CHAPTER 1: RELATIONS AND FUNCTIONS

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1 CHAPTER 1: RELATIONS AND FUNCTIONS Previous Years Board Exam (Important Questions & Answers) 1. If f(x) = x + 7 and g(x) = x 7, x R, find ( fog) (7) Given f(x) = x + 7 and g(x) = x 7, x R fog(x) = f(g(x)) = g(x) + 7 = (x 7) + 7 = x (fog) (7) = 7. MARKS WEIGHTAGE 06 marks 2. If f(x) is an invertible function, find the inverse of f (x) = 3 x 2 3x 2 Given f ( x) 3x 2 Let y y 2 3x 2 y x 3 1 x 2 f ( x) 3 3. Let T be the set of all triangles in a plane with R as relation in T given by R = {(T 1, T 2 ) :T 1 T 2 }. Show that R is an equivalence relation. (i) Reflexive R is reflexive if T 1 R T 1 Since T 1 T 1 R is reflexive. (ii) Symmetric R is symmetric if T 1 R T 2 T 2 R T 1 Since T 1 T 2 T 2 T 1 R is symmetric. (iii) Transitive R is transitive if T 1 R T 2 and T 2 R T 3 T 1 R T 3 Since T 1 T 2 and T 2 T 3 T 1 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 2, then find the value of 2 * 4. Given a *b = a + 3b 2 a, b z 2*4 = x 4 2 = = 0. Let * be a binary operation on N given by a * b = HCF (a, b) a, b N. Write the value of 22 * 4. Given a * b = HCF (a, b), a, b N 22 * 4 = HCF (22, 4) = 2 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 1 -

2 6. Let f : N N be defined by function f is bijective. n 1, if n is odd f ( n) 2 n for all n N. Find whether the, if n is even 2 n 1, if n is odd Given that f : N N be defined by f ( n) 2 n for all n N., if n is even 2 Let x, y N and let they are odd then x 1 y 1 f ( x) f ( y) x y 2 2 If x, y N are both even then also x y f ( x) f ( y) x y 2 2 If x, y N are such that x is even and y is odd then x 1 y f ( x) and f ( y) 2 2 Thus, x y for f(x) = f(y) Let x = 6 and y = 6 1 We get f (6) 3, f () f(x) = f(y) but x y...(i) So, f (x) is not one-one. Hence, f (x) is not bijective. 7. If the binary operation *, defined on Q, is defined as a * b = 2a + b ab, for all a, b Q, find the value of 3 * 4. Given binary operation is a*b = 2a + b ab 3* 4 = * 4 = 2 x 1 8. What is the range of the function f ( x) ( x 1)? x 1 We have given f ( x) ( x 1) ( x 1), if x 1 0 or x 1 x 1 ( x 1), if x 1 0 or x 1 ( x 1) (i) For x > 1, f ( x) 1 ( x 1) ( x 1) (ii) For x < 1, f ( x) 1 ( x 1) x 1 Range of f ( x) is { 1, 1}. ( x 1) Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 2 -

3 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.} Prove that R is an equivalence relation. We have provided R = {(a, b) : a, b Z, and(a b) is divisible by } (i) As (a a) = 0 is divisible by. (a, a) R a R Hence, R is reflexive. (ii) Let (a, b) R (a b) is divisible by. (b a) is divisible by. (b a) is divisible by. (b, a) R Hence, R is symmetric. (iii) Let (a, b) R and (b, c) Z Then, (a b) is divisible by and (b c) is divisible by. (a b) + (b c) is divisible by. (a c) is divisible by. (a, c) R R is transitive. Hence, R is an equivalence relation. 3ab 10. Let * be a binary operation on Q defined by a* b. Show that * is commutative as well as associative. Also find its identity element, if it exists. For commutativity, condition that should be fulfilled is a * b = b * a 3ab 3ba Consider a* b b* a a * b = b * a Hence, * is commutative. For associativity, condition is (a * b) * c = a * (b * c) 3ab 9ab Consider ( a* b)* c * c 2 3bc 9ab and a*( b* c) a* 2 Hence, (a * b) * c = a * (b * c) * is associative. Let e Q be the identity element, Then a * e = e * a = a 3ae 3ea a e If f : R R be defined by f(x) = (3 x 3 ) 1/ 3, then find fof(x). If f : R R be defined by f(x) = (3 x 3 ) 1/3 then ( fof) x = f( f(x)) = f [(3 x 3 ) 1/3 ] = [3 {(3 x 3 ) 1/3 } 3 ] 1/3 = [3 (3 x 3 )] 1/3 = (x 3 ) 1/3 = x Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 3 -

4 12. Let A = N N and * be a binary operation on A defined by (a, b) * (c, d) = (a + c, b + d). Show that * is commutative and associative. Also, find the identity element for * on A, if any. Given A = N N * is a binary operation on A defined by (a, b) * (c, d) = (a + c, b + d) (i) Commutativity: Let (a, b), (c, d) N N Then (a, b) * (c, d) = (a + c, b + d) = (c + a, d + b) (a, b, c, d N, a + c = c + a and b + d = d + c) = (c, d) * b Hence, (a, b) * (c, d) = (c, d) * (a, b) * is commutative. (ii) Associativity: let (a, b), (b, c), (c, d) Then [(a, b) * (c, d)] * (e, f) = (a + c, b + d) * (e, f) = ((a + c) + e, (b + d) + f) = {a + (c + e), b + (d + f)] ( set N is associative) = (a, b) * (c + e, d + f) = (a, b) * {(c, d) * (e, f)} Hence, [(a, b) * (c, d)] * (e, f) = (a, b) * {(c, d) * (e, f)} * is associative. (iii) Let (x, y) be identity element for on A, Then (a, b) * (x, y) = (a, b) (a + x, b + y) = (a, b) a + x = a, b + y = b x = 0, y = 0 But (0, 0) A For *, there is no identity element. 13. If f : R R and g : R R are given by f(x) = sin x and g(x) = x 2, find gof(x). Given f : R R and g : R R defined by f (x) = sin x and g(x) = x 2 gof(x) = g [f(x)] = g (sin x) = (sin x) 2 = sin 2 x 14. Consider the binary operation* on the set {1, 2, 3, 4, } defined by a * b = min. {a, b}. Write the operation table of the operation *. Required operation table of the operation * is given as * If f : R R is defined by f(x) = 3x + 2, define f[f(x)]. f (f (x)) = f (3x + 2) =3. (3x + 2) + 2 = 9x = 9x + 8 Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 4 -

5 16. Write fog, if f : R R and g : R R are given by f(x) = 8x 3 and g(x) = x 1/3. fog (x) = f (g(x)) = f (x 1/3 ) = 8(x 1/3 ) 3 = 8x 17. Let A = {1, 2, 3}, B = {4,, 6, 7} and let f = {(1, 4), (2,), (3, 6)} be a function from A to B. State whether f is one-one or not. f is one-one because f(1) = 4 ; f(2) = ; f(3) = 6 No two elements of A have same f image. 18. Let f : R R be defined as f(x) =10x +7. Find the function g : R R such that gof = fog =I R. gof = fog = IR fog = IR fog(x) = I (x) f (g(x)) = x [I(x) = x being identity function] 10(g(x)) + 7 = x [f(x) = 10x + 7] x 7 g( x) 10 x 7 i.e., g : R R is a function defined as g( x) Let A = R {3} and B = R {1}. Consider the function f : A B defined by Show that f is one-one and onto and hence find f 1. Let x 1, x 2 A. x1 2 x2 2 Now, f(x 1 ) = f(x 2 ) x1 3 x2 3 ( x 2)( x 3) ( x 3)( x 2) x x 3x 2x 6 x x 2x 3x x 2x 2x 3x x1 x2 x1 x2 Hence f is one-one function. For Onto x 2 Let y xy 3y x 2 x 3 xy x 3y 2 x( y 1) 3y 2 3y 2 x ( i) y 1 x 2 f ( x) x 3. Prepared by: M. S. KumarSwamy, TGT(Maths) Page - -

6 From above it is obvious that y except 1, i.e., y B R {1} x A Hence f is onto function. Thus f is one-one onto function. It f 1 1 3y 2 is inverse function of f then f ( y) [from (i)] y The binary operation * : R R R is defined as a * b = 2a + b. Find (2 * 3) * 4 (2 * 3) * 4 = (2 2 +3) * 4 = 7 * 4 = = 18 x 1, if x is odd 21. Show that f : N N, given by f ( x) is both one-one and onto. x 1, if x is even For one-one Case I : When x 1, x 2 are odd natural number. f(x 1 ) = f(x 2 ) x 1 +1 = x 2 +1 x 1, x 2 N x 1 = x 2 i.e., f is one-one. Case II : When x 1, x 2 are even natural number f(x 1 ) = f(x 2 ) x 1 1 = x 2 1 x 1 = x 2 i.e., f is one-one. Case III : When x 1 is odd and x 2 is even natural number f(x 1 ) = f(x 2 ) x 1 +1 = x 2 1 x 2 x 1 = 2 which is never possible as the difference of odd and even number is always odd number. Hence in this case f (x 1 ) f(x 2 ) i.e., f is one-one. Case IV: When x 1 is even and x 2 is odd natural number Similar as case III, We can prove f is one-one For onto: f(x) = x +1 if x is odd = x 1 if x is even For every even number y of codomain odd number y - 1 in domain and for every odd number y of codomain even number y +1 in Domain. i.e. f is onto function. Hence f is one-one onto function. 22. 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 - 6 -

7 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. 23. If the binary operation * on the set Z of integers is defined by a * b = a + b, then write the identity element for the operation * in Z. Let e Z be required identity a* e = a a Z a + e = a e = a a + e = 24. 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 = {, 6, 7, 8, 9} given by R = {(a, b) : a b, is divisible by 2}, 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 2} Reflexivity Here (a, a) R as a a = 0 = 0 divisible by 2 i.e., R is reflexive. Symmetry Let (a, b) R (a, b) R a b is divisible by 2 a b = ± 2m b a = 2m b a is divisible by 2 (b, a) R Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 7 -

8 Hence R is symmetric Transitivity Let (a, b), (b, c) R Now, (a, b), (b, c) R a b, b c are divisible by 2 a b = ±2m and b c = ±2n a b + b c = ± 2(m + n) (a c) = ± 2k [k = m + n] (a c) = 2k (a c) is divisible by 2 (a, c) R. Hence R is transitive. Therefore, R is an equivalence relation. The elements related to 6 are 6, 8. ab 26. Let * be a binary operation, on the set of all non-zero real numbers, given by a* b for all a, b R {0}. Find the value of x, given that 2 * (x * ) = 10. Given 2 * (x * ) = 10 x 2* 10 2* x 10 2 x x x Let A = {1, 2, 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 [(2, )]. Given, R is a relation in A A defined by (a, b)r(c, d) a + d = b + c (i) Reflexivity: a, b A Q a + b = b + a (a, b)r(a, b) So, R in reflexive. (ii) Symmetry: Let (a, b) R (c, d) Q (a, b)r(c, d) a + d = b + c b + c = d + a [Q a, b, c, d N and N is commutative under addition[ c + b = d + a (c, d)r(a, b) So, R is symmetric. (iii) Transitivity: Let (a, b)r(c, d) and (c, d)r(e, f) Now, (a, b)r(c, d) and (c, d)r(e, f) a + d = b + c and c + f = d + e a + d + c + f = b + c + d + e a + f = b + e (a, b)r(e, f). R is transitive. Hence, R is an equivalence relation. 2nd Part: Equivalence class: [(2, )] = {(a, b) A A: (a, b)r(2, )} = {(a, b) A A: a + = b + 2} = {(a, b) A A: b a = 3} = {(1, 4), (2, ), (3, 6), (4, 7), (, 8), (6, 9)} Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 8 -

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

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