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

• Last Updated : 30 Apr, 2021

### 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 = 1R.

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 = 1R.

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-ONE

For 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 even

f(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 number

f(p) = p-1

y = p-1

p = y+1

Hence, when p is odd y is even.

Case 2: When p is even number

f(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) = x2â€“ 3x + 2, find f (f(x)).

Solution:

f(x) = x2â€“ 3x + 2

f(f(x)) = f(x2â€“ 3x + 2)

= (x2â€“ 3x + 2)2 – 3(x2â€“ 3x + 2) + 2

= x4 + 9x2 + 4 -6x3 – 12x + 4x2 – 3x2 + 9x – 6 + 2

f(f(x)) = x4 – 6x3 + 10x2 – 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-ONE

For 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 negative

f(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 <0

Hence, 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) = x3 is injective.

Solution:

As, it is mentioned here

f : R â†’ R defined by f(x) = x3, x âˆˆ R

To prove f is injective (or one-one).

ONE-ONE

The function f is defined as

f(x) = x3

f(y) = y3

f(x) = f(y)

x3 = y3

x = y

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

### (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

• Reflexive

As, 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)

• Symmetric

As, 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.

• Transitive

When 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!.

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