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


NCERT Solutions for Class 12 Maths Exercise 1.2 Chapter 1 Relations and Functions – FREE PDF Download

NCERT Solutions class 12 Maths Relations and Functions



1. Show that the function  defined by  is one-one and onto, where  is the set of all non-zero real numbers. Is the result true, if the domain  is replaced by N with co-domain being same as ?

 

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=NRf=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.


2. Check the injectivity and surjectivity of the following functions:

 

(i)  given by 

(ii)  given by 

(iii)  given by 

(iv)  given by 

(v)  given by 

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.


3. Prove that the Greatest integer Function , given by  is neither one-one nor onto, where  denotes the greatest integer less than or equal to 

 

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.


4. Show that the Modulus Function , given by  is neither one-one nor onto, where is  if  is positive or 0 and  is  if  is negative.

 

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.


5. Show that the Signum Function , given by  is neither one-one nor 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.


6. Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let  = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that  is one-one.

 

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.


7. In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.

 

(i)  defined by 

(ii)  defined by 

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=±y1−−−−√x=±y−1

 f(y1−−−−√)=1+y1=yyf(y−1)=1+y−1=−y≠y

  is not onto.

Therefore,  is not bijective.


8. Let A and B be sets. Show that  : A x B  B x A such that  is a bijective function.

 

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.


9. Let  be defined by  for all  N.

 

State whether the function  is bijective. Justify your answer.

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.


10. Let A = R – {3} and B = R – {1}. Consider the function  : A  B defined by  Is  one-one and onto? Justify your answer.

 

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.


11. Let  be defined as  Choose the correct answer:

 

(A)  is one-one onto

(B)  is many-one onto

(C)  is one-one but not onto

(D)  is neither one-one nor onto

Ans.  and 

Let , then  and 

 

 

Therefore,  is not one-one function.

Now, 

 

  and 

Therefore,  is not onto function.

Therefore, option (D) is correct.


12. Let  be defined as  Choose the correct answer:

 

(A)  is one-one onto

(B)  is many-one onto

(C)  is one-one but not onto

(D)  is neither one-one nor onto

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.