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

NCERT Solutions for Class 12 Maths Exercise 1.3 Chapter 1 Relations and Functions – FREE PDF Download

NCERT Solutions class 12 Maths Relations and Functions

1. Let  be defined as  Find the function  such that 

Ans. Given:  

Now    and 

 

 

2. Let  be defined as  if  is odd and  if  is even. Show that  is invertible. Find the inverse of  Here, W is the set of all whole numbers.

 

Ans. Given:  defined as  

Injectivity: Let  be any two odd real numbers, then 

 

 

Again,  let  be any two even whole numbers, then 

 

 

Is  is even and  is odd, then 

Also, if  odd and  is even, then 

Hence, 

 

  is an injective mapping.

Surjectivity: Let  be an arbitrary whole number.

If  is an odd number, then there exists an even whole number  such that

If  is an even number, then there exists an odd whole number  such that

Therefore, every  W has its pre-image in W.

So,  is a surjective. Thus  is invertible and  exists.

For : 

  and   

 

Hence, 

3. If  is defined by  find 

 

Ans. Given:  

 

 

= x4+9x2+4−6x3−12x+4x2−3x2+9x−6+2x4+9×2+4−6×3−12x+4×2−3×2+9x−6+2

 

4. Show that the function  defined by   R is one-one and onto function.

 

Ans. It is given that f: R→{x∈R:−1<x<1}R→{x∈R:−1<x<1} is defined as f(x) = x1+|x|,x∈Rx1+|x|,x∈R
For one- one
Suppose f(x) = f(y), where x, y ∈∈ R
⇒x1+|x|=y1+|y|⇒x1+|x|=y1+|y|
It can be observed that if x is positive and y is negative.
Then, we have
x1+x=y1−yx1+x=y1−y⇒2xy=x−y⇒2xy=x−y
Since, x is positive and y is negative,
x > y⇒x−y>0⇒x−y>0, but, 2xy is negative
Then, 2xy ≠≠ x – y.
Thus, the case of x being positive and y being negative can be ruled out.
Under a similar argument, x being negative and y being positive can also be ruled out.
∴∴ x and y have to be either positive or negative.
When x and y are both positive,we have
f(x) = f(y)⇒x1+x=y1+y⇒x1+x=y1+y⇒x+xy=y+xy⇒x=y⇒x+xy=y+xy⇒x=y
When x and y are both negative,we have
f(x) =f(y) ⇒x1−x=y1−y⇒x1−x=y1−y⇒x−xy=y−xy⇒x=y⇒x−xy=y−xy⇒x=y
∴∴ f is one – one
For Onto
Now, let y∈∈R such that -1< y <1.
If y is negative then, there exists x = y1+y∈Ry1+y∈R such that
f(x) = f(y1+y)=(y1+y)1+∣∣y1+y∣∣f(y1+y)=(y1+y)1+|y1+y|=y1+y1+(−y1+y)=y1+y−y=y=y1+y1+(−y1+y)=y1+y−y=y
If y is positive, then, there exists x = y1−y∈Ry1−y∈R such that
f(x) = f(y1−y)=(y1−y)1+∣∣y1−y∣∣f(y1−y)=(y1−y)1+|y1−y|=y1−y1+(y1−y)=y1−y+y=y=y1−y1+(y1−y)=y1−y+y=y
Therefore, f is onto. Hence, f is one -one and onto. 

5. Show that the function  given by  is injective.

 

Ans. Let  R be such that  

 

 

Therefore,  is one-one function, hence  is injective.

6. Give examples of two functions  and  such that  is injective but  is not injective.

 

(Hint: Consider  and  )

Ans. Given: two functions  and  

Let  and   

Therefore,  is injective but  is not injective.

7. Give examples of two functions  and  such that  is onto but  is not onto.

 

(Hint: Consider  and  )

Ans. Let  

 

These are two examples in which  is onto but  is not onto.

8. Given a non empty set X, consider P (X) which is the set of all subsets of X.

 

Define the relation R in P (X) as follows:

For subsets A, B in P (X), ARB if and only if AB. Is R an equivalence relation on P (X)? Justify your answer.

Ans. (i) A  A   R is reflexive. 

(ii) A  B  B  A  R is not commutative.

(iii) If A  B, B  C, then A  C   R is transitive.

Therefore, R is not equivalent relation.

9. Given a non-empty set X, consider the binary operation * : P (X) x P (X)  P (X) given by A * B = A  B  A, B in P (X), where P (X) is the power set of X. Show that X is the identity element for this operation and X is the only invertible element in P (X) with respect to the operation *.

 

Ans. Let S be a non-empty set and P(S) be its power set. Let any two subsets A and B of S. 

 A  B  S

 A  B  P(S)

Therefore,  is an binary operation on P(S).

Similarly, if A, B  P(S) and A – B  P(S), then the intersection of sets  and difference of sets are also binary operation on P(S) and A  S = A = S  A for every subset A of sets

 A  S = A = S  A for all A  P(S)

 S is the identity element for intersection  on P(S).

10. Find the number of all onto functions from the set {1, 2, 3, ……., } to itself.

 

Ans. The number of onto functions that can be defined from a finite set A containing  elements onto a finite set B containing  elements =  

11. Let S =  and T = {1, 2, 3}. Find  of the following functions F from S to T, if it exists.

 

(i) F = 

(ii) F = 

Ans. S =  and T = {1, 2, 3} 

(i) F = 

 

 

 

(ii) 

F is not one-one function, since element  and  have the same image 1.

Therefore, F is not one-one function.

12. Consider the binary operation * : R x R  R and o = R x R  R defined as  and  R. Show that * is commutative but not associative, o is associative but not commutative. Further, show that  R,  [If it is so, we say that the operation * distributes over the operation o]. Does o distribute over *? Justify your answer.

 

Ans. Part I:    also   operation * is commutative. 

Now, 

And 

Here,   operation * is not associative.

Part II:  R

And, 

    operation  is not commutative.

Now  and 

Here     operation  is associative.

Part III: L.H.S.   = 

R.H.S.  =  = L.H.S.   Proved.

Now, another distribution law: 

L.H.S. 

R.H.S. 

As L.H.S.  R.H.S.

Therefore, the operation  does not distribute over.

13. Given a non-empty set X, let * : P (X) x P (X)  P (X) be defined as A * B = (A – B)  (B – A),  A, B  P (X). Show that the empty set  is the identity for the operation * and all the elements A of P (X) are invertible with A-1 = A. (Hint:  and )

 

Ans. For every A  P(X), we have 

 = 

And  = 

  is the identity element for the operation * on P(X).

Also A * A = (A – A)  (A – A) = 

 Every element A of P(X) is invertible with  = A.

14. Define binary operation * on the set {0, 1, 2, 3, 4, 5} as a∗b={a+b,ifa+b<6a+b−6ifa+b≥6a∗b={a+b,ifa+b<6a+b−6ifa+b≥6

 

Show that zero is the identity for this operation and each element  of the set is invertible with  being the inverse of 

Ans. A binary operation (or composition) * on a (non-empty) set is a function * : A x A A. We denote  by  for every ordered pair  A x A. 

 A binary operation on a no-empty set A is a rule that associates with every ordered pair of elements  (distinct or equal) of A some unique element  of A.

*012345
0012345
1123450
2234501
3345012
4450123
5501234

For all  A, we have  (mod 6) = 0

And  and 

 0 is the identity element for the operation.

Also on 0 = 0 – 0 = 0 *

2 * 1 = 3 = 1 * 2

   

15. Let A = {–1, 0, 1, 2}, B = {–4, –2, 0, 2} and  be the functions defined by   A and  A. Are  and  equal? Justify your answer.

 

(Hint: One may note that two functions  and  such that    A, are called equal functions).

Ans. When  then  and  

At   and 

At   and 

At   and 

Thus for each  A, 

Therefore,  and  are equal function.

16. Let A = {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is:

 

(A) 1

(B) 2

(C) 3

(D) 4

Ans. It is clear that 1 is reflexive and symmetric but not transitive. 

Therefore, option (A) is correct.

17. Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is:

 

(A) 1
(B) 2
(C) 3
(D) 4

Ans. 2
Therefore, option (B) is correct. 

18. Let  be the Signum Function defined as  and  be the Greatest Function given by  where  is greatest integer less than or equal to  Then, does  and  coincide in (0, 1)?

 

Ans. It is clear that  and  

Consider  which lie on (0, # 1)
Now, 
And 
  in (0, 1]
No, fog and gof don’t  coincide in (0, 1].

19. Number of binary operation on the set  are:

 

(A) 10
(b) 16
(C) 20
(D) 8

Ans. A = 
A x A = 
  = 4
Number of subsets =  = 16
Hence number of binary operation is 16.
Therefore, option (B) is correct.