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 06 7 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 etremely thankful to Honourable Shri D. Manivannan, Deputy Commissioner, KVS RO Hyderabad, who blessed and motivates me to complete this project work. This materials consists 8 easy Chapters out of 3 Chapters having overall weightage of 55 marks out of 00 marks. Each Chapter 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 maimum 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 ecellent form. I also etend my special thanks to respected sir Shri. P. S. Raju, Principal, KV Gachibowli, and respected sir Shri. E. Krishnamurthy, Principal, KV NFC Nagar Ghatkesar 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 net 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 NCERT Important Questions & Answers Page Relations and Functions Board Important Questions & Answers Page 3 0. Inverse Trigonometric Functions Important Concepts & Formulae Page 4 Inverse Trigonometric Functions NCERT Important Questions & Answers Page 5 34 Inverse Trigonometric Functions Board Important Questions & Answers Page Matrices NCERT Important Questions & Answers Page 4 58 Matrices Board Important Questions & Answers Page Determinants NCERT Important Questions & Answers Page Determinants Board Important Questions & Answers Page Continuity and Differentiability NCERT Important Questions & Answers Page 0 6 Continuity and Differentiability Board Important Questions Page Vector Algebra NCERT Important Questions & Answers Page 0 8 Vector Algebra Board Important Questions & Answers Page Linear Programming NCERT Important Questions & Answers Page Linear Programming Board Important Questions & Answers 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 NCERT Important Questions & Answers MARKS WEIGHTAGE 05 marks. Show that the relation R in the set R of real numbers, defined as R = {(a, b) : a b } is neither refleive nor symmetric nor transitive. We have R = {(a,b) :a b }, where a, b R For refleivity, we observe that is not true. So, R is not refleive 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 refleive, nor symmetric and nor transitive. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - -. Prove that the relation R in R defined by R = {(a, b): a b 3 } is neither refleive nor symmetric nor transitive. Given that R = {(a, b): a b 3 } It is observed that, R as So, R is not refleive. Now, (, ) (as < 3 =8) But (, ) R (as 3 > ) So, R is not symmetric We have 3,,, R as 3 and But 3, R as Therefore, R is not transitive. Hence, R is neither refleive nor symmetric nor transitive. 3. 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 refleive. Let (a, b) R. a b is even (a b) is even (a b) is even

7 (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. 4. Show that each of the relation R in the set A { Z : 0 }, 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 { Z : 0 } {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 refleive. 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 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 5. Let A = R {3} and B = R {}. Prove that the function f : A B defined by f() = one-one and onto? Justify your answer. 3 is f Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 3 -

8 Here, A = R { 3), B = R {} and f : A B is defined as f() = Let, y A such that f() = f(y) y ( )( y 3) ( y )( 3) 3 y 3 y 3 y 6 y 3y 6 3 y 3y 3 3y y y 3 Therefore, f is one- one. Let y B = R {}. Then, y The function f is onto if there eists A such that f() = y. Now, f() = y y y 3y 3 ( y) 3y 3y A [ y ] y Thus, for any y B, there eists 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 If f ( ),, show that (fof)() =, for all. What is the inverse of f? Given that f ( ), Then ( fof )( ) f ( f ( )) f Therefore (fof)() =, for all 3 Hence, the given function f is invertible and the inverse of f is itself. 7. Show that f :[,] R, given by f ( ),, is one-one. Find the inverse of the function f :[,] Range f. Given that f :[,] R, given by f ( ),, Let f() = f(y) Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 4 -

9 y y y y y y y Therefore, f is a one-one function. y Let y y y y So, for every y ecept in the range there eists in the domain such that f() = 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 eists. Let y be an arbitrary element of range f. Since, f :[,] Range f is onto, we have y = f() for some [,] y y y y, y y y Now, let us define g: Range f [,] as g( y), y y ( gof )( ) g( f ( )) g 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 8. Consider f : R R given by f () = Show that f is invertible. Find the inverse of f. Here, f :R R is given by f() = Let,y R, such that f() = f(y) = 4y = 4y = y Therefore, f is a one-one function. Let y = y 3 There eist, R, y R 4 y 3 Therefore, for any y R, there eist R such that 4 y 3 y 3 f ( ) f 4 3 y 4 4 Therefore, f is onto function. Thus, f is one-one and onto and therefore, f eists. 3 Let us define g : R R by g( ) 4 (4 3) 3 Now, ( gof )( ) g( f ( )) g(4 3) 4 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 5 -

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

11 y 6 6 y 6 6 y Therefore, gof I R and fog I[ 5, ) y 6 Hence, f is invertible and the inverse of f if given by f ( y) g( y) 3. 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 eists 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 *.. Let * be the binary operation on N defined by a*b =HCF of a and b. Is * commutative? Is * associative? Does there eist 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. 3. 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. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 7 -

12 It can be observed that for,3,4 Q, we have *3 = 3 = and 3* = 3 = *3 3* Thus, the operation is not commutative. 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 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 8 -

13 * 3 = 3 = 8 and 3* = 3 = Hence, *3 3* Thus, the operation is not commutative. 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. 4. 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. 5. Show that the function f :R { R : < <} defined by f( ) =, R onto function. is one-one and Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 9 -

14 It is given that f :R { R : < <} defined by f( ) =, R y Suppose, f() = f(y), where, y R y It can be observed that if is positive and y is negative, then we have y y y y Since, is positive and y is negative, then > y y > 0 But, y is negative. Then, y y. Thus, the case of being positive and y being negative can be ruled out. Under a similar argument, being negative and y being positive can also be ruled out. Therefore, and y have to be either positive or negative. y When and y are both positive, we have f ( ) f ( y) y y y y y y When and y are both negative, we have f ( ) f ( y) y y y y y Therefore, f is one-one. Now, let y R such that < y <. y If y is negative, then there eists R such that y y y y y y y f ( ) f y y y y y y y y y If y is positive, then there eists R such that y y y y y y y f ( ) f y y y y y y y y Therefore, f is onto. Hence, f is one-one and onto. 6. Show that the function f :R R given by f () = 3 is injective. Here, f :R R is given as f() = 3. Suppose, f() = f(y),where,y R 3 = y 3 (i) Now, we need to show that = y Suppose, y, their cubes will also not be equal. 3 y 3 However, this will be a contradiction to Eq. i). Therefore, = y. Hence, f is injective. a b, ifa b 6 7. 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} Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 0 -

15 The operation * on X is defined as a b, ifa b 6 a* b a b 6, ifa b 6 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 eists 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. 8. 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 refleive, 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. 9. Show that if f : A B and g : B C are one-one, then gof : A C is also one-one. Suppose gof ( ) = gof ( ) g (f ( )) = g(f ( )) f ( ) = f ( ), as g is one-one =, as f is one-one Hence, gof is one-one. 0. 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, Prepared by: M. S. KumarSwamy, TGT(Maths) Page - -

16 a b c a b a b c ( a* b)* c * c 4 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 3 4 f ( ) 5 7 and 3 7 g : R R 5 5 is 7 4 defined by g( ) 5 3, then fog = I 3 7 A and gof = I B, where, A R, B R 5 5 ; I A() =, A, I B () =, B are called identity functions on sets A and B, respectively We have gof ( ) g Similarly, fog( ) f Thus, gof () =, B and fog () =, A, which implies that gof = I B and fog = I A.. Let f : N R be a function defined as f () = 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 = , for some in N, which implies that y = ( + 3) y This gives, as y 6. y 6 3 Let us define g : S N by g( y) Now gof () = g(f ()) = g( ) = g (( + 3) + 6) and ( 3) ( 3 3) 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 - -

17 CHAPTER : RELATIONS AND FUNCTIONS Previous Years Board Eam (Important Questions & Answers). If f() = + 7 and g() = 7, R, find ( fog) (7) Given f() = + 7 and g() = 7, R fog() = f(g()) = g() + 7 = ( 7) + 7 = (fog) (7) = 7. MARKS WEIGHTAGE 05 marks. If f() is an invertible function, find the inverse of f () = Given f ( ) 5 3 Let y 5 5y 3 5y 3 5 f ( ) 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) Refleive R is refleive if T R T Since T T R is refleive. (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 = = + 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 - 3 -

18 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, y N and let they are odd then y f ( ) f ( y) y If, y N are both even then also y f ( ) f ( y) y If, y N are such that is even and y is odd then y f ( ) and f ( y) Thus, y for f() = f(y) Let = 6 and y = We get f (6) 3, f (5) 3 f() = f(y) but y...(i) So, f () is not one-one. Hence, f () 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 = 8. What is the range of the function f ( ) ( )? We have given f ( ) ( ) ( ), if 0 or ( ), if 0 or ( ) (i) For >, f ( ) ( ) ( ) (ii) For <, f ( ) ( ) Range of f ( ) is {, }. ( ) Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 4 -

19 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 refleive. (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 eists. 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() = (3 3 ) / 3, then find fof(). If f : R R be defined by f() = (3 3 ) /3 then ( fof) = f( f()) = f [(3 3 ) /3 ] = [3 {(3 3 ) /3 } 3 ] /3 = [3 (3 3 )] /3 = ( 3 ) /3 = Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 5 -

20 . 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 (, y) be identity element for on A, Then (a, b) * (, y) = (a, b) (a +, b + y) = (a, b) a + = a, b + y = b = 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() = sin and g() = 5, find gof(). Given f : R R and g : R R defined by f () = sin and g() = 5 gof() = g [f()] = g (sin ) = 5 (sin ) = 5 sin 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() = 3 +, define f[f()]. f (f ()) = f (3 + ) =3. (3 + ) + = = Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 6 -

21 6. Write fog, if f : R R and g : R R are given by f() = 8 3 and g() = /3. fog () = f (g()) = f ( /3 ) = 8( /3 ) 3 = 8 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() =0 +7. Find the function g : R R such that gof = fog =I R. gof = fog = IR fog = IR fog() = I () f (g()) = [I() = being identity function] 0(g()) + 7 = [f() = 0 + 7] 7 g( ) 0 7 i.e., g : R R is a function defined as g( ) 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, A. Now, f( ) = f( ) 3 3 ( )( 3) ( 3)( ) Hence f is one-one function. For Onto Let y y 3y 3 y 3y ( y ) 3y 3y ( i) y f ( ) 3. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 7 -

22 From above it is obvious that y ecept, i.e., y B R {} 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, if is odd. Show that f : N N, given by f ( ) is both one-one and onto., if is even For one-one Case I : When, are odd natural number. f( ) = f( ) + = +, N = i.e., f is one-one. Case II : When, are even natural number f( ) = f( ) = = i.e., f is one-one. Case III : When is odd and is even natural number f( ) = f( ) + = = which is never possible as the difference of odd and even number is always odd number. Hence in this case f ( ) f( ) i.e., f is one-one. Case IV: When is even and is odd natural number Similar as case III, We can prove f is one-one For onto: f() = + if is odd = if 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 - 8 -

23 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 } Refleivity Here (a, a) R as a a = 0 = 0 divisible by i.e., R is refleive. 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 - 9 -

24 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, given that * ( * 5) = 0. Given * ( * 5) = 0 5 * 0 * 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) Refleivity: a, b A Q a + b = b + a (a, b)r(a, b) So, R in refleive. (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 - 0 -

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

26 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) = f () = y. Clearly, domain of f = range of f and range of f = domain of f. The inverse of sine function is defined as sin = sinq =, where [ /, /] and [, ]. Thus, sin has infinitely many values for given [, ] 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 ) =, provided that and cos (cos) = and cos (cos ) =, provided that and 0 tan (tan) = and tan(tan ) =, provided that and cot (cot) = and cot(cot ) =, provided that < < and 0 < <. sec (sec) = and sec(sec ) = cosec (cosec) = and cosec(cosec ) =, sin cos ec or cos ec sin Prepared by: M. S. KumarSwamy, TGT(Maths) Page - -

27 cos s ec or sec cos tan cot or cot tan sin cos tan cot sec cosec cos sin tan cot cos ec s ec tan sin cos cot sec cos ec sin cos, where tan cot, where sec cos ec, where or y tan tan y tan, if y y y tan tan y tan, if y y y y tan tan y tan sin sin y sin y y, if, y 0, y sin sin y sin y y, if, y 0, y sin sin y sin y y, if, y 0, y sin sin y sin y y, if, y 0, y cos cos y cos y y, if, y 0, y Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 3 -

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

29 CHAPTER : INVERSE TRIGONOMETRIC FUNCTIONS NCERT Important Questions & Answers. Find the values of Let tan () cos sin tan tan tan where, tan MARKS WEIGHTAGE 05 marks 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 y z Prove that 3sin sin (3 4 ),, Let sin sin, then 3 We know that sin 3 3sin 4sin sin (3sin 4sin ) sin (3 4 ) 3 3sin sin (3 4 ) 7 3. Prove that tan tan tan 4 7 Given tan tan tan tan tan tan 4 y LHS 4 7 tan tan y tan.. y tan tan tan tan tan RHS Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 5 -

30 3 4. Prove that tan tan tan Given tan tan tan 7 7 LHS tan tan tan tan = tan tan tan tan tan tan 4 tan 3 7 y 4 tan tan y tan.. y tan tan tan RHS Simplify : tan, 0 Let = tan θ, then θ = tan (i) tan sec tan tan tan tan tan 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 [using (i)] cos 6. Simplify : tan, Let = sec θ, then θ = sec (i) tan tan tan sec tan tan tan (cot ) tan tan tan cot tan Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 6 -

31 sec [using (i)] cos sin 7. Simplify : tan,0 cos sin cos sin cos sin tan tan cos cos cos sin cos sin cos cos (inside the bracket divide numerator and denominator by cos ) tan tan tan tan tan tan tan 4 4 tan 4 y 8. Simplify : tan sin cos,, y 0 and y y y tan sin cos,, y 0 and y y y tan sin and tan y cos y tan ( tan tan y) tan.(tan tan ) tan(tan tan ) y y y y tan tan tan tan y tan. y. y y. y 9. If tan tan, find the value of. 4 Given that tan tan 4 tan y tan tan y tan 4. y ( )( ) ( )( ) ( )( ) tan 4 4 ( ) ( ) 4 tan Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 7 -

32 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 cos sin cos sin LHS cos sin sin sin sin sin sin y sin y y sin sin sin sin RHS Prove that tan sin cos RHS sin cos Let sin sin cos sin sin tan tan cos 3 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 8 -

33 3 3 Let cos y cos y sin y cos y sin tan y y tan cos then the equation becomes tan y tan tan tan y RHS tan tan tan tan tan y tan y tan tan tan LHS Prove that tan tan tan tan LHS = tan tan tan tan y tan tan tan tan y tan +... y tan tan tan tan tan tan tan tan tan tan tan tan RHS Prove that sin sin cot, 0, sin sin 4 sin sin Given cot, 0, sin sin 4 sin sin LHS cot sin sin Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 9 -

34 sin sin sin sin cot sin sin sin sin (by rationalizing the denominator) sin sin sin sin sin cot cot sin sin sin sin cos ( cos ) cos cot cot cot sin sin sin cos cot cos cos and sin sin cos sin cos cos cot cot cot RHS sin 5. Prove that tan cos 4 Let = cosy y cos y y cos sin cos y cos y LHS tan tan cos y cos y y y cos sin 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 tan 4 tan 6. Solve for : tan tan,( 0) Given tan tan,( 0) tan tan tan tan tan tan Prepared by: M. S. KumarSwamy, TGT(Maths) Page

35 tan tan tan tan ( ) ( ) ( ) ( )( ) tan ( ) ( ) ( ) tan ( ) tan tan tan tan given, so we do not take 3 7. Solve for : tan (cos ) tan (cos ec) Given tan (cos ) tan (cos ec) 3 cos tan tan (cos ec) tan tan cos cos tan tan sin sin cos cos sin sin sin cot cot cot Solve for : sin ( ) sin Given sin ( ) sin sin sin ( ) sin cos ( ) sin ( ) cos ( ) cos sin cos sin cos( ) cos sin sin cos sin sin sin 0 ( ) 0 0 or 0 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 3 -

36 0 or But does not satisfy the given equation, so = 0. y 9. Simplify: tan tan y y y y Given tan tan = tan tan y y y y tan tan tan tan y y tan y tan tan 4 0. Epress Given cos tan, in the simplest form. sin cos tan, sin y. y cos sin cos sin cos sin tan tan cos sin cos sin cos sin cos cos sin,sin cos and sin sin cos cos sin tan tan tan tan tan cos sin tan 4 4. Simplify : cot, Let = sec θ, then θ = sec (i) cot cot cot sec tan cot cot (cot ) sec tan. Prove that sin sin cos Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 3 -

37 Let sin and 5 3 sin y Therefore sin and sin y 5 7 Now, cos sin and cos y sin y We have cos( y) cos cos y sin sin y y cos sin sin cos Prove that sin cos tan Let sin, cos y and tan z Then sin, cos y and tan z Now, 44 5 cos sin and sin y cos y sin 3 sin y 5 3 tan and tan y cos 5 5 cos y tan tan y tan( y) tan z tan.tan y tan( y) tan z tan( z) tan( z) y z y z 4 63 sin cos tan a cos bsin a 4. Simplify: tan, if tan bcos asin b a cos bsin a cos bsin tan tan bcos bcos asin bcos asin bcos Prepared by: M. S. KumarSwamy, TGT(Maths) Page

38 a tan tan b a a tan tan (tan ) tan a tan b b b 5. Solve: tan tan 3 4 Given tan tan y tan tan tan y tan.3 4. y 5 5 tan tan (6 )( ) 0 or 6 Since = does not satisfy the equation, as the L.H.S. of the equation becomes negative, is the only solution of the given equation. 6 Prepared by: M. S. KumarSwamy, TGT(Maths) Page

39 CHAPTER : INVERSE TRIGONOMETRIC FUNCTIONS MARKS WEIGHTAGE 05 marks Previous Years Board Eam (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

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

41 6 y cos sin sin cos Prove that tan tan cos LHS tan tan tan tan tan tan tan y tan 7 tan tan tan 34 4 cos cos tan cos cos cos RHS Solve for : cos tan 3 cos tan 3 ( ) cos tan 3 cos tan 3 tan tan 4 tan tan 4 tan tan 3 3 tan tan tan tan tan.tan y. y Prepared by: M. S. KumarSwamy, TGT(Maths) Page

42 Prove that tan cos, (0,) LHS tan tan cos tan cos cos RHS Prove that sin sin LHS sin sin cos Let cos cos sin cos sin 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 and cot y Then sin and cot y Prepared by: M. S. KumarSwamy, TGT(Maths) Page

43 Now cos sin and sin y cot y cos y LHS cos( y) cos cos y sin sin y RHS Write the value of tan tan 5 Let tan tan tan tan tan tan tan 5 tan y 4. Find the value of the following: tan sin cos,, y 0 and y y Let tan and y tan tan, tan y y tan sin cos y tan tan tan sin cos tan tan tan tan tan sin (sin ) cos (cos ) sin and cos tan tan tan tan y tan tan tan.tan y 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

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

45 5 5 tan tan sec tan tan sec tan tan tan tan tan tan tan tan tan tan tan tan tan tan tan tan tan tan tan tan tan tan tan 4 7 y tan tan y tan.. y tan tan tan If tan tan y then write the value of + y + y. 4 tan tan y 4 y y tan tan. y 4. y 4 y y y y Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 4 -

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

47 CHAPTER 3: MATRICES MARKS WEIGHTAGE 03 marks NCERT Important Questions & Answers. If a matri has 8 elements, what are the possible orders it can have? What, if it has 5 elements? Since, a matri containing 8 elements can have any one of the following orders : 8, 8, 9, 9, 3 6,6 3 Similarly, a matri containing 5 elements can have order 5 or 5.. Construct a 3 4 matri, whose elements are given by: (i) a ij = 3i + j (ii) a ij = i j (i) The order of given matri is 3 4, so the required matri 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 matri 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 matri 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

48 Hence, the required matri 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 matri 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 and y from the following equation: y y y Prepared by: M. S. KumarSwamy, TGT(Maths) Page

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

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

51 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

52 3k k 4 4 4k k By definition of equality of matri 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 matri of order, show that tan 0 I + A = (I A) cos sin sin cos 0 Let A where tan 0 tan tan Now, cos and cos tan tan cos sin RHS ( I A) sin cos ( ) ( ) 3 3 ( ) 3 3 ( ) 0 0 LHS RHS Epress the matri matri. 4 B as the sum of a symmetric and a skew symmetric Prepared by: M. S. KumarSwamy, TGT(Maths) Page

53 4 B 3 4 B ' Let P ( B B ') Now P ' 3 3 P 3 3 Thus P ( B B ') is a symmetric matri Also, let Q ( B B ') Now Q ' 0 3 Q Thus Q ( B B ') is a skew symmetric matri Now, P Q B Thus, B is represented as the sum of a symmetric and a skew symmetric matri. 6. Epress the following matrices as the sum of a symmetric and a skew symmetric matri: ( 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

54 Thus P ( A A ') is a symmetric matri Now, Q ( A A') Q ' 0 Q Thus Q ( A A ') is a skew symmetric matri. 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 matri Now, Q ( A A') Q ' Q Thus Q ( A A ') is a skew symmetric matri. 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

55 Now, P ( A A') P ' P 5 5 Thus P ( A A ') is a symmetric matri Now, Q ( A A') Q ' 5 0 Q 3 0 Thus Q ( A A ') is a skew symmetric matri. 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 matri Now, Q ( A A') Q ' 3 0 Q Thus Q ( A A ') is a skew symmetric matri. Representing A as the sum of P and Q, Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 5 -

56 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 matri 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 - 5 -

57 A (applying R R + R 3 ) A (applying R R R 3 ) A Using elementary transformations, find the inverse of 6 3, if it eists. 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 matri on the LHS. Therefore, A does not eist. Note Suppose A = IA, after applying the elementary transformation, if any row or column of a matri on LHS is zero, then A does not eist. 0. Show that the matri B AB is symmetric or skew-symmetric according as A is symmetric or skew-symmetric. We suppose that A is a symmetric matri, 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 matri. Now, we suppose that A is a skew-symmetric matri. Then, A = A Consider (B AB) = [B (AB)] = (AB) (B ) [(AB) = B A and (A ) = A] Prepared by: M. S. KumarSwamy, TGT(Maths) Page

58 = (B A )B = B ( A)B= B AB [ A = A] (B AB) = B AB which shows that B AB is a skew-symmetric matri.. If A and B are symmetric matrices, prove that AB BA is a skew-symmetric matri. 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 matri Using elementary transformations, find the inverse of 3, if it eists 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

59 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? O 0 4. For what values of : Prepared by: M. S. KumarSwamy, TGT(Maths) Page

60 0 4 0 Since Matri multiplication is associative, therefore 0 0 O O 4 O 4 4 O If, show that A 5A + 7I = 0. 3 Given that A A A 5A 7I O Find, if O? 0 3 Given that O 0 3 Since Matri multiplication is associative, therefore O O 3 ( ) ( 5).9 ( )( 3) O 48 O Find the matri X so that X Prepared by: M. S. KumarSwamy, TGT(Maths) Page

61 Given that X The matri given on the RHS of the equation is a 3matri and the one given on the LHS of the equation is as a 3 matri. Therefore, X has to be a matri. 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 matri 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

62 Let A, B, C Find a matri 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 matri 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 matri 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

63 CHAPTER 3: MATRICES Previous Years Board Eam (Important Questions & Answers). Use elementary column operation C C C in the matri 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 matri such that A = A, then write the value of 7A (I + A) 3, where I is an identity matri. 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 y z 4 4. If y w 0 5, find the value of + y. y z 4 Given that y w 0 5 Equating, we get - y = - (i) - y = 0 (ii) z = 4, w = 5 (ii) (i) y + y = 0 + = and y = + y = + = 3. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

64 5. Solve the following matri equation for : 0 Given that O = 0 = 0 O If y , find ( y) Given that y y y Equating we get 8 + y = 0 and + = 5 y = 8 and = y = + 8 = For what value of, is the matri A 0 3 a skew-symmetric matri? 3 0 A will be skew symmetric matri if A = A' Equating, we get = 8. If matri 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

65 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 matri A. 9 4 Given that A A A 3 6. If A is a square matri 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 y 3 5, write the value of. 0 Given that y 3 5 y 0 3 y 5 y 0 3 y 5 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 6 -

66 Equating the corresponding elements we get. y = 0...(i) 3 + y = 5...(ii) Adding (i) and (ii), we get y y = = 5 = Find the value of + y from the following equation: 7 y Given that 7 y y y Equating the corresponding element we get + 3 = 7 and y 4 = and y = and y = 9 + 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 , write the value of Given that Equating the corresponding elements, we get = 3 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 6 -

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

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

69 CHAPTER 4: DETERMINANTS MARKS WEIGHTAGE 0 marks NCERT Important Questions & Answers 6. If, then find the value of Given that On epanding both determinants, we get 8 = = = 0 = 36 = ± 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 Epanding 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 Epanding along R, we obtain 0 c a b ( c) b b ( b) b c a c a b c a b c c

70 = c (a b + b bc) b (b c c ac) = abc + cb bc b c + bc + abc = 4 abc 4. If, y, z are different and We have 3 y y y 3 z z z y y y z z z 0 then show that + yz = 0 3 Now, we know that If some or all elements of a row or column of a determinant are epressed as sum of two (or more) terms, then the determinant can be epressed as sum of two (or more) determinants. 3 y y y y y 3 z z z z z 3 ( ) y y yz y y (Using C 3 C and then C C ) z z z z ( yz) y y z ( yz) 0 y y 0 z z z (Using R R R and R 3 R 3 R ) Taking out common factor (y ) from R and (z ) from R 3, we get ( yz)( y )( z ) 0 y 0 z = ( + yz) (y ) (z ) (z y) (on epanding along C ) Since Δ = 0 and, y, z are all different, i.e., y 0, y z 0, z 0, we get + yz = 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

71 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 epanding, prove that b c q r y z a p c a r p z b q y a b p q y c r z b c q r y z LHS c a r p z a b p q y b c c a a b q r r p p q (interchange row and column) y z z y b c c a c q r r p r [usingc 3 C 3 (C + C )] y z z z b c c a c ( ) q r r p r (taking common from C 3 ) y z z z Prepared by: M. S. KumarSwamy, TGT(Maths) Page

72 b a c ( ) q p r (using C C C 3 and C C C 3 ) y z a b c p q r (using y z C C ) a p b q y RHS (interchange row and column) c r z 7. Using the property of determinants and without epanding, 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 Epanding 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 epanding, 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 Prepared by: M. S. KumarSwamy, TGT(Maths) Page

73 Taking common factors (a c) and (b c) from R and R respectively, we get 0 ( a c) ( a c)( b c) 0 ( b c) c c Now, epanding 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 epanding, 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, epanding 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 epanding, prove that yz y y z y y z z y yz z z z y yz LHS y y z z z y ( )( )( )( ) Applying R R, R yr and R3 zr3, we have Prepared by: M. S. KumarSwamy, TGT(Maths) Page

74 3 yz 3 y y yz yz z z 3 yz yz y yz z 3 3 y (take out yz common from C 3 ) z 3 3 y y 3 3 z z 3 3 Epanding corresponding to C 3, we get 3 3 y y 3 3 z z 0 0 (using R R R and R3 R3 R ) ( y )( z ) ( z )( y ) = (y + ) (y ) (z ) (z + + z) (z + ) (z ) (y ) (y + + y) = (y ) (z ) [(y + ) (z + + z) (z + ) (y + + y)] = (y )(z )[yz + y + yz + z z zy z yz y 3 y] = (y )(z )[yz zy + z y ] = (y )(z )[yz(z y) + (z y )] = (y )(z )[yz(z y) + (z y)(z + y)] = (y ) (z ) [(z y) (y + yz + z)] = ( y) (y z) (z ) (y + yz + z) = RHS.. Using the property of determinants and without epanding, prove that 4 4 (5 4)(4 ) 4 4 LHS (5 4) 4 4 (5 4) Epanding along C, we get = (5 + 4) {(4 ) (4 )} (5 4)(4 ) = RHS. (using C C + C + C 3 ) [take out (5 + 4) common from C ]. (Using R R R and R 3 R 3 R ) Prepared by: M. S. KumarSwamy, TGT(Maths) Page

75 . Using the property of determinants and without epanding, prove that y k y y y y k y k (3 y k) y y y k 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 (5 + 4) common from C ]. (3 y k) 0 k 0 (Using R R R and R 3 R 3 R ) 0 0 Epanding 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 epanding, 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 Epanding along R, we get = (a + b + c) {( b c a) ( c a b)} = (a + b + c) [ (b + c + a) ( ) (c + a + b)] Prepared by: M. S. KumarSwamy, TGT(Maths) Page (Using C C C and C 3 C 3 C )

76 3 ( a b c)( a b c)( a b c) ( a b c) = RHS 4. Using the property of determinants and without epanding, prove that y z y z y z y ( y z) z z y y z y LHS z y z y z z y ( y z) y ( y z) y z y (using C C + C + C 3 ) ( y z) z y y ( y z) y z y [take out ( + y + z) common from C ]. z y y ( y z) 0 y z 0 (Using R R R and R 3 R 3 R ) 0 0 z y y ( y z)( y z)( y z) Epanding along R 3, we get ( y z)( y z)( y z) ( 0) 3 ( y z)( y z)( y z) ( y z) =RHS 3 5. Using the property of determinants and without epanding, prove that LHS (using C C + C + C 3 ) ( ) [take out ( ) ( ) common from C ]. (Using R R R and R 3 R 3 R ) Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 7 -

77 ( ) 0 ( ) 0 ( ) Take out ( ) common from R and same from R 3, we get ( )( )( ) 0 Epanding along C, we get ( )( )( ) 0 ( )( )( )( ) = RHS 6. Using the property of determinants and without epanding, 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 Epanding 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 epanding, 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

78 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 Epanding 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

79 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, epansion 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

80 A 5A + 7I = O 3 A A eists. 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 matri 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 matri A = 3, Show that A 3 6A + 5A + I = O. Hence, find A. 3 Prepared by: M. S. KumarSwamy, TGT(Maths) Page

81 Given that A = A A. A and A A. A A 6A 5A I O A 3 (6 3) (3 6) ( 4) A eist 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

82 A A A Solve system of linear equations, using matri method, + y + z = y z = 3 3y 5z = 9 The given system can be written as AX = B, where 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 eists. 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 Now, X A B y z T Prepared by: M. S. KumarSwamy, TGT(Maths) Page

83 y z Hence,, y and z 5. Solve system of linear equations, using matri method, y + z = 4 + y 3z = 0 + y + z = The given system can be written as AX = B, where 4 A 3, X y and B 0 z Here, A 3 = ( + 3) ( ) ( + 3) + ( ) = = 0 0 Thus, A is non-singular, Therefore, its inverse eists. 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 Now, X A B y z y z Hence, =, y = and z =. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

84 6. Solve system of linear equations, using matri method, + 3y +3 z = 5 y + z = 4 3 y z = 3 The given system can be written as AX = B, where 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 eists. 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 Now, X A B y z y z Hence, =, y = and z =. 7. Solve system of linear equations, using matri method, y + z = y 5z = 5 y + 3z = The given system can be written as AX = B, where Prepared by: M. S. KumarSwamy, TGT(Maths) Page

85 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 eists. 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 Now, X A B y z y z Hence, =, y = and z = If A = 3 4 find A. Using A,Solve system of linear equations: 3y + 5z = 3 + y 4z = 5 + y z = 3 The given system can be written as AX = B, where 3 5 A 3 4, X y and B 5 z 3 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 8 -

86 3 5 Here, A 3 4 = ( 4 + 4) ( 3) ( 6 + 4) + 5 (3 ) = = 0 Thus, A is non-singular. Therefore, its inverse eists. 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 Now, X A B y z y z Hence, =, 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 matri method. Let the prices (per kg) of onion, wheat and rice be Rs., Rs. y and Rs. z, respectively then 4 + 3y + z = 60, + 4y + 6z = 90, 6 + y + 3z = 70 This system of equations can be written as AX = B, where 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 eists. 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 - 8 -

87 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 Now, X A B y z y z = 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 epanding 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

88 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 Epanding 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

89 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 Epanding 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 ) ( )( )( )( ) Epanding 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

90 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 ) Epanding 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: y z y z y z Let p, q and r, then the given equations become 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 eists. 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

91 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 A B y z y z p, q, r 3 5,, y 3 z 5 =, y = 3 and z = If a, b, c, are in A.P, then find the determinant of 3 a Let A 3 4 b 4 5 c 3 a 0 0 ( b a c) 4 5 c T 3 a 3 4 b 4 5 c (using R R R R 3 ) But a,b, c are in AP. Using b = a + c, we get 3 a A [Since, all elements of R are zero] 4 5 c Prepared by: M. S. KumarSwamy, TGT(Maths) Page

92 3 37. Show that the matri A satisfies the equation A 4A + I = O, where I is identity matri and O is zero matri. 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 matri method. 3 y + 3z = 8 + y z = 4 3y + z = 4 The system of equation can be written as AX = B, where 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 eists. 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

93 5 8 X A B y z y 34 7 z 5 3 Hence =, y = and z = 3. Given that ( ) y z y z 39. Show that y ( z) yz 3 yz( y z) z yz ( y) ( y z) y z y ( z) yz z yz ( y) Applying R R, R yr,r 3 zr 3 to Δ and dividing by yz, we get ( y z) y z y y( z) y z yz z yz z( y) Taking common factors, y, z from C, C and C 3 respectively, we get ( y z) yz y ( z) y yz z z ( y) Applying C C C, C 3 C 3 C, we have ( y z) ( y z) ( y z) y ( z) y 0 z 0 ( y) z Taking common factor ( + y + z) from C and C 3, we have ( y z) ( y z) ( y z) ( y z) y ( z) y 0 z 0 ( y) z Applying R R (R + R 3 ), we have yz z y ( y z) y y z 0 z 0 y z Applying C (C + y C ) and C 3 C 3 + z C, we get Prepared by: M. S. KumarSwamy, TGT(Maths) Page

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

95 CHAPTER 3: DETERMINANTS Previous Years Board Eam (Important Questions & Answers) MARKS WEIGHTAGE 0 marks 9. Let A be a square matri of order 3 3. Write the value of A, where A = 4. Since A = n A where n is order of matri 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 matri and A =, then write the value of AA', where A' is the transpose of matri A. AA ' = A. A' = A. A = A = = 4. [since, AB = A. B and A = A', where A and B are square matrices.]. If A is a 3 3 matri, 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 3 4. If 3, find the value of. 5 4 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 9 -

96 3 Given that = 3 7 = 3 = 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 Epanding 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 Epanding 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 - 9 -

97 8. Write the value of the determinant Given determinant A = = 0 ( R = R 3 ) 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. 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, 3 + y + z = y + 3z = y + z = 00 The above system of equation may be written in matri form as AX = B 3 00 A 4 3, X y and B 300 z 00 3 Here, A 4 3 3( 3) (4 3) (4 ) A eists. 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

98 5 00 Now, X A B y z y z = 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 y z LHS a y z y a z Applying C C + C + C 3, we get a y z y z a y z a y z a y z y a z Apply R R R, we get y z ( a y z) a y z y a z 0 a 0 ( a y z) a y z y a z Epanding along R, we get = (a + + y + z) {0 + a (a + z z)} = a (a + + 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 matri 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, y, z respectively. From the statement we have + y+ z = 0 + y =3 + y 4z = 0 The above system of linear equations may be written in matri form as AX = B where 0 A 0, X y and B 3 4 z 0 Prepared by: M. S. KumarSwamy, TGT(Maths) Page

99 Here, A 0 ( 4 0) ( 8 0) ( ) A eists. 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 Now, X A B y z y z 0 0 = 5, y = 3, z = T 3. The management committee of a residential colony decided to award some of its members (say ) 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 matri 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 + y + z = + 3y + 3z = 33 y + z = 0 The above system of linear equation can be written in matri form as AX = B where A 3 3, X y and B 33 z 0 Here, A 3 3 Prepared by: M. S. KumarSwamy, TGT(Maths) Page

100 = (3 + 6) ( 3) + ( 4 3) = = 3 A eists. 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 Now, X A B y z y z = 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: 3 y z y 3y z y 3( y z)( y yz z) z y z 3z 3 y z LHS y 3y z y z y z 3z Applying C C + C + C 3 y z y z y z 3y z y y z y z 3z Taking out ( + y + z) along C, we get y z ( y z) 3y z y y z 3z Applying R R R ; R 3 R 3 R y z ( y z) 0 y y 0 z z Applying C C C 3 Prepared by: M. S. KumarSwamy, TGT(Maths) Page

101 y z z ( y z) 0 3y y 0 3z z Epanding along I column, we get D = ( + y + z)[(3y ( + z) + 3z ( y)] = 3( + y + z)[y + z + yz + z yz] = 3( + y + z)(y + yz + z) = 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 matri method. Apart from these values, namely, Honesty, Regularity and Hardwork, suggest one more value which the school must include for awards. Let, y and z be the awarded money for honesty, Regularity and hardwork. From the statement + y +z = 6000 (i) + 3z =000 (ii) +z = y y +z = 0 (iii) The above system of three equations may be written in matri form as AX = B, where 6000 A 0 3, X y and B z 0 Here, A 0 3 (0 6) ( 3) ( 0) Hence A eist 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 Now, X A B y z 3 0 Prepared by: M. S. KumarSwamy, TGT(Maths) Page

102 y z =500, y = 000, z = 3500 Ecept above three values, school must include discipline for award as discipline has great importance in student s life If, then write the value of Given that 3 3 ( +) ( + ) ( )( 3) = =3 7 =3 7 =4 = 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

103 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 Epanding along R we get = a 3.(7 6) = a Using matrices, solve the following system of equations: y + z = 4; + y 3z = 0; + y + z = Given equations y + z = 4 + y 3z = 0 + y + z = We can write this system of equations as AX = B where 4 A 3, X y and B 0 z Here, A 3 = ( + 3) - (- ) ( + 3) + ( - ) = = 0 A eists. 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 Now, X A B y z 3 Prepared by: M. S. KumarSwamy, TGT(Maths) Page

104 y z The required solution is =, y = -, z = If A = and B = 3 0, find (AB). 5 0 For B B i.e., B is invertible matri B eist. 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

105 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 0 -

106 CHAPTER 5: CONTINUITY AND DIFFERENTIABILITY NCERT Important Questions & Answers MARKS WEIGHTAGE 0 marks 3, if. Find all points of discontinuity of f, where f is defined by f ( ). 3, if Ans. 3, if Here, f ( ). 3, if lim lim At =, LHL ( ) ( 3) f Putting = h has when h 0 lim lim lim lim f ( ) (( h) 3) (4 h 3) (7 h) 7 h 0 h 0 h 0 lim lim At =, RHL ( ) ( 3) f Putting = + h as + when h 0 lim lim lim lim f ( ) (( h) 3) (4 h 3) ( h) h 0 h 0 h 0 LHL RHL. Thus, f() is discontinuous at =., if 0. Find all points of discontinuity of f, where f is defined by f ( ) 0, if 0, if 0 Ans. Here, f ( ) 0, if 0 lim lim LHL f ( ) 0 0 Putting = 0 h as 0 when h 0 lim lim 0 h lim h f ( ) 0 h 0 0 h h 0 h lim lim RHL f ( ) 0 0 Putting = 0 + h as 0 + ; h 0 lim lim 0 h lim h f ( ) 0 h 0 0 h h 0 h LHL RHL. Thus, f() is discontinuous at = 0. 3 if 3, 3. Find all points of discontinuity of f, where f is defined by f ( ), if Ans. For <, f() = 3 3 and for >, f() = + is a polynomial function, so f is continuous in the above interval. Therefore, we have to check the continuity at =. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 0 -

107 LHL lim lim 3 ( ) ( 3) f Putting = h has when h 0 lim lim lim f h h h h h 0 h ( ) (( ) 3) (8 6 3) lim (5 6 3 h h h ) 5 h 0 RHL lim lim ( ) ( ) Putting = + h as + when h 0 lim lim lim f ( ) (( h) ) (4 4h h ) h 0 h 0 lim (5 4 h h ) 5 h 0 Also, f() = () 3 3 = 8 3 = 5 [since f() = 3 3] LHL = RHL =f(). Thus, f() is continuous at =. Hence, there is no point of discontinuity for this function f(). a, if 3 4. Find the relationship between a and b so that the function f defined by f ( ) b 3, if 3 is continuous at = 3. Ans. a, if 3 Here, f ( ) b 3, if 3 LHL lim lim ( ) ( ) 3 f 3 a Putting = 3 h has 3 when h 0 lim lim lim f ( ) ( a(3 h) ) (3a ah ) 3a 3 h 0 h 0 RHL lim lim ( ) ( 3) 3 f 3 b Putting = 3 + h as 3 + when h 0 lim lim lim f ( ) ( b(3 h) 3) (3b bh 3) 3b 3 3 h 0 h 0 Also, f(3) =3 a + [since f() = a + ] Since, f() is continuous at = 3. LHL = RHL = f(3) 3a 3b 3 3a 3b a b 3 ( ), if 0 5. For what value of is the function defined by f ( ) continuous at = 4, if 0 0? What about continuity at =? Ans. ( ), if 0 Here, f ( ) 4, if 0 Prepared by: M. S. KumarSwamy, TGT(Maths) Page

108 lim lim At = 0, LHL ( ) ( ) 0 f 0 Putting = 0 h as 0 when h 0 lim lim lim f ( ) [(0 h) (0 h)] ( h h) 0 0 h 0 h 0 lim lim At = 0, RHL ( ) (4 ) 0 f 0 Putting = 0 + h as 0 + ; h 0 lim lim lim f ( ) [4(0 h) ] (4h ) 0 h 0 h 0 LHL RHL. Thus, f() is discontinuous at = 0 for any value of. lim lim At =, LHL ( ) (4 ) f Putting = h as when h 0 lim lim lim f ( ) (4( h) ) (4 4h ) 5 h 0 h 0 lim lim At =, RHL ( ) (4 ) f Putting = + h as + ; h 0 lim lim lim f ( ) [4( h) ] (4 4h ) 5 h 0 h 0 LHL = RHL. Thus, f() is continuous at = for any value of. sin, if 0 6. Find all points of discontinuity of f, where f ( ), if 0 sin, if 0 Ans. Here, f ( ), if 0 lim lim sin At = 0, LHL f ( ) 0 0 Putting = 0 h as 0 when h 0 lim lim sin(0 h) lim sin( h) lim sin h lim sin h f ( ) 0 h 0 0 h h 0 h h 0 h h 0 h lim lim sin At = 0, RHL f ( ) 0 0 Putting = 0 + h as 0 + ; h 0 lim lim sin(0 h) lim sin h f ( ) 0 h 0 0 h h 0 h Also, f(0) = 0 + = LHL = RHL = f(0). Thus, f() is continuous at = 0. When < 0, sin and both are continuous. Therefore, sin is also continuous. When > 0, f() = + is a polynomial. Therefore f is continuous. Hence, there is no point of discontinuity for this function f(). Prepared by: M. S. KumarSwamy, TGT(Maths) Page

109 sin, if 0 7. Determine if f defined by f ( ) is a continuous function? 0, if 0 Ans. sin, if 0 Here, f ( ) 0, if 0 lim lim At = 0, LHL f ( ) sin 0 0 Putting = 0 h as 0 when h 0 lim lim lim ( ) (0 ) sin sin 0 sin 0 f 0 h h 0 h h 0 h h = 0 value between and (since sin, for all values of R) lim lim At = 0, RHL f ( ) sin 0 0 Putting = 0 + h as 0 + ; h 0 lim lim lim ( ) (0 ) sin sin 0 sin 0 f 0 h h 0 h h 0 h h = 0 value between and (since sin, for all values of R) LHL = RHL = f(0). Thus, f() is continuous at = Find the values of k so that the function f Ans. k cos, if Here, f ( ) 3, if lim lim k cos LHL f ( ) k cos, if f ( ) is continuous at point 3, if Putting = h as when h 0 lim k cos lim h lim ksinh f ( ) h 0 h 0 h k lim sinh k k h 0 h h Since f() is continuous at, therefore LHL = f Also, f = 3 k 3 k 6 Prepared by: M. S. KumarSwamy, TGT(Maths) Page

110 k, if 5 9. Find the values of k so that the function f f ( ) 3 5, if 5 Ans. k, if 5 Here, f ( ) 3 5, if 5 lim lim At = 5, LHL ( ) ( ) 5 f 5 k Putting = 5 h has 5 when h 0 is continuous at point = 5. lim lim lim f ( ) ( k(5 h) ) (5k kh ) 5k 5 h 0 h 0 lim lim At = 5, RHL ( ) (3 5) 5 f 5 Putting = 5 + h as 5 + ; h 0 lim lim lim f ( ) (3(5 h) 5) (0 3 h) 0 5 h 0 h 0 Also, f(5) = 5k + Since f() is continuous at = 5, therefore LHL = RHL = f(5) 9 5k 0 5k 9 k 5 5, if 0. Find the values of a and b such that the function defined by f ( ) a b, if 0, if 0 continuous function. Ans. 5, if Here, f ( ) a b, if 0, if 0 lim lim At =, LHL f ( ) (5) 5 lim lim At =, RHL ( ) ( ) f a b Putting = + h as + when h 0 lim lim lim f ( ) ( a( h) b) ( a ah b) a b h 0 h 0 Also, f() = 5 Since f() is continuous at =, therefore LHL = RHL = f() a b () is a At =0, LHL lim lim ( ) ( ) 0 f 0 a b Putting = 0 h has 0 when h 0 lim lim lim f ( ) ( a(0 h) b) (0 a ah b) 0a b 0 h 0 h 0 Prepared by: M. S. KumarSwamy, TGT(Maths) Page

111 lim lim At =, RHL f ( ) () 0 0 Also, f(0) = Since f() is continuous at = 0, therefore LHL = RHL = f(0) Since, f() is continuous at = 0. LHL = RHL = f(0) 0a + b =. () Subtracting Eq. () from Eq. (), we get 8a =6 a = Put a = in Eq. (), we get + b = 5 b =. Prove that the function f given by f () =, R is not differentiable at =. Ans., if 0 Given, f ( ) ( ), if 0 We have to check the differentiability at = Here, f() = = 0 ' lim f ( h) f () lim ( h) 0) Lf () h 0 h h 0 h lim h h 0 h and ' lim f ( h) f () lim ( h) 0 Rf () h 0 h h 0 h ' ' Lf () Rf (). lim h 0 Hence, f() is not differentiable at = h h. Find dy d if y cos,0 Ans. Let = tan tan, then we have tan y cos cos cos cos tan y tan dy d 3. Find dy d if y sin,0 Ans. Let = tan tan, then we have tan y sin sin sin cos sin sin tan y y tan Prepared by: M. S. KumarSwamy, TGT(Maths) Page

112 dy 0 d 4. Find dy d if y cos, Ans. Let = tan tan, then we have tan y cos cos cos sin cos cos tan y y tan dy 0 d 5. Find dy d if y sec,0 Ans. Let cos cos, then we have y sec sec sec sec sec cos cos y cos 6. Differentiate sin (tan e ) with respect to. Ans. Let y = sin (tan e ) Differentiating both sides w.r.t., we get dy d sin(tan d e ) cos(tan e ) (tan e ) d d d d cos(tan e ) ( e ) ( e ) d e cos(tan e ) e e cos(tan ) ( ) e e 7. Differentiate log (cos e ) with respect to. Ans. Let y = log (cos e ) Differentiating both sides w.r.t., we get dy d d log(cos e ) (cos e ) d d cos e d d ( sin e ) ( e ) ( tan e ). e e tan e cos e d 8. Differentiate cos (log + e ), > 0 with respect to. Ans. Let y = cos (log + e ) Differentiating both sides w.r.t., we get dy d d cos(log e ) sin(log e ) (log e ) d d d Prepared by: M. S. KumarSwamy, TGT(Maths) Page

113 e sin(log e ) e sin(log e ) ( e ) sin(log e ) 9. Find dy d if y + y + = a b. Ans. Given that y + y + = a b Putting u = y, v = y and w =, we get u + v + w = a b du dv dw Therefore, () d d d Now, u = y. Taking logarithm on both sides, we have log u = log y Differentiating both sides w.r.t., we have du d (log y) log y d ( ). dy log y. u d d d y d du dy dy u log y y log y () d y d y d Also v = y Taking logarithm on both sides, we have log v = y log Differentiating both sides w.r.t., we have dv y d (log ) log dy y log dy v d d d d dv y dy y y dy v log log (3) d d d Again w = Taking logarithm on both sides, we have log w = log. Differentiating both sides w.r.t., we have dw d (log ) log d ( ) log. w d d d dw w log log (4) d From (), (), (3), (4), we have dy y y dy y log y log log 0 y d d y dy y (..log ) log. y y y log y d y dy log y. y log y y d. y.log 0. Differentiate sin with respect to. Ans. Let y = sin Let u = and v = sin then we have y = u v dy du dv Therefore, () d d d Now, u = Taking logarithm on both sides, we have log u = log. Differentiating both sides w.r.t., we have Prepared by: M. S. KumarSwamy, TGT(Maths) Page

114 du d (log ) log d ( ) log. u d d d du u log log () d Again v = sin Taking logarithm on both sides, we have log v = (sin) log. Differentiating both sides w.r.t., we have dv dv sin cos (log ) v[cos (log )] [cos (log )] v d d From (), () and (3), we get dy du dv sin log [cos (log )] d d d. Differentiate (log ) + log with respect to. Ans. Let y = (log ) + log Let u = (log ) and v = log then we have y = u + v dy du dv Therefore, () d d d Now, u = (log) Taking logarithm on both sides, we have log u = log(log). Differentiating both sides w.r.t., we have du d d log(log ) log(log ) ( ) log(log ) u d d d log du u log(log ) (log ) log(log ) d log log () Again v = log Taking logarithm on both sides, we have log v = (log) log = (log) Differentiating both sides w.r.t., we have dv d log (log ) log v d d dv log log log v d (3) From (), () and (3) dy log log (log ) log(log ) d log dy (log ) log log(log ) log.log d. Differentiate (sin ) Ans. Let y = (sin ) sin sin with respect to. Let u = (sin ), v = sin then we have y = u + v dy du dv Therefore, d d d () Now, u = (sin ) Taking logarithm on both sides, we have log u = log(sin ). Differentiating both sides w.r.t., we have Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 0 -

115 du d d log(sin ) log(sin ) ( ) cos log(sin ) u d d d sin du u cot log(sin ) (sin ) cot log(sin ) () d Again v = sin Differentiating both sides w.r.t., we have dv d ( ) (3) d d From (), () and (3) dy (sin ) cot log(sin ) d 3. Differentiate sin + (sin ) cos with respect to. Ans. Let y = sin + (sin ) cos Let u = sin, v = (sin ) cos then we have y = u + v dy du dv Therefore, () d d d Now, u = sin Taking logarithm on both sides, we have log u = (sin) log. Differentiating both sides w.r.t., we have du sin d log log d (sin ) sin log cos u d d d du sin sin sin u log cos log cos d () Again v = sin cos Taking logarithm on both sides, we have log v = (cos) log(sin) Differentiating both sides w.r.t., we have dv d d cos log(sin ) log(sin ) (cos ) cos cos log(sin )( sin ) v d d d sin dv cos vcot cos sin log(sin ) sin cot cos sin log(sin ) (3) d From (), () and (3) dy sin sin cos log cos sin cot cos sin log(sin ) d 4. Find dy if = a (θ + sin θ), y = a ( cos θ). d Ans. Given that = a (θ + sin θ), y = a ( cos θ) Differentiating w.r.t. θ, we get d dy a( cos ), a(sin ) d d dy sin cos sin dy Therefore, d a(sin ) sin tan d d a( cos ) cos cos cos d 5. Find dy if = cos θ cos θ, y = sin θ sin θ d Ans. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - -

116 Given that = cos θ cos θ, y = sin θ sin θ Differentiating w.r.t. θ, we get d sin ( sin ) sin sin d dy cos (cos ) cos cos d dy dy sin sin Therefore, d d d cos cos d 6. Find dy if = a (θ sin θ), y = a ( + cos θ) d Ans. Given that = a (θ sin θ), y = a ( + cos θ) Differentiating w.r.t. θ, we get d dy a( cos ), a(0 sin ) a sin d d dy sin cos cos dy sin sin Therefore, d a cot d d a( cos ) cos sin sin d 7. If Ans. a y a sin t cos t,, show that dy y sin t Given that a, y a Multiplying both we get, cos sin t cos t sin t cos t sin t cos t y a a a. a a a Differentiating both sides w.r.t., we get dy y 0 dy y dy y d d d 8. If y = 3e + e 3, prove that t d y d dy 5 6y 0 d d Ans. Given that y = 3e + e 3 Differentiating both sides w.r.t., we get dy 3 3 6e 6 6( e e ) d Again, Differentiating both sides w.r.t., we get d y 3 6(e 3 e ) d d y dy Now, 5 6y 6(e 3 e ) 5(6( e e )) 6(3e e ) d d e 8e 30e 30e 8e e 0 9. If y = 3 cos (log ) + 4 sin (log ), show that y + y + y = 0 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - -

117 Ans. Given that y = 3 cos (log ) + 4 sin (log ) Differentiating both sides w.r.t., we get dy y 3sin(log ) d (log ) 4cos(log ) d (log ) d d d 3sin(log ) 4 cos(log ) 3sin(log ) 4 cos(log ) y 3sin(log ) 4 cos(log ) Again, Differentiating both sides w.r.t., we get d d y y. 3cos(log ) (log ) 4sin(log ) (log ) d d y 3cos(log ) 4sin(log ) 3cos(log ) 4sin(log ) y y y y y y If e y ( + ) =, show that Ans. Given that e y ( + ) = y e d y d dy d Differentiating both sides w.r.t., we get y dy dy e d ( ) d ( ) dy d Again, Differentiating both sides w.r.t., we get d y dy d ( ) d 3. If y = (tan ), show that ( + ) y + ( + ) y = Ans. Given that y = (tan ) Differentiating both sides w.r.t., we get dy d tan (tan ) tan d d tan y ( ) y tan Again, Differentiating both sides w.r.t., we get ( ) y y ( ) y ( ) y 3. Verify Rolle s theorem for the function f () = + 8, [ 4, ]. Ans. Given function is f() = + 8, [ 4, ]. Since, a polynomial function is continuous and derivable on R, therefore (i) f() is continuous on [ 4,]. (ii) f() is derivable on ( 4,). Also, f( 4) = ( 4) + ( 4) 8 = 0 (since f () = + 8) Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 3 -

118 and f() = + 8 = 0 f( 4) = f() This means that all the conditions of Rolle's theorem are satisfied by f() in [ 4,]. Therefore, it eists at least one real c [ 4, ] such that f (c) = 0. Now, f () = ' d + 8 f ( ) ( 8) d Putting f (c) = 0 c + = 0 c =. Thus, f ( ) =0 and ( 4,). Hence, Rolle's theorem is verified with c =. 33. Verify Mean Value Theorem, if f () = 4 3 in the interval [a, b], where a = and b = 4. Ans. Here, f() = 4 3, [, 4] which is a polynomial function, so it is continuous and derivable at all R, therefore (i) f() is continuous on [, 4] (ii) f() is derivable on (, 4). Therefore, Conditions of Lagrange's theorem are satisfied on [, 4]. Hence, there is atleast one real number. c (, 4) such that Now, f '( ) d ( 4 3) 4 d f (4) 4 4(4) and f () ' f (4) f () 3 ( 6) 3 6 f ( c) c 4 c 4 5 c (, 4) Hence Mean Value Theorem is verified. 34. Verify Mean Value Theorem, if f () = in the interval [a, b], where a = and b = 3. Find all c (, 3) for which f (c) = 0. Ans. Given, f () = 3 5 3, (, 3), which is a polynomial function. Since, a polynomial function is continuous and derivable at all R, therefore (i) f() is continuous on [, 3]. (ii) f() is derivable on (, 3). Therefore, Condition of Lagrange's MVT are satisfied on [, 3]. Hence, there eists aleast one real c (, 3). 3 Now, f '( ) d ( 5 3 ) d 3 f (3) (3) 5(3) 3(3) and f () ' f (3) f () 7 ( 7) 7 7 f ( c) 0 3 3c 0c 3 0 3c 0c c 0c c, out of which 7 (,3) Hence Mean Value Theorem is verified. 35. If y y dy 0, for, < <, prove that d ( ) Ans. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 4 -

119 Given that y y 0 y y Squaring both sides, we get ( y) y ( ) y y y 0 ( y)( y) y( y) 0 ( y)( y y) 0 y 0 or y y 0 y or y( ) y or y But y = does not satisfy the given equation So, we consider, y Differentiating both sides w.r.t., we get d d ( ) ( ) ( ) dy d d d ( ) d d ( ) ( ) ( ) dy a y 36. If cos y = cos (a + y), with cos a ±, prove that d sin a Ans. cos y Given that cos y = cos (a + y) cos( a y) Differentiating both sides w.r.t. y, we get d d cos y cos( a y)( sin y) cos y[ sin( a y)] dy dy cos( a y) cos ( a y) sin( a y) cos y cos( a y)sin y sin( a y y) sin a cos ( a y) cos ( a y) cos ( a y) dy a y d sin a cos ( ) d y cos ( ) 37. If = a (cos t + t sin t) and y = a (sin t t cos t), find d Ans. Given that = a (cos t + t sin t) and y = a (sin t t cos t) Differentiating both sides w.r.t. t, we get d d d d a cost t sin t sin t ( t) a[ sin t ( tsin t sin t)] at cost dt dt dt dt dy d d d and a sin t t cost cos t ( t) a[cos t ( tsin t cos t)] atsin t dt dt dt dt dy dy at sin t Now, dt tan t d d at cost dt Again, Differentiating both sides w.r.t., we get d y d dt sec t 3 tan t sec t sec t d d d at cost at 38. Differentiate Ans. sin tan cos w.r.t. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 5 -

120 sin cos sin Let y = tan tan cos cos sin tan tan (tan ) cos Differentiating both sides w.r.t., we get dy d 39. Differentiate sin w.r.t. 4 Ans. Let y = sin sin 4 ( ) Let tan tan then we have tan y sin sin (sin ) tan tan Differentiating both sides w.r.t., we get dy d log ( ) log d ( ) d Differentiate sin w.r.t. e cos. Ans. Let u = sin and v = e cos Differentiating u and v w.r.t., we get du d sin (sin ) sin cos d d dv d and e (cos ) e ( sin ) ( sin ) e d d du du sin cos cos Now, d cos cos dv dv ( sin ) e e d cos cos cos Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 6 -

121 CHAPTER 5: CONTINUITY AND DIFFERENTIABILITY MARKS WEIGHTAGE 0 marks Previous Years Board Eam (Important Questions). For what value of k is the following function continuous at =? ; f ( ) k ; 3 ; 4 3, 0. Discuss the continuity of the following function at = 0 : f ( ) tan 0, 0 3 a b ; 3. If the function f () given by f ( ) ; is continuous at =, find the values of a and 5a b ; b. 4. Find the relationship between a and b so that the function f defined by: a, 3 f ( ) is continuous at = 3. b 3, 3 5. Show that the function f() = 3, R, is continuous but not differentiable at = 3. k k, 0 6. Find the value of k, for which f ( ) is continuous at = 0., 0 k, 7. Find the value of k so that the function f, defined by f ( ) is continuous at =. cos, 8. Find the value of a for which the function f defined as asin ( ), 0 f ( ) is continuous at = 0. tan sin 3, 0 9. Find all points of discontinuity of f, where f is defined as follows : 3 ; 3 f ( ) ; ; 3 0. Show that the function f defined as follows, is continuous at =, but not differentiable: 3 ; 0 f ( ) - ; 5 4 ;. Verify Lagrange s mean value theorem for the following function: f() = + + 3, for [4, 6]. sec. If f ( ) sec, then f (). Also find ' f. 3. Find dy d,if ( y ) y. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 7 -

122 4. If sin y = sin(a + y), prove that 5. If (cos ) y = (sin y), find dy d. 6. If y sin, then prove that dy a y d sin a sin ( ) ( ) d y dy 3 y 0 d d. 7. If y = e d y dy (sin + cos ), then show that y 0. d d d y 8. If = a(cost + t sin t) and y = a(sin t - t cost), then find d. y dy y 9. If log( y ) tan, then show that d y d y dy 0. If y = a cos(log ) + b sin(log ), then show that y d d. If y a( y), then show that. Find dy d,if y sin 3. Find dy d,if / y (cos ) (sin ) dy d 0 y 4. Differentiate the following with respect to : tan sin sin 5. If y cot sin sin, find dy d 6. Differentiate the following function w.r.t. : () cos + (sin ) tan 7. If asin,, then show that y e 3 4 ( ) d y dy a y 0 d d 8. If y cos, find dy 5 d 9. If y = cosec - d y dy, >. then show that ( ) ( ) 0 d d. 30. If y log tan dy then show that sec 4 d. Also find d y d at Differentiate the following function with respect to : f ( ) tan tan t 3. If a d y cost log tan, y a( sin t), then find d. d y 33. If a( sin ), y a( cos ), then find d. cos 34. Differentiate w.r.t. y y dy log 35. If e, then show that d log( e) Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 8 -

123 36. If tan log y a, then show that ( ) d y dy ( a) 0 d d. 37. Prove that d a sin a a d a 38. If y log then show that d y dy ( ) 0 d d 39. If y log a then show that ( ) d y dy a 0 d d 40. If y sin, then show that ( d y dy ) 0 d d. 4. Differentiate tan w.r.t.. d d y d y 4. If = a (cos t + t sin t) and y = a (sin t t cos t), 0 t, find,, dt dt d. m m m n 43. If y ( y), then show that dy y. d dy y 44. If y ( y), then show that. d t 45. If a sin t, y a cost log tan, then find d y d. 46. If y 47. If dy ( log y), then show that. d log y y dy log e, then show that d ( log ) y e y Differentiate the following with respect to : sin (36) 49. If = a cos 3 and y = a sin 3 d y, then find the value of d at. 6 dy sin ( a y) 50. If sin (a + y) + sin a cos(a + y) = 0, prove that. d sin a d y dy 5. If y = sin (log ), then prove that y 0. d d 5. Show that the function f() = is continuous but not differentiable at = Differentiate tan with respect to cos. 54. Differentiate tan with respect to sin. 55. If y, then show that d y dy y d y d 0 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 9 -

124 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 0 -

125 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) eternally 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 Eternally 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 - -

126 (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 : eternally] 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 - -

127 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 - 3 -

128 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 - 4 -

129 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 - 5 -

130 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 - 6 -

131 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 - 7 -

132 7. If with reference to the right handed system of mutually perpendicular unit vectors i, j and k, 3 i j, i j 3 k, then epress in the form, where is parallel to and is perpendicular to Let, is a scalar, i.e. 3 i j. Now, ( 3 ) i ( ) j 3k Now, since is perpendicular to, we should have 3( 3 ) ( ) Therefore, 3 i j and i j 3k 0. i.e., 8. 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 - 8 -

133 CHAPTER 0: VECTOR ALGEBRA Previous Years Board Eam (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 - 9 -

134 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

135 0. Find a vector a of magnitude5, making an angle of 4 with -ais., with y-ais and an acute angle with z-ais. 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 - 3 -

136 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 ( a)( a) 4, then write the value of. Given that ( a)( a) 4.. a a. a. a 4 a 4. a a 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 - 3 -

137 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

138 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

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

140 CHAPTER : LINEAR PROGRAMMING MARKS WEIGHTAGE 06 marks NCERT Important Questions & Answers. Determine graphically the minimum value of the objective function Z = y subject to the constraints: y 5; 3 + y 3 ; 3y ; 0, y 0 Given that Z = y... () y 5... () 3 + y 3... (3) 3y... (4) 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 y < 300 (see Step 3(ii) of corner Point Method.) i.e., 5 + 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 = y has no minimum value subject to the given constraints.. Solve the following Linear Programming Problems graphically: Maimise Z = 5 + 3y subject to 3 + 5y 5, 5 + y 0, 0, y 0. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

141 Our problem is to maimize Z = 5 + 3y (i) Subject to constraints 3 + 5y 5 (ii) 5 + y 0 (iii) 0, y 0 (iv) Firstly, draw the graph of the line 3 + 5y = 5 Secondly, draw the graph of the line 5 + y = 0 On solving given equations 3 + 5y = 5 and 5 + y = 0, we get = 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 maimum value of Z is at the point B, Show that the minimum of Z occurs at more than two points. Minimise and Maimise Z = + y subject to + y 00, y 0, + y 00;, y 0. Our problem is to minimize and maimize Z = + y (i) Subject to constraints are + y 00 (ii) y 0 (iii) + y 00 (iv) Prepared by: M. S. KumarSwamy, TGT(Maths) Page

142 0, y 0 (v) Firstly, draw the graph of the line + y = 00 Secondly, draw the graph of line y = 0 Thirdly, draw the graph of line + y = 00 On solving equations y = 0 and + y = 00, we get B(0, 40) and on solving the equations y = 0 and + 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 maimum 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 mi two types of foods in such a way that vitamin contents of the miture 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 miture. Let the miture contain kg of Food I and y kg of Food II. Clearly, 0, y 0. We make the following table from the given data: Prepared by: M. S. KumarSwamy, TGT(Maths) Page

143 Since the miture must contain at least 8 units of vitamin A and 0 units of vitamin C, we have the constraints: + y 8 + y 0 Total cost Z of purchasing kg of food I and y kg of Food II is Z = y Hence, the mathematical formulation of the problem is: Minimise Z = y... () subject to the constraints: + y 8... () + y 0... (3), 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 y < 380 i.e., 5 + 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 miing strategy for the dietician would be to mi kg of Food I and 4 kg of Food II, and with this strategy, the minimum cost of the miture will be Rs Reshma wishes to mi two types of food P and Q in such a way that the vitamin contents of the miture 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 miture. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

144 Let Reshma mies kg of food P and y kg of food Q. Construct the following table: The miture must contain atleast 8 units of vitamin A and units of Vitamin B. Total cost Z of purchasing food is Z = y The mathematical formulation of the given problem is Minimize Z = y (i) Subject to the constraints 3 + 4y 8 (ii) 5 + y (iii) 0, y 0 (iv) Firstly, draw the graph of the line 3 + 4y = 8 Secondly, draw the graph of the line 5 + y = On solving equations 3 + 4y = 8 and 5 + y =, we get B, 8 The corner points of the feasible region are A,0, B, and C 0, (see the below figure) 3 The values of Z at these points are as follows : Prepared by: M. S. KumarSwamy, TGT(Maths) Page

145 As the feasible region is unbounded, therefore 60 may or may not be the minimum value of Z. For this, we graph the inequality y < 60 or 3 + 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 3 + 4y < 8 therefore, the minimum cost of the miture will be Rs. 60 at the 8 line segment joining the points A,0 and B, 3 6. 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 eceed 50 units. Determine the number of units of each type of computers which the merchant should stock to get maimum 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 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 maimize Z = 5 + 3y (i) Subject to the constraints + y (ii) 3 + y 0 (iii) 0, y 0 (iv) Firstly, draw the graph of the line + y = Secondly, draw the graph of the line 3 + y = 0 On solving equations + y = and 3 + y = 0, we get B(4, 4). Feasible region is OABCO. (see below figure) Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 4 -

146 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 maimum value of Z is Rs. 3 at B(4, 4). Thus, the manufacturer should produce 4 pedestal lamps and 4 wooden shades to maimize his profits. 7. 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 miture of these two foods and also meets the minimal nutritional requirements. Let the diet contains 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 = 4 + 6y (i) Subject to constraints 3 + 6y 80 (ii) 4 + 3y 00 (iii) 0, y 0 (iv) Firstly, draw the graph of the line 3 + 6y = 80 Secondly, draw the graph of the line 4 + 3y = 00 4 On solving the equations 3 + 6y = 80 and4 + 3y = 00, we get 4, 3 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 4 -

147 It can be seen that the feasible region is unbounded. (see the below figure) 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, 4 + 6y < 04 or + 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 + 3y < 5 Therefore, the minimum cost of the miture 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 and y be the number of packets of food P and Q respectively. Obviously 0, y 0. Mathematical formulation of the given problem is as follows: Minimise Z = 6 + 3y (vitamin A) subject to the constraints + 3y 40 (constraint on calcium), i.e. 4 + y () 4 + 0y 460 (constraint on iron), i.e. + 5y 5... () 6 + 4y 300 (constraint on cholesterol), i.e. 3 + y (3) 0, y 0... (4) Prepared by: M. S. KumarSwamy, TGT(Maths) Page

148 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: 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. 9. 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 maimise her profit assuming that she can sell all the items that she produced? What will be the maimum profit? Let and y be the number of items M and N respectively. Total profit on the production = Rs ( y) Mathematical formulation of the given problem is as follows: Maimise Z = y subject to the constraints: + y (constraint on Machine I)... () + y (constraint on Machine II)... () + 5 y 5 (constraint on Machine III)... (3) 4 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 = y at these corner points. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

149 We see that the point (4, 4) is giving the maimum value of Z. Hence, the manufacturer has to produce 4 units of each item to get the maimum 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 units and y units of the commodity be transported from the factory at P to the depots at A and B respectively. Then (8 y) units will be transported to depot at C Hence, we have 0, y 0 and 8 y 0 i.e. 0, y 0 and + y 8 Prepared by: M. S. KumarSwamy, TGT(Maths) Page

150 Now, the weekly requirement of the depot at A is 5 units of the commodity. Since units are transported from the factory at P, the remaining (5 ) units need to be transported from the factory at Q. Obviously, 5 0, i.e. 5. Similarly, (5 y) and 6 (5 + 5 y) = + y 4 units are to be transported from the factory at Q to the depots at B and C respectively. Thus, 5 y 0, + y 4 0 i.e. y 5, + y 4 Total transportation cost Z is given by Z = y + 00 ( 5 ) + 0 (5 y) + 00 ( + y 4) + 50 (8 y) = 0 ( 7 y + 90) Therefore, the problem reduces to Minimise Z = 0 ( 7y + 90) subject to the constraints: 0, y 0... () + y 8... () 5... (3) y 5... (4) and + y 4... (5) 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 mies 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 mied in order to produce a miture having a minimum cost per bag? What is the minimum cost of the miture per bag? Let the farmer mies bags of brand P and y bags of brand Q. We construct the following table: Prepared by: M. S. KumarSwamy, TGT(Maths) Page

151 So, our problem is minimize Z = y (i) Subject to constraints 3 +.5y 8 + y (ii).5 +.5y y 36 (iii) + 3y 4 (iv) 0, y 0 (v) Firstly, draw the graph of the line 3 +.5y = 8 Secondly, draw the graph of the line.5 +.5y = 45 Thirdly, draw the graph of the line + 3y = 4 On solving equations 3 +.5y = 8 and + 3y = 4, we get C(3, 6). Similarly, on solving equations.5 +.5y = 45 and + 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: Prepared by: M. S. KumarSwamy, TGT(Maths) Page

152 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 y < 950 or 5 + 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 5 + 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 miture to minimize the cost to Rs A dietician wishes to mi together two kinds of food X and Y in such a way that the miture 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 miture which will produce the required diet? Let the dietician mies kg of food X and y kg of food Y. We construct the following table: So, our problem is to minimize Z = 6 + 0y (i) Subject to constraints + y 0 (ii) + y + y 6 (iii) 3 + y 8 (iv) 0, y 0 (v) Firstly, draw the graph of the line + y = 0 Secondly, draw the graph of the line line + y = 6 Thirdly, draw the graph of the line 3 + y = 8 On solving equations + y = 6 and + y = 0, we get B(, 4) Similarly, solving the equations 3 + y = 8 and + 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). Prepared by: M. S. KumarSwamy, TGT(Maths) Page

153 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,6 + 0y < or4 + 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 4 + 5y < 8. Therefore, the minimum value of Z is at B(, 4). Thus, the miture should contain kg of food X and 4 kg of food Y. The minimum cost of the miture is Rs.. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

154 CHAPTER : LINEAR PROGRAMMING Previous Years Board Eam (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 maimise his profit. Formulate an LPP and solve it graphically. Let the manufacturer produces padestal lamps and y wooden shades; then time taken by pedestal lamps and y wooden shades on grinding/cutting machines = ( + y) hours and time taken on the sprayer = (3 + y) hours. Since grinding/cutting machine is available for at the most hours. + y and sprayer is available for at most 0 hours. Thus, we have 3 + y 0 Now profit on the sale of lamps and y shades is, Z = 5 + 5y. So, our problem is to find and y so as to Maimise Z = 5 + 5y (i) Subject to the constraints: 3 + y 0 (ii) + y (iii) 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

155 Let us evaluate the objective function at each corner point as shown below: We find that maimum value of Z is Rs. 60 at B(4, 4). Hence, manufacturer should produce 4 lamps and 4 shades to get maimum 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 maimum 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 maimum profit? Make it as an LPP and solve graphically. What is the maimum profit per week? Let 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 = y We have to maimize Z, subject to the constraints 9 + y y 60...(i) + 3y 30...(ii) 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 - 5 -

156 Now the value of objective function is obtained at corner points as Hence, the company will get the maimum 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 maimise his profit? Make it as a LPP and solve it graphically. Suppose dealer purchase electronic sewing machines and y manually operated sewing machines. If Z denotes the total profit. Then according to question (Objective function) Z = + 8 y Also, + y y y 44 0, y 0. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 5 -

157 We have to maimise 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 maimum when = 8 and y =. Hence, dealer should purchase 8 electronic sewing machines and manually operated sewing machines to obtain the maimum profit ` 39 under given condition. 4. An aeroplane can carry a maimum of 00 passengers. A profit of `500 is made on each eecutive 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 eecutive class. However at least four times as many passengers prefer to travel by economy class than by the eecutive class. Determine how many tickets of each type must be sold in order to maimise 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 tickets of eecutive class and y tickets of economy class. Let Z be net profit of the airline. Here, we have to maimise z. Now Z = y Z = y...(i) According to question 0...(ii) Also + y 00...(iii) (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

158 Now value of Z is calculated at corner point as Hence, 40 tickets of eecutive class and 60 tickets of economy class should be sold to maimise 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 maimise 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, y unit of goods A and B are produced respectively. Let Z be total revenue Here Z = y...(i) Also + 3y 30...(ii) 3 + y 7...(iii) 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

159 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 maimum 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 maimize 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 and y hectare of land be allocated to crop A and B respectively. If Z is the profit then Z = y (i) We have to maimize Z subject to the constraints + y 50 (ii) 0 +0y y 80 (iii) 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 + y = 50 Secondly, draw the graph of the line + y = 80 Prepared by: M. S. KumarSwamy, TGT(Maths) Page

160 Feasible region is bounded. Now value of Z is calculated at corner point as Hence the co-operative society of farmers will get the maimum profit of Rs. 4,95,000 by allocating 30 hectares for crop A and 0 hectares for crop B. Yes, because ecess 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 mi packets from two different suppliers. Supplier S had a packet of mi of 4 units of A and units of B that costs Rs. 0. The supplier T has a packet of mi of unit of A and unit of B costs Rs.4. How many packets of mied from S and T should the company purchase to honour the contract requirement and yet minimize cost? Make a LPP and solve graphically. Let and y units of packet of mies are purchased from S and T respectively. If Z is total cost then Z = 0 + 4y...(i) is objective function which we have to minimize. Here constraints are. 4 + y 80...(ii) + y 60...(iii) Also, 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

161 Point of intersection of + y = 60...(vi) and 4 + y = 80...(vii) (vi) (vii) + y 4 y = = 0 = 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 y < 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 mies from S and 40 packets of mies 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 maimise 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 and y respectively. Here profit is the objective function Z. Z =5 + 3y (i) We have to maimise Z subject to the constrains + y (ii) 3 + y 0 (iii) 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

162 Since (0, 0) Satisfy 3 y 0 Graph of 3 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 y and 3 y 0 are solved as 3 ( ) y 8 4 Coordinate of B (4, 4) Now we evaluate objective function Z at each corner. Hence maimum 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 eceed 50 units. Determine the number of units of each type of computers which the merchant should stock to get maimum 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 and y respectively. Here, Profit is the objective function Z. Z = y (i) we have to maimise z subject to the constraints + y 50 (ii) (Demand Constraint) Prepared by: M. S. KumarSwamy, TGT(Maths) Page

163 y 70,00,000 (iii) (Investment constraint) 5 + 8y 400 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 + y = 50 and 5 + 8y = 400 are solved and we get = 00, y = 50 Now, we evaluate objective function Z at each corner Maimum 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 maimise profit? Make it as an L.P.P. and solve it graphically. Let the factory makes 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 maimized Z = 5 + 6y Prepared by: M. S. KumarSwamy, TGT(Maths) Page

164 Subject to 0, y y y 40 On plotting graph of above constraints or inequalities, we get shaded region. From figure, possible points for maimum value of z are at (4, 0), (8, 0), (0, 5). at (4, 0), z = 0 at (8, 0), z = = 60 (maimum) at (0, 5), z = 50 8 pieces of item A and 0 pieces of item B produce maimum 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 maimum 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 and y respectively Then, the given problem is Maimize, z = + y Prepared by: M. S. KumarSwamy, TGT(Maths) Page

165 Subjected to 0, y y y y 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 = + y = = 5 At (0, 0), z = + y = = 30 Maimum At (0, 0), z = = 0 As Z is maimum at (0, 0), i.e., = 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 maimum 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 maimum 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 Maimum profit on ring = Rs 300 Maimum profit on chain = Rs 90 Let gold rings manufactured per day = Chains manufactured per day = y L.P.P. is maimize Z = y Subject to 0, y 0, + y 4 and + y 6 On plotting graph of above constraints or inequalities, we get shaded region. Possible points for maimum Z are (6, 0), (8, 6) and (0, 4). At (6, 0), Z = = 4800 At (8, 6), Z = = 5440 Maimum At (0, 4), Z = = 4560 Z is maimum at (8, 6). 8 gold rings & 6 chains must be manufactured per day. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 6 -

166 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 6 -

167 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

168 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

169 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 bo 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) eactly 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 (eactly one of them solve the problem) = P(A) P(B ) + P(A ) P(B) Prepared by: M. S. KumarSwamy, TGT(Maths) Page

170 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 boes I, II and III, each containing two coins. In bo I, both coins are gold coins, in bo II, both are silver coins and in the bo III, there is one gold and one silver coin. A person chooses a bo at random and takes out a coin. If the coin is of gold, what is the probability that the other coin in the bo is also of gold? Let E, E and E 3 be the events that boes I, II and III are chosen, respectively. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

171 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 bo is of gold = the probability that gold coin is drawn from the bo 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 eclusive and ehaustive 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 eperience, 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

172 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 si. Find the probability that it is actually a si. Let E be the event that the man reports that si occurs in the throwing of the die and let S be the event that si occurs and S be the event that si does not occur. Then P(S ) = Probability that si occurs = 6 P(S ) = Probability that si does not occur = 5 6 P(E S ) = Probability that the man reports that si occurs when si has actually occurred on the die = Probability that the man speaks the truth = 3 4 P(E S ) = Probability that the man reports that si occurs when si 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 si has occurred is actually a si 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 eclusive and ehaustive. Moreover, P(E ) = P(E ) = Let E : ball drawn is red. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

173 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 eamination. 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 eclusive and ehaustive. 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 eclusive and ehaustive. 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

174 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 eclusive and ehaustive 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 eclusive and ehaustive 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

175 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 eclusive and ehaustive 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 eactly 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 eclusive and ehaustive events. and n(e ) =, n(e ) = 4 Also, n(s) = 6 P(E ) = and P(E ) = Let E : The event that eactly one head show up, P(E/E ) = P(eactly 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 - 7 -

176 The probability that the girl threw,,, 3 or 4 with the die, if she obtained eactly 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 eclusive and ehaustive 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 eclusive and ehaustive 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 - 7 -

177 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 eperiments 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

178 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 eperiment 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) = i p( i ) 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

179 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) = i p( i ) 0 i Also E(X ) = 0 Thus, Var (X) = E (X ) (E(X)) = () Therefore Var( ) 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

180 (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) si appears on at least one die When a die is tossed two times, we obtain (6 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 si appears on atleast one die. So, X may have values 0 or. P(X = 0) = P (si does not appear on any of the die) = Prepared by: M. S. KumarSwamy, TGT(Maths) Page

181 P(X = ) = P (si 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

182 = 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 0 k if P( X ) 3 k, if 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