# NCERT Solutions for Class 12 Maths Relations and Functions

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*NCERT Solutions*## NCERT Solutions for Class 12 Maths Relations and Functions Exercise Miscellaneous

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** Miscellaneous: Ques No 1.**

Let f : R → R be defined as f(x) = 10x + 7. Find the function g : R → R such that g o f = f o g = I_{R}.

**NCERT Solutions:**

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

Let a, b ∈ R such that f(a) = f(b). Then,

⇒ 10a + 7 = 10b + 7

⇒ 10a = 10b

⇒ a = b

Thus, the given function is one-one function.

Now, let y = 10x + 7 ⇒ x = (y – 7)/10. Put x = (y – 7)/10 in f(x) = 10x + 7.

Then, f((y – 7)/10) = 10((y – 7)/10) + 7 = y – 7 + 7 = y. Since f(x) = y, then the given function is on to function.

Therefore, the given function is one-one on to function. Hence, the given function f is an invertible function.

Let us define ?: R → R as g(?) = (y – 7)/10.

Now, we have

??(?) = ?(?(?)) = ?(10? + 7) = (10x + 7 – 7)/10 = x

and

??(?) = ?(?(?)) = f((y – 7)/10) = 10((y – 7)/10) + 7 = y – 7 + 7 = y

Thus, we get ??? = I_{?} and ??? = I_{R}

Hence, the required function ?: R → R is defined as g(?) = (y – 7)/10.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** Miscellaneous: Ques No 2.**

Let f : W → W be defined 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.

**NCERT Solutions:**

Given that f: W → W is defined as f(x) = n – 1 if n is odd and f(x) = n + 1 if n is even.

For one – one,

Let f(n) = f(m). If n is odd and m is even, then we have

n − 1 = m + 1.

⇒ n − m = 2

It is an impossible case.

Thus, 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

We see that in both cases the given function is one-one function.

For onto, it is clear that any odd number 2r + 1 in co-domain N is the image of 2r in domain N and any even number 2r in co-domain N is the image of 2r + 1 in domain N.

Thus, the given function f is onto function. Hence, the given function f is an invertible function.

Let us define ?: W → W as g(?) = m – 1 if m is odd and g(m) = m + 1 if m is even.

Now, when n is odd

??(?) = ?(?(?)) = ?(? − 1) = ? − 1 + 1 = ?

When n is even

??(?) = ?(?(?)) = ?(? + 1) = ? + 1 − 1 = ?

When m is odd

??(?) = ?(?(?)) = ?(? − 1) = ? − 1 + 1 = ?

When m is even

??(?) = ?(?(?)) = ?(? + 1) = ? + 1 − 1 = ?

Thus, ??? = I_{W} and ??? = I_{W}

Thus, the given function f is invertible and the inverse of f is given by f^{-1} = ?, which is the same as f. Hence, the inverse of f is f itself.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** Miscellaneous: Ques No 3.**

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

**NCERT Solutions:**

Given that f : R → R is defined by f(x) = x^{2} – 3x + 2. Thus, we get

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** Miscellaneous: Ques No 4.**

Show that the function f : R → {x ∈ R : – 1 < x < 1} defined by f(x) = x/(1 + |x|), x ∈ R is one-one and onto function.

**NCERT Solutions:**

Given that f : R → {x ∈ R : – 1 < x < 1} defined by f(x) = x/(1 + |x|), x ∈ R.

For one-one, f(x) = f(y), where x, y ∈ R.

If x is positive and y is negative. Then, we have

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. Thus, x and y have to be either positive or negative.

When x and y are both positive, we have

When x and y are both negative, we have

Thus, the given function is one-one function.

For onto

Now, let y ∈ R such that −1 < y < 1.

If y is negative, then, there exists x = y/(1 + y) ∈ R such that

If y is positive, then, there exists x = y/(1 – y) ∈ R such that

Thus, the given function is onto function.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** Miscellaneous: Ques No 5.**

Show that the function f : R → R given by f(x) = x^{3} is injective.

**NCERT Solutions:**

Given that f: R → R is given as f(x) = x^{3}.

For one – one

f(x) = f(y), where x, y ∈ R.

⇒ x^{3} = y^{3}

⇒ x = y.

Thus, the given function is one-one function or injective.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** Miscellaneous: Ques No 6.**

Give examples of two functions f : N → Z and g : Z → Z such that g o f is injective but g is not injective.

**NCERT Solutions:**

Let the function f: N → Z as f(x) = x and the function ?: Z → Z as ? (x) = |?|.

Now, we observed that (−1) =|−1| = 1 and (1) = |1| = 1. Thus, f(−1) = ?(1), but −1 ≠ 1.

Thus, the assumed function ? is not injective.

Now, ?of: N → Z is defined as ??(?) = ?(?(?)) = ?(?) = |?|.

Let x, y ∈ N such that ?of(x) = ?of(y).

⇒ |?| = |?|

Since x and y ∈ N, both are positive.

∴ |?| = |?| ⇒ ? = ?

Thus, the function gof(x) is injective.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** Miscellaneous: Ques No 7.**

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

**NCERT Solutions:**

Let the function f : N → N as f(x) = x + 1 and the function g : N → N as

Since 1 in co-domain N but this is not an image of any of domain N, the function f(x) = x + 1 is not onto.

Now, the function ?of: N → N is defined by

??(?) = ?(?(?)) = ?(? + 1) = ? + 1 − 1 = ? , where ? ∈ ? ⇒ ? + 1 > 1

Then, for y ∈ N, there exists x = y ∈ N such that ?of(x) = y.

Therefore, the function ?of(x) is an onto function.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** Miscellaneous: Ques No 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.

**NCERT Solutions:**

Since every set is a subset of itself, ARA for all A ∈ P(X). Thus, R is 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. Thus, R is not symmetric.

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

⇒ A ⊂ C

⇒ ARC

Thus, R is transitive.

Hence, R is not an equivalence relation as it is not symmetric.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** Miscellaneous: Ques No 9.**

Given a 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 ∗.

**NCERT Solutions:**

Given that the binary operation *: P(X) × P(X) → P(X) given by A * B = A ∩ B ∀ A, B in P(X)

We know that A∩X = A = X ∩ A for all A ∈ P(X).

Thus, A * X = A = X * A for all A ∈ P(X)

Hence, 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 or 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*.

Thus, X is the identity element for this operation and X is the only invertible element in P(X) with respect to the operation ∗.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** Miscellaneous: Ques No 10.**

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

**NCERT Solutions:**

The number of onto functions from the set {1, 2, 3, … , n} to itself is the permutation on n symbols 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 n symbols 1, 2, …, n, which is n.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** Miscellaneous: Ques No 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)}**

**NCERT Solutions:**

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

Thus, we have 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 = {(a, 2), (b, 1), (c, 1)}**

**NCERT Solutions:**

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

Thus, we have F (a) = 2, F (b) = 1, F(c) = 1

Since F (b) = F(c) = 1, the function F is not one-one. Hence, the function F is not an invertible function.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** Miscellaneous: Ques No 12.**

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

**NCERT Solutions:**

Given that *: R × R → ? ∗ ? = |? − ?|, ? ∗ ? = |? − ?|

For a, b ∈ R, we have ? ∗ ? = |? − ?| and ? ∗ ? = |? − ?| = |−(? − ?)| = |? − ?|

Thus, a * b = b * a. Hence, the operation * is commutative.

It can be observed that

(1 ∗ 2) ∗ 3 = (|1 − 2|) ∗ 3 = 1 ∗ 3 = |1 − 3| = 2

and

1 ∗ (2 ∗ 3) = 1 ∗ (|2 − 3|) = 1 ∗ 1 = |1 − 1| = 0

Thus, (1 ∗ 2) ∗ 3 ≠ 1 ∗ (2 ∗ 3) where 1, 2, 3 ∈ R. Hence, the operation * is not associative.

Now, given that o: R × R → R is defined as aob = a, ∀ a, b ∈ R.

Since 1 o 2 = 1 and 2 o 1 = 2, then 1 o 2 ≠ 2 o 1 where 1, 2 ∈ R. Hence, the operation o is not commutative.

Let a, b, c ∈ R. Then, we have

(a o b) o c = a o c = a

and

a o (b o c) = a o b = a

Since (a o b) o c = a o (b o c), where a, b, c ∈ R. Hence, the operation o is associative.

Now, let a, b, c ∈ R, then we have

a * (b o c) = a * b =|? − ?|

(a * b) o (a * c) = (|? − ?|) ? (|? − ?|) = |? − ?|

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

Now,

1 ? (2 ∗ 3) = 1 ? (|2 − 3|) = 1 ? 1 = 1

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

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

Hence, the operation o does not distribute over *.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** Miscellaneous: Ques No 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.

**NCERT Solutions:**

Given that *: P(X) × P(X) → P(X) is defined as A * B = (A − B) ∪ (B − A) ∀ 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

∴ A * Φ = A = Φ * A 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.

Now, we observed that

A * A = ( A – A) ∪ (A – A) = Φ ∪ Φ = Φ for all A ∈ P(X).

Hence, all the elements A of P(X) are invertible with A^{−1} = A.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** Miscellaneous: Ques No 14.**

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

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

**NCERT Solutions:**

Given that the binary operation ∗ on the set {0, 1, 2, 3, 4, 5} as

An element e ∈ X is the identity element for the operation *, if ? ∗ ? = ? = ? ∗ ? for all ? ∈ X.

For ? ∈ X, we have ? ∗ 0 = ? + 0 = ? and 0 ∗ ? = 0 + ? = ? ∴ ? ∗ 0 = ? = 0 ∗ ? for all ? ∈ 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.

Thus, a = −b or b = 6 – a

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

Thus, b = 6 − a is the inverse of a for all a ∈ X. Hence, the inverse of an element a ∈ X, a ≠ 0 is 6 − a i.e., a−1 = 6 – a.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** Miscellaneous: Ques No 15.**

Let A = {– 1, 0, 1, 2}, B = {– 4, – 2, 0, 2} and f, g : A → B be functions defined by f(x) = x^{2} – x, x ∈ A and g(x) = 2|x – 1/2|, x ∈ A. Are f and g equal? Justify your answer.

**NCERT Solutions:**

Given that A = {−1, 0, 1, 2}, B = {−4, −2, 0, 2} and f, ?: A → B are defined by

f(x) = x^{2} − x, x ∈ A and ?(?) = 2|x – 1/2| – 1**.**

It is observed that f(−1) = (−1)^{2} − (−1) = 1 + 1 = 2 and g(-1) = 2|-1 – 1/2| – 1 = 2**.**

Thus, f (−1) = ?(−1).

It is observed that f(0) = (0)^{2} − (0) = 0 and g(0) = 2|0 – 1/2| – 1 = 0.

Thus, we get f(0) = g(0).

It is observed that f(1) = (1)^{2} − (1) = 0 and g(1) = 2|1 – 1/2| – 1 = 0.

Thus, we get f(1) = g(1).

It is observed that f(2) = (2)^{2} − (2) = 2 and g(2) = 2|2 – 1/2| – 1 = 2.

Thus, we get f(2) = g(2).

It is observed that f(a) = g(a) for all a ∈ A. Hence, the functions f and ? are equal.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** Miscellaneous: Ques No 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

**NCERT Solutions:**

Given that A = {1, 2, 3}. Let the relation is R.

For reflexive, (1, 1), (2, 2), (3, 3) ∈ R.

For symmetric, (1, 2), (2, 1) ∈ R and (1, 3), (3, 1) ∈ R

For not transitive, (3, 1), (1, 2) ∈ R, but (3, 2) ∉ R.

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

Thus, the relation is R = {(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 1), (3, 1)}, which is reflexive and symmetric but not transitive.

Hence, the total number of desired relations is one. The correct answer is A.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** Miscellaneous: Ques No 17.**

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

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

**NCERT Solutions:**

Given that A = {1, 2, 3}. The smallest equivalence relation containing (1, 2) is given by, R = {(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 odd any one pair [say (2, 3)] to R, 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 R) is the universal relation.

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

The **correct answer is B**.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** Miscellaneous: Ques No 18.**

Let f : R → R be the Signum Function defined as and g : R → R 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]?

**NCERT Solutions:**

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** Miscellaneous: Ques No 19.**

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

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

**NCERT Solutions:**

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

Thus, the operation * 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 2^{4} = 16

The correct answer is B.