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

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

Page No 29:

Question 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 = 1_R.

Answer:

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

One-one:

Let f(x) = f(y), where x, y ∈R.

⇒ 10x + 7 = 10y + 7

⇒ x = y

∴ f is a one-one function.

Onto:

For y ∈ R, let y = 10x + 7.

Therefore, for any y ∈ R, there exists such that 

∴ f is onto.

Therefore, f is one-one and onto.

Thus, f is an invertible function.

Let us define g: R → R as

Now, we have:

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

Question 2:

Let f: W → W be defined as f(n) = n − 1, if 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.

Answer:

It is given that:

f: W → W is defined as

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 be ignored under a similar argument.

∴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

∴f is one-one.

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.

∴f is onto.

Hence, f is an invertible function.

Let us define g: W → W as:

Now, when n is odd:

And, when n is even:

Similarly, when m is odd:

When m is even:

∴

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 f itself.

Question 3:

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

Answer:

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

Question 4:

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

Answer:

It is given that f: R → {x ∈ R: −1 < x < 1} is defined as f(x) =, x ∈R.

Suppose f(x) = f(y), where x, y ∈ R.

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

Since x is positive and y is negative:

x > y ⇒ x − y > 0

But, 2xy is negative.

Then, .

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

 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:

∴ f is one-one.

Now, let y ∈ R such that −1 < y < 1.

If x is negative, then there existssuch that

If x is positive, then there existssuch that

∴ f is onto.

Hence, f is one-one and onto.

Question 5:

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

Answer:

f: R → R is given as f(x) = x³.

Suppose f(x) = f(y), where x, y ∈ R.

⇒ x³ = y³ … (1)

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 (1).

∴ x = y

Hence, f is injective.

Question 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.

(Hint: Consider f(x) = x and g(x) =)

Answer:

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

We first show that g is not injective.

It can be observed that:

g(−1) = 

g(1) = 

∴ g(−1) = g(1), but −1 ≠ 1.

∴ g is not injective.

Now, gof: N → Z is defined as.

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

⇒ 

Since x and y ∈ N, both are positive.

Hence, gof is injective

Question 7:

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

(Hint: Consider f(x) = x + 1 and

Answer:

Define f: N → N by,

f(x) = x + 1

And, g: N → N by,

We first show that g is not onto.

For this, consider element 1 in co-domain N. It is clear that this element is not an image of any of the elements in domain N.

∴ f is not onto.

Now, gof: N → N is defined by,

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

Hence, gof 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 ⊂ B. Is R an equivalence relation on P(X)? Justify you answer:

Answer:

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

∴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.

∴ R is not symmetric.

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

⇒ A ⊂ C

⇒ ARC

∴ R is transitive.

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

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Question 9:

Given a non-empty set X, consider the binary operation *: P(X) × P(X) → P(X) given by A * B = A ∩ B &mnForE; A, B in 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*.

Answer:

It is given that.

We know that.

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

Now, an elementis invertible if there existssuch that

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*.

Hence, the given result is proved.

Question 10:

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

Answer:

Onto functions from the set {1, 2, 3, … ,n} to itself is simply a 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!.

Question 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)}

Answer:

S = {a, b, c}, 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.

Question 12:

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

Answer:

It is given that *: R ×R → and o: R × R → R isdefined as

 and a o b = a, &mnForE;a, b ∈ R.

For a, b ∈ R, we have:

∴a * b = b * a

∴ The operation * is commutative.

It can be observed that,

∴The operation * is not associative.

Now, consider the operation o:

It can be observed that 1 o 2 = 1 and 2 o 1 = 2.

∴1 o 2 ≠ 2 o 1 (where 1, 2 ∈ R)

∴The operation o is not commutative.

Let a, b, c ∈ R. Then, we have:

(a o b) o c = a o c = a

a o (b o c) = a o b = a

⇒ a o b) o c = a o (b o c)

∴ 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 o (2 * 3) =

(1 o 2) * (1 o 3) = 1 * 1 =

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

The operation o does not distribute over *.

Question 13:

Given a non-empty set X, let *: P(X) × P(X) → P(X) be defined as A * B = (A − B) ∪ (B − A), &mnForE; 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. (Hint: (A − Φ) ∪ (Φ − A) = A and (A − A) ∪ (A − A) = A * A = Φ).

Answer:

It is given that *: P(X) × P(X) → P(X) is defined as

A * B = (A − B) ∪ (B − A) &mnForE; 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. &mnForE; 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.

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

Question 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 ≠ 0 of the set is invertible with 6 − a being the inverse of a.

Answer:

Let X = {0, 1, 2, 3, 4, 5}.

The operation * on X is defined as:

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

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.

i.e.,

a = −b or b = 6 − a

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

∴b = 6 − a is the inverse of a &mnForE; a ∈ X.

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

Question 15:

Let A = {−1, 0, 1, 2}, B = {−4, −2, 0, 2} and f, g: A → B be functions defined by f(x) = x² − x, x ∈ A and. Are f and g equal?

Justify your answer. (Hint: One may note that two function f: A → B and g: A → B such that f(a) = g(a) &mnForE;a∈A, are called equal functions).

Answer:

It is 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.

It is observed that:

Hence, the functions f and g are equal.

Question 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

Answer:

The given set is A = {1, 2, 3}.

The smallest relation containing (1, 2) and (1, 3) which is reflexive and symmetric, but not transitive is given by:

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

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

Relation 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 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.

The correct answer is A.

Question 17:

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

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

Answer:

It is given that A = {1, 2, 3}.

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 odd 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.

The correct answer is B.

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Question 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]?

Answer:

It is given that,

f: R → R is defined as

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.

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

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

Question 19:

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

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

Answer:

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.

The correct answer is B.