QUESTION BANK II PUC SCIENCE

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1 QUESTION BANK II PUC SCIENCE I. Very Short answer questions. (x9=9). Define Symmetric relation. Ans: A relation R on the set A is said to be symmetric if for all a, b, A, ar b Implies bra. i.e. (a, b) R (b, a) R. Let A = {4, 6, 8, 0} R = {(a, b); a + b = 5, a, b A} Show that the relation R Is empty. Ans: This is an empty set, as no pair (a, b) satisfies the condition a + b = 5.. Give an example of a relation defined on a suitable set which is i. reflexive, symmetric and transitive. ii. reflexive, symmetric but not transitive. iii. reflexive, transitive but not symmetric. iv. symmetric, transitive but not reflexive. vi. symmetric, but not reflexive and not transitive. vii. not reflexive, not symmetric, and not transitive. Solution: Consider a Set A = {a, b, c} i. Define a relation R, on A as. R = {(a, a), (b, b), (c, c)}. Clearly R is reflexive, symmetric and transitive ii. Consider the relation R on A as. R = {(a, a), (b, b), (c, c), (a, b), (b, a), (a, c), (c, a)}. R is symmetric, reflexive But R is not transitive for (b, a) R and (a, c) R but (b, c) R. iii. Consider the relation R on A as R = {(a, a), (b, b), (c, c), (a, b)} iv. R is reflexive and transitive but not symmetric for (a, b) R but (b, a) R Consider the relation R 4 on A as R 4 = {(a, a), (b, b), (c, c), (b, a)} is symmetric and transitive but not reflexive. Since (c, c) R 4. v. Consider the relation R 5 on A as R 5 = {(a, a), (b, b), (c, c), (a, b), (c, a)} R 5 is reflexive. R 5 is not symmetric because (a, b) R 5 but (b, a) R 5 Also R 5 is not transitive because (c, a) R 5, (a, b) R 5 but (c, b) R 5. vi. vii. Consider the relation R 6 on A by R 6 = {(a, a), (c, c), (a, b), (b, a)} R 6 is Symmetric but R 6 is not reflexive because (b, b) R 6. Also R 6 is not transitive for (b, a) R 6, (a, b) R 6 but (b, b) R 6. The relation R 7 defined on A as R 7 = {(a, b), (b, c) } is not reflexive, not symmetric and not transitive. * FUNCTION * (One mark question and answers) 4. Let A = {,, } B = {4, 5, 6, 7,} and let f = {(, 4), (, 5), (, 6)} be a function from A to B. show that f is not onto. Ans: 7 B has no pre image in A. so f is not onto. 5. If, f : R R is defined by f(x) = 4x- x R prove that f is one-one. Solution: For any two elements x, x R such that f(x ) = f (x ) we have 4x = 4x 4x = 4x x = x Hence f is one one. 6. Define transitive relation.

2 Ans: A relation R on the set S is said to be transitive relation if arb and brc arc. i.e. if (a, b) R and (b, c) R (a, c) R. 7. Let f : R R be the function defined by f (x) = x- x R. write Ans; y = f(x) = x x = (y + ) f ( y) = (y+) i.e. f ( y) = (x+) 8. Define a binary operation on a set. Ans: Let S be a non-empty set. The function *: S X S S which associates each ordered pair (a, b) of the elements of S to a unique element of S denotes by a*b is called a binary operation or a binary composition on S. 9. Determine whether or not each of the definition defined below is a binary operation justify. a) on Z + defined by a*b = a-b b) on Z + defined by a*b = a Solution: a) We have for a, b, Z + a*b = a b. We know the + a b is always positive. a, b Z, a*b = a b is a positive integer. Hence * is a binary operation on Z + b) Cleary a*b = a Z for all a, b Z Thus * is a binary operation on Z + * SHORT ANSINER QUESTIONS* (Answers to two marks questions) 0. Define an equivalence relation. Give on example of a relation which is transitive but not reflexive. Solution: A relation R on a Set S is called on equivalence relations. if R is reflexive, symmetric, and transitive. Ex. The relation < (less than) defined on the set R of all real numbers is transitive, but not reflexive.. Show that the signum function f : R R f defined by if x > 0 f ( x) 0 if x = 0 if x < 0 is neither one-one nor onto. Solution: From the graph of the function we have f() = and f() =, i.e. f() = f() = but f is not one-one Again for 4 R (co domain) there exists no x R (domain) Such that f(x) = 4 because f(x) = or for x 0. f is not onto.

3 . State whether the function f : R defined by f(x) = + x x R is one-one, onto or objective justify your answer. Solution: We have f(x) = + x for all x R. clearly we observe that f(x) = + = 5 and f( ) = + ( ) = 5 i.e, f() = f( ) but f is not one-one. Again for y there exist no real number x, such that f(x) = because if f(x) = + x = x = x = R f is not onto.. Show that the greatest integer function f : R R defined by f(x) = [x] is neither oneone nor onto. Solution: We have f(x) = [x] = greatest integer less than or equal to x. i.e. [.] =, [] = but.. Now we have [x] = for all x [, ] f is not one-one. Again R (codomain), but there existing no x R (domain) such that f(x) = because [x] is always an integer x R f is not onto. 4. Cheek the injectivity and surjectivity of the function f : Z Z defined by f(x) = x x Z Solution: We have f(x ) = f(x ) x x x x f is injective. Now 7 Z (codomain) and it has no pre-image in Z because if f(x) = 7 then x 7 x 7 Z f is not surjective. 5. If f(x) = x and g(x) = 5x then find (i) got and (ii) fog. Solution: We have f(x) = x and g(x) = 5x consider go f(x) = g( x ) = 5 x 5 x x 0 = 5 x x 0 Now fog(x) = f [g(x)] = f( 5x ) = 5x 6. If f ; R R be given by f ( x) x Solution: We have f ( x) x Now to find fo f(x). fof ( x) f f ( x) f x = f(y) where y = ( x fof ( x) f ( y) y = x = (x ) x. Hence fo f(x) = x.

4 7. Consider f : {,, } {a, b, c} given by f() = a, f() = b f() = C find f and show that f f Solution: Consider f : {,, } {a, b, c} given by f() = a, f() = b and f() = C f () a, f () b and f()=4 f ( a), f ( b) and f( c) f a,, b,, c, g( say) we observe that g is also objective.,,,,, g a b c f g f f f 4x 8. If f(x) =, x show that 6 x 4 (fof) (x) = x for all x. what is the inverse f? 4x 4 4 f( x) fof x f f ( x) = 6 x 4 6 f( x) 4 4x x 4 4 (4x ) (6 x 4) 4x x 6 (4x ) 4( x 4) 4 Solution: fof I f f 9. Let f : R R be defined by f(x) = 0x+7, Find the function g : R R such that gof = fog = Ig Solution: We have f(x) = 0x+7 By data, gof(x) = Ig(x) g [f(x)] = x. g [0x+7] = x. y 7 Let y = 0x+7 x 0 y 7 Then g(0x+7) = g0 7 y 0 0. Consider a function f : f : 0, R given by f(x) = Sinx. and g : f : 0, R given by g(x) = cosx. Show that f and g are one-one but (f + g) is not one-one. Solution: Since for any two distinct elements x and x in 0,, sin x sin x and cos x cos x Both f and g are one-one But (f + g) (o) = f(o) + g(o) = and 4

5 = f sin cos Therefore (f + g) is not one one. Examine whether the binary operation * defined below are commutative, associative a. On Q defined loy a * b = ab + b. On Z + defined loy a * b = a b c. On Q defined loy a * b = ab/ Solution: a) For every a, b, Q we have i) a * b = ab + Q Now a * b = ab + = ba + (Usual multiplication is commutative) = b * a * is commutative in Q. ii) consider 4, 5, 6 Q. 4 * ( 5 * 6) = 4 * (0 + ) = 4 * ) = 4 () + = 4 + = 5 and (4*5) * 6 = * 6 = ( x 6) + = 57 a * (b * c) = (a * b) * c is not true for all a, b, c, Q. Hence * is not associative. b) For all a, b, * Z +, clearly a * b = a b Z + i.e. * is a binary operation in Z + i) If a then a b b a i.e. a * b b * a for a b. Thus * is not commutative in Z + ii) Consider,, 4, Z + ( * ) * 4 = * 4 = 8 * 4 = 8 4 = and * ( * 4) = * 4 = () 4 = 8 ( * ) * 4 * ( * 4) f g g Thus (a*b)*c = a*(b*c) is not true a, b, c, Z Hence * is not associative in Z +. C. We have a * b = ab a,b Q i) Now a * b = ab = ba = b * a (Usual multiplication is commutative) * is commutative in Q. bc ii. Now consider a * (b * c) = a * bc a abc 4 ab abc ( a* b)* c * c 4 (a * b) * c = a * (b * c) for all a, b, c, Q. Hence * is associative in Q.. Write the multiplicative modulo table for the set A = {, 5, 7, }. Find the identity element w.r.t X X 5 7 Solution: The elements in the row corresponding 5 7 to the element, coincides with the elements of a above the horizontal line in the same order Thus acts as an identify element

6 * ANSIEWERS TO THREE MARKS QUESTIONS*. The relation R defined on the Set A = {,,, 4, 5} by R = {(a, b) ; a b is even}. Show that the relation R is on equivalence relation. (consider O as an even num ber) Proof: For any a A we have a a = o considered as even i.e. (a, a,) R for all a A Thus R is reflexive relation. Let (a, b) R a b is even b a is also even (b, a ) R. Thus the relation R is symmetric. Let (a, b) R and (b, c) R. a b is even and b c even a b k and b c =l for k, l Z. a b + b c = k + l = (k + l) a b + b c = (n) where n = (k+l) a c = n a c = even (a, c) R R Is transitive. Hence R Is an equivalence relation. 4. Let A = R {} and B = R {}. Consider the function f : A B defined by f(x) = x x x Solution: We have f(x) = for all x R x x x Now let f(x ) = f(x ) x x Is f is one-one and onto? Justify your answer. x x x x 6 x x x x 6 x x Thus f ( x ) f ( x) x x i.e. f is one-one. Let y B thus y l. x Now f ( x) y y x y ( x ) x y y x ( y ) x ( y ). y Also y for if y y y y y which is not true. Thus for every y Bthere exists y x A such that f ( x) y y i.e. f is onto. Hence f is bijective function. 5. If f and g are two functions defined by f(x) = x + 5 and g(x) = x + find gof(), ii) fog() and iii) gog(). Solution: We have f ( x) x 5, g( x) x. Now gof ( x) g[ f ( x)] g(x 5) 6

7 gof x ( ) (x 5) 4x 5 0x 4x 0x 6 gof () 4() 0() ii) fog( x) f ( x ) ( x ) 5 iii) fog x ( ) x 7 fog() () gog x ( ) g( x ) 4 ( x ) x x gog() Let * be the binary operation on N given by a * b = l.c.m of a and b Is * commutative? Is * associative? Find the identify in N w.r.t *. Solution: We have a * b = l.c.m. of a and b a, b N clearly l.c.m of two positive integers is a positive integer. Thus N is closed under the operation *. i) Clearly l.c.m of a and b = l.c.m. of b and a a*b = b*a for all a, b, N Thus * is commutative. ii) Let a, b, c, N be arbitrary. Now a*(b * c) = a*(l.c.m. of b and c) = l.c.m. of [ a and (l.c.m of b and c)] = l.c.m of (a,b,c)... () Again, (a * b) * c = (l.c.m of a and b) * c = l.c.m of [(l.c.m. of a and b) and c] = l.c.m of (a,b,c)...() Thus (a * b) *c = a* (b * c) a, b, c N Hence * is associative. iii) Let e be the identify element in N. Then for all a N, we have a * e = e * a = a a* e = l.c.m of a and e = a e* a = l.c.m. of e and a = a l.c.m of a and e = l.c.m. of e and a = a for all a N. implies e =. Thus acts as identify element in N. * FIVE MARKS QUESTIONS AND ANSWERS* 7. Consider f : R [ 5, ) defined by with y 6 f( y) Solution: We have f : R [ 5, ) by f ( x) 9x 5 Let a, a Such that f(a ) = f(a ) R 9a 6a 5 9a 6a 5 f ( x) 9x 5 show that f is invertible 7

8 9a 9a 6a 6a 0 a a a a a a a a 9a 9a a a 0 9a 9a 6 o as a, a R a a Thus f is one-one. Let y f ( x) 9x 5 9x 5 y ( y 5) x 8 y 5 y 6 x y 6 f( y) For every element y B, there exists a pre-image x in[ 5,. So f is onto Thus f is one-one and onto and therefore invertible Hence inverse function of f is given by ( ) y f y 6 4x 8. If f : A A defined by f( x) where AR. show that f is 4 f invertible and f 4x Solution: We have f( x) with x 4 4x 4x Now f ( x) f ( x) 4 4 4x x 8x 4x x 8x 4x 4x x x Thus f is one-one. Now let y A and f(x) f ( x) y yaand 4 x y 4 4y yaand x Thus for every y A, there exists 4 y x A y such that 6y f ( x) y. That is f is onto. Hence f is bijective. Now f ( x) y f ( y) x 8

9 4y 4x x y 4 4 Thus : f A A is defined by 4x f( x) 4 Clearly f f. 9. Let f; w f : w w be defined by n n is odd f( n) n if 'n' iseven Show that f is invertible and find The inverse of f. Hence w is the set of all whole numbers. Solution: Let x, x w be arbitrary Let x and x are even. f ( x ) f ( x ) x x x x Let x and x are odd f ( x ) f ( x ) x x x x Let x and x are odd f ( x ) f ( x) x x x x In both the cases, f ( x ) f ( x) x x Supposing x is odd and x is even then x x. Now f ( x) xand f ( x) x Also x x i.e. x x f ( x ) f ( x) Hence f is one-one Let m0 w (co domain) and f(n) = m n m if 'n' is odd Now f ( n) m f ( n) n m if 'n' is even. nm if 'n' is odd f( n) n m if 'n' is even. Thus for every m w (co domain) there exists n w (= m+ or m ) Such that f(n) = m. Also (considering O as even) Let f( o) 0 i.e. f () 0 Thus f is onto Hence f is a bijection. f ( n) m f ( m) n. m n m is odd f( m) m n m is even m n n is even f( m) m n n is odd 9

10 n n is even Thus f( n) n n is odd Is inverse function. 0

RELATIONS AND FUNCTIONS

For more important questions visit : www.4ono.com CHAPTER 1 RELATIONS AND FUNCTIONS IMPORTANT POINTS TO REMEMBER Relation R from a set A to a set B is subset of A B. A B = {(a, b) : a A, b B}. If n(a)