# Class 12 NCERT Solutions- Mathematics Part I – Chapter 1 Relations And Functions -Miscellaneous Exercise on Chapter 1 | Set 1

### Question 1. Let f : **R â†’ R** be defined as f(x) = 10x + 7. Find the function g : **R â†’ R** such that g o f = f o g = 1_{R}.

**Solution:**

As, it is mentioned here

f : R â†’ R be defined as f(x) = 10x + 7

To, prove the function one-one

Let’s take f(x) = f(y)

10x + 7 = 10y + 7

x = y

Hence f is one-one.To, prove the function onto

y âˆˆ R, y = 10x+7

So, it means for y âˆˆ R, there exists

Hence f is onto.As, f is one-one and onto. This f is invertible function.

Let’s say g : R â†’ R be defined as

Hence, g : R â†’ R such that g o f = f o g = 1

_{R}.g : R â†’ R is defined as

### Question 2. Let f : W â†’ W be defined as f(n) = n â€“ 1, if n is odd and f(n) = n + 1, if n is even. Show that f is invertible. Find the inverse of f. Here, W is the set of all whole numbers.

**Solution:**

The function f is defined as

As, we know f is invertible, if and only if f is one-one and onto.

ONE-ONEFor the pair of number, we will deal with three cases:

Case 1: When both numbers p and q are odd numbers.f(p) = p-1

f(q) = q-1

f(p) = f(q)

p-1 = q-1

p – q = 0

Case 2: When both numbers p and q are even numbers.f(p) = p+1

f(q) = q+1

f(p) = f(q)

p+1 = q+1

p – q = 0

Case 3:When p is odd and q is evenf(p) = p-1

f(q) = q+1

f(p) = f(q)

p-1 = q+1

p – q = 2

Subtracting an odd number and even always gives a odd number, not even. Hence, the case 3 result is impossible.

So, the function f is one-one, for case 1 and case 2 only.

ONTO

Case 1: When p is odd numberf(p) = p-1

y = p-1

p = y+1

Hence, when p is odd y is even.

Case 2: When p is even numberf(p) = p+1

y = p+1

p = y-1

Hence, when p is even y is odd.

So, it means for y âˆˆ W, there exists p = y+1 and y-1 for odd and even value of p respectively.

Hence f is onto.As, f is one-one and onto. This f is an invertible function.

Let’s say g : W â†’ W be defined as

f = g

Hence, The inverse of f is f itself

### Question 3. If f : **R â†’ R** is defined by f(x) = x^{2}â€“ 3x + 2, find f (f(x)).

**Solution:**

f(x) = x

^{2}â€“ 3x + 2f(f(x)) = f(x

^{2}â€“ 3x + 2)= (x

^{2}â€“ 3x + 2)2 – 3(x^{2}â€“ 3x + 2) + 2= x

^{4}+ 9x^{2}+ 4 -6x^{3}– 12x + 4x^{2}– 3x^{2}+ 9x – 6 + 2

f(f(x)) = x^{4}– 6x^{3}+ 10x^{2}– 3x

### Question 4. Show that the function f : R â†’ {x âˆˆ R : â€“ 1 < x < 1} defined by f(x) = , x âˆˆ R is one one and onto function.

**Solution:**

As, it is mentioned here

f : R â†’ {x âˆˆ R : â€“ 1 < x < 1} defined by , x âˆˆ R

As, we know f is invertible, if and only if f is one-one and onto.

ONE-ONEFor the pair of number, we will deal with three cases:

Case 1: When both numbers p and p are positive numbers.

The function f is defined as

Case 1: When both numbers p and q are positive numbers.f(p) = f(q)

p(1+q) = q(1+p)

p = q

Case 2: When number p and q are negative numbers.f(p) = f(q)

p(1-q) = q(1-p)

p = q

Case 3: When p is positive and q is negativef(p) = f(q)

p(1-q) = q(1+p)

p + q = 2pq

Here, RHS will be negative and LHS will be positive. Hence, the case 3 result is impossible.

So, the function f is one-one, for case 1 and case 2.

ONTO

Case 1: When p>0.

Case 2: When p <0Hence, p is defined for all the values of y, pâˆˆ R

Hence f is onto.As, f is one-one and onto. This f is an invertible function.

### Question 5. Show that the function f : **R â†’ R** given by f(x) = x^{3} is injective.

**Solution:**

As, it is mentioned here

f : R â†’ R defined by f(x) = x

^{3}, x âˆˆ RTo prove f is injective (or one-one).

ONE-ONEThe function f is defined as

f(x) = x

^{3}f(y) = y

^{3}f(x) = f(y)

x

^{3}= y^{3}x = y

The function f is one-one, so f is injective.

### Question 6. Give examples of two functions f : **N â†’ Z** and g :** Z â†’ Z** such that g o f is injective but g is not injective.

### (Hint : Consider f(x) = x and g (x) = | x |).

**Solution:**

Two functions, f : N â†’ Z and g : Z â†’ Z

Taking f(x) = x and g(x) = |x|

Let’s check, whether g is injective or not

g(5) = |5| = 5

g(-5) = |-5| = 5

As, we can see here that

Taking two integers, 5 and -5

g(5) = g(-5)

but, 5 â‰ -5

So, g is not an injective function.Now, g o f: N â†’ Z is defined as

g o f = g(f(x)) = g(x) = |x|

Now, as x,yâˆˆ N

g(x) = |x|

g(y) = |y|

g(x) = g(y)

|x| = |y|

x = y (both x and y are positive)

Hence, g o f is an injective.

### Question 7. Give examples of two functions f : **N â†’ N** and g : **N â†’ N** such that g o f is onto but f is not onto.

### (Hint : Consider f(x) = x + 1 and

**Solution:**

Two functions, f : N â†’ N and g : N â†’ N

Taking f(x) = x+1 and

As, f(x) = x+1

y = x+1

x = y-1

But, when y=1, x = 0. Which doesn’t satiny this relation f : N â†’ N.

Hence. f is not an onto function.Now, g o f: N â†’ N is defined as

g o f = g(f(x)) = g(x+1)

When x+1=1, we have

g(x+1) = 1 (1âˆˆ N)

And, when x+1>1, we have

g(x+1) = (x+1)-1 = x

y = x, which also satisfies x,yâˆˆ N

Hence, g o f is onto.

## Question 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 A âŠ‚ B. Is R an equivalence relation on P(X)? Justify your answer.

**Solution:**

Given, A and B are the subsets of P(x), AâŠ‚ B

To check the equivalence relation on P(X), we have to check

ReflexiveAs, we know that every set is the subset of itself.

Hence, AâŠ‚ A and BâŠ‚ B

ARA and BRB is reflexive for all A,Bâˆˆ P(X)

SymmetricAs, it is given that AâŠ‚ B. But it doesn’t make sure that BâŠ‚ A.

To be symmetric it has to be A = B

ARB is not symmetric.

TransitiveWhen AâŠ‚ B and BâŠ‚ C

Then of course, AâŠ‚ C

Hence, R is transitive.

So, as R is not symmetric.

R is not an equivalence relation on P(X).

### Question 9. Given a non-empty set X, consider the binary operation âˆ— : P(X) Ã— P(X) â†’ P(X) given by A âˆ— B = A âˆ© B âˆ€ 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 âˆ—.

**Solution:**

Given, P(X) Ã— P(X) â†’ P(X) is defined as A*B = Aâˆ©B âˆ€ A, B âˆˆ P(X)

This implies, AâŠ‚ X and B âŠ‚ X

So, Aâˆ©X = A and Bâˆ©X = B âˆ€ A, B âˆˆ P(X)

â‡’ A*X = A and B*X = B

Hence, X is the identity element for intersection of binary operator.

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

**Solution:**

Onto function from the set {1,2,3,…..,n} to itself is just same as the permutations of n.

1Ã—2Ã—3Ã—4Ã—…….Ã—n

Which is n!.

## Please

Loginto comment...