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Class 12, NCERT Chapter 1,  Exercise 1.2, Q12 f : R → R defined as  f ( x ) = 3 x . Let  x ,  y  ∈  R  such that  f ( x ) =  f ( y ). ⇒ 3 x  = 3 y ⇒  x  =  y ∴ f  is one-one. Also, for any real number ( y)  in co-domain  R , there exists  in  R  such that . ∴ f  is onto. Hence, function  f  is one-one and onto. The correct answer is A.

Class 12, NCERT Chapter 1,  Exercise 1.2, Q11 f : R → R is defined as  f (x) = x 4 Let  x ,  y  ∈  R  such that  f ( x ) =  f ( y ). ∴ does not imply that x 1 =x 2 . For instance, f (1) =  f (-1) = 1 ∴  f  is not one-one. Consider an element 2 in co-domain  R . It is clear that there does not exist any  x  in domain  R  such that  f ( x ) = 2. ∴  f  is not onto. Hence, function  f  is neither one-one nor onto. The correct answer is D

Class 12, NCERT Chapter 1,  Exercise 1.2, Q10 A =  R  - {3}, B =  R  - {1} f : A → B is defined as Let x,y∈A such that f(x)=f(y). ⇒ (x-2)/(x-3)=(y-2)/(y-3) ⇒ (x-2)(y-3)=(y-2)(x-3) ⇒ xy-3x-2y+6=xy-3y-2x+6 ⇒ -3x-2y=-3y-2x ⇒ 3x-2x=3y-2y ⇒ x=y ∴  f  is one-one. Let  y  ∈B =  R  - {1}. Then,  y  ≠ 1. The function  f  is onto if there exists  x  ∈A such that  f ( x ) =  y. Now, f(x)=y ⇒(x-2)(x-3)=y ⇒x-2=xy-3y ⇒x(1-y)=-3y+2 ⇒ x=2-3y/1-y  ∈A   Thus, for any  y  ∈ B, there exists  x=2-3y/1-y  ∈A     such that Hence, function  f  is one-one and onto.

Class 12, NCERT Chapter 1,  Exercise 1.2, Q9 Thus it is bijective.

Class 12, NCERT Chapter 1,  Exercise 1.2, Q8 f: A × B → B × A is defined as f(a, b) = (b, a). ∴ f is one-one. Now, let (b, a) ∈ B × A be any element. Then, there exists (a, b) ∈A × B such that f(a, b) = (b, a). [By definition of f] ∴ f is onto. Hence, f is bijective.

Class 12, NCERT Chapter 1,  Exercise 1.2, Q7 (i)  Here f(x) = 3 – 4x Then f(x 1 ) = 3 – 4x 1 and f (x 2 ) = 3 – 4x 2 Now f(x 1 ) = f (x 2 ) ⇒ 3 – 4x 1  = 3 – 4x 2 ⇒ – 4x 1  = –4x 2 ⇒ x 1  = x 2 ∴ f(x 1 ) = f(x 2 ) & x 1  = x 2 So, f is one-one function. Let f(x) = y ∈ R Then y=3 - 4x ⇒ x= 3-y/4 (ii) Let x 1  = 2 and x 2  = –2 ∈ R Here f(x) = 1 + x 2 Then f(x 1 ) = f (2) = 1 + (2)  2  = 5 and f (x 2 ) = f(–2) = 1 + (–2)  2  = 5 ∴ f (x 1 ) = f(x 2 ) but x 1  ≠ x 2 So, f is not one-one function. Let f(x) = – 2 ⇒R Then 1 + x 2  = –2 ⇒ x 2  = –3 = ± √–3 ∈R So, f is not onto function.

Class 12, NCERT Chapter 1,  Exercise 1.2, Q6 It is given that A={1,2,3} B={4,5,6,7} f : A →B defined as f (x)=  {(1, 4), (2, 5), (3, 6)}  Therefore  f(1)=4,   f(2)=5,   f(3)=6 It is seen that the images of distinct elements of A under  f  are distinct. Hence, function  f  is one-one

Signum 

Class 12, NCERT Chapter 1,  Exercise 1.2, Q4 f : R → R is given by, It is seen that f(- 1)=  |-1|=1, f (1)=|1|=1   ∴ f (-1) =  f (1), but - 1 ≠ 1. ∴  f  is not one-one. Now, consider -1 ∈  R . It is known that  f ( x ) = |x| is always non-negative. Thus, there does not exist any element  x  in domain  R  such that  f ( x ) = |x| = -1. ∴  f  is not onto. Hence, the modulus function is neither one-one nor onto.

Class 12, NCERT Chapter 1,  Exercise 1.2, Q3 f :R → R given by,  f (x) = [x]  It is seen that f (1.2) = [1.2] = 1, f (1.9) = [1.9] = 1. ,  Therefore f (1.2) = f (1.9), but 1.2 is not equal to 1.9. ,  therefore f is not one-one.  Now, consider 0.7  ∈    R .  It is known that f (x)= [x] is always an integer. Thus, there does not exist any element x  ∈  R such that f (x) = 0.7. , therefore  f is not onto.  Hence, the greatest integer functio n is neither one-one nor onto. 

Class 12, NCERT Chapter 1,  Exercise 1.2, Q2 (i)  f :  N  →  N  is given by, f ( x ) =  x 2 It is seen that for  x ,  y  ∈ N ,  f ( x ) =  f ( y )  ⇒  x 2  =  y 2   ⇒  x  =  y . ∴ f  is injective. Now, 2 ∈  N . But, there does not exist any  x  in  N  such that  f ( x ) =  x 2  = 2. ∴  f  is not surjective. Hence, function  f  is injective but not surjective. (ii)  f :  Z  →  Z  is given by, f ( x ) =  x 2 It is seen that  f (-1) =  f (1) = 1, but -1 ≠ 1. ∴  f  is not injective. Now,-2 ∈  Z . But, there does not exist any element  x  ∈ Z  such that  f ( x ) =  x 2  = -2. ∴  f  is not surjective. Hence, function  f  is neither injective nor surjective. (iii)  f :  R  →  R  is given by, f ( x ) =  x 2 It is seen that  f (-1) =  f (1) = 1, but -1 ≠ 1. ∴  f  is not injective. Now,-2 ∈  R . But, there does not exist any element  x  ∈  R  such that  f ( x ) =  x 2  = -2. ∴  f  is not surjective. Hence, function  f  is neither injective nor surjective. (iv)  f :

Class 12, NCERT Chapter 1,  Exercise 1.2, Q1 It is given that f: R∗ →R∗  is defined by f(x) =1/x For one – one: Let x, y ∈  R∗  such that f(x) = f(y) ⇒ 1/x =1/y ⇒ x = y Therefore,  f is one–one . For onto: For y ∈  R∗, there exists x =1/y ∈  R∗                       [as y ≠ 0]  such that ⇒ f(x) = 1/ [1/y] = y Therefore, f is onto . Thus, the given function f is one-one and onto. Now, consider function g:N→R∗  defined by g(x) =1/x We have, ⇒ g(x) = g(y) ⇒ 1/x=1/y ⇒ x = y Therefore, g is one–one. Further, it is clear that g is not onto as for 1.2 ∈ R∗,  there does not exit any x in N such that g(x) =1/1.2

Class 12, NCERT Chapter 1,  Example14 Since f is one-one, three elements of {1, 2, 3} must be taken to 3 different elements of the co-domain {1, 2, 3} under f. Hence, f has to be onto .

Class 12, NCERT Chapter 1,  Example13 Suppose f is not one-one. Then there exist two elements, say 1 and 2 in the domain whose image in the co-domain is the same. Also, the image of 3 under f can be only one element. Therefore, the range set can have at the most two elements of the co-domain {1, 2, 3}, showing that f is not onto, a contradiction. Hence, f must be one-one.

Class 12, NCERT Chapter 1,  Example12 Suppose f(x 1 ) = f(x 2 ). Note that if x 1 is odd and x 2 is even, then we will have x 1 + 1 = x 2 – 1, i.e., x 2 – x 1 = 2, which is impossible. Similarly, the possibility of x 1 being even and x 2 being odd can also be ruled out, using a similar argument. Therefore, both x 1 and x 2 must be either odd or even. Suppose both x 1 and x 2 are odd. Then f(x 1 ) = f(x 2 ) ⇒ x 1 + 1 = x 2 + 1 ⇒ x 1 = x 2 Similarly, if both x 1 and x 2 are even, then also f(x 1 ) = f(x 2 ) ⇒ x 1 – 1 = x 2 – 1 ⇒ x 1 = x 2 . Thus, f is one-one . Also, any odd number 2r + 1 in the co-domain N is the image of 2r + 2 in the domain N and any even number 2r in the co-domain N is the image of 2r – 1 in the domain N .  Thus, f is onto .

Class 12, NCERT Chapter 1,  Example11 Since f(– 1) = 1 = f(1), f is not one-one.  Also, the element – 2 in the co-domain R is not the image of any element x in the domain R. Therefore f is not onto.

Class 12, NCERT Chapter 1,  Example9 f is not one-one,  as f(1) = f(2) = 1.                [  it has two preimages ] But f is onto,  as given any y ∈ N, y ≠ 1,  we can choose x as y + 1  such that f(y + 1) = y + 1 – 1 = y.  Also for 1 ∈ N, we have f(1) = 1.

Class 12, NCERT Chapter 1,  Example10  f is one-one, as  f ( x 1 ) = f ( x 2 ) ⇒  2 x 1 = 2 x 2  ⇒ x 1 = x 2 Also, given any real number y in R,  there exists y/2 in R,  such that  f( y/2 ) = 2 . ( y/2 ) = y.  Hence, f is onto

Class 12, NCERT Chapter 1,  Example8 The function f is one-one,  for  f ( x 1 ) = f ( x 2 ) ⇒ 2 x 1 = 2 x 2 ⇒  x 1 = x 2 Further, f is not onto, as for 1 ∈ N ,  there does not exist any x in N ,  such that  f(x) = 2x = 1.