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On April 21, 2024, 11:35 AM

NCERT Solutions Class 12 Maths: Relations and Functions Revised Syllabus

Miscellaneous Exercise Question 1:

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 exists such that

If x is positive, then there exists such that

∴ f is onto.

Hence, f is one-one and onto.

Miscellaneous Exercise Question 2:

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.

Miscellaneous Exercise Question 3:

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.

Miscellaneous Exercise Question 4:

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!.

Miscellaneous Exercise Question 5:

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.

Miscellaneous Exercise Question 6:

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.

Miscellaneous Exercise Question 7:

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.