MATHEMATICS. MINIMUM LEVEL MATERIAL for CLASS XII Project Planned By. Honourable Shri D. Manivannan Deputy Commissioner,KVS RO Hyderabad

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1 MATHEMATICS MINIMUM LEVEL MATERIAL for CLASS XII 05 6 Project Planned By Honourable Shri D. Manivannan Deputy Commissioner,KVS RO Hyderabad Prepared by M. S. KUMARSWAMY, TGT(MATHS) M. Sc. Gold Medallist (Elect.), B. Ed. Kendriya Vidyalaya gachibowli

2 PREFACE It gives me great pleasure in presenting the Minimum Level Study Material in Mathematics for Class XII. It is in accordance with the latest CBSE syllabus of the session I am extremely thankful to Honourable Shri D. Manivannan, Deputy Commissioner, KVS RO Hyderabad, who blessed and motivates me to complete this project work. This materials consists 7 easy Chapters out of 3 Chapters having overall weightage of 45 marks out of 00 marks. Each Chapter has Important Concepts and Formulae along with all NCERT Most Important Questions & answers and previous years Important Board Questions with Answers. At the end, NCERT Most Important Questions from other left out chapters have been added to cover maximum portions. I avail this opportunity to convey my sincere thanks to respected sir, Shri U. N. Khaware, Additional Commissioner(Acad), KVS Headquarter, New Delhi, respected sir, Shri S. Vijay Kumar, Joint Commissioner(Admn), KVS Headquarter, New Delhi, respected sir Shri P. V. Sairanga Rao, Deputy Commissioner, KVS Headquarter, New Delhi, respected sir Shri. D. Manivannan, Deputy Commissioner, KVS RO Hyderabad, respected sir Shri Isampal, Deputy Commissioner, KVS RO Bhopal, respected sir Shri P. Deva Kumar, Deputy Commissioner, KVS RO Bangalore, respected sir Shri Nagendra Goyal, Deputy Commissioner, KVS RO Ranchi, respected sir Shri Y. Arun Kumar, Assistant Commissioner(Acad), KVS Headquarter, New Delhi, respected sir Shri Sirimala Sambanna, Assistant Commissioner, KVS RO Hyderabad, respected sir Shri. K. L. Nagaraju, Assistant Commissioner, KVS RO Bangalore, respected sir Shri.Gangadharaiah, Assistant Commissioner, KVS RO Bangalore and respected Shri M.K. Kulshreshtha, Assistant Commissioner, KVS RO Chandigarh for their blessings, motivation and encouragement in bringing out this project in such an excellent form. I also extend my special thanks to respected sir Shri. P. S. Raju, Principal, KV Gachibowli, respected madam Smt. Nirmala Kumari M., Principal, KV Mysore & respected sir Shri. M. Vishwanatham, Principal, KV Raichur for their kind suggestions and motivation while preparing this Question Bank. I would like to place on record my thanks to respected sir Shri. P. K. Chandran, Principal, presently working in KV Bambolim. I have started my career in KVS under his guidance, suggestions and motivation. Inspite of my best efforts to make this notes error free, some errors might have gone unnoticed. I shall be grateful to the students and teacher if the same are brought to my notice. You may send your valuable suggestions, feedback or queries through to kumarsir34@gmail.com that would be verified by me and the corrections would be incorporated in the next year Question Bank. M. S. KUMARSWAMY

3 DEDICATED TO MY FATHER LATE SHRI. M. S. MALLAYYA

4 INDEX S. NO. CONTENT PAGE NO.. Relations and Functions Important Concepts & Formulae Page 7 Relations and Functions NCERT Important Questions & Answers Page 8 9 Relations and Functions Board Important Questions & Answers Page 0 7. Inverse Trigonometric Functions Important Concepts & Formulae Page 8 3 Inverse Trigonometric Functions NCERT Important Questions & Answers Page 3 4 Inverse Trigonometric Functions Board Important Questions & Answers Page Matrices Important Concepts & Formulae Page Matrices NCERT Important Questions & Answers Page 56 7 Matrices Board Important Questions & Answers Page Determinants Important Concepts & Formulae Page 77 8 Determinants NCERT Important Questions & Answers Page Determinants Board Important Questions & Answers Page Vector Algebra Important Concepts & Formulae Page 9 6 Vector Algebra NCERT Important Questions & Answers Page 7 34 Vector Algebra Board Important Questions & Answers Page Linear Programming Important Concepts & Formulae Page 4 45 Linear Programming NCERT Important Questions & Answers Page Linear Programming Board Important Questions & Answers Page Probability Important Concepts & Formulae Page Probability NCERT Important Questions & Answers Page Probability Board Important Questions & Answers Page Other Chapters NCERT Most Important Questions Page 0 6

5 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - -

6 CHAPTER : RELATIONS AND FUNCTIONS QUICK REVISION (Important Concepts & Formulae) Relation Let A and B be two sets. Then a relation R from A to B is a subset of A B. R is a relation from A to B R A B. MARKS WEIGHTAGE 05 marks Total Number of Relations Let A and B be two nonempty finite sets consisting of m and n elements respectively. Then A B consists of mn ordered pairs. So, total number of relations from A to B is nm. Domain and range of a relation Let R be a relation from a set A to a set B. Then the set of all first components or coordinates of the ordered pairs belonging to R is called the domain of R, while the set of all second components or coordinates of the ordered pairs in R is called the range of R. Thus, Dom (R) = {a : (a, b) R} and Range (R) = {b : (a, b) R}. Inverse relation Let A, B be two sets and let R be a relation from a set A to a set B. Then the inverse of R, denoted by R, is a relation from B to A and is defined by R = {(b, a) : (a, b) R}. Types of Relations Void relation : Let A be a set. Then A A and so it is a relation on A. This relation is called the void or empty relation on A. It is the smallest relation on set A. Universal relation : Let A be a set. Then A A A A and so it is a relation on A. This relation is called the universal relation on A. It is the largest relation on set A. Identity relation : Let A be a set. Then the relation I A = {(a, a) : a A} on A is called the identity relation on A. Reflexive Relation : A relation R on a set A is said to be reflexive if every element of A is related to itself. Thus, R reflexive (a, a) R a A. A relation R on a set A is not reflexive if there exists an element a A such that (a, a) R. Symmetric relation : A relation R on a set A is said to be a symmetric relation iff (a, b) R (b, a) R for all a, b A. i.e. arb bra for all a, b A. A relation R on a set A is not a symmetric relation if there are atleast two elements a, b A such that (a, b) R but (b, a) R. Transitive relation : A relation R on A is said to be a transitive relation iff (a, b) R and (b, c) R (a, c) R for all a, b, c A. i.e. arb and brc arc for all a, b, c A. Antisymmetric relation : A relation R on set A is said to be an antisymmetric relation iff (a, b) R and (b, a) R a = b for all a, b A. Equivalence relation : A relation R on a set A is said to be an equivalence relation on A iff It is reflexive i.e. (a, a) R for all a A. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - -

7 It is symmetric i.e. (a, b) R (b, a) R for all a, b A. It is transitive i.e. (a, b) R and (b, c) R (a, c) R for all a, b, c A. Congruence modulo m Let m be an arbitrary but fixed integer. Two integers a and b are said to be congruence modulo m if a b is divisible by m and we write a b(mod m). Thus, a b (mod m) a b is divisible by m. Some Results on Relations If R and S are two equivalence relations on a set A, then R S is also an equivalence relation on A. The union of two equivalence relations on a set is not necessarily an equivalence relation on the set. If R is an equivalence relation on a set A, then R is also an equivalence relation on A. Composition of relations Let R and S be two relations from sets A to B and B to C respectively. Then we can define a relation SoR from A to C such that (a, c) SoR b B such that (a, b) R and (b, c) S. This relation is called the composition of R and S. Functions Let A and B be two empty sets. Then a function 'f ' from set A to set B is a rule or method or correspondence which associates elements of set A to elements of set B such that (i) All elements of set A are associated to elements in set B. (ii) An element of set A is associated to a unique element in set B. A function f from a set A to a set B associates each element of set A to a unique element of set B. If an element a A is associated to an element b B, then b is called 'the f image of a or 'image of a under f or 'the value of the function f at a'. Also, a is called the preimage of b under the function f. We write it as : b = f (a). Domain, CoDomain and Range of a function Let f : AB. Then, the set A is known as the domain of f and the set B is known as the codomain of f. The set of all f images of elements of A is known as the range of f or image set of A under f and is denoted by f (A). Thus, f (A) = {f (x) : x A} = Range of f. Clearly, f (A) B. Equal functions Two functions f and g are said to be equal iff (i) The domain of f = domain of g (ii) The codomain of f = the codomain of g, and (iii) f (x) = g(x) for every x belonging to their common domain. If two functions f and g are equal, then we write f = g. Types of Functions (i) Oneone function (injection) A function f : A B is said to be a oneone function or an injection if different elements of A have different images in B. Thus, f : A B is oneone a b f (a) f (b) for all a, b A f (a) = f (b) a = b for all a, b A. Algorithm to check the injectivity of a function Step I : Take two arbitrary elements x, y (say) in the domain of f. Step II : Put f (x) = f (y) Step III : Solve f (x) = f (y). If f (x) = f (y) gives x = y only, then f : A B is a oneone function (or an injection) otherwise not. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 3 -

8 Graphically, if any straight line parallel to x-axis intersects the curve y = f (x) exactly at one point, then the function f (x) is oneone or an injection. Otherwise it is not. If f : R R is an injective map, then the graph of y = f (x) is either a strictly increasing curve or a dy dy strictly decreasing curve. Consequently, 0 or 0 for all x. dx dx n Pm, if n m Number of oneone functions from A to B, 0, if n m where m = n(domain) and n = n(codomain) (ii) Ontofunction (surjection) A function f : AB is said to be an onto function or a surjection if every element of B is the fimage of some element of A i.e., if f (A) = B or range of f is the codomain of f. Thus, f : A B is a surjection iff for each b B, a A that f (a) = b. Algorithm for Checking the Surjectivity of a Function Let f : A B be the given function. Step I : Choose an arbitrary element y in B. Step II : Put f (x) = y. Step III : Solve the equation f (x) = y for x and obtain x in terms of y. Let x = g(y). Step IV : If for all values of y B, for which x, given by x = g(y) are in A, then f is onto. If there are some y B for which x, given by x = g(y) is not in A. Then, f is not onto. Number of onto functions :If A and B are two sets having m and n elements respectively such that n m, then number of onto functions from A to B is n r nr n ( ). C r (iii) Bijection (oneone onto function) A function f : A B is a bijection if it is oneone as well as onto. In other words, a function f : A B is a bijection if r m (i) It is oneone i.e. f (x) = f (y) x = y for all x, y A. (ii) It is onto i.e. for all y B, there exists x A such that f (x) = y. Number of bijections : If A and B are finite sets and f : A B is a bijection, then A and B have the same number of elements. If A has n elements, then the number of bijections from A to B is the total number of arrangements of n items taken all at a time i.e. n! (iv) Manyone function A function f : A B is said to be a manyone function if two or more elements of set A have the same image in B. f : AB is a manyone function if there exist x, y A such that x y but f (x) = f ( y). Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 4 -

9 (v) Into function A function f : AB is an into function if there exists an element in B having no preimage in A. In other words f : A B is an into function if it is not an onto function. (vi) Identity function Let A be a nonempty set. A function f : AA is said to be an identity function on set A if f associates every element of set A to the element itself. Thus f : A A is an identity function iff f (x) = x, for all x A. (vii) Constant function A function f : A B is said to be a constant function if every element of A has the same image under function of B i.e. f (x) = c for all x A, where c B. Composition of functions Let A, B and C be three nonvoid sets and let f : A B, g : B C be two functions. For each x A there exists a unique element g( f (x)) C. The composition of functions is not commutative i.e. fog gof. The composition of functions is associative i.e. if f, g, h are three functions such that (fog)oh and fo(goh) exist, then (fog)oh = fo(goh). The composition of two bijections is a bijection i.e. if f and g are two bijections, then gof is also a bijection. Let f : AB. The foi A = I B of = f i.e. the composition of any function with the identity function is the function itself. Inverse of an element Let A and B be two sets and let f : A B be a mapping. If a A is associated to b B under the function f, then b is called the f image of a and we write it as b = f (a). Inverse of a function If f : A B is a bijection, we can define a new function from B to A which associates each element y B to its preimage f (y) A. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 5 -

10 Algorithm to find the inverse of a bijection Let f : A B be a bijection. To find the inverse of f we proceed as follows : Step I : Put f (x) = y, where y B and x A. Step II : Solve f (x) = y to obtain x in terms of y. Step III : In the relation obtained in step II replace x by f (y) to obtain the inverse of f. Properties of Inverse of a Function (i) The inverse of a bijection is unique. (ii) The inverse of a bijection is also a bijection. (iii) If f : A B is a bijection and g : B A is the inverse of f, then fog = I B and gof = I A, where I A and I B are the identity functions on the sets A and B respectively. If in the above property, we have B = A. Then we find that for every bijection f : A A there exists a bijection g : A A such that fog = gof = I A. (iv) Let f : A B and g : B A be two functions such that gof = I A and fog = I B. Then f and g are bijections and g = f. Binary Operation Let S be a nonvoid set. A function f from S S to S is called a binary operation on S i.e. f : S S S is a binary operation on set S. Generally binary operations are represented by the symbols *,,. etc. instead of letters f, g etc. Addition on the set N of all natural numbers is a binary operation. Subtraction is a binary operation on each of the sets Z, Q, R and C. But, it is a binary operation on N. Division is not a binary operation on any of the sets N, Z, Q, R and C. However, it is not a binary operation on the sets of all nonzero rational (real or complex) numbers. Types of Binary Operations (i) Commutative binary operation A binary operation * on a set S is said to be commutative if a * b = b * a for all a, b S Addition and multiplication are commutative binary operations on Z but subtraction is not a commutative binary operation, since 3 3. Union and intersection are commutative binary operations on the power set P(S) of all subsets of set S. But difference of sets is not a commutative binary operation on P(S). (ii) Associative binary operation A binary operation * on a set S is said to be associative if (a * b) * c = a * (b * c) for all a, b, c S. (iii) Distributive binary operation Let * and o be two binary operations on a set S. Then * is said to be (i) Left distributive over o if a*(b o c) = (a * b) o (a * c) for all a, b, c S (ii) Right distributive over o if (b o c) * a = (b * a) o (c * a) for all a, b, c S. (iv) Identity element Let * be a binary operation on a set S. An element e S is said to be an identity element for the binary operation * if a * e = a = e * a for all a S. For addition on Z, 0 is the identity element, since 0 + a = a = a + 0 for all a R. For multiplication on R, is the identity element, since a = a = a for all a R. For addition on N the identity element does not exist. But for multiplication on N the identity element is. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 6 -

11 (v) Inverse of an element Let * be a binary operation on a set S and let e be the identity element in S for the binary operation *. An element a S is said to be an inverse of a S, if a * a= e = a* a. Addition on N has no identity element and accordingly N has no invertible element. Multiplication on N has as the identity element and no element other than is invertible. Let S be a finite set containing n elements. Then the total number of binary operations on S is Let S be a finite set containing n elements. Then the total number of commutative binary operation n( n ) on S is n. n n. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 7 -

12 CHAPTER : RELATIONS AND FUNCTIONS NCERT Important Questions & Answers MARKS WEIGHTAGE 05 marks. Determine whether each of the following relations are reflexive, symmetric and transitive : (iv) Relation R in the set Z of all integers defined as R = {(x, y): x y is an integer } For reflexive put y = x, x x = 0 which is an integer for all x Z. So, R is reflexive on Z. For symmetry let (x,y) R, then (x y) is an integer λ and also y x = λ [ λ Z λ Z] y x is an integer (y, x) R. So, R is symmetric. For transitivity let (x,y) R and (y, z) R x y = integer and y z = integers, then x z is also an integer (x, z) R. So, R is transitive.. Show that the relation R in the set R of real numbers, defined as R = {(a, b) : a b } is neither reflexive nor symmetric nor transitive. We have R = {(a,b) :a b }, where a, b R For reflexivity, we observe that is not true. So, R is not reflexive as, R For symmetry, we observe that 3 but 3 > ( ) (, 3) R but (3, ) R. So, R is not symmetric. For transitivity, we observe that ( 3 ) and 3 () but > () (, 3) R and ( 3,) R but (, ) R. So, R is not transitive. Hence, R is neither reflexive, nor symmetric and nor transitive. 3. Show that the relation R in R defined as R = {a, b) : a b}, is reflexive and transitive but not symmetric. We have R = {(a,b) : a b}. Let a,b R. Reflexive: for any a R we have a a. So, R is reflexive. Symmetric: we observe that (, 3) R but (3, ) R. So, R is not symmetric. Transitivity: (a,b) R and (b, c) R a b and b c a c (a,c) R So, R is transitive. Hence, R is reflexive and transitive but not symmetric. 4. Check whether the relation R in R defined by R = {(a, b): a b 3 } is reflexive, symmetric or transitive. Given that R = {(a, b): a b 3 } It is observed that, R as So, R is not reflexive. Now, (, ) (as < 3 =8) But (, ) R (as 3 > ) So, R is not symmetric. 3 8 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 8 -

13 We have 3,,, R as 3 and But 3, R as Therefore, R is not transitive. Hence, R is neither reflexive nor symmetric nor transitive. 5. Show that the relation R in the set A = {,, 3, 4, 5} given by R = {(a, b) : a b is even}, is an equivalence relation. Show that all the elements of {, 3, 5} are related to each other and all the elements of {, 4} are related to each other. But no element of {, 3, 5} is related to any element of, 4}. Given that A = {,, 3, 4, 5} and R = {(a, b) : a b is even} It is clear that for any element a A, we have (which is even). R is reflexive. Let (a, b) R. a b is even (a b) is even (a b) is even (b a) is even b a is even (b, a) R R is symmetric. Now, let (a, b) R and (b, c) R. a b is even and b c is even (a b) is even and (b c) is even (a c) = (a b) + (b c) is even (Since, sum of two even integers is even) a c is even (a, c) R R is transitive. Hence, R is an equivalence relation. Now, all elements of the set {,, 3} are related to each other as all the elements of this subset are odd. Thus, the modulus of the difference between any two elements will be even. Similarly, all elements of the set {, 4} are related to each other as all the elements of this subset are even. Also, no element of the subset {, 3, 5} can be related to any element of {, 4} as all elements of {, 3, 5} are odd and all elements of {, 4} are even. Thus, the modulus of the difference between the two elements (from each of these two subsets) will not be even. 6. Show that each of the relation R in the set A { x Z : 0 x }, given by R = {(a, b): a b is a multiple of 4} is an equivalence relation. Find the set of all elements related to. A { x Z : 0 x } {0,,,3, 4,5, 6, 7,8,9,0,,} and R = {(a, b): a b is a multiple of 4} For any element a A, we have (a, a) R a a = 0 is a multiple of 4. R is reflexive. Now, let (a, b) R a b is a multiple of 4. (a b) is a multiple of 4 b a is a multiple of 4. (b, a) R Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 9 -

14 R is symmetric. Now, let (a, b), (b, c) R. a b is a multiple of 4 and b c is a multiple of 4. (a b) is a multiple of 4 and (b c) is a multiple of 4. (a b + b c) is a multiple of 4 (a c) is a multiple of 4 a c is a multiple of 4 (a, c) R R is transitive. Hence, R is an equivalence relation. The set of elements related to is {, 5, 9} since = 0 is a multiple of 4 5 = 4 is a multiple of 4 9 = 8 is a multiple of 4 7. In each of the following cases, state whether the functions is one-one, onto or bijective. Justify answer. (i) f : R R defined by f (x) = 3 4x (ii) f : R R defined by f (x) = + x (i) Here, f :R R is defined by f(x) = 3 4x Let x, x R such that f(x ) = f(x ) 3 4 x = 3 4x 4 x = 4x x = x Therefore, f is one-one. For any real number y in R, there exists 3 y 3 y 3 y in R such that f 3 4 y Therefore, f is onto. Hence, f is bijective. (ii) Here f :R R is defined as f(x) = + x Let x, x R such that f(x ) = f(x ) + x = + x x = x x = x For instance, f() = f( ) = Therefore, f(x ) = f(x ) does not imply that x = x Therefore, f is not one-one. Consider an element in co-domain R. It is seen that f(x) = + x is positive for all x R. Thus, there does not exist any x in domain R such that f(x) =. Therefore, f is not onto. Hence, f is neither one-one nor onto. 8. Let A = R {3} and B = R {}. Consider the function f : A B defined by f(x) = one-one and onto? Justify your answer. Here, A = R { 3), B = R {} and f : A B is defined as f(x) = Let x, y A such that f(x) = f(y) x x 3 x x 3 is f Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 0 -

15 x y ( x )( y 3) ( y )( x 3) x 3 y 3 xy 3x y 6 xy 3y x 6 3x y 3y x 3x x 3y y x y Therefore, f is one- one. Let y B = R {}. Then, y The function f is onto if there exists x A such that f(x) = y. Now, f(x) = y x y x xy 3y x 3 x( y) 3y 3y x A [ y ] y Thus, for any y B, there exists 3 y A such that y 3y 3 y y 3y y y f y y 3y 3y 3 3y 3 y Therefore, f is onto. Hence, function f is one-one and onto. 4x 3 9. If f ( x), x, show that (fof)(x) = x, for all x. What is the inverse of f? 6x x 3 Given that f ( x), x 6x 4 3 4x 3 Then ( fof )( x) f ( f ( x)) f 6x 4 4x x 4 6x 8x 34x x 4x 3 4x 8 4x x 4 Therefore (fof)(x) = x, for all x 3 Hence, the given function f is invertible and the inverse of f is itself. x 0. Show that f :[,] R, given by f ( x), x, is one-one. Find the inverse of the x function f :[,] Range f. x Given that f :[,] R, given by f ( x), x, x Let f(x) = f(y) x y xy x xy y x y x y x y Therefore, f is a one-one function. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - -

16 x y Let y x xy y x x y So, for every y except in the range there exists x in the domain such that f(x) = y. Hence, function f is onto. Therefore, f :[,] Range f is one-one and onto and therefore, the inverse of the function f :[,] Range f exists. Let y be an arbitrary element of range f. Since, f :[,] Range f is onto, we have y = f(x) for some x [,] x y y x xy y x, y x y y Now, let us define g: Range f [,] as g( y), y y x x x x x ( gof )( x) g( f ( x)) g x x x x x x y y y y y and ( fog)( y) f ( g( y)) f y y y y y y Therefore, gof = fog = I R, Therefore, f = g y Therefore, f ( y), y y. Consider f : R R given by f (x) = 4x + 3. Show that f is invertible. Find the inverse of f. Here, f :R R is given by f(x) = 4x + 3 Let x,y R, such that f(x) = f(y) 4x + 3 = 4y + 3 4x = 4y x = y Therefore, f is a one-one function. Let y = 4x + 3 y 3 There exist, x R, y R 4 y 3 Therefore, for any y R, there exist x R such that 4 y 3 y 3 f ( x) f 4 3 y 4 4 Therefore, f is onto function. Thus, f is one-one and onto and therefore, f exists. x 3 Let us define g : R R by g( x) 4 (4x 3) 3 Now, ( gof )( x) g( f ( x)) g(4x 3) x 4 y 3 y 3 and ( fog)( y) f ( g( y)) f 4 3 y 4 4 Therefore, gof = fog = I R Prepared by: M. S. KumarSwamy, TGT(Maths) Page - -

17 Hence, f is invertible and the inverse of f is given by y 3 f ( y) g( y) 4. Consider f : R + [4, ) given by f (x) = x + 4. Show that f is invertible with the inverse f of given by f y = y 4, where R + is the set of all non-negative real numbers. Here, function f : R + [4, ] is given as f(x) = x + 4 Let x,y R +, such that f(x) = f(y) x + 4 = y + 4 x = y x = y [as x = y R + ] Therefore, f is a one-one function. For y [4, ), let y = x + 4 x = y 4 0 [as y 4] x = y 4 0 Therefore, for any y R +, There exists x = y 4 R + such that f(x) = f( y 4 ) = ( y 4 ) + 4 = y = y Therefore, f is onto. Thus, f is one-one and onto and therefore, f exists. Let us define g : [4, ) R + by g(y) = y 4 Now, gof(x) = g(f(x)) = g(x + 4) = and fog(y) = f(g(y)) = f( 4 ( 4) 4 y ) = Therefore, gof I R and fog I[4, ) x x x Hence, f is invertible and the inverse of f if given by y 4 4 ( y 4) 4 y f ( y) g( y) y 4 3. Consider f : R + [ 5, ) given by f (x) = 9x + 6x 5. Show that f is invertible with y 6 f ( y). 3 Here, function f : R + [ 5, ) is given as f(x) = 9x + 6x 5. Let y be any arbitrary element of [ 5, ). Let y = 9x + 6x 5 y = (3x + ) 5 = (3x + ) 6 (3x + ) = y + 6 (3x + ) = y 6 [as y 5 y + 6 0] y 6 x 3 Therefore, f is onto, thereby range f = [ 5, ). y 6 Let us define g : [ 5, ) R + as g( y) 3 Now, (gof)(x) = g(f(x)) = g(9x + 6x 5) = g((3x + ) 6) (3x ) 6 6 3x x 3 3 and (fog)(y) = f(g(y)) y 6 6 y 6 6 y Therefore, gof I R and fog I[ 5, ) y 6 y 6 f Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 3 -

18 y 6 Hence, f is invertible and the inverse of f if given by f ( y) g( y) 3 4. Let * be the binary operation on N given by a* b = LCM of a and b. (i) Find 5*7, 0*6 (ii) Is * commutative? (iii) Is * associative? (iv) Find the identity of * in N (v) Which elements of N are invertible for the operation *? The binary operation on N is defined as a*b = LCM of a and b. (i) We have 5 *7 = LCM of 5 and 7 = 35 and 0*6 = LCM of 0 and 6 = 80 (ii) It is known that LCM of a and b = LCM of b and a for a, b N. Therefore, a*b = b*a. Thus, the operation * is commutative. (iii) For a, b, c N, we have (a*b) *c = (LCM of a and b) *c = LCM of a, b, and c a* (b*c) = a * (LCM of b and c) = LCM of a, b, and c Therefore, (a*b) *c = a* (b*c). Thus, the operation is associative. (iv) It is known that LCM of a and = a = LCM of and a, a N. a* = a = *a, a N Thus, is the identity of * in N. (v) An element a in N is invertible with respect to the operation *, if there exists an element b in N such that a*b = e = b*a. Here, e =. This means that LCM of a and b = = LCM of b and a This case is possible only when a and b are equal to. Thus, is the only invertible element of N with respect to the operation *. 5. Let * be the binary operation on N defined by a*b =HCF of a and b. Is * commutative? Is * associative? Does there exist identity for this binary operation on N? The binary operation * on N is defined as a*b = HCF of a and b. It is known that HCF of a and b = HCF of b and a for a,b N. Therefore, a*b = b*a. Thus, the operation is commutative. For a,b,c N, we have (a*b)*c = (HCF of a and b)*c = HCF of a,b and c a*(b*c) = a*(hcf of b and c) = HCF of a,b, and c Therefore, (a*b)*c = a*(b*c) Thus, the operation * is associative. Now, an element e N will be the identity for the operation if a*e = a = e*a, a N. But this relation is not true for any a N. Thus, the operation * does not have identity in N. 6. Let * be a binary operation on the set Q of rational number as follows : (i) a*b = a b (ii) a*b = a + b (iii) a*b = a + ab (iv) a*b = (a b) ab (v) a*b = (vi) a*b = ab 4 Find which of the binary operation are commutative and which are associative? (i) On Q, the operation * is defined as a*b = a b. It can be observed that for,3,4 Q, we have *3 = 3 = and 3* = 3 = *3 3* Thus, the operation is not commutative. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 4 -

19 It can also be observed that (*3)*4 = ( )*4 = 4 = 5 and *(3*4) = *( ) = ( ) = 3 *(3*4) * (3*4) Thus, the operation * is not associative. (ii) On Q, the operation * is defined as a*b = a + b. For a,b Q, we have a*b = a + b = b + a = b*a Therefore, a*b = b*a Thus, the operation * is commutative. It can be observed that (* )* 3 = ( + ) * 3 ( + 4) * 4 = 5 * 4 = = 4 and *(* 3) = * ( + 3 ) = * (4 + 9) = *3 = + 3 = 70 (*)* 3 *(*3) where,, 3 Q Thus, the operation * is not associative. (iii) On Q, the operation is defined as a*b = a + ab It can be observed that * = + = + = 3, * = + = + = 4 * * where, Q Thus, the operation * is not commutative. It can also be observed that (*)*3 = ( + )*3 = 3*3 = = = and *(* 3) = *( + 3) = *8 = + 8 = 9 (*)* 3 *(*3) where,, 3 Q Thus, the operation * is not associative. (iv) On Q, the operation * is defined by a*b = (a b). For a,b Q, we have a*b = (a b) and b*a = (b a) = [ (a b)] = (a b) Therefore, a*b = b*a Thus, the operation * is commutative. It can be observed that (* ) * 3 = ( ) * 3 = ( ) * 3 = ( 3) = 4 and * ( *3) = * ( 3) = * () = ( ) = 0 (*)* 3 *(* 3) where,, 3 Q Thus, the operation * is not associative. ab (v) On Q, the operation * is defined as a*b = 4 ab ba For a,b Q, we have a * b = = =b*a 4 4 Therefore, a*b = b*a Thus, the operation * is commutative. ab ab. c For a,b,c Q, we have a*(b*c )= * 4 abc c and a*(b*c) = Therefore, (a*b)*c = a*(b*c). Thus, the operation * is associative. bc bc a. * 4 abc a (vi) On Q, the operation is defined as a*b = ab It can be observed that for 3 Q * 3 = x 3 = 8 and 3* = 3 x = Hence, *3 3* Thus, the operation is not commutative. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 5 -

20 It can also be observed that for,,3 Q (*)*3 = (. )*3 = 4 *3 = 4.3 = 36 and *(* 3) = *(. 3 ) = *8 =.8 = 34 (*)*3 *(*3) Thus, the operation * is not associative. Hence, the operations defined in parts (ii), (iv), (v) are commutative and the operation defined in part (v) is associative. 7. Show that none of the operation given in the above question has identity. An element e Q will be the identity element for the operation if a*e = a = e*a, a Q (i) a*b = a b If a*e = a,a 0 a e = a, a 0 e = 0 Also, e*a = a e a = a e = a e = 0 = a,a 0 But the identiry is unique. Hence this operation has no identity. (ii) a*b = a + b If a*e = a, then a + e = a For a =, ( ) + e = 4 + e Hence, there is no identity element. (iii) a*b = a + ab If a*e = a a + ae = a ae = 0 e = 0,a 0 a Also if e*a = a e + ea = a e, a a a e 0, a 0 a But the identity is unique. Hence this operation has no identify. (iv) a*b = (a b) If a*e = a, then (a e) = a. A square is always positive, so for a =,( e) Hence, there is no identity element. (v) a*b = ab /4 If a*e = a, then ae /4 = a. Hence, e = 4 is the identity element. a*4 =4 *a =4a/4 = a (vi) a*b = ab If a*e =a then ae = a e = e = ± But identity is unique. Hence this operation has no identity. Therefore only part (v) has an identity element. x 8. Show that the function f :R {x R : < x <} defined by f( x) =, x R x onto function. x It is given that f :R {x R : < x <} defined by f( x) =, x R x is one-one and Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 6 -

21 x y Suppose, f(x) = f(y), where x, y R x y It can be observed that if x is positive and y is negative, then we have x y xy x y x y Since, x is positive and y is negative, then x > y x y > 0 But, xy is negative. Then, xy x y. Thus, the case of x being positive and y being negative can be ruled out. Under a similar argument, x being negative and y being positive can also be ruled out. Therefore, x and y have to be either positive or negative. x y When x and y are both positive, we have f ( x) f ( y) x xy y xy x y x y x y When x and y are both negative, we have f ( x) f ( y) x xy y xy x y x y Therefore, f is one-one. Now, let y R such that < y <. y If y is negative, then there exists x R such that y y y y y y y f ( x) f y y y y y y y y y If y is positive, then there exists x R such that y y y y y y y f ( x) f y y y y y y y y Therefore, f is onto. Hence, f is one-one and onto. 9. Show that the function f :R R given by f (x) = x 3 is injective. Here, f :R R is given as f(x) = x 3. Suppose, f(x) = f(y),where x,y R x 3 = y 3 (i) Now, we need to show that x = y Suppose, x y, their cubes will also not be equal. x 3 y 3 However, this will be a contradiction to Eq. i). Therefore, x = y. Hence, f is injective. a b, ifa b 6 0. Define a binary operation * on the set {0,,, 3, 4, 5} as a* b. Show a b 6, ifa b 6 that zero is the identity for this operation and each element a 0 of the set is invertible with (6 a) being the inverse of a. Let X = {0,,, 3,4, 5} a b, ifa b 6 The operation * on X is defined as a* b a b 6, ifa b 6 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 7 -

22 An element e X is the identity element for the operation *, if a*e = a = e*a a X For a X, we observed that a*0 = a + 0 = a [a X a + 0 < 6] 0*a = 0 + a = a [a X 0 + a < 6] a*0 = a = 0*a a X Thus, 0 is the identity element for the given operation *. An element a X is invertible, if there exists b X such that a*b = 0 = b*a a b 0 b a, ifa b 6 i.e. a b 6 0 b a 6, ifa b 6 i.e., a = b or b = 6 a But X = {0,,, 3,4,5} and a, b X Then, a b Therefore, b = 6 a is the inverse of a, a X. Hence, the inverse of an element a X, a 0 is (6 a) i.e., a = 6 a.. Show that the relation R in the set Z of integers given by R = {(a, b) : divides a b} is an equivalence relation. R is reflexive, as divides (a a) for all a Z. Further, if (a, b) R, then divides a b. Therefore, divides b a. Hence, (b, a) R, which shows that R is symmetric. Similarly, if (a, b) R and (b, c) R, then a b and b c are divisible by. Now, a c = (a b) + (b c) is even. So, (a c) is divisible by. This shows that R is transitive. Thus, R is an equivalence relation in Z.. Show that if f : A B and g : B C are one-one, then gof : A C is also one-one. Suppose gof (x ) = gof (x ) g (f (x )) = g(f (x )) f (x ) = f (x ), as g is one-one x = x, as f is one-one Hence, gof is one-one. 3. Determine which of the following binary operations on the set N are associative and which are a b commutative. (a) a * b = a, b N (b) a * b = a, b N (a) Clearly, by definition a * b = b * a = a, b N. Also (a * b) * c = ( * c) = and a * (b * c) = a * () =, a, b, c N. Hence R is both associative and commutative. a b b a (b) a* b b* a, shows that * is commutative. Further, a b c a b a b c ( a* b)* c * c 4 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 8 -

23 b c a b c a b c a b c But a*( b* c) a* in general. 4 4 Hence, * is not associative Show that if f : R R 5 5 is defined by 3x 4 f ( x) 5x 7 and 3 7 g : R R 5 5 is 7x 4 defined by g( x) 5x 3, then fog = I 3 7 A and gof = I B, where, A R, B R 5 5 ; I A(x) = x, x A, I B (x) = x, x B are called identity functions on sets A and B, respectively. 3x x 4 5x 7 x 8 0x 8 4x We have gof ( x) g x 5x 7 3x 4 5x 0 5x x 7 7x x 4 5x 3 x 0x 4x Similarly, fog( x) f x 5x 3 7x 4 35x x 3 Thus, gof (x) = x, x B and fog (x) = x, x A, which implies that gof = I B and fog = I A. 5. Let f : N R be a function defined as f (x) = 4x + x + 5. Show that f : N S, where, S is the range of f, is invertible. Find the inverse of f. Let y be an arbitrary element of range f. Then y = 4x + x + 5, for some x in N, which implies that y = (x + 3) y This gives x, as y 6. y 6 3 Let us define g : S N by g( y) Now gof (x) = g(f (x)) = g(4x + x + 5) = g ((x + 3) + 6) and (x 3) (x 3 3) x y 6 3 y 6 3 fog( y) f 3 6 y y 6 6 y 6 6 y Hence, gof = I N and fog =I S. This implies that f is invertible with f = g. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 9 -

24 CHAPTER : RELATIONS AND FUNCTIONS Previous Years Board Exam (Important Questions & Answers). If f(x) = x + 7 and g(x) = x 7, x R, find ( fog) (7) Given f(x) = x + 7 and g(x) = x 7, x R fog(x) = f(g(x)) = g(x) + 7 = (x 7) + 7 = x (fog) (7) = 7. MARKS WEIGHTAGE 05 marks. If f(x) is an invertible function, find the inverse of f (x) = 3 x 5 3x Given f ( x) 5 3x Let y 5 5y 3x 5y x 3 5x f ( x) 3 3. Let T be the set of all triangles in a plane with R as relation in T given by R = {(T, T ) :T T }. Show that R is an equivalence relation. (i) Reflexive R is reflexive if T R T Since T T R is reflexive. (ii) Symmetric R is symmetric if T R T T R T Since T T T T R is symmetric. (iii) Transitive R is transitive if T R T and T R T 3 T R T 3 Since T T and T T 3 T T 3 R is transitive From (i), (ii) and (iii), we get R is an equivalence relation. 4. If the binary operation * on the set of integers Z, is defined by a *b = a + 3b, then find the value of * 4. Given a *b = a + 3b a, b z *4 = + 3 x 4 = + 48 = Let * be a binary operation on N given by a * b = HCF (a, b) a, b N. Write the value of * 4. Given a * b = HCF (a, b), a, b N * 4 = HCF (, 4) = Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 0 -

25 6. Let f : N N be defined by function f is bijective. n, if n is odd f ( n) n for all n N. Find whether the, if n is even n, if n is odd Given that f : N N be defined by f ( n) n for all n N., if n is even Let x, y N and let they are odd then x y f ( x) f ( y) x y If x, y N are both even then also x y f ( x) f ( y) x y If x, y N are such that x is even and y is odd then x y f ( x) and f ( y) Thus, x y for f(x) = f(y) Let x = 6 and y = We get f (6) 3, f (5) 3 f(x) = f(y) but x y...(i) So, f (x) is not one-one. Hence, f (x) is not bijective. 7. If the binary operation *, defined on Q, is defined as a * b = a + b ab, for all a, b Q, find the value of 3 * 4. Given binary operation is a*b = a + b ab 3* 4 = * 4 = x 8. What is the range of the function f ( x) ( x )? x We have given f ( x) ( x ) ( x ), if x 0 or x x ( x ), if x 0 or x ( x ) (i) For x >, f ( x) ( x ) ( x ) (ii) For x <, f ( x) ( x ) x Range of f ( x) is {, }. ( x ) Prepared by: M. S. KumarSwamy, TGT(Maths) Page - -

26 9. Let Z be the set of all integers and R be the relation on Z defined as R = {(a, b) ; a, b Z, and (a b) is divisible by 5.} Prove that R is an equivalence relation. We have provided R = {(a, b) : a, b Z, and(a b) is divisible by 5} (i) As (a a) = 0 is divisible by 5. (a, a) R a R Hence, R is reflexive. (ii) Let (a, b) R (a b) is divisible by 5. (b a) is divisible by 5. (b a) is divisible by 5. (b, a) R Hence, R is symmetric. (iii) Let (a, b) R and (b, c) Z Then, (a b) is divisible by 5 and (b c) is divisible by 5. (a b) + (b c) is divisible by 5. (a c) is divisible by 5. (a, c) R R is transitive. Hence, R is an equivalence relation. 3ab 0. Let * be a binary operation on Q defined by a* b. Show that * is commutative as well as 5 associative. Also find its identity element, if it exists. For commutativity, condition that should be fulfilled is a * b = b * a 3ab 3ba Consider a* b b* a 5 5 a * b = b * a Hence, * is commutative. For associativity, condition is (a * b) * c = a * (b * c) 3ab 9ab Consider ( a* b)* c * c 5 5 3bc 9ab and a*( b* c) a* 5 5 Hence, (a * b) * c = a * (b * c) * is associative. Let e Q be the identity element, Then a * e = e * a = a 3ae 3ea 5 a e If f : R R be defined by f(x) = (3 x 3 ) / 3, then find fof(x). If f : R R be defined by f(x) = (3 x 3 ) /3 then ( fof) x = f( f(x)) = f [(3 x 3 ) /3 ] = [3 {(3 x 3 ) /3 } 3 ] /3 = [3 (3 x 3 )] /3 = (x 3 ) /3 = x Prepared by: M. S. KumarSwamy, TGT(Maths) Page - -

27 . Let A = N N and * be a binary operation on A defined by (a, b) * (c, d) = (a + c, b + d). Show that * is commutative and associative. Also, find the identity element for * on A, if any. Given A = N N * is a binary operation on A defined by (a, b) * (c, d) = (a + c, b + d) (i) Commutativity: Let (a, b), (c, d) N N Then (a, b) * (c, d) = (a + c, b + d) = (c + a, d + b) (a, b, c, d N, a + c = c + a and b + d = d + c) = (c, d) * b Hence, (a, b) * (c, d) = (c, d) * (a, b) * is commutative. (ii) Associativity: let (a, b), (b, c), (c, d) Then [(a, b) * (c, d)] * (e, f) = (a + c, b + d) * (e, f) = ((a + c) + e, (b + d) + f) = {a + (c + e), b + (d + f)] ( set N is associative) = (a, b) * (c + e, d + f) = (a, b) * {(c, d) * (e, f)} Hence, [(a, b) * (c, d)] * (e, f) = (a, b) * {(c, d) * (e, f)} * is associative. (iii) Let (x, y) be identity element for on A, Then (a, b) * (x, y) = (a, b) (a + x, b + y) = (a, b) a + x = a, b + y = b x = 0, y = 0 But (0, 0) A For *, there is no identity element. 3. If f : R R and g : R R are given by f(x) = sin x and g(x) = 5x, find gof(x). Given f : R R and g : R R defined by f (x) = sin x and g(x) = 5x gof(x) = g [f(x)] = g (sin x) = 5 (sin x) = 5 sin x 4. Consider the binary operation* on the set {,, 3, 4, 5} defined by a * b = min. {a, b}. Write the operation table of the operation *. Required operation table of the operation * is given as * If f : R R is defined by f(x) = 3x +, define f[f(x)]. f (f (x)) = f (3x + ) =3. (3x + ) + = 9x = 9x + 8 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 3 -

28 6. Write fog, if f : R R and g : R R are given by f(x) = 8x 3 and g(x) = x /3. fog (x) = f (g(x)) = f (x /3 ) = 8(x /3 ) 3 = 8x 7. Let A = {,, 3}, B = {4, 5, 6, 7} and let f = {(, 4), (,5), (3, 6)} be a function from A to B. State whether f is one-one or not. f is one-one because f() = 4 ; f() = 5 ; f(3) = 6 No two elements of A have same f image. 8. Let f : R R be defined as f(x) =0x +7. Find the function g : R R such that gof = fog =I R. gof = fog = IR fog = IR fog(x) = I (x) f (g(x)) = x [I(x) = x being identity function] 0(g(x)) + 7 = x [f(x) = 0x + 7] x 7 g( x) 0 x 7 i.e., g : R R is a function defined as g( x) 0 9. Let A = R {3} and B = R {}. Consider the function f : A B defined by Show that f is one-one and onto and hence find f. Let x, x A. x x Now, f(x ) = f(x ) x 3 x 3 ( x )( x 3) ( x 3)( x ) x x 3x x 6 x x x 3x 6 3x x x 3x x x x x Hence f is one-one function. For Onto x Let y xy 3y x x 3 xy x 3y x( y ) 3y 3y x ( i) y x f ( x) x 3. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 4 -

29 From above it is obvious that y except, i.e., y B R {} x A Hence f is onto function. Thus f is one-one onto function. It f 3y is inverse function of f then f ( y) [from (i)] y 0. The binary operation * : R R R is defined as a * b = a + b. Find ( * 3) * 4 ( * 3) * 4 = ( +3) * 4 = 7 * 4 = = 8 x, if x is odd. Show that f : N N, given by f ( x) is both one-one and onto. x, if x is even For one-one Case I : When x, x are odd natural number. f(x ) = f(x ) x + = x + x, x N x = x i.e., f is one-one. Case II : When x, x are even natural number f(x ) = f(x ) x = x x = x i.e., f is one-one. Case III : When x is odd and x is even natural number f(x ) = f(x ) x + = x x x = which is never possible as the difference of odd and even number is always odd number. Hence in this case f (x ) f(x ) i.e., f is one-one. Case IV: When x is even and x is odd natural number Similar as case III, We can prove f is one-one For onto: f(x) = x + if x is odd = x if x is even For every even number y of codomain odd number y - in domain and for every odd number y of codomain even number y + in Domain. i.e. f is onto function. Hence f is one-one onto function.. Consider the binary operations * : R R R and o : R R R defined as a * b = a b and aob = a for all a, b R. Show that * is commutative but not associative, o is associative but not commutative. For operation * * : R R R such that a*b = a b a, b R Commutativity a*b = a b = b a = b * a i.e., * is commutative Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 5 -

30 Associativity a, b, c R (a * b) * c = a b * c = a b c a * (b * c) = a * b c = a b c But a b c a b c (a*b)* c a*( b * c) " a, b, c R * is not associative. Hence, * is commutative but not associative. For Operation o o : R R R such that aob = a Commutativity a, b R aob = a and boa = b a b aob boa o is not commutative. Associativity: " a, b, c R (aob) oc = aoc = a ao(boc) = aob = a (aob) oc = ao (boc) o is associative Hence o is not commutative but associative. 3. If the binary operation * on the set Z of integers is defined by a * b = a + b 5, then write the identity element for the operation * in Z. Let e Z be required identity a* e = a a Z a + e 5 = a e = a a + 5 e = 5 4. If the binary operation * on set R of real numbers is defined as a*b = 3 ab, write the identity 7 element in R for *. Let e R be identity element. a * e = a a R 3ae 7a a e e 7 3a Prove that the relation R in the set A = {5, 6, 7, 8, 9} given by R = {(a, b) : a b, is divisible by }, is an equivalence relation. Find all elements related to the element 6. Here R is a relation defined as R = {(a, b) : a b is divisible by } Reflexivity Here (a, a) R as a a = 0 = 0 divisible by i.e., R is reflexive. Symmetry Let (a, b) R (a, b) R a b is divisible by a b = ± m b a = m b a is divisible by (b, a) R Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 6 -

31 Hence R is symmetric Transitivity Let (a, b), (b, c) R Now, (a, b), (b, c) R a b, b c are divisible by a b = ±m and b c = ±n a b + b c = ± (m + n) (a c) = ± k [k = m + n] (a c) = k (a c) is divisible by (a, c) R. Hence R is transitive. Therefore, R is an equivalence relation. The elements related to 6 are 6, 8. ab 6. Let * be a binary operation, on the set of all non-zero real numbers, given by a* b for all 5 a, b R {0}. Find the value of x, given that * (x * 5) = 0. Given * (x * 5) = 0 x5 * 0 * x 0 5 x 05 0 x x Let A = {,, 3,, 9} and R be the relation in A A defined by (a, b) R (c, d) if a + d = b + c for (a, b), (c, d) in A A. Prove that R is an equivalence relation. Also obtain the equivalence class [(, 5)]. Given, R is a relation in A A defined by (a, b)r(c, d) a + d = b + c (i) Reflexivity: a, b A Q a + b = b + a (a, b)r(a, b) So, R in reflexive. (ii) Symmetry: Let (a, b) R (c, d) Q (a, b)r(c, d) a + d = b + c b + c = d + a [Q a, b, c, d N and N is commutative under addition[ c + b = d + a (c, d)r(a, b) So, R is symmetric. (iii) Transitivity: Let (a, b)r(c, d) and (c, d)r(e, f) Now, (a, b)r(c, d) and (c, d)r(e, f) a + d = b + c and c + f = d + e a + d + c + f = b + c + d + e a + f = b + e (a, b)r(e, f). R is transitive. Hence, R is an equivalence relation. nd Part: Equivalence class: [(, 5)] = {(a, b) A A: (a, b)r(, 5)} = {(a, b) A A: a + 5 = b + } = {(a, b) A A: b a = 3} = {(, 4), (, 5), (3, 6), (4, 7), (5, 8), (6, 9)} Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 7 -

32 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 8 -

33 CHAPTER : INVERSE TRIGONOMETRIC FUNCTIONS QUICK REVISION (Important Concepts & Formulae) MARKS WEIGHTAGE 05 marks Inverse Trigonometrical Functions A function f : A B is invertible if it is a bijection. The inverse of f is denoted by f and is defined as f (y) = x f (x) = y. Clearly, domain of f = range of f and range of f = domain of f. The inverse of sine function is defined as sin x = sinq = x, where [ /, /] and x [, ]. Thus, sin x has infinitely many values for given x [, ] There is one value among these values which lies in the interval [ /, /]. This value is called the principal value. Domain and Range of Inverse Trigonometrical Functions Properties of Inverse Trigonometrical Functions sin (sin) = and sin(sin x) = x, provided that x and cos (cos) = and cos (cos x) = x, provided that x and 0 tan (tan) = and tan(tan x) = x, provided that x and cot (cot) = and cot(cot x) = x, provided that < x < and 0 < <. sec (sec) = and sec(sec x) = x cosec (cosec) = and cosec(cosec x) = x, x x sin cos x ec or cos ec x sin Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 9 -

34 x x cos s x ec or sec x cos x x tan cot x or cot x tan x x sin x cos x tan cot sec cosec x x x x x x cos x sin x tan cot cos ec s ec x x x x x x tan x sin cos cot sec x cos ec x x x x sin x cos x, where x tan x cot x, where x sec x cos ec x, where x or x x y tan x tan y tan, if xy xy x y tan x tan y tan, if xy xy x y xy tan x tan y tan sin x sin y sin x y y x, if x, y 0, x y sin x sin y sin x y y x, if x, y 0, x y sin x sin y sin x y y x, if x, y 0, x y sin x sin y sin x y y x, if x, y 0, x y cos x cos y cos xy x y, if x, y 0, x y Prepared by: M. S. KumarSwamy, TGT(Maths) Page

35 cos x cos y cos xy x y, if x, y 0, x y cos x cos y cos xy x y, if x, y 0, x y cos x cos y cos xy x y, if x, y 0, x y sin ( ) sin, cos ( ) cos x x x x tan ( ) tan, cot ( ) cot x x x x sin x sin x x, cos x cos x x x x tan x tan sin cos x x x 3 3sin x sin 3x 4 x, 3cos x cos 4x 3 3x 3 3x x 3tan x tan 3x Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 3 -

36 CHAPTER : INVERSE TRIGONOMETRIC FUNCTIONS MARKS WEIGHTAGE 05 marks NCERT Important Questions & Answers. Find the principal values of sin. Let sin sin We know that the range of principal value of sin is, sin sin sin sin( ) sin 6 6, where, sin 6 6 Hence the principal value of sin is 6. Find the principal values of tan 3 Let tan 3 tan 3 We know that the range of principal value of tan is, tan 3 tan tan tan( ) tan 3 3, where, tan Hence the principal value of tan 3is 3 3. Find the principal values of cos Let cos cos cos is 0, We know that the range of principal value of cos cos cos 3 3 cos cos( ) cos 3, 0, cos where 3 3 Hence the principal value of cos is 3 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 3 -

37 4. Find the principal values of Let cos cos cos We know that the range of principal value of cos is 0, 3 cos cos cos cos cos( ) cos , 0, cos 3 where Hence the principal value of cos is 4 5. Find the principal values of cosec Let ec cos cos ec We know that the range of principal value of cos ec is, {0} cosec cos ec cos ec cos ec( ) cos ec 4 4, where, {0} cos ec 4 4 Hence the principal value of cosec is 4 6. Find the values of tan () cos sin Let tan x tan x tan x where x, 4 4 tan 4 Let cos y cos y cos cos cos cos( ) cos y where y 0, 3 Let sin z sin z sin sin z where z, sin 6 tan cos sin x y z Prepared by: M. S. KumarSwamy, TGT(Maths) Page

38 3 7. Prove that 3sin x sin (3x 4 x ), x, Let sin x x sin, then 3 We know that sin 3 3sin 4sin sin (3sin 4sin ) sin (3x 4 x ) 3 3sin x sin (3x 4 x ) 7 8. Prove that tan tan tan 4 7 Given tan tan tan tan tan tan 4 x y LHS 4 7 tan x tan y tan. x. y tan tan tan tan tan RHS Prove that tan tan tan Given tan tan tan 7 7 LHS tan tan tan x tan 7 tan x tan 7 x 4 = tan tan tan tan tan 3 7 x y 4 tan x tan y tan. x. y tan tan tan RHS x 0. Simplify : tan, x 0 x Let x = tan θ, then θ = tan x (i) x tan sec tan tan tan tan tan Prepared by: M. S. KumarSwamy, TGT(Maths) Page

39 cos sec cos tan tan tan cos tan sin sin cos cos sin cos sin cos tan tan sin sin cos and sin sin cos sin tan tan tan tan x [using (i)] cos. Simplify : tan, x x Let x = sec θ, then θ = sec x (i) tan tan tan x sec tan tan tan (cot ) tan tan tan cot tan sec x [using (i)] cos x sin x. Simplify : tan,0 x cos x sin x cos x sin x cos x sin x tan tan cos x cos x cos x sin x cos x sin x cos x cos x (inside the bracket divide numerator and denominator by cos x) tan x tan x tan tan tan x tan x tan x 4 4 tan x x 4 x y 3. Simplify : tan sin cos, x, y 0 and xy x y x y tan sin cos, x, y 0 and xy x y x y tan x sin and tan y cos x y tan ( tan x tan y) tan.(tan x tan ) tan(tan tan ) y x y x y x y tan tan tan x tan y tan x. y x. y Prepared by: M. S. KumarSwamy, TGT(Maths) Page

40 x y x. y x x 4. If tan tan, find the value of x. x x 4 x x Given that tan tan x x 4 x x tan x x x y tan x tan y tan x x 4 x. y x x ( x )( x ) ( x )( x ) ( x )( x ) tan x 4 x 4 ( x x x ) ( x x x ) x 4 tan 4 x x 4 x 4 x 4 x 4 3 x x 3 x 7 5. Find the value of cos cos cos cos cos cos where, [0, ] cos cos cos cos cos( ) cos Prove that cos sin sin Given cos sin sin Let cos x cos x sin x cos x x sin LHS cos sin sin sin Prepared by: M. S. KumarSwamy, TGT(Maths) Page

41 sin sin x sin y sin x y y x sin sin sin sin RHS Prove that tan sin cos RHS sin cos Let sin x sin x cos x sin x sin x tan x x tan cos x Let cos y cos y sin y cos y sin x tan y y tan cos x then the equation becomes tan x y tan tan tan x y RHS tan tan tan tan x tan y tan x. y tan tan tan LHS Prove that tan tan tan tan LHS = tan tan tan tan Prepared by: M. S. KumarSwamy, TGT(Maths) Page

42 x y tan tan tan x tan y tan +.. x. y tan tan tan tan tan tan tan tan tan tan tan tan RHS Prove that sin sin cot x x x, x 0, sin x sin x 4 sin sin Given cot x x x, x 0, sin x sin x 4 sin x sin x LHS cot sin x sin x sin sin sin sin cot x x x x sin x sin x sin x sin x (by rationalizing the denominator) sin x sin x sin x sin x sin x cot cot sin sin sin x sin x x x cos x ( cos x) cos x cot cot cot sin x sin x sin x x cos x x x cot cos x cos and sin x sin cos x x sin cos x cos x x cot cot cot RHS x sin x x 0. Prove that tan cos x x x 4 Let x = cosy y cos x LHS tan y y cos y cos y cos sin tan cos y cos y y y cos sin Prepared by: M. S. KumarSwamy, TGT(Maths) Page

43 y y cos y cos and cos y sin y y y cos sin tan y tan tan tan tan cos y y y cos sin tan 4 4 tan x tan x 4 tan x. Solve for x: tan x tan x,( x 0) x Given tan x tan x,( x 0) x x tan tan x x x x x tan tan x tan x tan x x x tan tan tan x x tan ( x) ( x) ( x) x ( x)( x) tan ( x) ( x) x ( x ) tan x x x x ( x ) x tan tan x tan tan x 4x x x x x x 3x x x x 3 3 x 0 given, so we do not take x x 3. Solve for x: tan (cos x) tan (cos ecx) Given tan (cos x) tan (cos ecx) Prepared by: M. S. KumarSwamy, TGT(Maths) Page x 3 cos x x tan tan (cos ecx) tan x tan cos x x cos x tan tan sin x sin x cos x cos x sin x sin x sin x x

44 cot x cot x cot x Solve for x: sin ( x) sin x Given sin ( x) sin x sin x sin ( x) sin x cos ( x) sin ( x) cos ( x) cos sin x x x x x x x x cos sin cos( ) cos sin sin x x cos x sin x sin sin x x x x 0 x(x ) 0 x 0 or x 0 x 0 or x But x does not satisfy the given equation, so x = 0. x x y 4. Simplify: Given tan tan tan y x y x x y tan = tan y x y x x y y tan 4 tan tan tan 5. Express Given x x y tan y x y tan x tan y tan cos x tan, x in the simplest form. sin x cos x tan, x sin x x y x. y tan x x x x x x cos sin cos sin cos sin tan x x x x cos sin cos sin x x cos sin Prepared by: M. S. KumarSwamy, TGT(Maths) Page

45 x x x x x x cos x cos sin,sin cos and sin x sin cos x x x cos sin tan x x tan tan tan tan x x x cos sin tan Simplify : cot, x x Let x = sec θ, then θ = sec x (i) cot cot cot x sec tan cot cot (cot ) sec x tan 7. Prove that sin sin cos Let sin x and 5 3 sin y Therefore sin x and sin y 5 7 Now, cos x sin x and cos y sin y We have cos( x y) cos x cos y sin xsin y x y cos sin sin cos Prove that sin cos tan Let sin x, cos y and tan z Then sin x, cos y and tan z Now, and 44 5 cos x sin x sin y cos y Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 4 -

46 3 sin x 3 sin y 5 3 tan x and tan y cos x 5 5 cos y tan x tan y tan( x y) tan z tan x.tan y tan( x y) tan z tan( z) tan( z) x y z x y z 4 63 sin cos tan a cos x bsin x a 9. Simplify: tan, if tan x bcos x asin x b a cos x bsin x a cos x bsin x tan tan bcos x bcos x asin x bcos x asin x bcos x a tan x tan b a a tan tan (tan x) tan x a tan x b b b 30. Solve: tan x tan 3x 4 Given tan x tan 3x 4 x 3x x y tan tan x tan y tan x.3x 4 x. y 5x 5x tan tan 6x 4 6x 4 6x 5x 0 (6x )( x ) 0 x or x 6 Since x = does not satisfy the equation, as the L.H.S. of the equation becomes negative, x is the only solution of the given equation. 6 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 4 -

47 CHAPTER : INVERSE TRIGONOMETRIC FUNCTIONS MARKS WEIGHTAGE 05 marks Previous Years Board Exam (Important Questions & Answers). Evaluate : sin sin 3 sin sin sin sin sin 3 6. Write the value of cot (tan a + cot a). cot tan a cot a cot cot a cot a 7 3. Find the principal values of cos cos 6. 7 cos cos cos cos cos cos cos cot Find the principal values of tan tan 4 3 tan tan tan tan 4 4 tan tan tan ( ) 4 4 a a b 5. Prove that: tan cos tan cos 4 b 4 b a a a LHS tan cos tan cos 4 b 4 b a a tan tan cos tan tan cos 4 b 4 b a a tan tan cos tan tan cos 4 b 4 b a a tan cos tan cos b b a a tan cos tan cos b b Prepared by: M. S. KumarSwamy, TGT(Maths) Page

48 a a tan cos tan cos b b a tan cos b a a tan cos tan cos b b a a tan cos tan cos b b b = RHS a a a a cos cos cos cos b b b 6. Solve: x x tan tan tan 3 8 x x tan tan tan 3 ( x ) ( x ) 8 tan tan ( x ).( x ) 3 x 8 x 8 tan tan tan tan ( x ) 3 x 3 x 8 6x 6 8x 8x 6x 6 0 x 3 4x 3x 8 0 (4x )( x 8) 0 x and x 8 4 As x = 8 does not satisfy the equation Hence x is only solution Prove that sin sin sin sin sin sin sin sin sin cos Let sin x and sin y Therefore sin x and 5 Now, 5 5 sin y cos x sin x and cos y sin y We have cos( x y) cos x cos y sin xsin y Prepared by: M. S. KumarSwamy, TGT(Maths) Page

49 6 x y cos sin sin cos Prove that tan tan cos LHS tan tan tan tan tan tan x tan y tan 7 tan tan tan 34 4 x cos cos tan x cos x cos cos RHS x x 9. Solve for x: cos tan x x 3 x x cos tan x x 3 ( x ) x cos tan x x 3 x x cos tan x x 3 tan x tan x 4 tan x tan x 4 tan x tan x 3 3 tan tan tan x x tan tan.tan x 3 3 x y x. y Prepared by: M. S. KumarSwamy, TGT(Maths) Page

50 x x 0. Prove that tan x cos, x (0,) x LHS tan tan x x x x cos tan x cos x x x cos RHS x Prove that sin sin LHS sin sin cos Let cos x cos x sin x cos x sin x x sin cos sin cos sin RHS Find the principal value of tan 3 sec ( ) tan 3 sec ( ) tan tan sec sec 3 3 sec sec sec sec Prove that : cossin cot Let sin x and cot y Then sin x and cot y Prepared by: M. S. KumarSwamy, TGT(Maths) Page

51 Now cos x sin x and sin y cot y cos y LHS cos( x y) cos x cos y sin xsin y RHS Write the value of tan tan 5 x x Let tan x tan tan x tan tan tan tan x 5 x tan x y 4. Find the value of the following: tan sin cos, x, y 0 and xy x y Let x tan and y tan tan x, tan y x y tan sin cos x y tan tan tan sin cos tan tan tan tan tan sin (sin ) cos (cos ) sin and cos tan tan tan tan x y tan tan tan.tan xy 5. Write the value of tan sin cos tan 3 sin cos 3 Prepared by: M. S. KumarSwamy, TGT(Maths) Page tan sin tan sin tan 6 3 tan 3 3

52 6. Prove that tan sin Let sin sin 4 4 tan 3 tan sin tan 4 tan 3 3tan 8tan 3tan 8tan tan tan tan tan sin If y cot cos x tan cos x, then prove that sin y tan cos tan cos tan cos y cot cos x tan cos x y x x x cos x x y cos tan x cos cos x x cos x y sin cos x x sin cos x x sin y tan cos x x cos 8. If sin sin cos x, then find the value of x. 5 sin sin cos x sin cos x sin 5 5 sin cos x sin cos x sin x 5 5 x Prove that tan sec tan LHS tan sec tan x tan Prepared by: M. S. KumarSwamy, TGT(Maths) Page

53 5 5 tan tan sec tan tan sec tan tan tan tan tan tan tan tan tan tan tan x tan tan x tan x tan tan tan tan tan tan tan tan tan 4 7 x y tan x tan y tan. x. y tan tan tan If tan x tan y then write the value of x + y + xy. 4 tan x tan y 4 x y x y tan tan x. y 4 x. y 4 x y xy x y xy Prepared by: M. S. KumarSwamy, TGT(Maths) Page

54 Prepared by: M. S. KumarSwamy, TGT(Maths) Page

55 CHAPTER 3: MATRICES QUICK REVISION (Important Concepts & Formulae) MARKS WEIGHTAGE 03 marks Matrix A matrix is an ordered rectangular array of numbers or functions. The numbers or functions are called the elements or the entries of the matrix. We denote matrices by capital letters. Order of a matrix A matrix having m rows and n columns is called a matrix of order m n or simply m n matrix (read as an m by n matrix). In general, an m n matrix has the following rectangular array: a a... a n a a... a n am am... a mn or A = [a ij ] m n, i m, j n i, j N Thus the ith row consists of the elements a i, a i, a i3,..., a in, while the j th column consists of the elements a j, a j, a 3j,..., a mj, In general a ij, is an element lying in the i th row and j th column. We can also call it as the (i, j) th element of A. The number of elements in an m n matrix will be equal to mn. x or x y y We can also represent any point (x, y) in a plane by a matrix (column or row) as, Types of Matrices (i) Column matrix A matrix is said to be a column matrix if it has only one column. In general, A = [a ij ] m is a column matrix of order m. (ii) Row matrix A matrix is said to be a row matrix if it has only one row. In general, B = [b ij ] n is a row matrix of order n. (iii) Square matrix A matrix in which the number of rows are equal to the number of columns, is said to be a square matrix. Thus an m n matrix is said to be a square matrix if m = n and is known as a square matrix of order n. In general, A = [a ij ] m m is a square matrix of order m. If A = [a ij ] is a square matrix of order n, then elements (entries) a, a,..., a nn are said to constitute the diagonal, of the matrix A. (iv) Diagonal matrix A square matrix B = [b ij ] m m is said to be a diagonal matrix if all its non diagonal elements are zero, that is a matrix B = [b ij ] m m is said to be a diagonal matrix if b ij = 0, when i j. (v) Scalar matrix Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 5 -

56 A diagonal matrix is said to be a scalar matrix if its diagonal elements are equal, that is, a square matrix B = [b ij ] n n is said to be a scalar matrix if b ij = 0, when i j b ij = k, when i = j, for some constant k. (vi) Identity matrix A square matrix in which elements in the diagonal are all and rest are all zero is called an identity if i j matrix. In other words, the square matrix A = [aij] n n is an identity matrix, if aij 0if i j We denote the identity matrix of order n by I n. When order is clear from the context, we simply write it as I. Observe that a scalar matrix is an identity matrix when k =. But every identity matrix is clearly a scalar matrix. (vii) Zero matrix A matrix is said to be zero matrix or null matrix if all its elements are zero. We denote zero matrix by O. Equality of matrices Two matrices A = [a ij ] and B = [b ij ] are said to be equal if (i) they are of the same order (ii) each element of A is equal to the corresponding element of B, that is a ij = b ij for all i and j. Operations on Matrices Addition of matrices The sum of two matrices is a matrix obtained by adding the corresponding elements of the given matrices. Furthermore, the two matrices have to be of the same order. a a a3 b b b3 Thus, if A = is a 3 matrix and B = is another 3 matrix. Then, a a a3 b b b3 we define a b a b a3 b3 A + B =. a b a b a3 b3 In general, if A = [a ij ] and B = [b ij ] are two matrices of the same order, say m n. Then, the sum of the two matrices A and B is defined as a matrix C = [c ij ] m n, where c ij = a ij + b ij, for all possible values of i and j. If A and B are not of the same order, then A + B is not defined. Multiplication of a matrix by a scalar If A = [a ij ] m n is a matrix and k is a scalar, then ka is another matrix which is obtained by multiplying each element of A by the scalar k. In other words, ka = k [a ij ] m n = [k (a ij )] m n, that is, (i, j)th element of ka is ka ij for all possible values of i and j. Negative of a matrix The negative of a matrix is denoted by A. We define A = ( ) A. Difference of matrices If A = [a ij ], B = [b ij ] are two matrices of the same order, say m n, then difference A B is defined as a matrix D = [d ij ], where d ij = a ij b ij, for all value of i and j. In other words, D = A B = A + ( ) B, that is sum of the matrix A and the matrix B. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 5 -

57 Properties of matrix addition (i) Commutative Law If A = [a ij ], B = [b ij ] are matrices of the same order, say m n, then A + B = B + A. (ii) Associative Law For any three matrices A = [a ij ], B = [b ij ], C = [c ij ] of the same order, say m n, (A + B) + C = A + (B + C). (iii) Existence of additive identity Let A = [a ij ] be an m n matrix and O be an m n zero matrix, then A + O = O + A = A. In other words, O is the additive identity for matrix addition. (iv) The existence of additive inverse Let A = [a ij ] m n be any matrix, then we have another matrix as A = [ a ij ] m n such that A + ( A) = ( A) + A= O. So A is the additive inverse of A or negative of A. Properties of scalar multiplication of a matrix If A = [a ij ] and B = [b ij ] be two matrices of the same order, say m n, and k and l are scalars, then (i) k(a +B) = k A + kb, (ii) (k + l)a = k A + l A Multiplication of matrices The product of two matrices A and B is defined if the number of columns of A is equal to the number of rows of B. Let A = [a ij ] be an m n matrix and B = [b jk ] be an n p matrix. Then the product of the matrices A and B is the matrix C of order m p. To get the (i, k) th element c ik of the matrix C, we take the ith row of A and k th column of B, multiply them elementwise and take the sum of all these products. In other words, if A = [a ij ] m n, B = [b jk ] n b k b k p, then the ith row of A is [a i a i... a in ] and the kth column of B is. then. b nk c ik = a i b k + a i b k + a i3 b 3k a in b nk = The matrix C = [c ik ] m p is the product of A and B. If AB is defined, then BA need not be defined. In the above example, AB is defined but BA is not defined because B has 3 column while A has only (and not 3) rows. If A, B are, respectively m n, k l matrices, then both AB and BA are defined if and only if n = k and l = m. In particular, if both A and B are square matrices of the same order, then both AB and BA are defined. Non-commutativity of multiplication of matrices Now, we shall see by an example that even if AB and BA are both defined, it is not necessary that AB = BA. Zero matrix as the product of two non zero matrices We know that, for real numbers a, b if ab = 0, then either a = 0 or b = 0. If the product of two matrices is a zero matrix, it is not necessary that one of the matrices is a zero matrix. Properties of multiplication of matrices The multiplication of matrices possesses the following properties: The associative law For any three matrices A, B and C. We have (AB) C = A (BC), whenever both sides of the equality are defined. The distributive law For three matrices A, B and C. A (B+C) = AB + AC (A+B) C = AC + BC, whenever both sides of equality are defined. The existence of multiplicative identity For every square matrix A, there exist an identity matrix of same order such that IA = AI = A. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

58 Transpose of a Matrix If A = [a ij ] be an m n matrix, then the matrix obtained by interchanging the rows and columns of A is called the transpose of A. Transpose of the matrix A is denoted by A or (A T ). In other words, if A = [a ij ] m n, then A = [a ji ] n m. Properties of transpose of the matrices For any matrices A and B of suitable orders, we have (i) (A ) = A, (ii) (ka) = ka (where k is any constant) (iii) (A + B) = A + B (iv) (A B) = B A Symmetric and Skew Symmetric Matrices A square matrix A = [a ij ] is said to be symmetric if A = A, that is, [a ij ] = [a ji ] for all possible values of i and j. A square matrix A = [a ij ] is said to be skew symmetric matrix if A = A, that is aji = aij for all possible values of i and j. Now, if we put i = j, we have a ii = a ii. Therefore a ii = 0 or a ii = 0 for all i s. This means that all the diagonal elements of a skew symmetric matrix are zero. Theorem For any square matrix A with real number entries, A + A is a symmetric matrix and A A is a skew symmetric matrix. Theorem Any square matrix can be expressed as the sum of a symmetric and a skew symmetric matrix. Elementary Operation (Transformation) of a Matrix There are six operations (transformations) on a matrix, three of which are due to rows and three due to columns, which are known as elementary operations or transformations. (i) The interchange of any two rows or two columns. Symbolically the interchange of ith and jth rows is denoted by R i R j and interchange of ith and jth column is denoted by C i C j. (ii) The multiplication of the elements of any row or column by a non zero number. Symbolically, the multiplication of each element of the ith row by k, where k 0 is denoted by R i k R i. The corresponding column operation is denoted by C i kc i (iii) The addition to the elements of any row or column, the corresponding elements of any other row or column multiplied by any non zero number. Symbolically, the addition to the elements of i th row, the corresponding elements of j th row multiplied by k is denoted by R i R i + kr j. The corresponding column operation is denoted by C i C i + kc j. Invertible Matrices If A is a square matrix of order m, and if there exists another square matrix B of the same order m, such that AB = BA = I, then B is called the inverse matrix of A and it is denoted by A. In that case A is said to be invertible. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

59 A rectangular matrix does not possess inverse matrix, since for products BA and AB to be defined and to be equal, it is necessary that matrices A and B should be square matrices of the same order. If B is the inverse of A, then A is also the inverse of B. Theorem 3 (Uniqueness of inverse) Inverse of a square matrix, if it exists, is unique. Theorem 4 If A and B are invertible matrices of the same order, then (AB) = B A. Inverse of a matrix by elementary operations If A is a matrix such that A exists, then to find A using elementary row operations, write A = IA and apply a sequence of row operation on A = IA till we get, I = BA. The matrix B will be the inverse of A. Similarly, if we wish to find A using column operations, then, write A = AI and apply a sequence of column operations on A = AI till we get, I = AB. In case, after applying one or more elementary row (column) operations on A = IA (A = AI), if we obtain all zeros in one or more rows of the matrix A on L.H.S., then A does not exist. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

60 CHAPTER 3: MATRICES MARKS WEIGHTAGE 03 marks NCERT Important Questions & Answers. If a matrix has 8 elements, what are the possible orders it can have? What, if it has 5 elements? Since, a matrix containing 8 elements can have any one of the following orders : 8, 8, 9, 9, 3 6,6 3 Similarly, a matrix containing 5 elements can have order 5 or 5.. Construct a 3 4 matrix, whose elements are given by: (i) a ij = 3i + j (ii) a ij = i j (i) The order of given matrix is 3 4, so the required matrix is a a a3 a4 A a a a3 a 4, where a ij = 3i + j a3 a3 a33 a Putting the values in place of i and j, we will find all the elements of matrix A. a 3, a 3, a a4 3 4, a 6, a 6 3 a3 6 3, a4 6 4, a a3 9, a , a Hence, the required matrix is A a a a3 a4 (ii) Here, A a a a3 a 4,where a ij = i j a3 a3 a33 a a = =, a = = 0, a 3 = 3 =, a 4 = 4 =, a = 4 = 3, a = 4 =, a 3 = 4 3 =, a 4 = 4 4 = 0 a 3 = 6 = 5, a 3 = 6 = 4, a 33 = 6 3 = 3 and a 34 = 6 4 = Prepared by: M. S. KumarSwamy, TGT(Maths) Page

61 Hence, the required matrix is 0 A a b a c 5 3. Find the value of a, b, c and d from the equation: a b 3c d 0 3 a b a c 5 Given that a b 3c d 0 3 By definition of equality of matrix as the given matrices are equal, their corresponding elements are equal. Comparing the corresponding elements, we get a b = (i) a b = 0 (ii) a + c = 5 (iii) and 3c + d = 3 (iv) Subtracting Eq.(i) from Eq.(ii), we get a = Putting a = in Eq. (i) and Eq. (iii), we get b = and + c = 5 b = and c = 3 Substituting c = 3 in Eq. (iv), we obtain d = 3 d = 3 9 = 4 Hence, a =,b =, c = 3 and d = Find X and Y, if X + Y = 0 9 and X Y = ( X Y ) ( X Y ) X X X Now,( X Y) ( X Y) Y X Y Find the values of x and y from the following equation: x y x y x y Prepared by: M. S. KumarSwamy, TGT(Maths) Page

62 x y or x + 3 = 7 and y 4 = 4 or x = 7 3 and y = 8 or x = 4 and y = 8 i.e. x = and y = Find AB, if A 0 and B We have AB Thus, if the product of two matrices is a zero matrix, it is not necessary that one of the matrices is a zero matrix If A 3, then show that A 3 3A 40 I = O A A. A So, A A. A Now, A 3A 40I If x y 3 5, find the values of x and y. 0 x y x y x y 5 Prepared by: M. S. KumarSwamy, TGT(Maths) Page

63 By definition of equality of matrix as the given matrices are equal, their corresponding elements are equal. Comparing the corresponding elements, we get x y = 0 (i) and 3x + y = 5 (ii) Adding Eqs. (i) and (ii), we get 5x = 5 x = 3 Substituting x = 3 in Eq. (i), we get 3 y = 0 y = 6 0 = 4 x y x 6 4 x y 9. Given 3 z w w z w 3, find the values of x, y, z and w. By definition of equality of matrix as the given matrices are equal, their corresponding elements are equal. Comparing the corresponding elements, we get 3x = x + 4 x = 4 x = and 3y = 6 + x + y y = 6 + x y= 6 x Putting the value of x, we get 6 8 y 4 Now, 3z = + z + w, z = + w w z (i) Now, 3w = w + 3 w = 3 Putting the value of w in Eq. (i), we get 3 z Hence, the values of x, y, z and w are, 4, and 3. cos x sin x 0 0. If F( x) sin x cos x 0, show that F(x) F(y) = F(x + y). 0 0 cos x sin x 0 cos y sin y 0 LHS F( x) F( y) sin x cosx 0 sin y cos y cos x cos y sin xsin y sin y cos x sin x cos y 0 sin x cos y cos xsin y sin xsin y cos x cos y cos( x y) sin( x y) 0 sin( x y) cos( x y) 0 F( x y) RHS Find A 5A + 6I, if A 3 0 Prepared by: M. S. KumarSwamy, TGT(Maths) Page

64 0 0 5 A A. A A 5A 6I If A 0, prove that A 3 6A + 7A + I = A A. A A A. A A 6A 7A I O If A 4 and I 0, find k so that A = ka I Given than A = ka I k k k k k 0 Prepared by: M. S. KumarSwamy, TGT(Maths) Page

65 3k k 4 4 4k k By definition of equality of matrix as the given matrices are equal, their corresponding elements are equal. Comparing the corresponding elements, we get 3k = k = k = k = 4k = 4 k = 4 = k k = Hence, k = 0 tan 4. If A and I is the identity matrix of order, show that tan 0 I + A = (I A) cos sin sin cos 0 x Let A where x tan x 0 tan tan Now, x x cos and cos tan x tan x cos sin RHS ( I A) sin cos x x 0 0 x x x 0 x 0 x x x x x x x x x x( x ) x x x x x x x x x( x ) x x x x x x x 3 3 x x x x x x x( x ) x x x x x 3 3 x x x x x x x( x ) x x x x x 0 0 x x LHS RHS 0 x 0 x 5. Express the matrix matrix. 4 B as the sum of a symmetric and a skew symmetric Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 6 -

66 4 B 3 4 B ' Let P ( B B ') Now P ' 3 3 P 3 3 Thus P ( B B ') is a symmetric matrix Also, let Q ( B B ') Now Q ' 0 3 Q Thus Q ( B B ') is a skew symmetric matrix Now, P Q B Thus, B is represented as the sum of a symmetric and a skew symmetric matrix. 6. Express the following matrices as the sum of a symmetric and a skew symmetric matrix: ( i) ( ii) 3 ( iii) ( ) iv (i) 3 5 Let A, then A P Q where, P ( A A') and Q ( A A') Now, P ( A A') P ' 3 P 3 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 6 -

67 Thus P ( A A ') is a symmetric matrix Now, Q ( A A') Q ' 0 Q Thus Q ( A A ') is a skew symmetric matrix. Representing A as the sum of P and Q, P Q 3 A 0 (ii) 6 Let A 3, then A P Q 3 where, P ( A A') and Q ( A A') Now, P ( A A') P ' 3 3 P 3 3 Thus P ( A A ') is a symmetric matrix Now, Q ( A A') Q ' Q Thus Q ( A A ') is a skew symmetric matrix. Representing A as the sum of P and Q, P Q A (iii) 3 3 Let A, then A P Q 4 5 where, P ( A A') and Q ( A A') Prepared by: M. S. KumarSwamy, TGT(Maths) Page

68 Now, P ( A A') P ' P 5 5 Thus P ( A A ') is a symmetric matrix Now, Q ( A A') Q ' 5 0 Q 3 0 Thus Q ( A A ') is a skew symmetric matrix. Representing A as the sum of P and Q, P Q 5 0 A (iv) 5 Let A, then A P Q where, P ( A A') and Q ( A A') 5 4 Now, P ( A A') P ' P Thus P ( A A ') is a symmetric matrix Now, Q ( A A') Q ' 3 0 Q Thus Q ( A A ') is a skew symmetric matrix. Representing A as the sum of P and Q, Prepared by: M. S. KumarSwamy, TGT(Maths) Page

69 0 3 5 P Q A 3 0 cos sin 7. If A = sin cos and A + A = I, find the value of α. cos sin cos sin A A sin cos sin cos Now, A A' I cos sin cos sin 0 sin cos sin cos 0 cos cos 0 Comparing the corresponding elements of the above matrices, we have cos cos cos Obtain the inverse of the following matrix using elementary operations: Write A = I A, i.e., A A (applying R R ) A (applying R 3 R 3 3R ) A (applying R R R ) A (applying R 3 R 3 + 5R ) A (applying R R 3) 0 A 3 3 Prepared by: M. S. KumarSwamy, TGT(Maths) Page

70 A (applying R R + R 3 ) A (applying R R R 3 ) A Using elementary transformations, find the inverse of 6 3, if it exists. 6 3 Let A We know that A = IA A 0 6 A Using R R A Using R R R Now, in the above equation, we can see all the elements are zero in the second row of the matrix on the LHS. Therefore, A does not exist. Note Suppose A = IA, after applying the elementary transformation, if any row or column of a matrix on LHS is zero, then A does not exist. 0. Show that the matrix B AB is symmetric or skew-symmetric according as A is symmetric or skew-symmetric. We suppose that A is a symmetric matrix, then A = A Consider (B AB) ={B (AB)} = (AB) (B ) [ (AB) = B A ] = B A (B) [ (B ) = B] = B (A B) = B (AB) [ A = A] (B AB) = B AB which shows that B AB is a symmetric matrix. Now, we suppose that A is a skew-symmetric matrix. Then, A = A Consider (B AB) = [B (AB)] = (AB) (B ) [(AB) = B A and (A ) = A] Prepared by: M. S. KumarSwamy, TGT(Maths) Page

71 = (B A )B = B ( A)B= B AB [ A = A] (B AB) = B AB which shows that B AB is a skew-symmetric matrix.. If A and B are symmetric matrices, prove that AB BA is a skew-symmetric matrix. Here, A and Bare symmetric matrices, then A = A and B = B Now, (AB BA) = (AB) (BA) ((A B) = A B and (AB) = B A ) = B A A B = BA AB ( B = Band A = A) = (AB BA) (AB BA) = (AB BA) Thus, (AB BA) is a skew-symmetric matrix Using elementary transformations, find the inverse of 3, if it exists Let A 3. We know that A = IA A A (Using R R + R R 3 ) A (Using R R R and R 3 R 3 3R ) A (Using R R 3 ) A (Using R R and R3 R 3 ) A (Using R R R 3 and R R 4R 3 ) Prepared by: M. S. KumarSwamy, TGT(Maths) Page

72 A (Using R 5 5 R R ) A n n 4n 3. If A, then prove that A n n, where n is any positive integer. We are required to prove that for all n N n 4n P( n) n n () 4() 3 4 Let n =, then P()...( i) () which is true for n =. Let the result be true for n = k. k k 4k P( k) A...( ii) k k Let n = k + k ( k ) 4( k ) k 3 4k 4 P( k ) A k ( k ) k k k k k 4k 3 4 Now, LHS A A A k k ( k).3 ( 4 k). ( k).( 4) ( 4 k)( ) k.3 ( k). k.( 4) ( k)( ) 3 k 4 4k ( k ) 4( k ) k k k ( k ) Therefore, the result is true for n = k + whenever it is true for n = k. So, by principle of mathematical induction, it is true for all n N O? 0 x O 0 x 4. For what values of x : Prepared by: M. S. KumarSwamy, TGT(Maths) Page

73 0 4 0 Since Matrix multiplication is associative, therefore 0 0 x O 0 0 x 4 x O x 4 x x O 4 4x O 4 4x 0 4x 4 x 3 5. If, show that A 5A + 7I = 0. 3 Given that A A A 5A 7I O Find x, if x 5 0 x 0 4 O? 0 3 Given that x 5 0 x 0 4 O 0 3 Since Matrix multiplication is associative, therefore x 5 x O x 0 3 x 5 x 9 O x 3 x( x ) ( 5).9 ( )( x 3) O x 48 O x 48 0 x 48 x Find the matrix X so that X Prepared by: M. S. KumarSwamy, TGT(Maths) Page

74 Given that X The matrix given on the RHS of the equation is a 3matrix and the one given on the LHS of the equation is as a 3 matrix. Therefore, X has to be a matrix. a c Now, let X b d a 4c a 5c 3a 6c b 4d b 5d 3b 6d 4 6 Equating the corresponding elements of the two matrices, we have a + 4c = 7, a + 5c = 8, 3a + 6c = 9 b + 4d =, b + 5d = 4, 3b + 6d = 6 Now, a + 4c = 7 a = 7 4c a + 5c = 8 4 8c + 5c = 8 3c = 6 c = a = 7 4( ) = = Now, b + 4d = b = 4d and b + 5d = 4 4 8d + 5d = 4 3d = 0 d = 0 b = 4(0) = Thus, a =, b =, c =, d = 0 Hence, the required matrix X is 0 cos sin n cos n sin n 8. If A sin cos, then prove that A, n N sin n cos n We shall prove the result by using principle of mathematical induction. cos sin n cos n sin n We have P(n) : If A sin cos, then A, n N sin n cos n cos sin Let n =, then P() A sin cos Therefore, the result is true for n =. Let the result be true for n = k. So k cos k sin k P( k) A sin k cos k Now, we prove that the result holds for n = k + k cos( k ) sin( k ) i.e. P( k ) A sin( k ) cos( k ) k k cos k sin k cos sin Now, P( k ) A A. A sin k cos k sin cos cos cos k sin sin k cos sin k sin cos k sin cos k cos sin k sin sin k cos cosk cos( k ) sin( k ) cos( k ) sin( k ) sin( k ) cos( k ) sin( k ) cos( k ) Therefore, the result is true for n = k +. Thus by principle of mathematical induction, we have n cos n sin n A, n N sin n cos n Prepared by: M. S. KumarSwamy, TGT(Maths) Page

75 Let A, B, C Find a matrix D such that CD AB = O. Since A, B, C are all square matrices of order, and CD AB is well defined, D must be a square matrix of order. a b 5 a b 5 Let D, c d then CD AB 0 O 3 8 c d a 5c b 5d a 8c 3b 8d a 5c 3 b 5d 0 0 3a 8c 43 3b 8d 0 0 By equality of matrices, we get a + 5c 3 = 0... () 3a + 8c 43 = 0... () b + 5d = 0... (3) and 3b + 8d = 0... (4) Solving () and (), we get a = 9, c = 77. Solving (3) and (4), we get b = 0, d = 44. a b 9 0 Therefore D c d By using elementary operations, find the inverse of the matrix A In order to use elementary row operations we may write A = IA. 0 0 A, then A (applying R R R ) 0 A 0 (applying R R ) A 0 (applying R R R ) 5 5 Thus A 5 5 = 5 5 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 7 -

76 CHAPTER 3: MATRICES Previous Years Board Exam (Important Questions & Answers). Use elementary column operation C C C in the matrix equation Given that Applying C C C, we get a 4 3b a b. If a 8b write the value of a b. a 4 3b a b Give that a 8b On equating, we get a + 4 = a +, 3b = b +, a 8b = 6 a =, b = Now the value of a b = ( ) = - = 0 MARKS WEIGHTAGE 03 marks 3. If A is a square matrix such that A = A, then write the value of 7A (I + A) 3, where I is an identity matrix. 7A (I + A) 3 = 7A - {I 3 + 3I A + 3I.A + A 3 } = 7A {I + 3A + 3A + A A} [ I 3 = I = I, A = A] = 7A {I + 6A + A } = 7A {I + 6A + A} = 7A {I + 7A} = 7A I 7A = I x y z 4 4. If x y w 0 5, find the value of x + y. x y z 4 Given that x y w 0 5 Equating, we get x - y = - (i) x - y = 0 (ii) z = 4, w = 5 (ii) (i) x y x + y = 0 + x = and y = x + y = + = 3. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 7 -

77 5. Solve the following matrix equation for x : x 0 Given that x O 0 x x = 0 x = 0 O If y 5 x 0 0 5, find (x y) Given that y 5 x y x y x 0 5 Equating we get 8 + y = 0 and x + = 5 y = 8 and x = x y = + 8 = For what value of x, is the matrix A 0 3 a skew-symmetric matrix? x 3 0 A will be skew symmetric matrix if A = A' 0 0 x 0 x x Equating, we get x = 8. If matrix A and A = ka, then write the value of k. Given A = ka k k k k = Prepared by: M. S. KumarSwamy, TGT(Maths) Page

78 a b a c 5 9. Find the value of a if a b 3c d 0 3 a b a c 5 Given that a b 3c d 0 3 Equating the corresponding elements we get. a b = (i) a + c = 5 (ii) a b = 0 (iii) 3c + d =3 (iv) From (iii) a = b b a b Putting in (i) we get b b b a = (ii) c =5 - =5 - = 3 (iv) d =3 3 (3) =3 9 = 4 i.e. a =, b =, c = 3, d = If A , then find the matrix A. 9 4 Given that A A A 3 6. If A is a square matrix such that A = A, then write the value of (I + A) 3A. (I + A) 3A = I + A + A 3A = I + A - A = I + A - A [ A = A] = I = I. I = I 0. If x y 3 5, write the value of x. 0 Given that x y 3 5 x y 0 3x y 5 x y 0 3x y 5 Prepared by: M. S. KumarSwamy, TGT(Maths) Page

79 Equating the corresponding elements we get. x y = 0...(i) 3x + y = 5...(ii) Adding (i) and (ii), we get x y + 3x + y = x = 5 x = Find the value of x + y from the following equation: x 7 y Given that x 7 y x y x y Equating the corresponding element we get x + 3 = 7 and y 4 = x and y x = and y = 9 x + y = + 9 = 3 4 T 4. If A and B 3, then find A T B T. 0 T Given that B 3 B T T Now, A B If x, write the value of x Given that x x x Equating the corresponding elements, we get x = 3 Prepared by: M. S. KumarSwamy, TGT(Maths) Page

80 cos sin sin cos 6. Simplify: cos sin sin cos cos sin cos sin sin cos cos sin sin cos cos sin cos sin cos sin sin cos sin cos cos sin cos sin cos sin cos sin 0 7. Write the values of x y + z from the following equation: By definition of equality of matrices, we have x + y + z = 9... (i) x + z =5... (ii) y + z =7... (iii) (i) (ii) x + y + z x z = 9 5 y = 4... (iv) (ii) (iv) x y + z = 5 4 x y + z = x y z 9 x z 5 y z 7 y x If x 3 3, then find the value of y. y x Given that x 3 3 By definition of equality of matrices, we have y + x = 7 - x = - x = y + () = 7 y = 3 Prepared by: M. S. KumarSwamy, TGT(Maths) Page

81 Prepared by: M. S. KumarSwamy, TGT(Maths) Page

82 CHAPTER 4: DETERMINANTS QUICK REVISION (Important Concepts & Formulae) Determinant a b If A = c d, then determinant of A is written as A = a b = det (A) or Δ c d (i) For matrix A, A is read as determinant of A and not modulus of A. (ii) Only square matrices have determinants. MARKS WEIGHTAGE 0 marks Determinant of a matrix of order one Let A = [a ] be the matrix of order, then determinant of A is defined to be equal to a Determinant of a matrix of order two a a Let A = a a be a matrix of order, then the determinant of A is defined as: a a det (A) = A = Δ = = aa aa a a Determinant of a matrix of order 3 3 Determinant of a matrix of order three can be determined by expressing it in terms of second order determinants. This is known as expansion of a determinant along a row (or a column). There are six ways of expanding a determinant of order 3 corresponding to each of three rows (R, R and R 3 ) and three columns (C, C and C 3 ) giving the same value as shown below. Consider the determinant of square matrix A = [a ij ] 3 3 a a a i.e., A = 3 a a a 3 a a a Expansion along first Row (R ) Step Multiply first element a of R by ( ) ( + ) [( ) sum of suffixes in a ] and with the second order determinant obtained by deleting the elements of first row (R ) and first column (C ) of A as a lies in R and C, a a 3 i.e., ( ) a a a 3 33 Step Multiply nd element a of R by ( ) + [( ) sum of suffixes in a ] and the second order determinant obtained by deleting elements of first row (R ) and nd column (C ) of A as a lies in R and C, a a 3 i.e., ( ) a a a 3 33 Step 3 Multiply third element a3 of R by ( ) + 3 [( ) sum of suffixes in a3 ] and the second order determinant obtained by deleting elements of first row (R ) and third column (C 3 ) of A as a 3 lies in R and C 3, Prepared by: M. S. KumarSwamy, TGT(Maths) Page

83 i.e., ( ) a a 3 a3 a 3 a 3 Step 4 Now the expansion of determinant of A, that is, A written as sum of all three terms obtained in steps, and 3 above is given by a a3 a a3 a 3 a A ( ) a ( ) a ( ) a3 a a a a a a or A = a (a a 33 a 3 a 3 ) a (a a 33 a 3 a 3 ) + a 3 (a a 3 a 3 a ) Expansion along second row (R ) A = a a a 3 a a a 3 a a a Expanding along R, we get A ( ) a a a ( ) a a a ( ) a a a Expansion along first Column (C ) a3 a33 a3 a33 a3 a3 A = a a a 3 a a a 3 a a a By expanding along C, we get a a a a a a 3 A ( ) a ( ) a ( ) a3 a a a a a a For easier calculations, we shall expand the determinant along that row or column which contains maximum number of zeros. While expanding, instead of multiplying by ( ) i + j, we can multiply by + or according as (i + j) is even or odd. If A = kb where A and B are square matrices of order n, then A = k n B, where n =,, 3 Properties of Determinants Property The value of the determinant remains unchanged if its rows and columns are interchanged. if A is a square matrix, then det (A) = det (A ), where A = transpose of A. If R i = i th row and C i = i th column, then for interchange of row and columns, we will symbolically write C i R i Property If any two rows (or columns) of a determinant are interchanged, then sign of determinant changes. We can denote the interchange of rows by R i R j and interchange of columns by C i C j. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

84 Property 3 If any two rows (or columns) of a determinant are identical (all corresponding elements are same), then value of determinant is zero. Property 4 If each element of a row (or a column) of a determinant is multiplied by a constant k, then its value gets multiplied by k. o By this property, we can take out any common factor from any one row or any one column of a given determinant. o If corresponding elements of any two rows (or columns) of a determinant are proportional (in the same ratio), then its value is zero. Property 5 If some or all elements of a row or column of a determinant are expressed as sum of two (or more) terms, then the determinant can be expressed as sum of two (or more) determinants. Property 6 If, to each element of any row or column of a determinant, the equimultiples of corresponding elements of other row (or column) are added, then value of determinant remains the same, i.e., the value of determinant remain same if we apply the operation R i R i + kr j or C i C i + k C j. If Δ is the determinant obtained by applying R i kr i or C i kc i to the determinant Δ, then Δ = kδ. If more than one operation like R i R i + kr j is done in one step, care should be taken to see that a row that is affected in one operation should not be used in another operation. A similar remark applies to column operations. Area of triangle Area of a triangle whose vertices are (x, y ), (x, y ) and (x 3, y 3 ), is given by the expression x y x y () x y 3 3 Since area is a positive quantity, we always take the absolute value of the determinant in (). If area is given, use both positive and negative values of the determinant for calculation. The area of the triangle formed by three collinear points is zero. Minors and Cofactors Minor of an element a ij of a determinant is the determinant obtained by deleting its i th row and j th column in which element a ij lies. Minor of an element a= is denoted by M ij. Minor of an element of a determinant of order n(n ) is a determinant of order n. Cofactor of an element a ij, denoted by A ij is defined by Aij = ( ) i + j M ij, where M ij is minor of a ij. If elements of a row (or column) are multiplied with cofactors of any other row (or column), then their sum is zero. Adjoint and Inverse of a Matrix The adjoint of a square matrix A = [a ij ] n n is defined as the transpose of the matrix [A ij ] n n, where A ij is the cofactor of the element a ij. Adjoint of the matrix A is denoted by adj A. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

85 a a For a square matrix of order, given by a a The adj A can also be obtained by interchanging a and a and by changing signs of a and a, i.e., Theorem If A be any given square matrix of order n, then A(adj A) = (adj A) A = A I, where I is the identity matrix of order n A square matrix A is said to be singular if A = 0. A square matrix A is said to be non-singular if A 0 Theorem If A and B are nonsingular matrices of the same order, then AB and BA are also nonsingular matrices of the same order. Theorem 3 The determinant of the product of matrices is equal to product of their respective determinants, that is, AB = A B, where A and B are square matrices of the same order If A is a square matrix of order n, then adj(a) = A n. Theorem 4 A square matrix A is invertible if and only if A is nonsingular matrix. Let A is a non-singular matrix then we have A 0 A is invertible and A adja A Applications of Determinants and Matrices Application of determinants and matrices for solving the system of linear equations in two or three variables and for checking the consistency of the system of linear equations: Consistent system A system of equations is said to be consistent if its solution (one or more) exists. Inconsistent system A system of equations is said to be inconsistent if its solution does not exist. Solution of system of linear equations using inverse of a matrix Consider the system of equations a x + b y + c z = d a x + b y + c z = d a 3 x + b 3 y + c 3 z = d 3 a b c x d Let A a b c, X y and B d a3 b3 c 3 z d 3 Then, the system of equations can be written as, AX = B, i.e., Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 8 -

86 a b c x d a b c y d a3 b3 c 3 z d 3 Case I If A is a nonsingular matrix, then its inverse exists. Now AX = B or A (AX) = A B (premultiplying by A ) or (A A) X = A B (by associative property) or I X = A B or X = A B This matrix equation provides unique solution for the given system of equations as inverse of a matrix is unique. This method of solving system of equations is known as Matrix Method. Case II If A is a singular matrix, then A = 0. In this case, we calculate (adj A) B. If (adj A) B O, (O being zero matrix), then solution does not exist and the system of equations is called inconsistent. If (adj A) B = O, then system may be either consistent or inconsistent according as the system have either infinitely many solutions or no solution. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 8 -

87 CHAPTER 4: DETERMINANTS MARKS WEIGHTAGE 0 marks NCERT Important Questions & Answers x 6. If, then find the value of x. 8 x 8 6 x 6 Given that 8 x 8 6 On expanding both determinants, we get x x 8 = x 36 = x 36 = 0 x = 36 x = ± 6. Prove that a a b a b c a 3a b 4a 3b c a 3a 6a 3b 0a 6b 3c Prepared by: M. S. KumarSwamy, TGT(Maths) Page Applying operations R R R and R 3 R 3 3R to the given determinant Δ, we have a a b a b c 0 a a b 0 3a 7a 3b Now applying R 3 R 3 3R, we get a a b a b c 0 a a b 0 0 a Expanding along C, we obtain a a b a 0 0 a( a 0) a( a ) a 0 a b c a a 3 3. Prove that b c a b 4abc c c a b b c a a Let b c a b c c a b Applying R R R R 3 to Δ, we get 0 c b b c a b c c a b Expanding along R, we obtain 0 c a b ( c) b b ( b) b c a c a b c a b c c

88 = c (a b + b bc) b (b c c ac) = abc + cb bc b c + bc + abc = 4 abc 4. If x, y, z are different and We have x x x 3 y y y 3 z z z 3 x x x 3 3 y y y z z z 0 then show that + xyz = 0 3 Now, we know that If some or all elements of a row or column of a determinant are expressed as sum of two (or more) terms, then the determinant can be expressed as sum of two (or more) determinants. 3 x x x x x y y y y y 3 z z z z z 3 x x x x ( ) y y xyz y y (Using C 3 C and then C C ) z z z z ( xyz) y y x z ( xyz) 0 y x y x 0 x x z x z x z x (Using R R R and R 3 R 3 R ) Taking out common factor (y x) from R and (z x) from R 3, we get x x ( xyz)( y x)( z x) 0 y x 0 z x = ( + xyz) (y x) (z x) (z y) (on expanding along C ) Since Δ = 0 and x, y, z are all different, i.e., x y 0, y z 0, z x 0, we get + xyz = 0 a 5. Show that b abc abc bc ca ab a b c c LHS a b c Taking out factors a,b,c common from R, R and R 3, we get Prepared by: M. S. KumarSwamy, TGT(Maths) Page

89 abc a a a b b b c c c Applying R R + R + R 3, we have a b c a b c a b c abc b b b c c c Now applying C C C, C 3 C 3 C, we get 0 0 abc a b c b 0 c 0 abc ( 0) a b c abc abc bc ca ab = RHS a b c 6. Using the property of determinants and without expanding, prove that b c q r y z a p x c a r p z x b q y a b p q x y c r z b c q r y z LHS c a r p z x a b p q x y b c c a a b q r r p p q (interchange row and column) y z z x x y b c c a c q r r p r [usingc 3 C 3 (C + C )] y z z x z b c c a c ( ) q r r p r (taking common from C 3 ) y z z x z Prepared by: M. S. KumarSwamy, TGT(Maths) Page

90 b a c ( ) q p r (using C C C 3 and C C C 3 ) y x z a b c p q r (using x y z C C ) a p x b q y RHS (interchange row and column) c r z 7. Using the property of determinants and without expanding, prove that a ab ac ba b bc 4a b c ca cb c a ab ac a b c LHS ba b bc abc a b c R 3 ] ca cb c a b c ( abc)( abc) a b c Expanding corresponding to first row R, we get 0 a b c [taking out factors a from R, b from R and c from (taking out factors a from C, b from C and c from C 3 ) (using R R + R and R R R 3 ) a b c (0 ) 4a b c RHS 8. Using the property of determinants and without expanding, prove that a a b b ( a b)( b c)( c a) c c LHS b b a c a c Applying R R R3 and R R R3, we get 0 a c a c 0 a c ( a c)( a c) 0 b c b c 0 b c ( b c)( b c) c c c c Taking common factors (a c) and (b c) from R and R respectively, we get Prepared by: M. S. KumarSwamy, TGT(Maths) Page

91 0 ( a c) ( a c)( b c) 0 ( b c) c c Now, expanding corresponding to C, we get = (a c) (b c) (b + c a c) = (a b) (b c) (c a) = RHS 9. Using the property of determinants and without expanding, prove that a b c ( a b)( b c)( c a)( a b c) a b c LHS a b c a b c Applying C C C and C C C3, we get 0 0 a b b c c a b b c c a b b c c ( a b)( a ab b ) ( b c)( b bc c ) c 3 Taking common (a b) from C and (b c) from C, we get 0 0 ( a b)( b c) a b b c c ( a ab b ) ( b bc c ) c 3 Now, expanding along R, we get = (a b) (b c) [ (b + bc + c ) (a + ab + b )] = (a b) (b c) [b + bc + c a ab b ] = (a b) (b c) (bc ab + c a ) = (a b) (b c) [b(c a) + (c a) (c + a)] = (a b) (b c) (c a) (a + b + c)= RHS. 0. Using the property of determinants and without expanding, prove that x x yz y y zx x y y z z x xy yz zx z z xy x x yz LHS y y zx z z xy ( )( )( )( ) Applying R xr, R yr and R3 zr3, we have 3 x x xyz 3 y y xyz xyz z z 3 xyz Prepared by: M. S. KumarSwamy, TGT(Maths) Page

92 x xyz y xyz z x 3 3 y (take out xyz common from C 3 ) x z 3 x 3 y x y x 3 3 z x z x 3 3 Expanding corresponding to C 3, we get 3 3 y x y x 3 3 z x z x 0 0 (using R R R and R3 R3 R ) ( y x )( z x ) ( z x )( y x ) = (y + x) (y x) (z x) (z + x + xz) (z + x) (z x) (y x) (y + x + xy) = (y x) (z x) [(y + x) (z + x + xz) (z + x) (y + x + xy)] = (y x)(z x)[yz + yx + xyz + xz + x 3 + x z zy zx xyz xy x 3 x y] = (y x)(z x)[yz zy + xz xy ] = (y x)(z x)[yz(z y) + x(z y )] = (y x)(z x)[yz(z y) + x(z y)(z + y)] = (y x) (z x) [(z y) (xy + yz + zx)] = (x y) (y z) (z x) (xy + yz + zx) = RHS.. Using the property of determinants and without expanding, prove that x 4 x x x x 4 x (5x 4)(4 x) x x x 4 x 4 x x LHS x x 4 x x x x 4 5x 4 x x 5x 4 x 4 x 5x 4 x x 4 x x (5x 4) x 4 x x x 4 x x (5x 4) 0 x x 4 Expanding along C, we get = (5x + 4) {(4 x) (4 x)} (5x 4)(4 x) = RHS. (using C C + C + C 3 ) [take out (5x + 4) common from C ]. (Using R R R and R 3 R 3 R ). Using the property of determinants and without expanding, prove that y k y y y y k y k (3 y k) y y y k Prepared by: M. S. KumarSwamy, TGT(Maths) Page

93 y k y y LHS y y k y y y y k 3y k y y 3y k y k y 3y k y y k (3 y k) y k y y y y y y k (using C C + C + C 3 ) [take out (5x + 4) common from C ]. (3 y k) 0 k 0 (Using R R R and R 3 R 3 R ) 0 0 Expanding along C 3, we get (3 y k) ( k 0) k (3 y k) = RHS y k 3. Using the property of determinants and without expanding, prove that a b c a a b b c a b ( a b c) c c c a b a b c a a LHS b b c a b c c c a b a b c a b c a b c b b c a b (Using R R R R ) c c c a b Take out (a + b + c) common from R, we get ( a b c) b b c a b c c c a b 0 0 ( a b c) b b c a 0 c 0 c a b Expanding along R, we get = (a + b + c) {( b c a) ( c a b)} = (a + b + c) [ (b + c + a) ( ) (c + a + b)] 3 ( a b c)( a b c)( a b c) ( a b c) = RHS 3 3 (Using C C C and C 3 C 3 C ) Prepared by: M. S. KumarSwamy, TGT(Maths) Page

94 4. Using the property of determinants and without expanding, prove that x y z x y z y z x y ( x y z) z x z x y x y z x y LHS z y z x y z x z x y ( x y z) x y ( x y z) y z x y (using C C + C + C 3 ) ( x y z) x z x y x y ( x y z) y z x y [take out (x + y + z) common from C ]. x z x y x y ( x y z) 0 y z x 0 (Using R R R and R 3 R 3 R ) 0 0 z x y x y ( x y z)( x y z)( x y z) Expanding along R 3, we get ( x y z)( x y z)( x y z) ( 0) 3 ( x y z)( x y z)( x y z) ( x y z) =RHS 3 5. Using the property of determinants and without expanding, prove that x x x x x x LHS x x x x x x x x x x x (using C C + C + C 3 ) x x x x x ( x x ) x [take out x x ( x x ) 0 x x x x 0 x x x 3 ( x x ) common from C ]. (Using R R R and R 3 R 3 R ) x x x Prepared by: M. S. KumarSwamy, TGT(Maths) Page

95 ( x x ) 0 x x( x) x x 0 x( x ) x Take out ( x) common from R and same from R 3, we get x x ( x x )( x)( x) 0 x Expanding along C, we get ( x x )( x)( x) x 0 x x x x ( x x )( x)( x)( x x ) x 3 x 3 x 3 = RHS 6. Using the property of determinants and without expanding, prove that a b ab b ab a b a ( a b ) 3 b a a b a b ab b LHS ab a b a b a a b 0 a b b 0 a b a b( a b ) a( a b ) a b 0 b ( a b ) 0 a b a a b 0 b ( a b ) 0 a 0 0 a b Expanding along R, we get ( a b ) ( a b ) 3 ( a b ) RHS ( R R br ar ) 3 3 (Using C C bc3 and C C ac3 ) 7. Using the property of determinants and without expanding, prove that a ab ac ab b bc a b c ca cb c a ab ac LHS ab b bc ca cb c Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 9 -

96 Taking out common factors a, b and c from R, R and R 3 respectively, we get a a b c a b b c a b c c a a b c a b 0 (Using R R R and R 3 R 3 R ) 0 a c Multiply and divide C by a, C by b and C 3 by c and then take common out from C, C and C 3 respectively, we get a b c a b c abc 0 0 abc 0 Expanding along R 3, we get ( c ) ( a ) ( b ) a b c RHS 0 8. Find values of k if area of triangle is 4 sq. units and vertices are (i) (k, 0), (4, 0), (0, ) (ii) (, 0), (0, 4), (0, k) k 0 (i) We have Area of triangle = k(0 ) + (8 0) = 8 k(0 ) + (8 0) = ± 8 On taking positive sign k + 8 = 8 k = 0 k = 0 On taking negative sign k + 8 = 8 k = 6 k = 8 k =0, 8 0 (ii) We have Area of triangle = k (4 k) + (0 0) = 8 (4 k) + (0 0) = ± 8 [ 8 + k] = ± 8 On taking positive sign, k 8 = 8 k = 6 k = 8 On taking negative sign, k 8 = 8 k = 0 k = 0 k =0, 8 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 9 -

97 9. If area of triangle is 35 sq units with vertices (, 6), (5, 4) and (k, 4). Then find the value of k. 6 We have Area of triangle = k 4 (4 4) + 6(5 k) + (0 4k) = 70 (4 4) + 6 (5 k) + (0 4k) = ± k + 0 4k = ± 70 On taking positive sign, 0k + 50 = 70 0k = 0 k = On taking negative sign, 0k + 50 = 70 0k = 0 k = k =, 0. Using Cofactors of elements of second row, evaluate Given that Cofactors of the elements of second row 3 8 A ( ) (9 6) 7 3 A ( ) (5 8) and A3 ( ) (0 3) 7 Now, expansion of Δ using cofactors of elements of second row is given by aa a A a3 A3 = ( 7) = 4 7 = 7 Prepared by: M. S. KumarSwamy, TGT(Maths) Page If A =, show that A 5A + 7I = O. Hence find A. 3 Given that A = Now, A 5A + 7I = O A A. A O

98 A 5A + 7I = O 3 A A exists. Now, A.A 5A = 7I Multiplying by A on both sides, we get A.A (A ) 5A(A ) = 7I(A ) AI 5I = 7A (using AA = I and IA = A ) A ( A 5 I) 5I A A For the matrix A =, find the numbers a and b such that A + aa + bi = O. 3 Given that A = A A. A Now, A aa bi O a b O a a b 0 O 4 3 a a 0 b 3a b 8 a a 3 a b 0 0 If two matrices are equal, then their corresponding elements are equal. + 3a + b = 0 (i) 8 + a = 0 (ii) 4 + a = 0 (iii) and 3 + a + b = 0 (iv) Solving Eqs. (iii) and (iv), we get 4 + a = 0 a = 4 and 3 + a + b = b = 0 b = Thus, a = 4 and b = 3. For the matrix A = 3, Show that A 3 6A + 5A + I = O. Hence, find A. 3 Prepared by: M. S. KumarSwamy, TGT(Maths) Page

99 Given that A = A A. A and A A. A A 6A 5A I O A 3 (6 3) (3 6) ( 4) A exist 3 Now, A 6A 5A I O AA( AA ) 6 A( AA ) 5( AA ) ( IA ) O AAI 6AI 5I A O A 6A 5I A ( A A 6 A 5 I ) ( A A 6 A 5 I ) A Prepared by: M. S. KumarSwamy, TGT(Maths) Page

100 A A A Solve system of linear equations, using matrix method, x + y + z = x y z = 3 3y 5z = 9 The given system can be written as AX = B, where x A 4, X y and B z 9 A 4 (0 6) ( 0 0) (6 0) = = 68 0 Thus, A is non-singular, Therefore, its inverse exists. Therefore, the given system is consistent and has a unique solution given by X = A B. Cofactors of A are A = = 6, A = ( 0 + 0) = 0, A 3 = = 6 A = ( 5 3) = 8, A = 0 0 = 0, A 3 = (6 0) = 6 A 3 = ( + 4) =, A 3 = ( 4 ) = 6, A 33 = 8 = adj( A) A ( adja) A x 6 8 Now, X A B y z T Prepared by: M. S. KumarSwamy, TGT(Maths) Page

101 x y z Hence, x, y and z 5. Solve system of linear equations, using matrix method, x y + z = 4 x + y 3z = 0 x + y + z = The given system can be written as AX = B, where x 4 A 3, X y and B 0 z Here, A 3 = ( + 3) ( ) ( + 3) + ( ) = = 0 0 Thus, A is non-singular, Therefore, its inverse exists. Therefore, the given system is consistent and has a unique solution given by X = A B. Cofactors of A are A = + 3 = 4, A = ( + 3) = 5, A 3 = =, A = ( ) =, A = = 0, A 3 = ( + ) =, A 3 = 3 =, A 3 = ( 3 ) = 5, A 33 = + = 3 T adj( A) A ( adja) A 0 3 x 4 4 Now, X A B y z 3 x y z Hence, x =, y = and z =. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

102 6. Solve system of linear equations, using matrix method, x + 3y +3 z = 5 x y + z = 4 3x y z = 3 The given system can be written as AX = B, where 3 3 x 5 A, X y and B 4 3 z Here, A 3 = (4 + ) 3 ( 3) + 3 ( + 6) = = 40 0 Thus, A is non-singular. Therefore, its inverse exists. Therefore, the given system is consistent and has a unique solution given by X = A B Cofactors of A are A = 4 + = 5, A = ( 3) = 5, A 3 = ( + 6) = 5, A = ( 6 + 3) = 3, A = ( 4 9) = 3, A 3 = ( 9) =, A 3 = = 9, A 3 = ( 3) =, A 33 = 4 3 = 7 T adj( A) A ( adja) 5 3 A x Now, X A B y z x y z Hence, x =, y = and z =. 7. Solve system of linear equations, using matrix method, x y + z = 7 3x + 4y 5z = 5 x y + 3z = The given system can be written as AX = B, where Prepared by: M. S. KumarSwamy, TGT(Maths) Page

103 x 7 A 3 4 5, X y and B 5 3 z Here, A = ( 5) ( ) (9 + 0) + ( 3 8) 3 = = 4 0 Thus, A is non-singular. Therefore, its inverse exists. Therefore, the given system is consistent and has a unique solution given by X = A B Cofactors of A are A = 5 = 7, A = (9 + 0) = 9, A 3 = 3 8 =, A = ( 3 + ) =, A = 3 4 =, A 3 = ( + ) =, A 3 = 5 8 = 3, A 3 = ( 5 6) =, A 33 = = 7 T adj( A) A ( adja) 9 A 4 7 x Now, X A B y z 7 3 x y z Hence, x =, y = and z = If A = 3 4 find A. Using A,Solve system of linear equations: x 3y + 5z = 3x + y 4z = 5 x + y z = 3 The given system can be written as AX = B, where 3 5 x A 3 4, X y and B 5 z 3 Prepared by: M. S. KumarSwamy, TGT(Maths) Page

104 3 5 Here, A 3 4 = ( 4 + 4) ( 3) ( 6 + 4) + 5 (3 ) = = 0 Thus, A is non-singular. Therefore, its inverse exists. Therefore, the given system is consistent and has a unique solution given by X = A B Cofactors of A are A = = 0, A = ( 6 + 4) =, A 3 = 3 =, A = (6 5) =, A = 4 5 = 9, A 3 = ( + 3) = 5, A 3 = ( 0) =, A 3 = ( 8 5) = 3, A 33 = = 3 T 0 0 adj( A) A ( adja) A x 0 Now, X A B y z x y z Hence, x =, y = and z = The cost of 4 kg onion, 3 kg wheat and kg rice is Rs 60. The cost of kg onion, 4 kg wheat and 6 kg rice is Rs 90. The cost of 6 kg onion kg wheat and 3 kg rice is Rs 70. Find cost of each item per kg by matrix method. Let the prices (per kg) of onion, wheat and rice be Rs. x, Rs. y and Rs. z, respectively then 4x + 3y + z = 60, x + 4y + 6z = 90, 6x + y + 3z = 70 This system of equations can be written as AX = B, where 4 3 x 60 A 4 6, X y and B z Here, A 4 6 = 4( ) 3(6 36) + (4 4) 6 3 = = 50 0 Thus, A is non-singular. Therefore, its inverse exists. Therefore, the given system is consistent and has a unique solution given by X = A B Cofactors of A are, Prepared by: M. S. KumarSwamy, TGT(Maths) Page

105 A = = 0, A = (6 36) = 30, A 3 = 4 4 = 0, A = (9 4) = 5, A = = 0, A 3 = (8 8) = 0, A 3 = (8 8) = 0, A 3 = (4 4) = 0, A 33 = 6 6 = 0 T adj( A) A ( adja) A x Now, X A B y z x y z x = 5, y = 8 and z = 8. Hence, price of onion per kg is Rs. 5, price of wheat per kg is Rs. 8 and that of rice per kg is Rs Without expanding the determinant, prove that a a bc LHS b b ca c c ab Applying R ar, R br and R3 cr3, we get 3 a a abc 3 b b abc abc c c 3 abc a abc b abc c 3 3 b [Taking out factor abc from C 3 ] a c 3 3 a a bc a a 3 b b ca b b 3 c c ab c c 3 ( ) b b (using C C 3 and C C 3 ) a c a c 3 3 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 0 -

106 3 a a 3 b b RHS c c 3 b c c a a b 3. If a, b and c are real numbers, and c a a b b c 0. Show that either a + b + c = 0 or a = b = c. b c c a a b c a a b b c a b b c c a ( a b c) c a a b ( a b c) a b b c ( a b c) b c c a ( a b c) a b b c c a a b b c c a ( a b c) 0 b c c a 0 c a a b b a c b Expanding along C, we get ( ) b a b c c c a b a c b ( a b c) ( b c)( c b) ( c a)( b a) a b b c c a ( a b c) bc b c bc ( bc ac ab a ) ( a b c) bc b c bc bc ac ab a ( a b c) ab bc ac a b c It is given that Δ= 0, ( a b c) ab bc ac a b c 0 Either a b c 0 or ab bc ac a b c 0 ab bc ac a b c 0 ab bc ac a b c 0 a b c ab bc ac 0 a b ab b c bc c a ac 0 ( a b) ( b c) ( c a) 0 (using C C + C + C 3 ) [take out ( a b c) common from C ]. (Using R R R and R 3 R 3 R ) ( a b) ( b c) ( c a) 0 [since square of any real number is never negative] ( a b) ( b c) ( c a) 0 a b, b c, c a a b c Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 0 -

107 3. Prove that a bc ac c a ab b ac 4a b c ab b bc c a bc ac c LHS a ab b ac ab b bc c Taking out a from C, b from C and c from C 3, we get a b a c abc a b b a 0 b b c c c a c abc b b a b b c c 0 c a c abc 0 c a c b b c c Expanding along C,we get = (abc) [ ( b) { c(a c) + c (a + c) } ] = (ab c) (ac) = 4 a b c = RHS. [Using C C + C C 3] [Using R R R 3] 33. Using properties of determinants, prove that LHS (using C 3 C 3 + C ) ( ) (Taking out (α + β + γ) common from C ) ( ) 0 0 (Using R R R and R 3 R 3 R ) ( )( )( )( ) Expanding along C 3, we get = (α + β + γ) [(β α)(γ α ) (γ α)(β α )] = (α + β + γ) [(β α)(γ α)(γ + α) (γ α)(β α)(β + α)] = (α + β + γ) (β α)(γ α)[γ + α β α] = (α + β + γ) (β α)(γ α)(γ β) = (α + β + γ) (α β)(β γ)(γ α) = RHS Prepared by: M. S. KumarSwamy, TGT(Maths) Page

108 34. Using properties of determinants, prove that 3a a b a c LHS b a 3b b c c a c b 3c a b c a b a c a b c 3b b c a b c c b 3c a b a c ( a b c) 3b b c c b 3c Now applying R R R, R 3 R 3 R, we get a b a c ( a b c) 0 b a a b 0 a c c a 3a a b a c b a 3b b c 3( a b c)( ab bc ca) c a c b 3c (using C C + C + C 3 ) (Taking out (a + b + c) common from C ) Expanding along C, we get = (a + b + c)[(b + a) (c + a) (a b) (a c)] = (a + b + c)[4bc + ab + ac + a a + ac + ba bc] = (a + b + c) (3ab + 3bc + 3ac) = 3(a + b + c)(ab + bc + ca) = RHS 35. Solve the system of equations: x y z x y z x y z Let p, q and r, then the given equations become x y z p + 3q + 0r = 4, 4p 6q + 5r =, 6p + 9q 0r = This system can be written as AX = B, where 3 0 p 4 A 4 6 5, X q and B r 3 0 Here, A (0 45) 3( 80 30) 0(36 36) = = 00 0 Thus, A is non-singular. Therefore, its inverse exists. Therefore, the above system is consistent and has a unique solution given by X = A B Cofactors of A are A = 0 45 = 75, Prepared by: M. S. KumarSwamy, TGT(Maths) Page

109 A = ( 80 30) = 0, A 3 = ( ) = 7, A = ( 60 90) = 50, A = ( 40 60) = 00, A 3 = (8 8) = 0, A 3 = = 75, A 3 = (0 40) = 30, A 33 = = adj( A) A ( adja) A x X A B y z x y z p, q, r 3 5,, x y 3 z 5 x =, y = 3 and z = If a, b, c, are in A.P, then find the determinant of x x 3 x a Let A x 3 x 4 x b x 4 x 5 x c x x 3 x a 0 0 ( b a c) x 4 x 5 x c T x x 3 x a x 3 x 4 x b x 4 x 5 x c (using R R R R 3 ) But a,b, c are in AP. Using b = a + c, we get x x 3 x a A [Since, all elements of R are zero] x 4 x 5 x c Prepared by: M. S. KumarSwamy, TGT(Maths) Page

110 3 37. Show that the matrix A satisfies the equation A 4A + I = O, where I is identity matrix and O is zero matrix. Using this equation, find A. 3 Given that A A AA Hence, A 4A I O 0 0 Now, A 4A I O AA 4A I AA( A ) 4AA IA (Post multiplying by A because A 0) A( AA ) 4I A AI 4I A A 4I A A 38. Solve the following system of equations by matrix method. 3x y + 3z = 8 x + y z = 4x 3y + z = 4 The system of equation can be written as AX = B, where 3 3 x 8 A, X y and B 4 3 z Here, A 4 3 = 3 ( 3) + (4 + 4) + 3 ( 6 4) = 7 0 Hence, A is nonsingular and so its inverse exists. Now A =, A = 8, A 3 = 0 A = 5, A = 6, A 3 = A 3 =, A 3 = 9, A 33 = adj( A) A ( adja) A T Prepared by: M. S. KumarSwamy, TGT(Maths) Page

111 x 5 8 X A B y z x 7 y 34 7 z 5 3 Hence x =, y = and z = 3. Given that ( ) y z xy zx 39. Show that xy ( x z) yz 3 xyz( x y z) xz yz ( x y) ( y z) xy zx xy ( x z) yz xz yz ( x y) Applying R xr, R yr,r 3 zr 3 to Δ and dividing by xyz, we get x( y z) x y x z xy y( x z) y z xyz xz yz z( x y) Taking common factors x, y, z from C, C and C 3 respectively, we get ( y z) x x xyz y ( x z) y xyz z z ( x y) Applying C C C, C 3 C 3 C, we have ( y z) x ( y z) x ( y z) y ( x z) y 0 z 0 ( x y) z Taking common factor (x + y + z) from C and C 3, we have ( y z) x ( y z) x ( y z) ( x y z) y ( x z) y 0 z 0 ( x y) z Applying R R (R + R 3 ), we have yz z y ( x y z) y x y z 0 z 0 x y z Applying C (C + y C ) and C 3 C 3 + z C, we get Prepared by: M. S. KumarSwamy, TGT(Maths) Page

112 yz 0 0 ( ) x y z y x z z y y z z x y Finally expanding along R, we have Δ = (x + y + z) (yz) [(x + z) (x + y) yz] = (x + y + z) (yz) (x + xy + xz) = (x + y + z) 3 (xyz) Use product to solve the system of equations x y + z = y 3z = 3x y + 4z = 0 Consider the product Hence, Now, given system of equations can be written, in matrix form, as follows x 0 3 y 3 4 z x 0 y z Hence x = 0, y = 5 and z = 3 Prepared by: M. S. KumarSwamy, TGT(Maths) Page

113 CHAPTER 3: DETERMINANTS Previous Years Board Exam (Important Questions & Answers) MARKS WEIGHTAGE 0 marks 9. Let A be a square matrix of order 3 3. Write the value of A, where A = 4. Since A = n A where n is order of matrix A. Here A = 4 and n = 3 A = 3 4 = 3 0. Write the value of the following determinant: Given that Applying R R 6R (Since R is zero) If A is a square matrix and A =, then write the value of AA', where A' is the transpose of matrix A. AA ' = A. A' = A. A = A = x = 4. [since, AB = A. B and A = A', where A and B are square matrices.]. If A is a 3 3 matrix, A 0 and 3A = k A, then write the value of k. Here, 3A = k A 3 3 A = k A [ ka = kn A where n is order of A] 7 A = k A k = 7 a ib c id 3. Evaluate: c id a ib a ib c id ( a ib)( a ib) ( c id)( c id) c id a ib ( a ib)( a ib) ( c id)( c id) a i b c i d a b c d x 3 4. If 3, find the value of x. x 5 4 Prepared by: M. S. KumarSwamy, TGT(Maths) Page

114 x 3 Given that 3 x 5 4 4x + 8 3x 5 = 3 x 7 = 3 x = 0 5. If = , write the minor of the element a 3. 3 Minor of a 3 = 5 3 = 0 3 = Evaluate: cos5 sin 75 sin5 0 0 cos Expanding the determinant, we get cos 5. cos 75 - sin 5. sin 75 = cos ( ) = cos 90 = 0 [since cos (A + B) = cos A. cos B sina. sin B] 7. Using properties of determinants, prove the following: a a b a b Let a b a a b a b a b a Applying R R + R + R 3, we have 3( a b) 3( a b) 3( a b) a b a a b a b a b a Taking out 3(a + b) from st row, we have 3( a b) a b a a b a b a b a a a b a b a b a a b b a b 9 ( ) a b a b a Applying C C C and C C C ( a b) b b a b b b a Expanding along first row, we have D = 3(a + b) [. (4b b )] = 3 (a + b) 3b = 9b (a + b) Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 0 -

115 8. Write the value of the determinant Given determinant A = x = 0 ( R = R 3 ) 3 4 6x 9x x x 9x x 9. Two schools P and Q want to award their selected students on the values of Tolerance, Kindness and Leadership. The school P wants to award Rs. x each, Rs. y each and Rs. z each for the three respective values to 3, and students respectively with a total award money of Rs.,00. School Q wants to spend Rs. 3,00 to award its 4, and 3 students on the respective values (by giving the same award money to the three values as school P). If the total amount of award for one prize on each value is Rs.,00, using matrices, find the award money for each value. Apart from these three values, suggest one more value that should be considered for award. According to question, 3x + y + z = 00 4x + y + 3z = 300 x + y + z = 00 The above system of equation may be written in matrix form as AX = B 3 x 00 A 4 3, X y and B 300 z 00 3 Here, A 4 3 3( 3) (4 3) (4 ) A exists. Now, A = ( 3) =, A = (4 3) =, A 3 = (4 ) = 3, A = ( ) =, A = (3 ) =, A 3 = (3 ) = A 3 = (6 ) = 5, A 3 = (9 4) = 5, A 33 = (3 8) = 5 T 3 5 adj( A) A ( adja) 5 5 A Prepared by: M. S. KumarSwamy, TGT(Maths) Page - -

116 x 5 00 Now, X A B y z x y z x = 300, y = 400, z = 500 i.e., Rs. 300 for tolerance, Rs. 400 for kindness and Rs. 500 for leadership are awarded. One more value like punctuality, honesty etc may be awarded. 30. Using properties of determinants, prove that a x y z LHS x a y z x y a z Applying C C + C + C 3, we get a x y z y z a x y z a y z a x y z y a z Apply R R R, we get y z ( a x y z) a y z y a z 0 a 0 ( a x y z) a y z y a z Expanding along R, we get = (a + x + y + z) {0 + a (a + z z)} = a (a + x + y + z) = RHS 3. 0 students were selected from a school on the basis of values for giving awards and were divided into three groups. The first group comprises hard workers, the second group has honest and law abiding students and the third group contains vigilant and obedient students. Double the number of students of the first group added to the number in the second group gives 3, while the combined strength of first and second group is four times that of the third group. Using matrix method, find the number of students in each group. Apart from the values, hard work, honesty and respect for law, vigilance and obedience, suggest one more value, which in your opinion, the school should consider for awards. Let no. of students in Ist, nd and 3rd group to x, y, z respectively. From the statement we have x + y+ z = 0 x + y =3 x + y 4z = 0 The above system of linear equations may be written in matrix form as AX = B where x 0 A 0, X y and B 3 4 z 0 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - -

117 Here, A 0 ( 4 0) ( 8 0) ( ) A exists. Now, A = 4 0 = 4 A = ( 8 0) = 8 A 3 = = A = ( 4 ) = 5 A = 4 = 5 A 3 = ( ) = 0 A 3 = 0 = A 3 = (0 ) = A 33 = = adj( A) A ( adja) 8 5 A 5 0 x Now, X A B y z 0 0 x y z 0 0 x = 5, y = 3, z = T 3. The management committee of a residential colony decided to award some of its members (say x) for honesty, some (say y) for helping others and some others (say z) for supervising the workers to keep the colony neat and clean. The sum of all the awardees is. Three times the sum of awardees for cooperation and supervision added to two times the number of awardees for honesty is 33. If the sum of the number of awardees for honesty and supervision is twice the number of awardees for helping others, using matrix method, find the number of awardees of each category. Apart from these values, namely, honesty, cooperation and supervision, suggest one more value which the management of the colony must include for awards. According to question x + y + z = x + 3y + 3z = 33 x y + z = 0 The above system of linear equation can be written in matrix form as AX = B where x A 3 3, X y and B 33 z 0 Here, A 3 3 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 3 -

118 = (3 + 6) ( 3) + ( 4 3) = = 3 A exists. A = 9, A =, A 3 = 7 A = 3, A = 0, A 3 = 3 A 3 = 0, A 3 =, A 33 = T adj( A) A ( adja) 0 A x Now, X A B y z x y z x = 3, y = 4, z = 5 No. of awards for honesty = 3 No. of awards for helping others = 4 No. of awards for supervising = 5. The persons, who work in the field of health and hygiene should also be awarded. 33. Using properties of determinants, prove the following: 3x x y x z x y 3y z y 3( x y z)( xy yz zx) x z y z 3z 3x x y x z LHS x y 3y z y x z y z 3z Applying C C + C + C 3 x y z x y x z x y z 3y z y x y z y z 3z Taking out (x + y + z) along C, we get x y x z ( x y z) 3y z y y z 3z Applying R R R ; R 3 R 3 R x y x z ( x y z) 0 y x x y 0 x z x z Applying C C C 3 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 4 -

119 y z x z ( x y z) 0 3y x y 0 3z x z Expanding along I column, we get D = (x + y + z)[(3y (x + z) + 3z (x y)] = 3(x + y + z)[xy + z + yz + xz yz] = 3(x + y + z)(xy + yz + zx) = R.H.S. 34. A school wants to award its students for the values of Honesty, Regularity and Hard work with a total cash award of Rs. 6,000. Three times the award money for Hardwork added to that given for honesty amounts to `,000. The award money given for Honesty and Hardwork together is double the one given for Regularity. Represent the above situation algebraically and find the award money for each value, using matrix method. Apart from these values, namely, Honesty, Regularity and Hardwork, suggest one more value which the school must include for awards. Let x, y and z be the awarded money for honesty, Regularity and hardwork. From the statement x + y +z = 6000 (i) x + 3z =000 (ii) x +z = y x y +z = 0 (iii) The above system of three equations may be written in matrix form as AX = B, where x 6000 A 0 3, X y and B z 0 Here, A 0 3 (0 6) ( 3) ( 0) Hence A exist If Aij is co-factor of aij then A = = 6 A = ( 3) = ; A 3 = ( 0) = A = ( + ) = 3 A = 0 A 3 = ( ) = 3 A 3 = 3 0 = 3 A 3 = (3 ) = ; A 33 = 0 = T adj( A) A ( adja) 0 A 6 3 x Now, X A B y z 3 0 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 5 -

120 x y z x =500, y = 000, z = 3500 Except above three values, school must include discipline for award as discipline has great importance in student s life. x x If, then write the value of x. x 3 x 3 x x 4 Given that x 3 x 3 (x +) (x + ) (x )(x 3) = + x + x + x + x + 3x + x 3 =3 7x =3 7x =4 x = 36. Using properties of determinants, prove that a a b a b c LHS a 3a b 4a 3b c 3a 6a 3b 0a 6b 3c a a a b c a b a b c a 3a 4a 3b c a b 4a 3b c 3a 6a 0a 6b 3c 3a 3b 0a 6b 3c a b c a b c a 3 4a 3b c ab 4a 3b c 3 6 0a 6b 3c 3 3 0a 6b 3c a b c a 3 4a 3b c ab a 6b 3c a b c a 3 4a 3b c 3 6 0a 6b 3c a b c a 3 4a 3 3b 3 c 3 6 0a 3 6 6b 3 6 3c a. a 3 4 b 3 3 c a a b a b c a 3a b 4a 3b c a 3a 6a 3b 0a 6b 3c [since C = C in second determinant] Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 6-3

121 a. a 3 4 b.0 c.0 [since C = C 3 in second determinant and C = C 3 in third determinant] a Applying C C C and C 3 C 3 C we get 0 0 a Expanding along R we get = a 3.(7 6) = a Using matrices, solve the following system of equations: x y + z = 4; x + y 3z = 0; x + y + z = Given equations x y + z = 4 x + y 3z = 0 x + y + z = We can write this system of equations as AX = B where x 4 A 3, X y and B 0 z Here, A 3 = ( + 3) - (- ) ( + 3) + ( - ) = = 0 A exists. A = 4, A = 5, A 3 = A =, A = 0, A 3 = A 3 =, A 3 = 5, A 33 = 3 T adj( A) A ( adja) A 0 3 x 4 4 Now, X A B y z 3 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 7 -

122 x y z The required solution is x =, y = -, z = If A = and B = 3 0, find (AB). 5 0 For B B i.e., B is invertible matrix B exist. A = 3, A =, A 3 = A =, A =, A 3 = A 3 = 6, A 3 =, A 33 = 5 = (3 0) ( 0) ( 0 ) = = 0 T adj( B) B ( adjb) B 5 5 Now (AB) = B. A Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 8 -

123 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 9 -

124 CHAPTER 0: VECTOR ALGEBRA QUICK REVISION (Important Concepts & Formulae) MARKS WEIGHTAGE 06 marks Vector The line l to the line segment AB, then a magnitude is prescribed on the line l with one of the two directions, so that we obtain a directed line segment. Thus, a directed line segment has magnitude as well as direction. A quantity that has magnitude as well as direction is called a vector. A directed line segment is a vector, denoted as AB or simply as a, and read as vector AB or vector a. The point A from where the vector AB starts is called its initial point, and the point B where it ends is called its terminal point. The distance between initial and terminal points of a vector is called the magnitude (or length) of the vector, denoted as AB or a. The arrow indicates the direction of the vector. The vector OP having O and P as its initial and terminal points, respectively, is called the position vector of the point P with respect to O. Using distance formula, the magnitude of vector OP is given by OP x y z The position vectors of points A, B, C, etc., with respect to the origin O are denoted by a, b and c, etc., respectively Direction Cosines Consider the position vector OP( or r) of a point P(x, y, z) in below figure. The angles α, β, γ made by the vector r with the positive directions of x, y and z-axes respectively, are called its direction angles. The cosine values of these angles, i.e., cosα, cosβ and cos γ are called direction cosines of the vector r, and usually denoted by l, m and n, respectively. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 0 -

125 x The triangle OAP is right angled, and in it, we have cos (r stands for r ). Similarly, from the r y z right angled triangles OBP and OCP, we may write cos and cos. Thus, the coordinates r r of the point P may also be expressed as (lr, mr, nr). The numbers lr, mr and nr, proportional to the direction cosines are called as direction ratios of vector r, and denoted as a, b and c, respectively. l + m + n = but a + b + c, in general. Types of Vectors Zero Vector A vector whose initial and terminal points coincide, is called a zero vector (or null vector), and denoted as 0. Zero vector can not be assigned a definite direction as it has zero magnitude. Or, alternatively otherwise, it may be regarded as having any direction. The vectors AA, BB represent the zero vector, Unit Vector A vector whose magnitude is unity (i.e., unit) is called a unit vector. The unit vector in the direction of a given vector a is denoted by a Coinitial Vectors Two or more vectors having the same initial point are called coinitial vectors. Collinear Vectors Two or more vectors are said to be collinear if they are parallel to the same line, irrespective of their magnitudes and directions. Equal Vectors Two vectors a and b are said to be equal, if they have the same magnitude and direction regardless of the positions of their initial points, and written as a b. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - -

126 Negative of a Vector A vector whose magnitude is the same as that of a given vector (say, AB ), but direction is opposite to that of it, is called negative of the given vector. For example, vector BA is negative of the vector AB, and written as BA = AB. The vectors defined above are such that any of them may be subject to its parallel displacement without changing its magnitude and direction. Such vectors are called free vectors. Addition of Vectors Triangle law of vector addition If two vectors a and b are represented (in magnitude and direction) by two sides of a triangle taken in order, then their sum (resultant) is represented by the third side c ( a b ) taken in the opposite order. Subtraction of Vectors : To subtract b from a, reverse the direction of b and add to a. Geometrical Representation of Addition and Subtraction : Parallelogram law of vector addition If we have two vectors a and b represented by the two adjacent sides of a parallelogram in magnitude and direction, then their sum a b is represented in magnitude and direction by the diagonal of the parallelogram through their common point. This is known as the parallelogram law of vector addition. Properties of vector addition Property For any two vectors a and b, a b b a (Commutative property) Property For any three vectors a, b and c a b c a b c (Associative property), Property 3 For any vector a, we have a 0 0 a a, Here, the zero vector 0 is called the additive identity for the vector addition. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - -

127 Property 4 For any vector a, we have a a a a 0 Here, the vector a is called the additive inverse for the vector addition. Multiplication of a Vector by a Scalar Let a be a given vector and λ a scalar. Then the product of the vector a by the scalar λ, denoted as λ a, is called the multiplication of vector a by the scalar λ. Note that, λ a is also a vector, collinear to the vector a. The vector λ a has the direction same (or opposite) to that of vector a according as the value of λ is positive (or negative). Also, the magnitude of vector λ a is λ times the magnitude of the vector a, i.e., λ a = λ a Unit vector in the direction of vector a is given by a. a a Properties of Multiplication of Vectors by a Scalar For vectors a, b and scalars m, n, we have (i) m ( a ) = ( m) a = (m a ) (ii) ( m) ( a ) = m a (iii) m(n a ) = (mn) a = n(m a ) (iv) (m + n) a = m a + n a (v) m( a + b ) = m a + mb. Vector joining two points If P (x, y, z ) and P (x, y, z ) are any two points, then the vector joining P and P is the vector. PP PP = Position vector of head Position vector of tail = OP OP = ( x x) i ( y y ) j ( z z) k The magnitude of vector PP is given by ( x x ) ( y y ) ( z z ) SECTION FORMULA Internal Division : Position vector of a point C dividing a vector AB internally in the ratio of m : n mb na is OC = m n If C is the midpoint of AB, then OC divides AB in the ratio :. Therefore, position vector of C. b. a a b is Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 3 -

128 a b The position vector of the midpoint of AB is Position vector of any point C on AB can always be taken as c b a where + =.. n OA m. OB ( n m). OC, where C is a point on AB dividing it in the ratio m : n. External Division : Let A and B be two points with position vectors a and b respectively and let C be a point dividing AB externally in the ratio m : n. Then the position vector of C is given by mb na OC = m n Two vectors a and b are collinear if and only if there exists a nonzero scalar λ such that b = λa. If the vectors a and b are given in the component form, i.e. a a i a j a3k and b b i b j b k, then the two vectors are collinear if and only if 3 b i b j b k a i a j a k 3 3 b b b a a a 3 3 If a a i a j a3k, then a, a, a 3 are also called direction ratios of a. In case if it is given that l, m, n are direction cosines of a vector, then li mj nk (cos ) i (cos ) j (cos ) k is the unit vector in the direction of that vector, where α, β and γ are the angles which the vector makes with x, y and z axes respectively. Product of Two Vectors Scalar (or dot) product of two vectors The scalar product of two nonzero vectors a and b, denoted by a. b, is defined as a. b a b cos where, θ is the angle between a and b, 0 θ π If either a 0 or b 0, then θ is not defined, and in this case, we define a. b 0 Observations. a b is a real number. Let a and b be two nonzero vectors, then a. b 0 if and only if a and b are perpendicular to each other. i.e. a. b 0 a b If θ = 0, then a. b a b. In particular, a. a a, as θ in this case is 0. If θ = π, then a. b a b In particular, a.( a) a, as θ in this case is π. In view of the Observations and 3, for mutually perpendicular unit vectors i, j and k, we have i. i j. j k. k i. j j. k k. i 0 The angle between two nonzero vectors a and b, is given by a. b a. b cos, or cos a b a b The scalar product is commutative. i.e. a. b b. a Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 4 -

129 Property (Distributivity of scalar product over addition) Let a, b and c be any three vectors, then a.( b c) a. b a. c Property Let a and b be any two vectors, and be any scalar. Then ( a). b ( a. b) a.( b) Projection of a vector on a line If p is the unit vector along a line l, then the projection of a vector a on the line l is given by a. p. Projection of a vector a on other vector b b, is given by a. b or a. or ( a. b) b b If θ = 0, then the projection vector of AB will be AB itself and if θ = π, then the projection vector of AB will be BA 3 If or, then the projection vector of AB will be zero vector. If α, β and γ are the direction angles of vector a a i a j a3k, then its direction cosines may be a. i a a a3 given as cos,cos cos a i a a a Vector (or cross) product of two vectors The vector product of two nonzero vectors a and b, is denoted by a b and defined as ab a b sin n, where, θ is the angle between a and b, 0 θ π and n is a unit vector perpendicular to both a and b, such that a, b and n form a right handed system. If either a 0 or b 0, then θ is not defined and in this case, we define ab 0. Observations a b is a vector. Let a and b be two nonzero vectors. Then ab 0 if and only if and a and b are parallel (or collinear) to each other, i.e., ab 0 a b In particular, a a 0 and a( a) 0, since in the first situation, θ = 0 and in the second one, θ = π, making the value of sin θ to be 0. If then ab a b In view of the Observations and 3, for mutually perpendicular unit vectors i, j and k, we have i i j j k k 0 i j k, j k i, k i j In terms of vector product, the angle between two vectors a and b may be given as ab sin a b The vector product is not commutative, as ab b a In view of the above observations, we have j i k, k j i, i k j If a and b represent the adjacent sides of a triangle then its area is given as a b. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 5 -

130 If a and b represent the adjacent sides of a parallelogram, then its area is given by ab. The area of a parallelogram with diagonals a and b is a b ab is a unit vector perpendicular to the plane of a and b. ab a b is also a unit vector perpendicular to the plane of a and b. a b Property 3 (Distributivity of vector product over addition): If a, b and c are any three vectors and λ be a scalar, then (i) a( b c) ab a c (ii) ( ab) ( a) b a ( b) Let a and b be two vectors given in component form as a a i a j a3k and b b i b j b3 k, respectively. Then their cross product may be given by i j k ab a a a 3 b b b 3 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 6 -

131 CHAPTER 0: VECTOR ALGEBRA NCERT Important Questions & Answers MARKS WEIGHTAGE 06 marks. Find the unit vector in the direction of the sum of the vectors, a i j 5k and b i j 3k The sum of the given vectors is a b ( c, say) 4 i 3j k and c 4 3 ( ) 9 Thus, the required unit vector is 4 3 c c (4i 3 j k) i j k c Show that the points are A( i j k ), B( i 3j 5 k ), C(3 i 4j 4 k ) the vertices of a right angled triangle. We have AB ( ) i ( 3) j ( 5 ) k i j 6k BC (3 ) i ( 4 3) j ( 4 5) k i j k and CA ( 3) i ( 4) j ( 4) k i 3j 5k Then AB 4, BC 6, CA 35 AB BC CA Hence, the triangle is a right angled triangle. 3. Find the direction cosines of the vector joining the points A(,, 3) and B(,, ), directed from A to B. The given points are A(,, 3) and B(,,). Then AB ( ) i ( ) j ( ( 3)) k i 4j 4k Now, AB unit vector along AB = ( 4 4 ) AB 6 i j AB k 3 i 3 j 3 k Hence direction cosines are,, Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are i j k and i j k respectively, in the ratio : (i) internally (ii) externally The position vector of a point R divided the line segment joining two points P and Q in the ratio m: n is given by mb na Case I Internally m n mb na Case II Externally m n Position vectors of P and Q are given as OP i j k, OQ i j k Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 7 -

132 (i) Position vector of R [dividing (PQ ) in the ratio : internally] moq nop ( i j k ) ( i j k ) i 4j k 4 i j k m n (i) Position vector of R [dividing (PQ ) in the ratio : externally] moq nop ( i j k ) ( i j k ) 3i 0j 3k 3 i 3k m n 5. Find the position vector of the mid point of the vector joining the points P(, 3, 4) and Q(4,, ). Position vectors of P and Q are given as OP i 3j 4 k, OQ 4 i j k The position vector of the mid point of the vector joining the points P(, 3, 4) and Q(4,, ) is given by Position Vector of the mid-point of (PQ) = OQ OP 4 i j k i 3 j 4 k 6 i 4 j k 3 i j k 6. Show that the points A, B and C with position vectors, a 3i 4j 4k, b i j k and c i 3j 5k respectively form the vertices of a right angled triangle. Position vectors of points A, B and C are respectively given as a 3i 4j 4 k, b i j k and c i 3j 5k Now, AB b a i j k 3i 4j 4k i 3j 5k AB BC c b i 3j 5k i j k i j 6k BC CA a c 3i 4j 4k i 3j 5k i j k CA 4 6 BC AB CA Hence it form the vertices of a right angled triangle. 7. Find angle θ between the vectors a i j k and b i j k The angle θ between two vectors a and b is given by a. b cos a b Now, a. b ( i j k ).( i j k ) Therefore, we have cos cos If a 5 i j 3k and b i 3j 5k, then show that the vectors a b and a b are perpendicular. We know that two nonzero vectors are perpendicular if their scalar product is zero. Here, a b 5i j 3k i 3j 5k 6i j 8k and a b 5i j 3k i 3j 5k 4 i 4j k Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 8 -

133 Now, ( a b).( a b) (6i j 8 k ).(4 i 4j k ) Hence a b and a b are perpendicular. 9. Find a b, if two vectors a and b are such that a, b 3 and a. b = 4. We have a b ( a b).( a b) a. a a. b b. a b. b a ( a. b) b (4) a b 5 0. Show that the points A( i 3j 5 k ), B( i j 3 k ), C(7 i k ) are collinear. We have AB ( ) i ( 3) j (3 5) k 3 i j k BC (7 ) i (0 ) j ( 3) k 6i j 4k CA (7 ) i (0 3) j ( 5) k 9 i 3j 6k Now, AB 4, BC 56, CA 6 AB 4, BC 4, CA 3 4 CA AB BC Hence the points A, B and C are collinear.. If a, b, c are unit vectors such that a b c 0, find the value of a. b b. c c. a Given that a, b, c, a b c 0 ( a b c) ( a b c).( a b c) 0 a. a a. b a. c b. b b. c b. a c. a c. b c. c 0 a b c ( a. b b. c c. a) 0 ( a. b b. c c. a) 0 ( a. b b. c c. a) 3 3 a. b b. c c. a. If the vertices A, B, C of a triangle ABC are (,, 3), (, 0, 0), (0,, ), respectively, then find ABC. We are given the points A(,, 3), B(, 0, 0) and C(0,, ). Also, it is given that ABC is the angle between the vectors BA and BC Now, BA ( i j 3 k ) ( i 0j 0 k ) i j 3k BA and BC (0i j k ) ( i 0j 0 k ) i j k BC 4 6 BA. BC (i j 3 k ).( i j k ) 6 0 BA. BC 0 0 cos cos ABC BA BC ( 7)( 6) 0 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 9 -

134 0 ABC cos 0 3. Show that the points A(,, 7), B(, 6, 3) and C(3, 0, ) are collinear. The given points are A(,, 7), B (, 6, 3) and C(3, 0, ). AB (i 6j 3 k ) ( i j 7 k ) i 4j 4k AB BC (3i 0 j k ) (i 6j 3 k ) i 4j 4k BC and AC (3i 0 j k ) ( i j 7 k ) i 8j 8k AC AC AB BC Hence, the given points A, B and C are collinear. 4. Show that the vectors i j k, i 3 j 5k and 3 i 4 j 4k form the vertices of a right angled triangle. Let A= i j k B = i 3 j 5k and C = 3 i 4 j 4k AB ( i 3j 5 k ) ( i j k ) i j 6k AB BC (3i 4j 4 k ) ( i 3j 5 k ) i j k BC 4 6 and AC (3i 4j 4 k ) ( i j k ) i 3j 5k AC AB AC BC Hence, ABC is a right angled triangle. 5. Find a unit vector perpendicular to each of the vectors ( a b) and ( a b), where a i j k, b i j 3k. We have a b i 3j 4k and a b j k A vector which is perpendicular to both ( a b) and ( a b) is given by i j k ( a b) ( a b) 3 4 i 4j k ( c, say) 0 Now, c Therefore, the required unit vector is c c ( i 4 j k) i j k c Find the area of a triangle having the points A(,, ), B(,, 3) and C(, 3, ) as its vertices. We have AB j k and AC i j. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

135 The area of the given triangle is AB AC i j k Now, AB AC 0 4 i j k 0 Therefore, AB AC 6 4 Thus, the required area is AB AC 7. Find the area of a parallelogram whose adjacent sides are given by the vectors a 3i j 4k and b i j k. The area of a parallelogram with a and b as its adjacent sides is given by ab i j k Now, ab 3 4 5i j 4k Therefore, ab and hence, the required area is Find the area of the triangle with vertices A(,, ), B(, 3, 5) and C(, 5, 5). AB (i 3j 5 k ) ( i j k ) i j 3k AC ( i 5j 5 k ) ( i j k ) 4j 3k i j k Now, AB AC 3 6 i 3j 4k AB AC Area of triangle ABC = AB AC 6 sq. units. 9. Find the area of the parallelogram whose adjacent sides are determined by the vectors a i j 3k and b i 7j k. Adjacent sides of parallelogram are given by the vectors a i j 3k and b i 7j k. i j k Now, ab 3 0i 5j 5k 7 ab Hence, the area of the given parallelogram is 5 sq. units. 0. Let the vectors a and b be such that a 3 and b, then ab is a unit vector, find the 3 angle between a and b. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 3 -

136 Given that vectors a and b be such that a 3 and b. 3 Also, ab is a unit vector ab a. b sin 3 sin 3 sin 4. If i j k, i 5 j, 3 i j 3k and i 6 j k are the position vectors of points A, B, C and D respectively, then find the angle between AB and CD. Deduce that AB and CD are collinear. Note that if θ is the angle between AB and CD, then θ is also the angle between AB and CD. Now AB = Position vector of B Position vector of A = (i 5 j ) ( i j k ) i 4j k Therefore, AB Similarly, CD i 8j k CD AB. CD ( ) 4( 8) ( )() Thus, cos AB CD (3 )(6 ) Since 0 θ π, it follows that θ = π. This shows that AB and CD are collinear.. Let a, b and c be three vectors such that a 3, b 4, c 5 and each one of them being perpendicular to the sum of the other two, find a b c. Given that each one of them being perpendicular to the sum of the other two. Therefore, a.( b c) 0, b.( c a) 0, c.( a b) 0 Now, a b c ( a b c) ( a b c).( a b c) a. a a.( b c) b. b b.( c a) c.( a b) c. c a b c Therefore, a b c Find a vector of magnitude 5 units, and parallel to the resultant of the vectors a i 3j k and b i j k. Given vectors a i 3j k and b i j k. Let c be the resultant vector a and b then c (i 3 j k ) ( i j k ) 3 i j 0k c Unit vector in the direction of c = c c (3 i j ) c 0 Hence, the vector of magnitude 5 units and parallel to the resultant of vectors a and b is c 5 (3 ) 0 i j i j Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 3 -

137 4. The two adjacent sides of a parallelogram are i 4j 5k and i j 3k. Find the unit vector parallel to its diagonal. Also, find its area. Two adjacent sides of a parallelogram are given by a i 4j 5k and b i j 3k Then the diagonal of a parallelogram is given by c a b c a b i 4j 5k i j 3k 3 i 6j k c Unit vector parallel to its diagonal = c 3 6 (3 6 ) c 7 i c j k 7 i 7 j 7 k i j k Now, ab 4 5 i j 0k 3 Then the area of a parallelogram = ab sq. units. 5. Let a i 4j k, b 3 i j 7k and c i j 4k. Find a vector d which is perpendicular to both a and b and c. d 5. The vector which is perpendicular to both a and b must be parallel to ab. i j k Now, ab 4 3 i j 4k 3 7 Let d ( ab) (3 i j 4 k ) Also c. d 5 (i j 4 k ). (3 i j 4 k ) Required vector d 5 (3 i j 4 k ) 3 6. The scalar product of the vector i j k with a unit vector along the sum of vectors i 4j 5k and i j 3k is equal to one. Find the value of λ. Let a = i j k, b = i 4 j 5k and c = i j 3k Now, b c i 4j 5k i j 3 k ( ) i 6j k b c ( ) Unit vector along b c b c ( ) i 6 j k is b c 4 44 The scalar product of i j k with this unit vector is. b c ( ) i 6 j k ( i j k). ( i j k). b c 4 44 ( ) ( 6) Prepared by: M. S. KumarSwamy, TGT(Maths) Page

138 7. If a, b and c are mutually perpendicular vectors of equal magnitudes, show that the vector a b c is equally inclined to a, b and c. Given that a, b and c are mutually perpendicular vectors. a. b b. c c. a 0 It is also given that a b c Let vector a b c be inclined to a, b and c at angles andrespectively. ( a b c). a a. a b. a c. a a 0 0 cos a b c a a b c a a b c a a a a b c a a b c ( a b c). b a. b b. b c. b 0 b 0 cos a b c b a b c a a b c a b b a b c a a b c ( a b c). c a. c b. c c. c 0 0 c cos a b c c a b c a a b c a c c a b c a a b c Now as a b c, therefore, cos = cos = cos Hence, the vector a b c is equally inclined to a, b and c.. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

139 CHAPTER 0: VECTOR ALGEBRA Previous Years Board Exam (Important Questions & Answers). Write the projection of vector i j k along the vector j. ( i j k ). j 0 0 Required projection j 0 0 MARKS WEIGHTAGE 06 marks. Find a vector in the direction of vector i 3j 6k which has magnitude units. i 3j 6k i 3j 6k Required vector i 3j 6k 3(i 3j 6 k ) 6 i 9j 8k 7 3. Show that the vectors a b, b c and c a are coplanar if a, b, c are coplanar. Let a, b, c are coplanar then we have a b c 0 a.( bc) b.( c a) c.( a b) 0 Now, a b b c c a ( a b).{( b c) ( c a)} ( a b).{ bc b a c c c a} ( a b).{ bc b a c a} a.( b c) a.( b a) a.( c a) b.( b c) b.( b a) b.( c a) a b c b c a a b c a b c a b c 0 0 Hence, a, b, c are coplanar 4. Show that the vectors a, b, c are coplanar if a b, b c and c a are coplanar. Let a b, b c, c a are coplanar ( a b).{( b c) ( c a)} 0 ( a b).{ b c b a c c c a} 0 ( a b).{ bc b a c a} 0 a.( bc) a.( b a) a.( c a) b.( b c) b.( b a) b.( c a) 0 a b c a b c 0 a, b, c are coplanar Prepared by: M. S. KumarSwamy, TGT(Maths) Page

140 5. Write a unit vector in the direction of vector PQ, where P and Q are the points (, 3, 0) and (4, 5, 6) respectively. PQ (4 ) i (5 3) j (6 0) k 3 i j 6k 3i j 6k 3i j 6k 3i j 6k 3 Required unit vector = 6 i j k Write the value of the following : i ( j k ) j ( k i ) k ( i j ) i ( j k ) j ( k i ) k ( i j ) i j i k j k j i k i k j k j i k j i 0 7. Find the value of 'p' for which the vectors 3 i j 9k and i p j 3k are parallel. Since given two vectors are parallel p 3 p 6 p p 3 8. Find a.( b c), if a i j 3 k, b i j k and c 3 i j k. Given that a i j 3 k, b i j k and c 3 i j k 3 a.( b c) (4 ) ( 3) 3( 6) Show that the four points A, B, C and D with position vectors 4 i 5 j k, j k,3 i 9j 4k and 4( i j k ) are coplanar. Position vectors of A, B, C and D are Position vector of A = 4 i 5j k Position vector of B = j k Position vector of C = 3 i 9j 4k Position vector of D = 4( i j k ) = 4 i 4 j 4k AB 4i 6j k, AC i 4j 3 k, AD 8i j 3k Now, 4 6 AB.( AC AD) 4 3 4( 3) 6( 3 4) ( 3) AB.( AC AD) 0 Hence AB, AC and AD are coplanar i.e. A, B, C and D are coplanar. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

141 0. Find a vector a of magnitude5, making an angle of 4 with x-axis., with y-axis and an acute angle with z-axis. Direction cosines of required vector a are l cos, m cos 0 and n cos 4 l m n 0 cos cos cos n Unit vector in the direction of a i 0 j k a 5 0 i j k 5i 5k. If a and b are perpendicular vectors, a b = 3 and a = 5 find the value of b. Given a b = 3 a b 69 ( a b).( a b) 69 a a. b b 69 a b 69 a b a. b 0 b 69 a b. Find the projection of the vector i 3j 7k on the vector i 3 j 6k. Let a = i 3j 7k and b = i 3 j 6k Projection of the vector a on b a. b ( i 3j 7 k ).(i 3j 6 k ) = b i 3j 6 k If a and b are two unit vectors such that a b is also a unit vector, then find the angle between a and b. Given that a b is also a unit vector a b = a b ( a b).( a b) a a. b b a. b a, b a. b a. b a b cos Prepared by: M. S. KumarSwamy, TGT(Maths) Page

142 cos cos cos cos Prove that, for any three vectors a, b, c a b b c c a a b c a b b c c a ( a b).{( b c) ( c a)} ( a b).{ bc b a c c c a} ( a b).{ bc b a c a} a.( b c) a.( b a) a.( c a) b.( b c) b.( b a) b.( c a) a b c b c a a b c a b c a b c 5. Vectors a, b, c are such that a b c = 0 and a 3, b 5 and c 7. Find the angle between a and b. a b c 0 a b c ( a b) ( c) ( a b).( a b) c. c a a. b b c 9 a. b 5 49 a. b a. b a b cos 5 35 cos cos cos cos If a is a unit vector and ( x a)( x a) 4, then write the value of x. Given that ( x a)( x a) 4 x. x x. a a. x a. a 4 x a 4 x. a a. x x 4 x 5 x 5 7. For any three vectors a, b and c, write the value of the following: a( b c) b ( c a) c( a b) a( b c) b ( c a) c( a b) ab a c bc b a c a cb ab a c bc a b a c b c 0 Prepared by: M. S. KumarSwamy, TGT(Maths) Page

143 8. The magnitude of the vector product of the vector i j k with a unit vector along the sum of vectors i 4j 5k and i j 3k is equal to. Find the value of. Let a = i j k, b = i 4 j 5k and c = i j 3k Now, b c i 4j 5k i j 3 k ( ) i 6j k b c ( ) The vector product of i j k with this unit vector is. b c a ( b c) a b c b c i j k Now, a( b c) ( 6) i ( ) j (6 ) k 6 8 i (4 ) j (4 ) k a ( b c) 8 i (4 ) j (4 ) k b c (4 ) (4 ) (4 ) (4 ) ( 4 44) Find a unit vector perpendicular to each of the vectors a b and a b, where a 3i j k and b i j k. Ans. Given that a 3i j k and b i j k a b 3i j k ( i j k ) 3i j k i 4j 4k 5 i 6j k and a b (3i j k ) i j k 6i 4j 4k i j k 7 i 6j k Now, perpendicular vector of a b and a b i j k = 5 6 ( ) i (0 4) j (30 4) k 7 6 4i 4j k (i j k ) (i j k ) i j k Required unit vector = i j k i j k Prepared by: M. S. KumarSwamy, TGT(Maths) Page

144 0. If a i j 7k and b 5 i j k, then find the value of, so that a b and a b are perpendicular vectors. Given that a i j 7k and b 5 i j k a b i j 7k 5i j k 6 i j (7 ) k and a b i j 7k 5i j k 4 i (7 ) k Now, a b and a b are perpendicular vectors ( a b).( a b) 0 (6i j (7 ) k ).( 4 i (7 ) k (7 )(7 ) Prepared by: M. S. KumarSwamy, TGT(Maths) Page

145 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 4 -

146 CHAPTER : LINEAR PROGRAMMING QUICK REVISION (Important Concepts & Formulae) MARKS WEIGHTAGE 06 marks A half-plane in the xy-plane is called a closed half-plane if the points on the line separating the halfplane are also included in the half-plane. The graph of a linear inequation involving sign or is a closed half-plane. A half-plane in the xy-plane is called an open half-plane if the points on the line separating the halfplane are not included in the half-plane. The graph of linear inequation involving sign < or > is an open half-plane. Two or more linear inequations are said to constitute a system of linear inequations. The solution set of a system of linear inequations is defined as the intersection of solution sets of linear inequations in the system. A linear inequation is also called a linear constraint as it restricts the freedom of choice of the values x and y. LINEAR PROGRAMMING In linear programming we deal with the optimization (maximization or minimization) of a linear function of a number of variables subject to a number of restrictions (or constraints) on variables, in the form of linear inequations in the variable of the optimization function. A Linear Programming Problem is one that is concerned with finding the optimal value (maximum or minimum value) of a linear function (called objective function) of several variables (say x and y), subject to the conditions that the variables are non-negative and satisfy a set of linear inequalities (called linear constraints). The term linear implies that all the mathematical relations used in the problem are linear relations while the term programming refers to the method of determining a particular programme or plan of action. Objective function Linear function Z = ax + by, where a, b are constants, which has to be maximised or minimized is called a linear objective function. Variables x and y are called decision variables. Constraints The linear inequalities or equations or restrictions on the variables of a linear programming problem are called constraints. The conditions x 0, y 0 are called non-negative restrictions. Optimisation problem A problem which seeks to maximise or minimise a linear function (say of two variables x and y) subject to certain constraints as determined by a set of linear inequalities is called an optimisation problem. Linear programming problems are special type of optimisation problems. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 4 -

147 GRAPHICAL METHOD OF SOLVING LINEAR PROGRAMMING PROBLEMS Feasible region The common region determined by all the constraints including non-negative constraints x, y 0 of a linear programming problem is called the feasible region (or solution region) for the problem. The region other than feasible region is called an infeasible region. Feasible solutions Points within and on the boundary of the feasible region represent feasible solutions of the constraints. Any point outside the feasible region is called an infeasible solution. Optimal (feasible) solution: Any point in the feasible region that gives the optimal value (maximum or minimum) of the objective function is called an optimal solution. Theorem Let R be the feasible region (convex polygon) for a linear programming problem and let Z = ax + by be the objective function. When Z has an optimal value (maximum or minimum), where the variables x and y are subject to constraints described by linear inequalities, this optimal value must occur at a corner point (vertex) of the feasible region. A corner point of a feasible region is a point in the region which is the intersection of two boundary lines. Theorem Let R be the feasible region for a linear programming problem, and let Z = ax + by be the objective function. If R is bounded, then the objective function Z has both a maximum and a minimum value on R and each of these occurs at a corner point (vertex) of R. A feasible region of a system of linear inequalities is said to be bounded if it can be enclosed within a circle. Otherwise, it is called unbounded. Unbounded means that the feasible region does extend indefinitely in any direction. If R is unbounded, then a maximum or a minimum value of the objective function may not exist. However, if it exists, it must occur at a corner point of R. (By Theorem ). The method of solving linear programming problem is referred as Corner Point Method. The method comprises of the following steps:. Find the feasible region of the linear programming problem and determine its corner points (vertices) either by inspection or by solving the two equations of the lines intersecting at that point.. Evaluate the objective function Z = ax + by at each corner point. Let M and m, respectively denote the largest and smallest values of these points. 3. (i) When the feasible region is bounded, M and m are the maximum and minimum values of Z. (ii) In case, the feasible region is unbounded, we have: 4. (a) M is the maximum value of Z, if the open half plane determined by ax + by > M has no point in common with the feasible region. Otherwise, Z has no maximum value. (b) Similarly, m is the minimum value of Z, if the open half plane determined by ax + by < m has no point in common with the feasible region. Otherwise, Z has no minimum value. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

148 DIFFERENT TYPES OF LINEAR PROGRAMMING PROBLEMS A few important linear programming problems are listed below:. Manufacturing problems In these problems, we determine the number of units of different products which should be produced and sold by a firm when each product requires a fixed manpower, machine hours, labour hour per unit of product, warehouse space per unit of the output etc., in order to make maximum profit.. Diet problems In these problems, we determine the amount of different kinds of constituents/nutrients which should be included in a diet so as to minimise the cost of the desired diet such that it contains a certain minimum amount of each constituent/nutrients. 3. Transportation problems In these problems, we determine a transportation schedule in order to find the cheapest way of transporting a product from plants/factories situated at different locations to different markets. SOLUTION OF LINEAR PROGRAMMING PROBLEMS There are two types of linear programming problems-. When linear constraints and objective functions are given.. When linear constraints and objective functions are not given.. When linear constraints and objective functions are given. Working Rule (i) (ii) (iii) (iv) (v) (vi) Consider the linear equations of their corresponding linear inequations. Draw the graph of each linear equation. Check the solution region of each linear inequations by testing the points and then shade the common region of all the linear inequations. Determine the corner points of the feasible region. Find the value of objective function at each of the corner points obtained in above step. The maximum or minimum value out of all the values obtained in above step is the maximum or minimum value of the objective function.. When linear constraints and objective functions are not given. Working Rule (i) (ii) Identify the unknown variables in the given Linear programming problems. Denote them by x and y. Formulate the objective function in terms of x and y. Also, observe it is maximized or minimized. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

149 (iii) Write the linear constraints in the form of linear inequations formed by the given conditions. (iv) Consider the linear equations of their corresponding linear inequations. (v) Draw the graph of each linear equation. (vi) Check the solution region of each linear inequations by testing the points and then shade the common region of all the linear inequations. (vii) Determine the corner points of the feasible region. (viii) Evaluate the value of objective function at each corner points obtained in the above step. (ix) As the feasible region is unbounded, therefore the value may or may not be minimum or maximum value of the objective function. For this draw a graph of the inequality by equating the objective function with the above value to form linear inequation i.e. < for minimum or > for maximum. And check whether the resulting half plane has points in common with the feasible region or not. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

150 CHAPTER : LINEAR PROGRAMMING MARKS WEIGHTAGE 06 marks NCERT Important Questions & Answers. Determine graphically the minimum value of the objective function Z = 50x + 0y subject to the constraints: x y 5; 3x + y 3 ; x 3y ; x 0, y 0 Given that Z = 50x + 0y... () x y 5... () 3x + y 3... (3) x 3y... (4) x 0, y 0... (5) First of all, let us graph the feasible region of the system of inequalities () to (5). The feasible region (shaded) is shown in the below figure. Observe that the feasible region is unbounded. We now evaluate Z at the corner points. From this table, we find that 300 is the smallest value of Z at the corner point (6, 0). We know that if the region would have been bounded, this smallest value of Z is the minimum value of Z. But here we see that the feasible region is unbounded. Therefore, 300 may or may not be the minimum value of Z. To decide this issue, we graph the inequality 50x + 0y < 300 (see Step 3(ii) of corner Point Method.) i.e., 5x + y < 30 and check whether the resulting open half plane has points in common with feasible region or not. If it has common points, then 300 will not be the minimum value of Z. Otherwise, 300 will be the minimum value of Z. As shown in the above figure, it has common points. Therefore, Z = 50 x + 0 y has no minimum value subject to the given constraints.. Solve the following Linear Programming Problems graphically: Maximise Z = 5x + 3y subject to 3x + 5y 5, 5x + y 0, x 0, y 0. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

151 Our problem is to maximize Z = 5x + 3y (i) Subject to constraints 3x + 5y 5 (ii) 5x + y 0 (iii) x 0, y 0 (iv) Firstly, draw the graph of the line 3x + 5y = 5 Secondly, draw the graph of the line 5x + y = 0 On solving given equations 3x + 5y = 5 and 5x + y = 0, we get x = 0 9, y = 45 9 Feasible region is OABCO (see the below figure) The corner points of the feasible region are O(0, 0), A(, 0), B, 9 9 at these points are as follows: and C(0, 3) The values of Z Therefore, the maximum value of Z is at the point B, Show that the minimum of Z occurs at more than two points. Minimise and Maximise Z = x + y subject to x + y 00, x y 0, x + y 00; x, y 0. Our problem is to minimize and maximize Z = x + y (i) Subject to constraints are x + y 00 (ii) x y 0 (iii) x + y 00 (iv) Prepared by: M. S. KumarSwamy, TGT(Maths) Page

152 x 0, y 0 (v) Firstly, draw the graph of the line x + y = 00 Secondly, draw the graph of line x y = 0 Thirdly, draw the graph of line x + y = 00 On solving equations x y = 0 and x + y = 00, we get B(0, 40) and on solving the equations x y = 0 and x + y = 00, we get C(50, 00). Feasible region is ABCDA (see below figure) The corner points of the feasible region are A(0, 50), B(0, 40), C(50, 00) and D(0, 00). The values of Z at these points are as follows: The maximum value of Z is 400 at D(0, 00) and the minimum value of Z is 00 at all the points on the line segment joining A(0, 50) and B(0, 40). 4. A dietician wishes to mix two types of foods in such a way that vitamin contents of the mixture contain atleast 8 units of vitamin A and 0 units of vitamin C. Food I contains units/kg of vitamin A and unit/kg of vitamin C. Food II contains unit/kg of vitamin A and units/kg of vitamin C. It costs Rs 50 per kg to purchase Food I and Rs 70 per kg to purchase Food II. Formulate this problem as a linear programming problem to minimise the cost of such a mixture. Let the mixture contain x kg of Food I and y kg of Food II. Clearly, x 0, y 0. We make the following table from the given data: Prepared by: M. S. KumarSwamy, TGT(Maths) Page

153 Since the mixture must contain at least 8 units of vitamin A and 0 units of vitamin C, we have the constraints: x + y 8 x + y 0 Total cost Z of purchasing x kg of food I and y kg of Food II is Z = 50x + 70y Hence, the mathematical formulation of the problem is: Minimise Z = 50x + 70y... () subject to the constraints: x + y 8... () x + y 0... (3) x, y 0... (4) Let us graph the inequalities () to (4). The feasible region determined by the system is shown in the below figure. Here again, observe that the feasible region is unbounded. Let us evaluate Z at the corner points A(0,8), B(,4) and C(0,0). In the table, we find that smallest value of Z is 380 at the point (,4). We know that the feasible region is unbounded. Therefore, we have to draw the graph of the inequality 50x + 70y < 380 i.e., 5x + 7y < 38 to check whether the resulting open half plane has any point common with the feasible region. From the above figure, we see that it has no points in common. Thus, the minimum value of Z is 380 attained at the point (, 4). Hence, the optimal mixing strategy for the dietician would be to mix kg of Food I and 4 kg of Food II, and with this strategy, the minimum cost of the mixture will be Rs A cooperative society of farmers has 50 hectare of land to grow two crops X and Y. The profit from crops X and Y per hectare are estimated as Rs 0,500 and Rs 9,000 respectively. To control weeds, a liquid herbicide has to be used for crops X and Y at rates of 0 litres and 0 litres per hectare. Further, no more than 800 litres of herbicide should be used in order to Prepared by: M. S. KumarSwamy, TGT(Maths) Page

154 protect fish and wild life using a pond which collects drainage from this land. How much land should be allocated to each crop so as to maximise the total profit of the society? Let x hectare of land be allocated to crop X and y hectare to crop Y. Obviously, x 0, y 0. Profit per hectare on crop X = Rs 0500 Profit per hectare on crop Y = Rs 9000 Therefore, total profit = Rs (0500x y) The mathematical formulation of the problem is as follows: Maximise Z = 0500 x y subject to the constraints: x + y 50 (constraint related to land)... () 0x + 0y 800 (constraint related to use of herbicide) i.e. x + y () x 0, y 0 (non negative constraint)... (3) Let us draw the graph of the system of inequalities () to (3). The feasible region OABC is shown (shaded) in the below figure. Observe that the feasible region is bounded. The coordinates of the corner points O, A, B and C are (0, 0), (40, 0), (30, 0) and (0, 50) respectively. Let us evaluate the objective function Z = 0500 x y at these vertices to find which one gives the maximum profit. Hence, the society will get the maximum profit of Rs 4,95,000 by allocating 30 hectares for crop X and 0 hectares for crop Y. 6. A manufacturing company makes two models A and B of a product. Each piece of Model A requires 9 labour hours for fabricating and labour hour for finishing. Each piece of Model B requires labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available are 80 and 30 respectively. The company makes a profit of Rs 8000 on each piece of model A and Rs 000 on each piece of Model B. How many pieces of Model A and Model B should be manufactured per week to realise a maximum profit? What is the maximum profit per week? Suppose x is the number of pieces of Model A and y is the number of pieces of Model B. Then Total profit (in Rs) = 8000 x y Let Z = 8000 x y We now have the following mathematical model for the given problem. Maximise Z = 8000 x y... () subject to the constraints: 9x + y 80 (Fabricating constraint) i.e. 3x + 4y () Prepared by: M. S. KumarSwamy, TGT(Maths) Page

155 x + 3y 30 (Finishing constraint)... (3) x 0, y 0 (non-negative constraint)... (4) The feasible region (shaded) OABC determined by the linear inequalities () to (4) is shown in the below figure. Note that the feasible region is bounded. Let us evaluate the objective function Z at each corner point as shown below: We find that maximum value of Z is,68,000 at B (, 6). Hence, the company should produce pieces of Model A and 6 pieces of Model B to realise maximum profit and maximum profit then will be Rs,68, Reshma wishes to mix two types of food P and Q in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and units of vitamin B. Food P costs Rs 60/kg and Food Q costs Rs 80/kg. Food P contains 3 units/kg of Vitamin A and 5 units / kg of Vitamin B while food Q contains 4 units/kg of Vitamin A and units/kg of vitamin B. Determine the minimum cost of the mixture. Let Reshma mixes x kg of food P and y kg of food Q. Construct the following table: The mixture must contain atleast 8 units of vitamin A and units of Vitamin B. Total cost Z of purchasing food is Z = 60x + 80y The mathematical formulation of the given problem is Minimize Z = 60x + 80y (i) Subject to the constraints 3x + 4y 8 (ii) 5x + y (iii) x 0, y 0 (iv) Firstly, draw the graph of the line 3x + 4y = 8 Secondly, draw the graph of the line 5x + y = On solving equations 3x + 4y = 8 and 5x + y =, we get B, 8 The corner points of the feasible region are A,0, B, and C 0, (see the below figure) 3 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 5 -

156 The values of Z at these points are as follows : As the feasible region is unbounded, therefore 60 may or may not be the minimum value of Z. For this, we graph the inequality 60x + 80y < 60 or 3x + 4y < 8 and check whether the resulting half plane has points in common with the feasible region or not. It can be seen that the feasible region has no common point with 3x + 4y < 8 therefore, the minimum cost of the mixture will be Rs. 60 at the 8 line segment joining the points A,0 and B, 3 8. A factory makes tennis rackets and cricket bats. A tennis racket takes.5 hours of machine time and 3 hours of craftman s time in its making while a cricket bat takes 3 hour of machine time and hour of craftman s time. In a day, the factory has the availability of not more than 4 hours of machine time and 4 hours of craftsman s time. (i) What number of rackets and bats must be made if the factory is to work at full capacity? (ii) If the profit on a racket and on a bat is Rs 0 and Rs 0 respectively, find the maximum profit of the factory when it works at full capacity. Let the number of rackets and the number of cricket bats to be made in a day be x and y respectively. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 5 -

157 Construct the following table: The machine time is not available for more than 4 h..5x + 3y 4 The craftman s time is not available for more than 4 h. 3x + y 4 The profit on rackets is Rs. 0 and on bats is Rs. 0. Maximum Z = 0x + 0y (i) Subject to constraints.5x + 3y 4 (ii) 3x + y 4 (iii) x 0, y 0 (iv) Firstly, draw the graph of the line.5x + 3y = 4 Secondly, draw the graph of the line 3x + y = 4 On solving equations.5x + 3y = 4 and 3x + y = 4, we get B(4, ). Feasible region is OABCO(See below figure). The corner points of the feasible region are O (0, 0), A(8, 0), B(4, ) and C (0, 4). The values of Z at these points are as follows: Thus, the maximum profit of the factory when it works to its full capacity is Rs. 00. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

158 9. A manufacturer produces nuts and bolts. It takes hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and hour on machine B to produce a package of bolts. He earns a profit of Rs7.50 per package on nuts and Rs 7.00 per package on bolts. How many packages of each should be produced each day so as to maximise his profit, if he operates his machines for at the most hours a day? Let the manufacturer produces x nuts and y bolts. We construct the following table : Total profit, Z = 7.5x + 7y i.e.,maximize Z = 7.5x + 7y (i) Subject to constraints x + 3y (ii) 3x + y (iii) x 0, y 0 (iv) Firstly, draw the graph of the lines x + 3y = Secondly, draw the graph of the lines 3x + y = On solving equations x + 3y = and 3x + y =, we get B(3, 3). Feasible region is OABCO.(See below figure) The corner points of the feasible region are O (0, 0) A(4, 0), B(3, 3) and C(0,4). The values of Z at these points are as follows: The maximum value of Z is Rs at (3, 3). Thus, 3 packages of nuts and 3 packages of bolts should be produced each day to get the maximum profit Rs Prepared by: M. S. KumarSwamy, TGT(Maths) Page

159 0. A factory manufactures two types of screws, A and B. Each type of screw requires the use of two machines, an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on hand operated machines to manufacture a package of screws A, while it takes 6 minutes on automatic and 3 minutes on the hand operated machines to manufacture a package of screws B. Each machine is available for at the most 4 hours on any day. The manufacturer can sell a package of screws A at a profit of Rs 7 and screws B at a profit of Rs 0. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximise his profit? Determine the maximum profit. Let the manufacturer produces x package of screws A and y package of screws B. We construct the following table: The profits on a package of screws A is Rs. 7 and on the package of screws B is Rs. 0. Our problemi s to maximize Z = 7x + 0y (i) Subject to constraints 4x + 6y 40 x + 3y 0 (ii) 6x + 3y 40 x + y 80 (iii) and x 0, y 0 (iv) Firstly, draw the graph of the line x + 3y = 0 Secondly, draw the graph of the line x + y = 80 On solving equations x + 3y = 0 and x + y = 80, we get B(30, 0). The corner points of the feasible region are O (0, 0), A(40, 0), B(30, 0) and C(0, 40). Feasible region is OABCO.(See below figure) The values of Z at these points are as follows: Prepared by: M. S. KumarSwamy, TGT(Maths) Page

160 The maximum value of Z is Rs. 40 at B(30, 0). Thus, the factory should produce 30 packages of screws A and 0 packages of screws B to get the maximum profit of Rs A merchant plans to sell two types of personal computers a desktop model and a portable model that will cost Rs 5000 and Rs respectively. He estimates that the total monthly demand of computers will not exceed 50 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than Rs 70 lakhs and if his profit on the desktop model is Rs 4500 and on portable model is Rs Let the manufacturer produces x pedestal lamps and y wooden shades everyday. We construct the following table : The profit on a lamp is Rs. 5 and on the shades is Rs. 3. Our problem is to maximize Z = 5x + 3y (i) Subject to the constraints x + y (ii) 3x + y 0 (iii) x 0, y 0 (iv) Firstly, draw the graph of the line x + y = Secondly, draw the graph of the line 3x + y = 0 On solving equations x + y = and 3x + y = 0, we get B(4, 4). Feasible region is OABCO. (see below figure) Prepared by: M. S. KumarSwamy, TGT(Maths) Page

161 The corner points of the feasible region are O(0, 0), A(6, 0), B(4, 4) and C(0, 0). at these points are as follows: The values of Z The maximum value of Z is Rs. 3 at B(4, 4). Thus, the manufacturer should produce 4 pedestal lamps and 4 wooden shades to maximize his profits.. A diet is to contain at least 80 units of vitamin A and 00 units of minerals. Two foods F and F are available. Food F costs Rs 4 per unit food and F costs Rs 6 per unit. One unit of food F contains 3 units of vitamin A and 4 units of minerals. One unit of food F contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem. Find the minimum cost for diet that consists of mixture of these two foods and also meets the minimal nutritional requirements. Let the diet contains x unit of food F and y units of food F. We construct the following table : The cost of food F is Rs. 4 per unit and of food F is Rs. 6 per unit. So, our problem is to minimize Z = 4x + 6y (i) Subject to constraints 3x + 6y 80 (ii) 4x + 3y 00 (iii) x 0, y 0 (iv) Firstly, draw the graph of the line 3x + 6y = 80 Secondly, draw the graph of the line 4x + 3y = 00 4 On solving the equations 3x + 6y = 80 and4x + 3y = 00, we get B 4, 3 It can be seen that the feasible region is unbounded. (see the below figure) Prepared by: M. S. KumarSwamy, TGT(Maths) Page

162 The corner points of the feasible region are of Z at these points are as follows: A,0 and B 4, and C 0,. The values As the feasible is unbounded therefore, 04 may or may not be the minimum value of Z. For this, we draw a graph of the inequality, 4x + 6y < 04 or x + 3y < 5 and check, whether the resulting half plane has points in common with the feasible region or not. It can be seen that the feasible region has no common point with x + 3y < 5 Therefore, the minimum cost of the mixture will be Rs A dietician has to develop a special diet using two foods P and Q. Each packet (containing 30 g) of food P contains units of calcium, 4 units of iron, 6 units of cholesterol and 6 units of vitamin A. Each packet of the same quantity of food Q contains 3 units of calcium, 0 units of iron, 4 units of cholesterol and 3 units of vitamin A. The diet requires atleast 40 units of calcium, atleast 460 units of iron and at most 300 units of cholesterol. How many packets of each food should be used to minimise the amount of vitamin A in the diet? What is the minimum amount of vitamin A? Let x and y be the number of packets of food P and Q respectively. Obviously x 0, y 0. Mathematical formulation of the given problem is as follows: Minimise Z = 6x + 3y (vitamin A) subject to the constraints x + 3y 40 (constraint on calcium), i.e. 4x + y () 4x + 0y 460 (constraint on iron), i.e. x + 5y 5... () 6x + 4y 300 (constraint on cholesterol), i.e. 3x + y (3) x 0, y 0... (4) Let us graph the inequalities () to (4). The feasible region (shaded) determined by the constraints () to (4) is shown in below figure and note that it is bounded. The coordinates of the corner points L, M and N are (, 7), (5, 0) and (40, 5) respectively. Let us evaluate Z at these points: Prepared by: M. S. KumarSwamy, TGT(Maths) Page

163 From the table, we find that Z is minimum at the point (5, 0). Hence, the amount of vitamin A under the constraints given in the problem will be minimum, if 5 packets of food P and 0 packets of food Q are used in the special diet. The minimum amount of vitamin A will be 50 units. 4. A manufacturer has three machines I, II and III installed in his factory. Machines I and II are capable of being operated for at most hours whereas machine III must be operated for atleast 5 hours a day. She produces only two items M and N each requiring the use of all the three machines. The number of hours required for producing unit of each of M and N on the three machines are given in the following table: Number of hours required on Items machines I II III M N.5 She makes a profit of Rs 600 and Rs 400 on items M and N respectively. How many of each item should she produce so as to maximise her profit assuming that she can sell all the items that she produced? What will be the maximum profit? Let x and y be the number of items M and N respectively. Total profit on the production = Rs (600 x y) Mathematical formulation of the given problem is as follows: Maximise Z = 600 x y subject to the constraints: x + y (constraint on Machine I)... () x + y (constraint on Machine II)... () x + 5 y 5 (constraint on Machine III)... (3) 4 x 0, y 0... (4) Let us draw the graph of constraints () to (4). ABCDE is the feasible region (shaded) as shown in below figure determined by the constraints () to (4). Observe that the feasible region is bounded, coordinates of the corner points A, B, C, D and E are (5, 0) (6, 0), (4, 4), (0, 6) and (0, 4) respectively. Let us evaluate Z = 600 x y at these corner points. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

164 We see that the point (4, 4) is giving the maximum value of Z. Hence, the manufacturer has to produce 4 units of each item to get the maximum profit of Rs There are two factories located one at place P and the other at place Q. From these locations, a certain commodity is to be delivered to each of the three depots situated at A, B and C. The weekly requirements of the depots are respectively 5, 5 and 4 units of the commodity while the production capacity of the factories at P and Q are respectively 8 and 6 units. The cost of transportation per unit is given below: From/To Cost (in Rs.) A B C P Q How many units should be transported from each factory to each depot in order that the transportation cost is minimum. What will be the minimum transportation cost? Let x units and y units of the commodity be transported from the factory at P to the depots at A and B respectively. Then (8 x y) units will be transported to depot at C Hence, we have x 0, y 0 and 8 x y 0 i.e. x 0, y 0 and x + y 8 Now, the weekly requirement of the depot at A is 5 units of the commodity. Since x units are transported from the factory at P, the remaining (5 x) units need to be transported from the factory at Q. Obviously, 5 x 0, i.e. x 5. Similarly, (5 y) and 6 (5 x + 5 y) = x + y 4 units are to be transported from the factory at Q to the depots at B and C respectively. Thus, 5 y 0, x + y 4 0 i.e. y 5, x + y 4 Total transportation cost Z is given by Z = 60 x + 00 y + 00 ( 5 x) + 0 (5 y) + 00 (x + y 4) + 50 (8 x y) = 0 (x 7 y + 90) Therefore, the problem reduces to Minimise Z = 0 (x 7y + 90) subject to the constraints: x 0, y 0... () x + y 8... () x 5... (3) y 5... (4) and x + y 4... (5) Prepared by: M. S. KumarSwamy, TGT(Maths) Page

165 The shaded region ABCDEF represented by the constraints () to (5) is the feasible region (see below figure). Observe that the feasible region is bounded. The coordinates of the corner points of the feasible region are (0, 4), (0, 5), (3, 5), (5, 3), (5, 0) and (4, 0). Let us evaluate Z at these points. From the table, we see that the minimum value of Z is 550 at the point (0, 5). Hence, the optimal transportation strategy will be to deliver 0, 5 and 3 units from the factory at P and 5, 0 and units from the factory at Q to the depots at A, B and C respectively. Corresponding to this strategy, the transportation cost would be minimum, i.e., Rs A farmer mixes two brands P and Q of cattle feed. Brand P, costing Rs 50 per bag, contains 3 units of nutritional element A,.5 units of element B and units of element C. Brand Q costing Rs 00 per bag contains.5 units of nutritional element A,.5 units of element B, and 3 units of element C. The minimum requirements of nutrients A, B and C are 8 units, 45 units and 4 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag? What is the minimum cost of the mixture per bag? Let the farmer mixes x bags of brand P and y bags of brand Q. We construct the following table: So, our problem is minimize Z = 50x + 00y (i) Subject to constraints 3x +.5y 8 x + y (ii).5x +.5y 45 x + 9y 36 (iii) x + 3y 4 (iv) x 0, y 0 (v) Firstly, draw the graph of the line 3x +.5y = 8 Secondly, draw the graph of the line.5x +.5y = 45 Thirdly, draw the graph of the line x + 3y = 4 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 6 -

166 On solving equations 3x +.5y = 8 and x + 3y = 4, we get C(3, 6). Similarly, on solving equations.5x +.5y = 45 andx + 3y = 4, we get B(9, ). The corner points of the feasible region are A(8, 0), B(9, ), C(3, 6) and D (0, ). (See below figure) The values of Z at these points are as follows: As the feasible region is unbounded, therefore 950 may or may not be the minimum value of Z. For this, we draw a graph of the inequality 50x + 00y < 950 or 5x + 4y < 39 and check, whether the resulting half plane has points in common with the feasible region or not. It can be seen that the feasible region has no common point with 5x + 4y < 39. Therefore, the minimum value of Z is 950 at C(3, 6). Thus, 3 bags of brand P and 6 bags of brand Q should be used in the mixture to minimize the cost to Rs A dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains at least 0 units of vitamin A, units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg food is given below: Food Vitamin A Vitamin B Vitamin C X 3 Y One kg of food X costs Rs 6 and one kg of food Y costs Rs 0. Find the least cost of the mixture which will produce the required diet? Let the dietician mixes x kg of food X and y kg of food Y. We construct the following table: Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 6 -

167 So, our problem is to minimize Z = 6x + 0y (i) Subject to constraints x + y 0 (ii) x + y x + y 6 (iii) 3x + y 8 (iv) x 0, y 0 (v) Firstly, draw the graph of the line x + y = 0 Secondly, draw the graph of the line line x + y = 6 Thirdly, draw the graph of the line 3x + y = 8 On solving equations x + y = 6 and x + y = 0, we get B(, 4) Similarly, solving the equations 3x + y = 8 and x + y = 6, we get C(, 5). The corner points of the feasible region are A(0, 0), B(, 4), C(, 5) and D(0, 8). The values of Z at these points are as follows: As the feasible region is unbounded, therefore may or may not be the minimum value of Z. For this,we drawa graph of the inequality,6x + 0y < or4x + 5y < 8 and check, whether the resulting half plane has points in common with the feasible region or not. It can be seen that the feasible region has no common point with 4x + 5y < 8. Therefore, the minimum value of Z is at B(, 4). Thus, the mixture should contain kg of food X and 4 kg of food Y. The minimum cost of the mixture is Rs.. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

168 8. A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given below: Types of Machines Toys I II III A 8 6 B Each machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is Rs 7.50 and that on each toy of type B is Rs 5, show that 5 toys of type A and 30 of type B should be manufactured in a day to get maximum profit. Let the manufacturer makes x toys of type A and y toys of type B. We construct the following table: Our problem is to maximize Z = 7.50x + 5y (i) Subject to constraints x + 6y 360 x + y 60 (ii) 8x 360 x 0 (iii) 6x + 9y 360 x + 3y 0 (iv) x 0, y 0 (v) Firstly, draw the graph of the line x + y = 60 Secondly, draw the graph of the line x + 3y = 0 Thirdly, draw the graph of the line x = 0 On solving equations x + y = 60 and x + 3y = 0, we get C(5, 30) Similarly, solving the equations x = 0 and x + y = 60, we get B(0, 0). Feasible region is OABCDO. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

169 The corner points of the feasible region are A(0, 0), B(0, 0), C(5, 30) and D(0, 40). The values of Z at these points are as follows: Thus, the maximum value of Z is 7.5 C(5, 30). Thus, the manufacturer should manufacture 5 toys of type A and 30 toys of type B to maximize the profit. 9. An aeroplane can carry a maximum of 00 passengers. A profit of Rs 000 is made on each executive class ticket and a profit of Rs 600 is made on each economy class ticket. The airline reserves at least 0 seats for executive class. However, at least 4 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximise the profit for the airline. What is the maximum profit? Let x passengers travel by executive class and y passengers travel by economy class. We construct the following table : So, our problem is to maximize Z = 000x + 600y (i) Subject to constraints x + y 00 (ii) x 0 (iii) y 4x 0 y 4x (iv) x 0, y 0 (v) Firstly, draw the graph of the line x + y = 00. Secondly, draw the graph of the line y = 4x Thirdly, draw the graph of the line x = 0 On solving the equations, we get A(0, 80), B(40, 60) and C(0, 80). Feasible region is ABCA.(See the below figure) Prepared by: M. S. KumarSwamy, TGT(Maths) Page

170 The corner points of the feasible region are A(0, 80), B(40, 60) and C(0, 80). The value of Z at these points are as follows: Thus, the maximum value of Z is at B(40, 60). Thus, 40 tickets of executive class and 60 tickets of economy class should be sold to maximize the profit and the maximum profit is Rs A toy company manufactures two types of dolls, A and B. Market tests and available resources have indicated that the combined production level should not exceed 00 dolls per week and the demand for dolls of type B is at most half of that for dolls of type A. Further, the production level of dolls of type A can exceed three times the production of dolls of other type by at most 600 units. If the company makes profit of Rs and Rs 6 per doll respectively on dolls A and B, how many of each should be produced weekly in order to maximise the profit? Let the company manufactures x dolls of type A and y dolls of type B, then x 0, y 0 (i) x + y 00 (ii) y x / x y 0 (iii) x 3y x 3y 600 (iv) Firstly, draw the graph of the line x + y = 00 Secondly, draw the graph of the line x y = 0 Thirdly, draw the graph of the line x 3y = 600 The point of intersection of lines x 3y = 600 and x + y = 00 is B(050,50); of lines x = y and x + y = 00 is C (800,400). Let Z be the total profit, then Z = x + 6y Feasible region is OABCO. (See the below figure) Prepared by: M. S. KumarSwamy, TGT(Maths) Page

171 The corner points of the feasible region are A(600, 0), B(050, 50) and C(800, 400). The values of Z at these points are as follows: The maximum value of Z is 6000 at C(800, 400). Thus, 800 and 400 dolls of types A and type B should be produced respectively to get the maximum profit of Rs Prepared by: M. S. KumarSwamy, TGT(Maths) Page

172 CHAPTER : LINEAR PROGRAMMING Previous Years Board Exam (Important Questions & Answers) MARKS WEIGHTAGE 06 marks. A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding/cutting machine and a sprayer. It takes hours on the grinding/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp. It takes hour on the grinding/cutting machine and hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at the most 0 hours and the grinding/cutting machine for at the most hours. The profit from the sale of a lamp is Rs. 5 and that from a shade is Rs. 5. Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximise his profit. Formulate an LPP and solve it graphically. Let the manufacturer produces x padestal lamps and y wooden shades; then time taken by x pedestal lamps and y wooden shades on grinding/cutting machines = (x + y) hours and time taken on the sprayer = (3x + y) hours. Since grinding/cutting machine is available for at the most hours. x + y and sprayer is available for at most 0 hours. Thus, we have 3x + y 0 Now profit on the sale of x lamps and y shades is, Z = 5x + 5y. So, our problem is to find x and y so as to Maximise Z = 5x + 5y (i) Subject to the constraints: 3x + y 0 (ii) x + y (iii) x 0 (iv) y 0 (v) The feasible region (shaded) OABC determined by the linear inequalities (ii) to (v) is shown in the figure. The feasible region is bounded. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

173 Let us evaluate the objective function at each corner point as shown below: We find that maximum value of Z is Rs. 60 at B(4, 4). Hence, manufacturer should produce 4 lamps and 4 shades to get maximum profit of Rs A manufacturing company makes two types of teaching aids A and B of Mathematics for class XII. Each type of A requires 9 labour hours of fabricating and labour hour for finishing. Each type of B requires labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available per week are 80 and 30 respectively. The company makes a profit of Rs. 80 on each piece of type A and Rs. 0 on each piece of type B. How many pieces of type A and type B should be manufactured per week to get a maximum profit? Make it as an LPP and solve graphically. What is the maximum profit per week? Let x and y be the number of pieces of type A and B manufactured per week respectively. If Z be the profit then, Objective function, Z = 80x + 0y We have to maximize Z, subject to the constraints 9x + y 80 3x + 4y 60...(i) x + 3y 30...(ii) x 0, y 0...(iii) The graph of constraints are drawn and feasible region OABC is obtained, which is bounded having corner points O(0, 0), A(0, 0), B(, 6) and C(0, 0) Prepared by: M. S. KumarSwamy, TGT(Maths) Page

174 Now the value of objective function is obtained at corner points as Hence, the company will get the maximum profit of Rs.,680 by making pieces of type A and 6 pieces of type B of teaching aid. Yes, teaching aid is necessary for teaching learning process as (i) it makes learning very easy. (ii) it provides active learning. (iii) students are able to grasp and understand concept more easily and in active manner. 3. A dealer in rural area wishes to purchase a number of sewing machines. He has only Rs. 5,760 to invest and has space for at most 0 items for storage. An electronic sewing machine cost him Rs. 360 and a manually operated sewing machine Rs. 40. He can sell an electronic sewing machine at a profit of Rs. and a manually operated sewing machine at a profit of Rs. 8. Assuming that he can sell all the items that he can buy, how should he invest his money in order to maximise his profit? Make it as a LPP and solve it graphically. Suppose dealer purchase x electronic sewing machines and y manually operated sewing machines. If Z denotes the total profit. Then according to question (Objective function) Z = x + 8 y Also, x + y 0 360x + 40y x + 6y 44 x 0, y 0. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

175 We have to maximise Z subject to above constraint. To solve graphically, at first we draw the graph of line corresponding to given inequations and shade the feasible region OABC. The corner points of the feasible region OABC are O(0, 0), A(6, 0), B(8, ) and C(0, 0). Now the value of objective function Z at corner points are obtained in table as From table, it is obvious that Z is maximum when x = 8 and y =. Hence, dealer should purchase 8 electronic sewing machines and manually operated sewing machines to obtain the maximum profit ` 39 under given condition. 4. An aeroplane can carry a maximum of 00 passengers. A profit of `500 is made on each executive class ticket out of which 0% will go to the welfare fund of the employees. Similarly a profit of `400 is made on each economy ticket out of which 5% will go for the improvement of facilities provided to economy class passengers. In both cases, the remaining profit goes to the airline s fund. The airline reserves at least 0 seats for executive class. However at least four times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximise the net profit of the airline. Make the above as an LPP and solve graphically. Do you think, more passengers would prefer to travel by such an airline than by others? Let there be x tickets of executive class and y tickets of economy class. Let Z be net profit of the airline. Here, we have to maximise z. Now Z = 500x 80 + y Z = 400x + 300y...(i) According to question x 0...(ii) Also x + y 00...(iii) x + 4x 00 5x 00 x 40...(iv) Shaded region is feasible region having corner points A (0, 0), B (40,0) C (40, 60), D (0,80) Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 7 -

176 Now value of Z is calculated at corner point as Hence, 40 tickets of executive class and 60 tickets of economy class should be sold to maximise the net profit of the airlines. Yes, more passengers would prefer to travel by such an airline, because some amount of profit is invested for welfare fund. 5. A manufacturer considers that men and women workers are equally efficient and so he pays them at the same rate. He has 30 and 7 units of workers (male and female) and capital respectively, which he uses to produce two types of goods A and B. To produce one unit of A, workers and 3 units of capital are required while 3 workers and unit of capital is required to produce one unit of B. If A and B are priced at Rs. 00 and Rs. 0 per unit respectively, how should he use his resources to maximise the total revenue? Form the above as an LPP and solve graphically. Do you agree with this view of the manufacturer that men and women workers are equally efficient and so should be paid at the same rate? Let x, y unit of goods A and B are produced respectively. Let Z be total revenue Here Z = 00x + 0y...(i) Also x + 3y 30...(ii) 3x + y 7...(iii) x 0...(iv) y 0...(v) On plotting graph of above constants or inequalities (ii), (iii), (iv) and (v). We get shaded region as feasible region having corner points A, O, B and C. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 7 -

177 For co-ordinate of 'C' Two equations (ii) and (iii) are solved and we get coordinate of C = (3, 8) Now the value of Z is evaluated at corner point as: Therefore maximum revenue is Rs.,60 when workers and 8 units capital are used for production. Yes, although women workers have less physical efficiency but it can be managed by her other efficiency. 6. A cooperative society of farmers has 50 hectares of land to grow two crops A and B. The profits from crops A and B per hectare are estimated as `0,500 and `9,000 respectively. To control weeds, a liquid herbicide has to be used for crops A and B at the rate of 0 litres and 0 litres per hectare, respectively. Further not more than 800 litres of herbicide should be used in order to protect fish and wildlife using a pond which collects drainage from this land. Keeping in mind that the protection of fish and other wildlife is more important than earning profit, how much land should be allocated to each crop so as to maximize the total profit? Form an LPP from the above and solve it graphically. Do you agree with the message that the protection of wildlife is utmost necessary to preserve the balance in environment? Let x and y hectare of land be allocated to crop A and B respectively. If Z is the profit then Z =0500x y (i) We have to maximize Z subject to the constraints x + y 50 (ii) 0x +0y 800 x + y 80 (iii) x 0, y 0 (iv) The graph of system of inequalities (ii) to (iv) are drawn, which gives feasible region OABC with corner points O (0, 0), A (40, 0), B (30, 0) and C (0, 50). Firstly, draw the graph of the line x + y = 50 Secondly, draw the graph of the line x + y = 80 Prepared by: M. S. KumarSwamy, TGT(Maths) Page

178 Feasible region is bounded. Now value of Z is calculated at corner point as Hence the co-operative society of farmers will get the maximum profit of Rs. 4,95,000 by allocating 30 hectares for crop A and 0 hectares for crop B. Yes, because excess use of herbicide can make drainage water poisonous and thus it harm the life of water living creature and wildlife. 7. A company produces soft drinks that has a contract which requires that a minimum of 80 units of the chemical A and 60 units of the chemical B go into each bottle of the drink. The chemicals are available in prepared mix packets from two different suppliers. Supplier S had a packet of mix of 4 units of A and units of B that costs Rs. 0. The supplier T has a packet of mix of unit of A and unit of B costs Rs.4. How many packets of mixed from S and T should the company purchase to honour the contract requirement and yet minimize cost? Make a LPP and solve graphically. Let x and y units of packet of mixes are purchased from S and T respectively. If Z is total cost then Z = 0x + 4y...(i) is objective function which we have to minimize. Here constraints are. 4x + y 80...(ii) x + y 60...(iii) Also, x 0...(iv) y 0...(v) On plotting graph of above constraints or inequalities (ii), (iii), (iv) and (v) we get shaded region having corner point A, P, B as feasible region. For coordinate of P Prepared by: M. S. KumarSwamy, TGT(Maths) Page

179 Point of intersection of x + y = 60...(vi) and 4x + y = 80...(vii) (vi) (vii) x + y 4x y = x = 0 x = 0 y = 40 Since co-ordinate of P = (0, 40) Now the value of Z is evaluated at corner point in the following table Since feasible region is unbounded. Therefore we have to draw the graph of the inequality. 0x + 4y < 60...(viii) Since the graph of inequality (viii) does not have any point common. So the minimum value of Z is 60 at (0, 40). i.e., minimum cost of each bottle is Rs. 60 if the company purchases 0 packets of mixes from S and 40 packets of mixes from supplier T. 8. A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of grinding/cutting machine and a sprayer. It takes hours on the grinding/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp. It takes one hour on the grinding/cutting machine and hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at the most 0 hours and the grinding/cutting machine for at the most hours. The profit from the sale of a lamp is Rs. 5 and that from a shade is Rs. 3. Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximise his profit? Make an L.P.P. and solve it graphically. Let the number of padestal lamps and wooden shades manufactured by cottage industry be x and y respectively. Here profit is the objective function Z. Z =5x + 3y (i) We have to maximise Z subject to the constrains x + y (ii) 3x + y 0 (iii) x 0 and y 0 (iv) On plotting graph of above constraints or inequalities (ii), (iii) and (iv) we get shaded region having corner point A, B, C as feasible region. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

180 Since (0, 0) Satisfy 3x y 0 Graph of 3x y 0 is that half plane in which origin lies. The shaded area OABC is the feasible region whose corner points are O, A, B and C. For coordinate B. Equation x y and 3x y 0 are solved as 3x ( x) 0 3x 4 4x 0 x 4 y 8 4 Coordinate of B (4, 4) Now we evaluate objective function Z at each corner. Hence maximum profit is ` 3 when manufacturer produces 4 lamps and 4 shades. 9. A merchant plans to sell two types of personal computers a desktop model and a portable model that will cost Rs. 5,000 and Rs. 40,000 respectively. He estimates that the total monthly demand of computers will not exceed 50 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than Rs. 70 lakhs and his profit on the desktop model is Rs. 4,500 and on the portable model is Rs. 5,000. Make an L.P.P. and solve it graphically. Let the number of desktop and portable computers to be sold be x and y respectively. Here, Profit is the objective function Z. Z = 4500x y (i) we have to maximise z subject to the constraints x + y 50 (ii) (Demand Constraint) Prepared by: M. S. KumarSwamy, TGT(Maths) Page

181 5000x y 70,00,000 (iii) (Investment constraint) 5x + 8y 400 x 0, y 0 (iv) (Non-negative constraint) On plotting graph of above constraints or inequalities, we get shaded region having corner point A, B, C as feasible region. For coordinates of C, equation x + y = 50 and 5x + 8y = 400 are solved and we get x = 00, y = 50 Now, we evaluate objective function Z at each corner Maximum profit is Rs.,50,000 when he plan to sell 00 unit desktop and 50 portable computers. 0. A factory makes two types of items A and B, made of plywood. One piece of item A requires 5 minutes for cutting and 0 minutes for assembling. One piece of item B requires 8 minutes for cutting and 8 minutes for assembling. There are 3 hours and 0 minutes available for cutting and 4 hours for assembling. The profit on one piece of item A is Rs 5 and that on item B is Rs 6. How many pieces of each type should the factory make so as to maximise profit? Make it as an L.P.P. and solve it graphically. Let the factory makes x pieces of item A and B by pieces of item. Time required by item A (one piece) cutting = 5 minutes, assembling = 0 minutes Time required by item B (one piece) cutting = 8 minutes, assembling = 8 minutes Total time cutting = 3 hours & 0 minutes, assembling = 4 hours Profit on one piece item A = Rs 5, item B = Rs 6 Thus, our problem is maximized Z = 5x + 6y Prepared by: M. S. KumarSwamy, TGT(Maths) Page

182 Subject to x 0, y 0 5x + 8y 00 0x + 8y 40 On plotting graph of above constraints or inequalities, we get shaded region. From figure, possible points for maximum value of z are at (4, 0), (8, 0), (0, 5). at (4, 0), z = 0 at (8, 0), z = = 60 (maximum) at (0, 5), z = 50 8 pieces of item A and 0 pieces of item B produce maximum profit of Rs 60.. One kind of cake requires 300 g of flour and 5 g of fat, another kind of cake requires 50 g of flour and 30 g of fat. Find the maximum number of cakes which can be made from 7 5 kg of flour and 600 g of fat, assuming that there is no shortage of the other ingredients used in making the cakes. Make it as an L.P.P. and solve it graphically. Let number of first kind and second kind of cakes that can be made be x and y respectively Then, the given problem is Maximize, z = x + y Prepared by: M. S. KumarSwamy, TGT(Maths) Page

183 Subjected to x 0, y 0 300x + 50y 7500 x + y 50 5x + 30y 600 x + y 40 On plotting graph of above constraints or inequalities, we get shaded region. From graph, three possible points are (5, 0), (0, 0), (0, 0) At (5, 0), z = x + y = = 5 At (0, 0), z = x + y = = 30 Maximum At (0, 0), z = = 0 As Z is maximum at (0, 0), i.e., x = 0, y = 0. 0 cakes of type I and 0 cakes of type II can be made.. A small firm manufactures gold rings and chains. The total number of rings and chains manufactured per day is atmost 4. It takes hour to make a ring and 30 minutes to make a chain. The maximum number of hours available per day is 6. If the profit on a ring is Rs. 300 and that on a chain is Rs 90, find the number of rings and chains that should be manufactured per day, so as to earn the maximum profit. Make it as an L.P.P. and solve it graphically. Total no. of rings & chain manufactured per day = 4. Time taken in manufacturing ring = hour Time taken in manufacturing chain = 30 minutes One time available per day = 6 Maximum profit on ring = Rs 300 Maximum profit on chain = Rs 90 Let gold rings manufactured per day = x Chains manufactured per day = y L.P.P. is maximize Z = 300x + 90y Subject to x 0, y 0, x + y 4 and x + y 6 On plotting graph of above constraints or inequalities, we get shaded region. Possible points for maximum Z are (6, 0), (8, 6) and (0, 4). At (6, 0), Z = = 4800 At (8, 6), Z = = 5440 Maximum At (0, 4), Z = = 4560 Z is maximum at (8, 6). 8 gold rings & 6 chains must be manufactured per day. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

184 Prepared by: M. S. KumarSwamy, TGT(Maths) Page

185 CHAPTER 3: PROBABILITY QUICK REVISION (Important Concepts & Formulae) MARKS WEIGHTAGE 0 marks Trial and Elementary Events : Let a random experiment be repeated under identical conditions. Then the experiment is called a trial and the possible outcomes of the experiment are known as elementary events or cases. Elementary events are also known as indecomposable events. Decomposable Events/Compound Events : Events obtained by combining together two or more elementary events are known as the compound events or decomposable events. Exhaustive Number of Cases : The total number of possible outcomes of a random experiment in a trial is known as the exhaustive number of cases. The total number of elementary events of a random experiment is called the exhaustive number of cases. Mutually Exclusive Events : Events are said to be mutually exclusive or incompatible if the occurrence of anyone of them prevents the occurrence of all the others, i.e., if no two or more of them can occur simultaneously in the same trial. Equally Likely Events : Events are equally likely if there is no reason for an event to occur in preference to any other events. The number of cases favourable to an events in a trial is the total number of elementary events such that the occurrence of any one of them ensures the happening of the event. Independent Events : Events are said to be independent if the happening (or non-happening) of one event is not affected by the happening (or non-happening) of others. Classical Definition of Probability of An Event : If there are n elementary events associated with a random experiment and m of them are favourable to an event A, then the probability of happening of A is denoted by P(A) and is defined as the ratio m n. m ( P A) n 0 P(A). A denotes not happening of A P( A ) = P(A) P(A) + P(A) = If P(A) =, then A is called certain event and A is called an impossible event if P(A) = 0. m The odds in favour of occurrence of the event A are defined by i.e., P(A) : P( A ) and the odds n m against the occurrence of A are defined by n m, i.e., P( A ) : P(A). m Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 8 -

186 Sample Space : The set of all possible outcomes of a random experiment is called the sample space associated with it and it is generally denoted by S. If E, E,..., E n are the possible outcomes of a random experiment, then S = {E, E,..., E n }. Each element of S is called a sample point. Event : A subset of the sample space associated with a random experiment is called an event. Elementary Events : Single element subsets of the sample space associated with a random experiment are known as the elementary events or indecomposable events. Compound Events : Those subsets of the sample space S associated to an experiment which are disjoint union of single element subsets of the sample space S are known as the compound or decomposable events. Impossible and Certain Event : Let S be the sample space associated with a random experiment. Then f and S, being subsets of S are events.the event f is called an impossible event and the event S is known as a certain event. Occurence or Happening of An Event : Let S be the sample space associated with a random experiment and let A be an event. If w is an outcome of a trial such that w A, then we say that the event A has occurred. If w A, we say that the event A has not occured. Algebra of Events Mutually Exclusive Events : Let S be the sample space associated with a random experiment and let A and A be two events.then A and A are mutually exclusive events if A A =. Mutually Exclusive and Exhaustive System of Events : Let S be the sample space associated with a random experiment. Let A, A,..., A n be subsets of S such that (i) A i A j = for i j, and (ii) A A... A n = S. If E, E,..., E n are elementary events associated with a random experiment. Then (i) E i E j = f for i j and (ii) E E.. E n = S. Favourable Events : Let S be the sample space associated with a random experiment and let A S. Then the elementary events belonging to A are known as the favourable events to A. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 8 -

187 Experimentally Probability : Let S be the sample space associated with a random experiment, and let A be a subset of S representing an event. Then the probability of the event A is defined as Number of elements in A n( A) P( A) Number of elements in S n( s) P() = 0, P (S) =. Addition theorem for two events : If A and B are two events associated with a random experiment, then P(A B) = P (A) + P(B) P(A B). If A and B are mutually exclusive events, then P(A B) = 0, therefore P(A B) = P(A) + P(B). This is the addition theorem for mutually exclusive events. Addition theorem for three events : If A, B, C are three events associated with a random experiment then, P(A B C) = P(A) + P(B) + P(C) P(A B) P(B C) P(A C) + P(A B C). If A, B, C are mutually exclusive events, then P (A B) = P (B C) = P (A C) = P(A B C) = 0. P(A B C) = P(A) + P(B) + P(C) Let A and B be two events associated with a random experiment.then (i) P( A B) = P(B) P(A B) (ii) P(A B ) = P(A) P(A B ) P ( A B) is known as the probability of occurence of B only. P(A B ) is known as the probability of occurence of A only. If B A, then (i) P(A B ) = P(A) P(B) (ii) P(B) P(A). Conditional Probability P (A/B) = Probability of occurrence of A given that B has already happened. P (B/A) = Probability of occurrence of B given that A has already happened. Properties of conditional probability Property P (S/F) = P(F/F) = Property If A and B are any two events of a sample space S and F is an event of S such that P(F) 0, then P((A B)/F) = P(A/F) + P(B/F) P ((A B)/F) Property 3 P(E /F) = P(E/F) Multiplication Theorem : If A and B are two events, then P(A B) = P(A)P(B / A), if P(A) 0 P(A B) = P(B)P(A/ B), if P(B ) 0. ( ) ( ) ( / ) P A B ( / ) P A P B A and P A B B P( A) P( B) Extension of multiplication theorem : If A, A,..., A n are n events related to a random experiment, then A A 3 A n P( A A A3... An ) P( A ) P P... P A A A A A... An where P(A i / A A... A i ) represents the conditional probability of the event A i, given that the events A, A,..., A i have already happened. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

188 Independent Events : Event are said to be independent, if the occurrence or nonoccurrence of one does not affect the probability of the occurrence or nonoccurrence of the other. If A and B are two independent events associated with a random experiment then, P(A/B) = P(A) and P(B/A) = P(B) and viceversa. If A and B are independent events associated with a random experiment, then P(A B) = P(A) P(B) i.e., the probability of simultaneous occurrence of two independent events is equal to the product of their probabilities. If A, A,..., A n are independent events associated with a random experiment, then P(A A A 3... A n ) = P (A ) P (A )... P(A n ) If A, A,..., A n are n independent events associated with a random experiment, then P( A A A... A ) P( A ) P( A )... P( A ) 3 n Prepared by: M. S. KumarSwamy, TGT(Maths) Page n Events A, A,..., A n are independent or mutually independent if the probability of the simultaneous occurence of (any) finite number of them is equal to the product of their separate probabilities while these events are pair wise independent if P(A i A j ) = P(A j ) P(A i ) for all i j. The Law of Total Probability : Let S be the sample space and let E, E,..., E n be n mutually exclusive and exhaustive events associated with a random experiment. If A is any event which occurs with E or E or... or E n, then A A A P( A) P( E ) P P( E ) P... P( En) P E E En Baye's Rule : Let S be the sample space and let E, E,..., E n be n mutually exclusive and exhaustive events associated with a random experiment. If A is any event which occurs with E or E or... or E n, then Ei P( Ei ) P( A/ Ei ) P, i,,... n n A P( E ) P( A / E ) i i The events E, E,..., E n are usually referred to as 'hypothesis' and the probabilities P(E ), P(E ),..., P(E n ) are known as the 'priori' probabilities as they exist before we obtain any information from the experiment. The probabilities P(A/E i ); i =,,..., n are called the likelihood probabilities as they tell us how likely the event A under consideration occur, given each and every priori probabilities. The probabilities P (E i /A); i =,,..., n are called the posterior probabilities as they are determined after the result of the experiment are known. Random Variable : A random variable is a real valued function having domain as the sample space associated with a given random experiment. A random variable associated with a given random experiment associates every event to a unique real number. Probability Distribution : If a random variable X takes values x, x,..., x n with respective, probabilities p, p,..., p n, then X : x x... x n i

189 P(X) : p p... p n is called the probability distribution of X. Binomial Distribution : A random variable X which takes values 0,,,..., n is said to follow n r n r binomial distribution if its probability distribution function is given by P( X r) Cr p q, r = 0,,,..., n where p, q > 0 such that p + q =. If n trials constitute an experiment and the experiment is repeated N times, then the frequencies of 0,,,..., n successes are given by N P(X = 0), N P (X=), N P (X=),..., N P (X=n). If X is a random variable with probability distribution X : x x... x n P(X) : p p... p n then the mean or expectation of X is defined as as and the variance of X is defined as n n X E( X ) p x and the variance of X is defined i n Var( X ) p x E( X ) p x E( X ) i i i i i i or Var (X) = E (X ) [E(X)] The mean of the binomial variate X ~ B (n, p) is n p. The variance of the binomial variate X ~ B (n, p) is npq, where p + q = i i The standard deviation of a binomial variate X ~ B (n, p) is Var( X ) mean > variance. npq Maximum Value of P(X = r) for given Values of n and p for a Binomial Variate X If (n + ) p is an integer, say m, then P( X r) n C r ( ) n r r p p is maximum when r = m or r = m. If (n + )p is not an integer, then P(X = r) is maximum when r = [(n + )p]. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

190 CHAPTER 3: PROBABILITY MARKS WEIGHTAGE 0 marks NCERT Important Questions & Answers. A die is thrown three times. Events A and B are defined as below: A : 4 on the third throw; B : 6 on the first and 5 on the second throw. Find the probability of A given that B has already occurred. The sample space has 6 outcomes. Now A = (,,4) (,,4)... (,6,4) (,,4) (,,4)... (,6,4) (3,,4) (3,,4)... (3,6,4) (4,,4) (4,,4)...(4,6,4) (5,,4) (5,,4)... (5,6,4) (6,,4) (6,,4)...(6,6,4) B = {(6,5,), (6,5,), (6,5,3), (6,5,4), (6,5,5), (6,5,6)} and A B = {(6,5,4)}. 6 Now, P( B) and P( A B) 6 6 P( A B) Then P( A / B) 6 P( B) A die is thrown twice and the sum of the numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once? Let E be the event that number 4 appears at least once and F be the event that the sum of the numbers appearing is 6. Then, E = {(4,), (4,), (4,3), (4,4), (4,5), (4,6), (,4), (,4), (3,4), (5,4), (6,4)} and F = {(,5), (,4), (3,3), (4,), (5,)} 5 We have P( E) and P( E) Also E F = {(,4), (4,)} Therefore P( E F) 36 P( E F) Hence, the required probability, P( E / F) 36 P( F) A black and a red dice are rolled. (a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5. (b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4. Let the first observation be from the black die and second from the red die. When two dice (one black and another red) are rolled, the sample space S = 6 6 = 36 (equally likely sample events) (i) Let E : set of events in which sum greater than 9 and F : set of events in which black die resulted in a 5 E = {(6,4), (4,6), (5, 5), (5,6), (6, 5), (6,6)} n(e) = 6 and F = {(5, ), (5,), (5, 3), (5, 4), (5, 5), (5,6)} n(f) = 6 E F = {(5, 5), (5,6)} n(e F) = Prepared by: M. S. KumarSwamy, TGT(Maths) Page

191 The conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5, is given by P(E/F) 6 6 P( E) and P( F) Also, P( E F) 36 P( E F) P( E / F) 36 P( F) (ii) Let E : set of events having 8 as the sum of the observations, F : set of events in which red die resulted in a (in any one die) number less than 4 E = {(,6), (3, 5), (4,4), (5, 3), (6,)} n(e) = 5 and F = {(, ), (, ),.. (3, ), (3, ) (5, ), (5, ),.} n(f) = 8 E F = {(5, 3), (6, )} n(e F) = The conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5, is given by P(E/F) 5 8 P( E) and P( F) Also, P( E F) 36 P( E F) P( E / F) 36 P( F) An instructor has a question bank consisting of 300 easy True / False questions, 00 difficult True / False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question? Total number of questions = = 400. Let E be the event that selected question is an easy question Then, n(e) = = P( E) 400 Let F be the event that selected question is a multiple choice question. Then, n(f) = = P( F) Also, P( E F) P( E F) P( E / F) 400 P( F) An urn contains 0 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls are black? Let E and F denote respectively the events that first and second ball drawn are black. We have to find P(E F) or P (EF). Prepared by: M. S. KumarSwamy, TGT(Maths) Page

192 Now P(E) = P (black ball in first draw) = 0 5 Also given that the first ball drawn is black, i.e., event E has occurred, now there are 9 black balls and five white balls left in the urn. Therefore, the probability that the second ball drawn is black, given that the ball in the first draw is black, is nothing but the conditional probability of F given that E has occurred. i.e. P(F E) = 9 4 By multiplication rule of probability, we have P (E F) = P(E) P(F E) = Two balls are drawn at random with replacement from a box containing 0 black and 8 red balls. Find the probability that (i) both balls are red. (ii) first ball is black and second is red. (iii) one of them is black and other is red. Total number of balls = 8, number of red balls = 8 and number of black balls = 0 Probability of drawing a red ball = 8 8 Similarly, probability of drawing a black ball = 0 8 (i) Probability of getting both red balls = P (both balls are red) = P (a red ball is drawn at first draw and again a red ball at second draw) = (ii) P (probability of getting first ball is black and second is red) = (iii) Probability of getting one black and other red ball = P(first ball is black and second is red) + P (first ball is red and second is black) = Probability of solving specific problem independently by A and B are and respectively. If 3 both try to solve the problem independently, find the probability that (i) the problem is solved (ii) exactly one of them solves the problem. Probability of solving the problem by A, P(A) = Probability of solving the problem by B, P(B) = 3 Probability of not solving the problem by A = P(A ) = P(A) = and probability of not solving the problem by B = P(B ) = P(B) = 3 3 (i) P (the problem is solved) = P(none of them solve the problem) = P( A' B ') P( A') P( B ') (since A and B are independent A and B are independent) (ii) P (exactly one of them solve the problem) = P(A) P(B ) + P(A ) P(B) Prepared by: M. S. KumarSwamy, TGT(Maths) Page

193 8. In a hostel, 60% of the students read Hindi news paper, 40% read English news paper and 0% read both Hindi and English news papers. A student is selected at random. (a) Find the probability that she reads neither Hindi nor English news papers. (b) If she reads Hindi news paper, find the probability that she reads English news paper. (c) If she reads English news paper, find the probability that she reads Hindi news paper. Let H : Set of students reading Hindi newspaper and E : set of students reading English newspaper. Let n(s) = 00 Then, n(h) = 60 n(e) = 40 and n(h E) = P( H ), P( E) and P( H E) (i) Required probability = P (student reads neither Hindi nor English newspaper) = P( H ' E ') P( H E)' P( H E) 3 4 P( H ) P( E) P( H E) (ii) Required probability = P(a randomly chosen student reads English newspaper, if he/she reads P( E H ) Hindi newspaper) = P( E / H ) 5 P( H ) (iii) Required probability = P (student reads Hindi newspaper when it is given that reads English P( H E) newspaper) = P( H / E) 5 P( E) 5 9. Bag I contains 3 red and 4 black balls while another Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from Bag II. Let E be the event of choosing the bag I, E the event of choosing the bag II and A be the event of drawing a red ball. Then P(E ) = P(E ) = Also P(A E ) = P(drawing a red ball from Bag I) = 3 7 and P(A E ) = P(drawing a red ball from Bag II) = 5 Now, the probability of drawing a ball from Bag II, being given that it is red, is P(E A) By using Bayes' theorem, we have 5 P( E) P( A / E) 35 P( E / A) P( E 3 5 ) P( A / E ) P( E ) P( A / E ) Given three identical boxes I, II and III, each containing two coins. In box I, both coins are gold coins, in box II, both are silver coins and in the box III, there is one gold and one silver coin. A person chooses a box at random and takes out a coin. If the coin is of gold, what is the probability that the other coin in the box is also of gold? Let E, E and E 3 be the events that boxes I, II and III are chosen, respectively. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

194 Then P(E ) = P(E ) = P(E 3 ) = 3 Also, let A be the event that the coin drawn is of gold Then P(A E ) = P(a gold coin from bag I) = / = P(A E ) = P(a gold coin from bag II) = 0 P(A E 3 ) = P(a gold coin from bag III) = Now, the probability that the other coin in the box is of gold = the probability that gold coin is drawn from the box I. = P(E A) By Bayes' theorem, we know that P( E) P( A / E ) P( E / A) P( E ) P( A / E ) P( E) P( A / E) P( E3 ) P( A/ E3) In a factory which manufactures bolts, machines A, B and C manufacture respectively 5%, 35% and 40% of the bolts. Of their outputs, 5, 4 and percent are respectively defective bolts. A bolt is drawn at random from the product and is found to be defective. What is the probability that it is manufactured by the machine B? Let events B, B, B 3 be the following : B : the bolt is manufactured by machine A B : the bolt is manufactured by machine B B 3 : the bolt is manufactured by machine C Clearly, B, B, B 3 are mutually exclusive and exhaustive events and hence, they represent a partition of the sample space. Let the event E be the bolt is defective. The event E occurs with B or with B or with B 3. Given that, P(B ) = 5% = 0.5, P (B ) = 0.35 and P(B 3 ) = 0.40 Again P(E B ) = Probability that the bolt drawn is defective given that it is manufactured by machine A = 5% = 0.05 Similarly, P(E B ) = 0.04, P(E B 3 ) = 0.0. Hence, by Bayes' Theorem, we have P( B ) P( E / B ) P( B / E) P( B ) P( E / B ) P( B ) P( E / B ) P( B3 ) P( E / B3 ) A doctor is to visit a patient. From the past experience, it is known that the probabilities that he will come by train, bus, scooter or by other means of transport are respectively 3 0, 5, 0 and. The probabilities that he will be late are, and, if he comes by train, bus and scooter respectively, but if he comes by other means of transport, then he will not be late. When he arrives, he is late. What is the probability that he comes by train? Let E be the event that the doctor visits the patient late and let T, T, T 3, T 4 be the events that the doctor comes by train, bus, scooter, and other means of transport respectively. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

195 Then P(T ) = 3 0, P(T ) = 5, P(T 3) = 0 and P(T 4) = 5 P(E T ) = Probability that the doctor arriving late comes by train = 4 Similarly, P(E T ) = 3, P(E T 3) = and P(E T 4) = 0, since he is not late if he comes by other means of transport. Therefore, by Bayes' Theorem, we have P(T E) = Probability that the doctor arriving late comes by train P( T ) P( E / T ) P( T / E) P( T ) P( E / T ) P( T ) P( E / T ) P( T3 ) P( E / T3 ) P( T4 ) P( E / T4 ) Hence, the required probability is 3. A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six. Let E be the event that the man reports that six occurs in the throwing of the die and let S be the event that six occurs and S be the event that six does not occur. Then P(S ) = Probability that six occurs = 6 P(S ) = Probability that six does not occur = 5 6 P(E S ) = Probability that the man reports that six occurs when six has actually occurred on the die = Probability that the man speaks the truth = 3 4 P(E S ) = Probability that the man reports that six occurs when six has not actually occurred on the die 3 = Probability that the man does not speak the truth = 4 4 Thus, by Bayes' theorem, we get P(S E) = Probability that the report of the man that six has occurred is actually a six 3 P( S) P( E / S) 4 3 P( S 6 4 / E) P( S 3 5 ) P( E / S) P( S) P( E / S) Hence, the required probability is A bag contains 4 red and 4 black balls, another bag contains red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag. Let E : first bag is selected, E : second bag is selected Then, E and E are mutually exclusive and exhaustive. Moreover, P(E ) = P(E ) = Let E : ball drawn is red. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 9 -

196 P(E/E ) = P(drawing a red ball from first bag) = 4 8 P(E/E ) = P(drawing a red ball from second bag) = 8 4 By using Baye s theorem, P( E ) P( E / E ) Required probability = P( E / E) P( E ) P( E / E ) P( E) P( E / E) Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 0% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hostlier? Let E : the event that the student is residing in hostel and E : the event that the student is not residing in the hostel. Let E : a student attains A grade, Then, E and E are mutually exclusive and exhaustive. Moreover, P(E ) = 60% = 60 3 and P(E ) = 40% = Then P(E/E ) = 30% = 30 3 and P(E/E ) = 0% = By using Baye s theorem, we obtain 3 3 P( E ) P( E / E ) 9 9 P( E / E) 5 0 P( E ) P( E / E ) P( E) P( E / E) In answering a question on a multiple choice test, a student either knows the answer or guesses. Let 3 4 be the probability that he knows the answer and 4 be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability 4. What is the probability that the student knows the answer given that he answered it correctly? Let E : the event that the student knows the answer and E : the event that the student guesses the answer. Then, E and E are mutually exclusive and exhaustive. Moreover, P(E ) = 3 4 and P(E ) = 4 Let E : the answer is correct. The probability that the student answered correctly, given that he knows the answer, is i.e., P P(E/E ) = Probability that the students answered correctly, given that the he guessed, is 4 i.e., P(E/E ) = 4 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 9 -

197 By using Baye s theorem, we obtain P( E ) P( E / E ) P( E / E) P( E ) P( E / E ) P( E) P( E / E) There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin? Let E : the event that the coin chosen is two headed, E : the event that the coin chosen is biased and E 3 : the event that the coin chosen is unbiased E, E, E 3 are mutually exclusive and exhaustive events. Moreover, P(E ) = P(E ) = P(E 3 ) = 3 Let E : tosses coin shows up a head, P(E/E ) = P(coin showing heads, given that it is a two headed coin) = 75 3 P(E/E ) = P(coin showing heads, given that it is a biased coin) = 75% 00 4 P(E/E3) = P(coin showing heads, given that it is an unbiased coin) = The probability that the coin is two headed, given that it shows head, is given by P(E /E) By using Baye s theorem, we obtain P( E) P( E / E ) P( E / E) P( E ) P( E / E) P( E) P( E / E ) P( E3) P( E / E3) An insurance company insured 000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accidents are 0.0, 0.03 and 0.5 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver? There are 000 scooter drivers, 4000 car drivers and 6000 truck drivers. Total number of drivers = = 000 Let E : the event that insured person is a scooter driver, E : the event that insured person is a car driver and E 3 : the event that insured person is a truck driver. Then, E, E, E 3 are mutually exclusive and exhaustive events. Moreover, P(E ) = 000, P(E ) = 4000 and P(E 3 ) = Let E : the events that insured person meets with an accident, P(E/E ) = P(scooter driver met with an accident) = P(E/E ) = P(car driver met with an accident) = P(E/E3) = P(truck driver met with an accident) = The probability that the driver is a scooter driver, given he met with an accident, is given by P(E /E) Prepared by: M. S. KumarSwamy, TGT(Maths) Page

198 By using Baye s theorem, we obtain P( E) P( E / E ) P( E / E) P( E ) P( E / E) P( E) P( E / E ) P( E3) P( E / E3) A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further, % of the items produced by machine A and % produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B? Let E : the event that the item is produced by machine A and E : the event that the item is produced by machine B. Then, E and E are mutually exclusive and exhaustive events. Moreover, 60 3 P(E ) = 60% and P(E ) = 40% = Let E : the event that the item chosen is defective, P(E/E ) = P(machine A produced defective items) = P(E/E ) = P(machine B produced defective items) = % 00 % 00 The probability that the randomly selected item was from machine B, given that it is defective, is given by P(E /E) By using Baye s theorem, we obtain P( E) P( E / E) P( E / E) P( E ) P( E / E ) P( E) P( E / E) Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets,, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw,, 3 or 4 with the die? Let E : the event that 5 or 6 is shown on die and E : the event that,, 3, or 4 is shown on die. Then, E and E are mutually exclusive and exhaustive events. and n(e ) =, n(e ) = 4 Also, n(s) = 6 P(E ) = and P(E ) = Let E : The event that exactly one head show up, P(E/E ) = P(exactly one head show up when coin is tossed thrice) = P{HTT, THT, TTH} = 3 8 P(E/E ) = P(head shows up when coin is tossed once) = Prepared by: M. S. KumarSwamy, TGT(Maths) Page

199 The probability that the girl threw,,, 3 or 4 with the die, if she obtained exactly one head, is given by P(E /E) By using Baye s theorem, we obtain P( E) P( E / E) P( E / E) P( E ) P( E / E ) P( E) P( E / E) A manufacturer has three machine operators A, B and C. The first operator A produces % defective items, where as the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on the job for 30% of the time and C is on the job for 0% of the time. A defective item is produced, what is the probability that it was produced by A? Let E : the event that item is produced by machine A, E : the event that item is produced by machine B and E 3 : the event that item is produced by machine C Here, E, E and E 3 are mutually exclusive and exhaustive events. Moreover, P(E ) = 50% P(E ) = 30% = and P(E 3 ) = 0% = 0 00 Let E : The event that item chosen is found to be defective, P(E/E ) = 00, P(E/E 5 ) = 00, P(E/E 3) = 7 00 By using Baye s theorem, we obtain P( E) P( E / E ) P( E / E) P( E ) P( E / E ) P( E ) P( E / E ) P( E ) P( E / E ) A card from a pack of 5 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond. Let E : the event that lost cards is a diamond n(e ) = 3 E : lost cards is not a diamond n(e ) = 5 3 = 39 And, n(s) = 5 Then, E and E are mutually exclusive and exhaustive events. P(E ) = 3 and P(E ) = Let E : the events that two cards drawn from the remaining pack are diamonds, When one diamond card is lost, there are diamond cards out of 5 cards. The cards can be drawn out of diamond cards in C ways. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

200 Similarly, diamond cards can be drawn out of 5 cards in 5 C ways. The probability of getting two cards, when one diamond card is lost, is given by P(E/E ) C 3 P(E/E ) = 5 C C 3 56 and P(E/E ) = 5 C By using Baye s theorem, we obtain P( E ) P( E / E ) P( E / E) P( E ) P( E / E ) P( E) P( E / E) Two cards are drawn successively with replacement from a well-shuffled deck of 5 cards. Find the probability distribution of the number of aces. The number of aces is a random variable. Let it be denoted by X. Clearly, X can take the values 0,, or. Now, since the draws are done with replacement, therefore, the two draws form independent experiments Therefore, P(X = 0) = P(non-ace and non-ace) = P(non-ace) P(non-ace) P(X = ) = P(ace and non-ace or non-ace and ace) = P(ace and non-ace) + P(non-ace and ace) = P(ace). P(non-ace) + P (non-ace). P(ace) and P(X = ) = P (ace and ace) Thus, the required probability distribution is 4. Find the probability distribution of number of doublets in three throws of a pair of dice. Let X denote the number of doublets. Possible doublets are (,), (,), (3,3), (4,4), (5,5), (6,6) Clearly, X can take the value 0,,, or 3. 6 Probability of getting a doublet Probability of not getting a doublet 6 6 Now P(X = 0) = P (no doublet) Prepared by: M. S. KumarSwamy, TGT(Maths) Page

201 P(X = ) = P (one doublet and two non-doublets) P(X = ) = P (two doublets and one non-doublet) and P(X = 3) = P (three doublets) Thus, the required probability distribution is 5. Find the variance of the number obtained on a throw of an unbiased die. The sample space of the experiment is S = {,, 3, 4, 5, 6}. Let X denote the number obtained on the throw. Then X is a random variable which can take values,, 3, 4, 5, or 6. Also P() = P() = P(3) = P(4) = P(5) = P(6) = 6 Therefore, the Probability distribution of X is n Now E(X) = xi p( xi ) i Also E(X 9 ) Thus, Var (X) = E (X ) (E(X)) Two cards are drawn simultaneously (or successively without replacement) from a well shuffled pack of 5 cards. Find the mean, variance and standard deviation of the number of kings. Let X denote the number of kings in a draw of two cards. X is a random variable which can assume the values 0, or. 48! 48 C!(48 )! Now P(X = 0) = P (no king) 5 C 5! 55!(5 )! P(X = ) = P (one king and one non-king) C C C Prepared by: M. S. KumarSwamy, TGT(Maths) Page

202 4 C 43 and P(X = ) = P (two kings) 5 C 55 Thus, the probability distribution of X is n Now Mean of X = E(X) = xi p( xi ) 0 i Also E(X ) = 0 Thus, Var (X) = E (X ) (E(X)) = () Therefore x Var( x) 0.37 () 7. Find the probability distribution of (i) number of heads in two tosses of a coin. (ii) number of tails in the simultaneous tosses of three coins. (iii) number of heads in four tosses of a coin. (i) When one coin is tossed twice, the sample space is S = {HH,HT,TH,TT}. Let X denotes, the number of heads in any outcome in S, X (HH) =, X (HT) =, X (TH) = and X (TT) = 0 Therefore, X can take the value of 0, or. It is known that P(HH) = P(HT) = P(TH) = P(TT) = 4 P(X = 0) = P (tail occurs on both tosses) = P({TT}) = 4 P(X = ) = P (one head and one tail occurs) = P({TH,HT}) = 4 and P(X = ) = P (head occurs on both tosses) = P({HH}) = 4 Thus, the required probability distribution is as follows (ii) When three coins are tossed thrice, the sample space is S = {HHH,HHT,HTH,HTT,THH,THT,TTH,TTT} which contains eight equally likely sample points. Let X represent the number of tails. Then, X can take values 0,, and 3. P(X = 0) = P (no tail) = P({HHH}) = 8, P (X = ) = P (one tail and two heads show up) = P({HHT,HTH,THH}) = 3 8, P (X = ) = P (two tails and one head show up) = P({HTT,THT,TTH}) = 3 8 and P(X = 3) = P (three tails show up) = P({TTT}) = 8 Thus, the probability distribution is as follows Prepared by: M. S. KumarSwamy, TGT(Maths) Page

203 (iii) When a coin is tossed four times, the sample space is S = {HHHH, HHHT, HHTH, HTHT, HTTH, HTTT, THHH, HTHH, THHT, THTH, HHTT, TTHH, TTHT, TTTH, THTT, TTTT} which contains 6 equally likely sample points. Let X be the random variable, which represents the number of heads. It can be seen that X can take the value of 0,,, 3 or 4. P (X = 0 ) = P(no head shows up) = P {TTTT} = 6, P(X = ) = P (one head and three tails show up) = P(HTTT,THTT,TTHT,TTTH) = 4, 6 4 P(X = ) = P (two heads and two tails show up) = P({HHTT,HTHT,HTTH,THHT,THTH,TTHH}) 6 3 =, 6 8 P(X = 3) = P (three heads and one tail show up) = P({HHHT,HHTH,HTHH,THHH}) = and P (X = 4) = P (four heads show up) = P ({HHHH}) = 6 Thus, the probability distribution is as follows: 8. Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as (i) number greater than 4 (ii) six appears on at least one die When a die is tossed two times, we obtain (6 x 6) = 36 number of sample points. (i) Let X be the random variable which denotes the number greater than 4 in two tosses of a die. So X may have values 0, or. Now, P(X = 0) = P (number less than or equal to 4 on both the tosses) = , P(X = ) = P (number less than or equal to 4 on first toss and greater than 4 on second toss) + P(number greater than 4 on first toss and less than or equal to 4 on second toss) = P(X = ) = P (number greater than 4 on both the tosses) = Probability distribution of X, i.e., number of successes is (ii) Let X be the random variable which denotes the number of six appears on atleast one die. So, X may have values 0 or. P(X = 0) = P (six does not appear on any of the die) = Prepared by: M. S. KumarSwamy, TGT(Maths) Page

204 P(X = ) = P (six appears on atleast one of the die) = 36 Thus, the required probability distribution is as follows 9. From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs. It is given that out of 30 bulbs, 6 are defective. Number of non-defective bulbs = 30 6 = 4 4 bulbs are drawn from the lot with replacement. Let p = P(obtaining a defective bulb when a bulb is drawn) = and q = P(obtaining a non-defective bulb when a bulb is drawn) = Using Binomial distribution, we have P(X = 0) = P (no defective bulb in the sample) = P(X = ) = P (one defective bulb in the sample) = C p q C p q P(X = ) = P (two defective and two non-defective bulbs are drawn) = P(X = 3) = P (three defective and one non-defective bulbs are drawn) = C p q P(X = 4) = P (four defective bulbs are drawn) = C4 p q 5 65 Therefore, the required probability distribution is as follows. 4 C p q A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails. Let X denotes the random variable which denotes the number of tails when a biased coin is tossed twice. So, X may have value 0, or. Since, the coin is biased in which head is 3 times as likely to occur as a tail. P{H} = 3 4 and P{T} = P (X = ) = P{HH} = 4 6 P(X = ) = P (one tail and one head) = P{HT,TH} = P{HT} + P{TH} + P{H}P{T} + P {T} P{H} Prepared by: M. S. KumarSwamy, TGT(Maths) Page

205 = P (X = ) = P (two tails) = P{TT} = P{T} P{T} = 4 6 Therefore, the required probability distribution is as follows 3. A random variable X has the following probability distribution: Determine (i) k (ii) P(X < 3) (iii) P(X > 6) (iv) P(0 < X < 3) (i) It is known that the sum of a probability distribution of random variable is one i.e., P( X ) =, therefore P(0) + P() + P() + P(3) + P(4) + P(5) + P(6) + P(7) = 0 k k k 3k k k 7k k 0k 9k 0 ( k )(0 k ) 0 k or k 0 k = is not possible as the probability of an event is never negative. k 0 3 (ii) P(X < 3) = P(0) + P() + P() = 0 k k 3k (iii) P(X > 6) = P(7) = 7k k (iv) P(0 < X < 3) = P() + P() = k k 3k 0 3. The random variable X has a probability distribution P(X) of the following form, where k is some number : k if x 0 k if x P( X ) 3 k, if x 0, otherwise (a) Determine the value of k. (b) Find P (X < ), P (X ), P(X ). Given distribution of X is Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 0 -