NCERT Solutions class 12 Maths Exercise 1.2 Ch 1 Relations and Functions

Ans.  

Part I:  and 

If  then 

  is one-one.

     is onto.

Part II:  When domain R is replaced by N, co-domain R remaining the same, then, f=N→Rf=N→R

If 

  where  N

  is one-one.

But every real number belonging to co-domain may not have a pre-image in N.

e.g.  N    is not onto.

Ans. (i)  given by 

If  then 

  is injective.

There are such numbers of co-domain which have no image in domain N.

e.g. 3 co-domain N, but there is no pre-image in domain of 

therefore  is not onto.   is not surjective.

(ii)  given by 

Since,  Z =  therefore, 

  and 1 have same image.     is not injective.

There are such numbers of co-domain which have no image in domain Z.

e.g. 3 co-domain, but  domain of     is not surjective.

(iii)  given by 

As 

  and 1 have same image.    is not injective.

e.g.  co-domain, but  domain R of        is not surjective.

(iv)  given by 

If  then 

i.e., for every  N, has a unique image in its co-domain.     is injective.

There are many such members of co-domain of  which do not have pre-image in its domain e.g., 2, 3, etc.

Therefore  is not onto.   is not surjective.

(v) given by 

If  then 

i.e., for every  Z, has a unique image in its co-domain.    is injective.

There are many such members of co-domain of  which do not have pre-image in its domain.

Therefore  is not onto. ff is not surjective.

Ans. Function , given by 

  and 

  is not one-one.

All the images of  R belong to its domain have integers as the images in co-domain. But no fraction proper or improper belonging to co-domain of  has any pre-image in its domain.

Therefore, ff is not onto.

Ans. Modulus Function , given by 

Now    

  contains 

Thus negative integers are not images of any element.       is not one-one.

Also second set R contains some negative numbers which are not images of any real number.

  is not onto.

Ans. Signum Function , given by 

  for 

       is not one-one.

Except  no other members of co-domain of  has any pre-image its domain.

  is not onto.

Therefore, ff is neither one-one nor onto.

Ans. A = {1, 2, 3}, B = {4, 5, 6, 7} and  = {(1, 4), (2, 5), (3, 6)}

Here,    and 

Here, also distinct elements of A have distinct images in B.

Therefore, is a one-one function.

Ans. (i)  defined by 

Now,   if  R, then  and 

And     if , then     is one-one.

Again, if every element of Y (– R) is image of some element of X (R) under  i.e., for every  Y, there exists an element  in X such that 

Now 

  is onto or bijective function.

(ii)  defined by 

Now,   if  R, then  and 

And     if , then 

    is not one-one.

Again, if every element of Y (– R) is image of some element of X (R) under  i.e., for every  Y, there exists an element  in X such that 

Now,    x=±y−1−−−−√x=±y−1

 f(y−1−−−−√)=1+y−1=−y≠yf(y−1)=1+y−1=−y≠y

  is not onto.

Therefore,  is not bijective.

Ans. Injectivity: Let  and  A x B such that 

  and 

  = 

  =  for all  A x B

So,  is injective.

Surjectivity: Let  be an arbitrary element of B x A. Then  B and  A.

  A x B

Thus, for all  B x A, their exists  A x B such that     

So,  A x B  B x A is an onto function, therefore  is bijective.

Ans.  be defined by 

(a)  and 

The elements 1, 2, belonging to domain of  have the same image 1 in its co-domain.

So,  is not one-one, therefore,  is not injective.

(b) Every number of co-domain has pre-image in its domain e.g., 1 has two pre-images 1 and 2.

So,  is onto, therefore,  is not bijective.

Ans. A = R – {3} and B = R – {1} and 

Let  A, then  and 

Now, for 

    is one-one function.

Now 

  = 

Therefore,  is an onto function.

Ans.  and 

Let , then  and 

Therefore,  is not one-one function.

Now, 

  and 

Therefore,  is not onto function.

Therefore, option (D) is correct.

Ans.  Let  R such that 

Therefore,  is one-one function.

Now, consider  R (co-domain of ) certainly  R (domain of )

Thus for all  R (co-domain of ) there exists  R (domain of ) such that

Therefore,  is onto function.

Therefore, option (A) is correct.