# NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions - Miscellaneous Exercise

### Access Exercises of Class 12 Maths Chapter 1 -  Relations and Functions

Exercise 1.1 Solutions : 16 Questions (14 Short Answers, 2 MCQ)

Exercise 1.2 Solutions : 12 Questions (10 Short Answers, 2 MCQ)

Exercise 1.3 Solutions : 14 Questions (12 Short Answers, 2 MCQ)

Exercise 1.4 Solutions : 13 Questions (12 Short Answers, 1 MCQ)

Miscellaneous Exercise Solutions: 19 Questions (7 Long answers, 9 Short Answer Type, 3 MCQs)

Miscellaneous Exercise

1. Let f : R → R be defined as f(x) = 10x + 7. Find the function g : R → R such that gof = fog = IR.

Sol. It is given that f : R → R is defined as f(x) = 10x + 7.

For one-one,

Let f(x) = f(y), where x, y ∈ R

⇒ 10x + 7 = 10y + 7 ⇒ x = y

Therefore, f is a one-one function.

For onto,

For y ∈ R, let y = 10x + 7

$$\Rarr\space x =\frac{y-7}{10}\epsilon \text{R}$$

Therefore, for any y ∈ R, there exists

$$x =\frac{y-7}{10}\epsilon\text{R}\\\text{such that}\\\text{f(x)} = f\bigg(\frac{y-7}{10}\bigg)\\=10\bigg(\frac{y-7}{10}\bigg) + 7$$

= y- 7+7 =y

Therefore, f is onto. Therefore, f is one-one and onto.

Thus, f is an invertible function.

Let us define g : R → R as

$$g(y) =\frac{y-7}{10}$$

Now, we have gof(x) = g(f(x)) = g(10x + 7)

$$=\frac{(10x +7)-7}{10} =\frac{10x}{10}=x$$

$$\text{and fog(y) = f(g(y)) =}\\\text{f}\bigg(\frac{y-7}{10}\bigg)\\= 10\bigg(\frac{y-7}{10}\bigg) + 7$$

y -7+7 =y

∴ gof = IR and fog = IR

Hence, the required function g : R → R is defined

$$\text{as g(y) =}\space\frac{y-7}{10}.$$

2. Let f : W → W be defind as f(n) = n – 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.

Sol. It is given that

$$\text{f : W → W is defined as}\\\text{f(n) =}\begin{cases}n-1,\space\text{if n is odd}\\n+1,\space\text{if n is even}\end{cases}$$

For one-one,

Let f(n) = f(m)

It can be observed that if n is odd and m is even, then we will have

n – 1 = m + 1 ⇒ n – m = 2

However, this is impossible. Similarly, the possibility of n being even and m being odd can also ignored under a similar argument.

Therefore, both n and m must be either odd or even.

Now, if both n and m are odd, then we have

f(n) = f(m) ⇒ n – 1 = m – 1 ⇒ n = m

Again, if both n and m are even, then we have

f(n) = f(m) ⇒ n + 1 = m + 1 ⇒ n = m

Therefore, f is one-one.

For onto,

It is clear that any odd number 2r + 1 in co-domain W is the image of 2r in domain W and any even number 2r in co-domain W is the image of 2r + 1 in domain W.

Therefore, f is onto. Hence, f is an invertible function.

Let us define g :

$$\text{W}\xrightarrow{}\text{W}\space\text{as g(m)} =\\\begin{cases}m+1,\space\text{if m is even}\\m-1,\space\text{if m is odd}\end{cases}$$

Now, when n is odd

gof(n) = g(f(n)) = g(n – 1) = n – 1 + 1 = n

( when n is odd, n – 1 is even)

and when n is even

gof(n) = g(f(n)) = g(n + 1) = n + 1 – 1 = n

( when n is odd, n + 1 is odd)

Similarly, when m is odd

fog(m) = f(g(m)) = f(m – 1) = m – 1 + 1 = m

when m is even

fog(m) = f(g(m)) = f(m + 1) = m + 1 – 1 = m

∴ gof = Iw and fog = Iw

Thus, f is invertible and the inverse of f is given by f–1 = g, which is the same as f. Hence, the inverse of f is itself.

3. If f : R → R is defined by f(x) = x2 – 3x + 2, find f(f(x)).

Sol. It is given that f : R → R is defined as f(x) = x2 – 3x + 2.

f(f(x)) = f(x2 – 3x + 2) = (x2 – 3x + 2)2 – 3(x2 – 3x + 2) + 2

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

= x4 – 6x3 + 10x2 – 3x

4. Show that the function f : R → {x ∈ R : – 1 < x < 1} defined by

$$\textbf{f(x) =}\frac{\textbf{x}}{\textbf{1 + |x|}}\textbf{,}\space\textbf{x}\epsilon \textbf{R}\space\textbf{is one-one}$$

and onto function.

Sol. It is given that f : R → {x ∈ R : – 1 < x < 1} is defined

$$\text{as f(x) =}\frac{x}{x + |x|},\space x\epsilon\text{R}$$

Suppose, f(x) = f(y),

where x, y ∈ R

$$\Rarr\space\frac{x}{1 + |x|} =\frac{x}{1 + |y|}$$

It can be observed that if x is positive and y is negative, then we have

$$\frac{x}{1 +x} = \frac{y}{1-y}\\\Rarr\space\text{2xy = x-y}$$

Since, x is positive and y is negative, then

x > y ⇒ 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 ruld out. Therefore, x and y have to be either positive or negative.

When x and y are both positive, we have

$$\text{f(x) = f(y)}\\\Rarr\space \frac{x}{1 +x} =\frac{y}{1 +y}$$

⇒ x – xy = y – yx ⇒ x = y

Therefore, f is one-one. Now, let y ∈ R such that – 1 < y < 1.

If y is negative, then there exists

$$x =\frac{y}{1+y}\epsilon \text{R}\\\text{such that}\\\text{f(x) = }f\bigg(\frac{y}{1 +y}\bigg)\\=\frac{\bigg(\frac{y}{1 +y}\bigg)}{1 + \frac{y}{1+y}} =\frac{\frac{y}{1+y}}{1 + \bigg(\frac{\normalsize-y}{1 + y}\bigg)}\\=\frac{y}{1 +y-y} = y$$

If y is positive, then there exists

$$x =\frac{y}{1-y}\epsilon\space\text{R}\\\text{such that}\\\text{f(x) = f}\bigg(\frac{y}{1-y}\bigg)\\=\frac{\bigg(\frac{y}{1-y}\bigg)}{1 + \begin{vmatrix}\bigg(\frac{y}{1-y}\bigg)\end{vmatrix}} =\frac{\frac{y}{1-y}}{1 + \frac{y}{1-y}}\\=\frac{y}{1-y+y} = y$$

Therefore, f is onto. Hence, f is one-one and onto.

5. Show that the function f : R → R given by
f(x) = x3 is injective.

Sol. Here, f : R → R is given as f(x) = x3.

Suppose, f(x) = f(y), where x, y ∈ R ⇒ x3 = y3 ...(i)

Now, we need to show that x = y

Suppose, x ≠ y, their cubes will also not be equal.

x3 ≠ y3

However, this will be a contradiction to Eq. (i).

Therefore, x = y. Hence, f is injective.

6. Give example of two functions f : N → Z and g : Z → Z such that gof is injective but g is not injective.

Sol. Define f : N → Z as f(x) = x and g : Z → Z as g(x) = |x|.

We first show that g is not injective. It can be observed that

g(– 1) = |– 1| = 1, g(1) = |1| = 1

Therefore, g(– 1) = g(1) but – 1 ≠ 1. Therefore, g is not injective.

Now, gof : N → Z is defined as gof(x) = g(f(x)) = g(x) = |x|

let x, y ∈ N such that gof(x) = gof(y) ⇒ |x| = |y|

Since, x and y ∈ N, both are positive.

∴ |x| = |y| ⇒ x = y

Hence, gof is injective.

7. Give an examples of two functions f : N → N and g : N → N such that gof is onto but f is not onto.

Sol. Define f : N → N by f(x) = x + 1

$$\text{and\space}\text{g : N → N by}\\\text{g(x) =}\begin{cases}x-1,\space \text{if x}\gt 1\\1,\space \text{if x = 1}\end{cases}$$

We first show that f is not onto.

For this, consider element 1 in co-domain N.

Let f(x) = 1 ⇒ x + 1 = 1 ⇒ x = 0, which is not a
natral number.

Therefore, f is not onto. Now, gof : N → N is
defined by,

gof(x) = g(f(x)) = g(x + 1) = (x + 1) – 1

[ x ∈ N ⇒ (x + 1) > 1]

= x

Then, it is clear that for y ∈ N, there exists x = y ∈ N such that gof(x) = y.

Hence, gof is 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 A ⊂ B. Is R an equivalence relation on P(X) ? Justify your answer.

Sol. Since, every set is a subset of itself, ARA for all A ∈ P(X).

Threfore, R is not reflexive.

Let ARB ⇒ A ⊂ B

This cannot be implied to B ⊂ A.

For instance, if A = {1, 2} and B = {1, 2, 3}, then it cannot be implied that B is related to A.

Therefore, R is not symmetric.

Further, if ARB and BRC, then A ⊂ B and B ⊂ C.

⇒ A ⊂ C ⇒ ARC. Therefore, R is transitive. Hence, R is not an equivalence relation since, it is not symmetric.

9. Given non-empty set X, consider the binary operation * : P(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 *.

Sol. Given that

* : P(X) × P(X) → P(X) is defind as A * B =

A ∩ B ∀A, B ∈ P(X)

We know that

A * X = A ∩ X = A =X ∩ A = X* A ∀ A ∈ P(X)

Thus, X is the identity element for the given binary operation *.

Now, an element A ∈ P(X) is invertible if there exists B ∈ P(X) such that

A * B = X = B * A

(As X is the identity element)

i.e., A ∩ B = X = B ∩ A

This case is possible only when A = X = B. Thus, X is the only invertible element in P(X) with respect to the given operation *

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

Sol. Onto function from the set {1, 2, 3, ...., n) to itself is simply a permutation on n symbol 1, 2, ....., n.

Thus, the total number of onto maps from {1, 2,....., n} to itself is the same as the total number of permutations on symbols 1, 2, ....., n, which is n!.

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

(i) F = {(a, 3), (b, 2), (c, 1)}

(ii) F = {(a, 2), (b, 1), (c, 1)}

Sol. Given, S = {a, b, c) and T = {1, 2, 3}

(i) F : S → T is defined as F = {(a, 3), (b, 2), (c, 1)}

⇒ F(a) = 3, F(b) = 2, F(c) = 1

Therefore, F–1 : T → S is given by F–1 = {(3,a), (2, b), (1, c)}

(ii) F : S → T is defined as F = {(a, 2), (b, 1), (c, 1)}

Since, F(b) = F(c) = 1, F is not one-one

Hence, F is not invertible i.e., F–1 does not exist.

12. Consider the binary operations * : R × R → R and o : R × R → R defined as a * b = |a – b| and aob = a ∀ a, b ∈ R. Show that * is commutative but not associative o is associative but not commulative. Further, show that ∀ a, b, c ∈ R, a * (boc) = (a * b)o(a * c). [If it is so, we say that the operation * distributes over the operation]. Does o distribute over * ? Justify your answer.

Sol. Given that * : R × R → R and o : R × R → R defined as

a * b = |a – b| and aob = a ∀ a, b ∈ R.

For a, b ∈ R, we have

a * b = |a – b|, b * a = |b – a|

= |– (a – b)| = |a – b|

Therefore, a * b = b * a

Therefore, the operation * is commutative.

It can be observed that,

(1 * 2) * 3 = (|1 – 2|) * 3 = 1 * 3 = |1 – 3| = 2

1 * (2 * 3) = 1 * (|2 – 3|) = 1 * 1 = |1 – 1| = 0

∴ (1 * 2) * 3 ≠ 1 * (2 * 3) (where, 1, 2, 3 ∈ R)

Therefore, the operation * is not associative.

Now, consider the operation o. It can be observed that 1 o 2 = 1 and 2 o 1 = 2.

Therefore, the 1 o 2 ≠ 2 o 1 where 1, 2 ∈ R.

Therefore, the operation o is not cummutative.

Let a, b, c ∈ R. Then, we have

(a o b) oc = a o c = a, a o (b o c) = a o b = a

⇒ (a o b)o c = a o(b o c)

Therefore, the operation o is associative.

Now, let a, b, c ∈ R then, we have

a * (b o c) = a * b = |a – b|

(a * b) o (a * c) = (|a – b|) o (|a – c|) = |a – b|

Hence, a * (b o c) = (a * b) o (a * c)

Now, 1 o (2 * 3) = 1 o (|2 – 3|) = 1 o 1 = 1

(1 o 2) * (1 o 3) = 1 * 1 = |1 – 1| = 0

Therefore, 1 o (2 * 3) ≠ (1 o 2) * (1 o 3) where 1, 2, 3 ∈ R.

Therefore, the operation o does not distribute over *.

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

Sol. Given that * : P(X) × P(X) → P(X) is defined as

A * B = (A – B) ∩ (B – A) for A, B ∈ P(X)

Let A ∈ P(X). Then, we have

A * φ = (A – φ) ∪ (φ – A) = A ∪ φ = A

φ * A = (φ – A) ∪ (A – φ) = φ ∪ A = A

Therefore, A * φ = A = φ* for all A ∈ P(X)

Thus, φ is the identity element for the given operation *.

Now, an element A ∈ P(X) will be invertible, if there exists B ∈ P(X) such that A * B = φ = B * A.

As φ is the identity element.

Now, we observed that A * A = (A – A) ∪ (A – A) = f ∪ f = f ∀ A ∈ P(X).

Hence, all the elements A of P(X) are invertible with A–1 = A.

14. Define a binary operation * on the set {0, 1, 2, 3,4, 5} as

$$\textbf{as a*b = }\begin{cases}\textbf{a+b,\space}\textbf{if a+b}\lt \textbf{6}\\\textbf{a+b-6,}\space\textbf{if a+b}\geq\textbf{6}\end{cases}$$

Show that zero is the identity for this operation and each element a ≠ 0 of the set is invertible with (6 – a) being the inverse of a.

Sol. Let X = {0, 1, 2, 3, 4, 5}

The operation * on X is defined as

$$a*b =\begin{cases}a+b,\space\text{if a +b}\lt 6\\a+b-6,\space\text{if a+b}\geq6\end{cases}$$

An element e ∈ X is the identity element for the
operation *, if

a * e = a = e * a ∀ a ∈ X

For a ∈ X, we observed that

a * 0 = a + 0 = a [ a ∈ X ⇒ a + 0 < 6]

0 * a = 0 + a = a [ a ∈ X ⇒ 0 + a < 6]

∴ a * 0 = a = 0 * a ∀ a ∈ X

Thus, 0 is the identity element for the given operation *.

An element a ∈ X is invertible, if there exists b ∈ X such that

a * b = 0 = b * a

$$\text{i.e.,\space}\\\begin{cases}a+b =0 =b+a\space\text{if a+b}\lt 6\\a+b-6 =0 =b+a-6,\space\text{if a+b}\geq6\end{cases}$$

i.e., a = – b or b = 6 – a

But X = {0, 1, 2, 3, 4, 5} and a, b ∈ X

Then, a ≠ – b

Therefore, b = 6 – a is the inverse of a, a ∈ X.

Hence, the inverse of an element a ∈ X, a ≠ 0 is (6 – a) i.e., a–1 = 6 – a.

15. Let A = {– 1, 0, 1, 2}, B = {– 4, – 2, 0, 2} and f, g : A → B be the functions defined by f(x) = x2 – x,

$$\textbf{x}\epsilon \textbf{A}\space\textbf{g(x)} \textbf{= 2}\begin{vmatrix}\textbf{x -}\frac{\textbf{1}}{\textbf{2}}\end{vmatrix}\textbf{-1}\textbf{,}\space\textbf{x}\epsilon \textbf{A.}$$

Sol. Given that A = {– 1, 0, 1, 2}, B = {– 4, – 2, 0, 2}.

Also, it is given that f, g : A → B are defined by

f(x) = x2 – x, x ∈ A and

$$\textbf{g(x)} = 2\begin{vmatrix}\textbf{x -}\frac{\textbf{1}}{\textbf{2}}\end{vmatrix}\textbf{-1,}\space\textbf{x}\epsilon \textbf{A.}$$

Sol. Given that A = {– 1, 0, 1, 2}, B = {– 4, – 2, 0, 2}.

Also, it is given that f, g : A → B are defined by

f(x) = x2 – x, x ∈ A and

$$g(x) = 2\begin{vmatrix}x + \frac{1}{2}\end{vmatrix}-1,\space x\epsilon A.$$

It is observed that f(– 1) = (– 1)2 – (– 1) = 1 + 1 = 2

$$\text{and\space g(\normalsize-1)} = 2\begin{vmatrix}(\normalsize-1) -\frac{1}{2}\end{vmatrix}\\ = 2\bigg(\frac{3}{2}\bigg) -1 = 3-1=2$$

⇒ f(– 1) = g(– 1)

Next f(0) = (0)2 – 0 = 0

$$\text{and}\space g(0) = 2\begin{vmatrix}0-\frac{1}{2}\end{vmatrix}-1\\2\bigg(\frac{1}{2}\bigg)-1$$

= 1 – 1 = 0

⇒ f(0) = g(0)

Next, f(1) = (1)2 – 1= 1 – 1 = 0

$$\text{and \space g(1) = 2}\begin{vmatrix}1 -\frac{1}{2}\end{vmatrix}-1\\=2\bigg(\frac{1}{2}\bigg)-1$$

= 1 – 1 = 0

⇒ f(1) = g(1)

Next, f(2) = (2)2 – 2 = 4 – 2 = 2

$$\text{and\space} g(2)= 2\begin{vmatrix}1 -\frac{1}{2}\end{vmatrix} \\= 2\bigg(\frac{3}{2}\bigg)-1$$

= 3 – 1 = 2

⇒ f(2) = g(2)

∴ f(a) = g(a) ∀ a ∈ A.

Hence, the functions f and g are equal.

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

Sol. (A) 1

This is because relation R is reflexive as (1, 1), (2, 2), (3, 3) ∈ R.

Relations R is symmetric since (1, 2), (2, 1) ∈ R and (1, 3), (3, 1) ∈ R.

But relation R is not transitive as (3, 1), (1, 2) ∈ R but (3, 2) ∉ R.

Now, if we add any one of the two pairs (3, 2) and (2, 3) (or both) to relation R, then relation R will become transitive.

Hence, the total number of desired relations is one.

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

(A) 1

(B) 2

(C) 3

(D) 4

Sol. (B) 2

It is given that A = {1, 2, 3}.

An equivalence relation is reflexive, symmetric and transitive.

The smallest equivalence relation containing (1, 2) is given by

R1 = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}

Now, we are left with only four pairs i.e., (2, 3), (3, 2), (1, 3) and (3, 1).

If we add any one pair [say (2, 3)] to R1, then for symmetry we must add (3, 2).

Also, for transitivity we are required to add (1, 3) and (3, 1).

Hence, the only equivalence relation (bigger than R1) is the universal relation.

This shows that the total number of equivalence relations containing (1, 2) is two.

18. Let f : R → R be the Signum Function defined as

$$\textbf{f(x) =}\begin{cases}\textbf{1,}\space\textbf{x}\geq\textbf{0}\\\textbf{0,}\space\textbf{x = 0}\\\textbf{\normalsize-1},\textbf{x}\lt \textbf{0}\end{cases}\space\textbf{and g : R}\xrightarrow{}\textbf{R}\space\\\textbf{be the Greatest}$$

Integer Function given by g(x) = [x], where [x] is greatest integer less than or equal to x. Then, does fog and gof coincide in (0, 1]?

Sol. It is given that f : R → R is defined as

$$\text{f(x) = }\begin{cases}1,\space x\geq0\\0,\space x =0\\-1,\space x\lt 0\end{cases}$$

Also, g : R → R is defined as g(x) = [x], where [x] is the greatest integer less than or equal to x.

Now, let x ∈ (0, 1]. Then, we have

[x] = 1 if x = 1 and [x] = 0 if 0 < x < 1

∴ fog(x) = f(g(x)) = f([x])

$$=\begin{cases}f(1),\space \text{if x = 1}\\f(0),\space\text{if x}\epsilon(0,1)\end{cases}\\ =\begin{cases}1,\space\text{if x = 1}\\0,\space x\epsilon(0,1)\end{cases}$$

gof(x) = g(f(x)) = g(1) = [1] = 1

[ ∵ x > 0]

Thus, when x ∈ (0, 1), we have fog(x) = 0 and
gof(x) = 1.

But fog(1) ≠ gof(1)

Hence, fog and gof do not coincide in (0, 1].

19. Number of binary operations on the set {a, b} are :

(A) 10

(B) 16

(C) 20

(D) 8

Sol. (B) 16

A binary operation * on {a, b} is a function from {a, b} × {a, b} → {a, b}.

i.e., * is a function from {(a, a), (a, b), (b, a), (b, b)} → {a, b}.

Hence, the total number of binary operations on the set {a, b} is 24 i.e., 16.
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