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

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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 = I_R.

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 = I_R and fog = I_R

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 = I_w and fog = I_w

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,

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

Are f and g equal? Justify your answer.

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

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

Are f and g equal? Justify your answer.

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}.