Chapter 1: Relations and Functions

Given, A = [–1, 1]

(i) f: [-1, 1] → [-1, 1], f (x) = x/2

Let f (x₁) = f(x₂)

x₁/ 2 = x₂

So, f (x) is one-one.

Also x ∈ [-1, 1]

x/2 = f (x) = [-1/2, 1/2]

Hence, the range is a subset of co-domain ‘A’

So, f (x) is not onto.

Therefore, f (x) is not bijective.

(ii) g (x) = |x|

Let g (x₁) = g (x₂)

|x₁| = |x₂|

x₁ = ± x₂

So, g (x) is not one-one

Also g (x) = |x| ≥ 0, for all real x

Hence, the range is [0, 1], which is subset of co-domain ‘A’

So, f (x) is not onto.

Therefore, f (x) is not bijective.

(iii) h (x) = x|x|

Let h (x₁) = h (x₂)

x₁|x₁| = x₂|x₂|

If x₁, x₂ > 0

x₁² = x₂²

x₁² – x₂² = 0

(x₁ – x₂)(x₁ + x₂) = 0

x₁ = x₂ (as x₁ + x₂ ≠ 0)

Similarly for x₁, x₂ < 0, we have x₁ = x₂

It’s clearly seen that for x₁ and x₂ of opposite sign, x₁ ≠ x₂.

Hence, f (x) is one-one.

For x ∈ [0, 1], f (x) = x² ∈ [0, 1]

For x < 0, f (x) = -x² ∈ [-1, 0)

Hence, the range is [-1, 1].

So, h (x) is onto.

Therefore, h (x) is bijective.

(iv) k (x) = x²

Let k (x₁) = k (x₂)

x₁² = x₂²

x₁ = ± x₂

Therefore, k (x) is not one-one.

(i) Given, x is greater than y; x, y ∈ N

If (x, x) ∈ R, then x > x, which is not true for any x ∈ N.

Thus, R is not reflexive.

Let (x, y) ∈ R

⇒ xRy

⇒ x > y

So, y > x is not true for any x, y ∈ N

Hence, R is not symmetric.

Let xRy and yRz

⇒ x > y and y > z

⇒ x > z

⇒ xRz

Hence, R is transitive.

(ii) x + y = 10; x, y ∈ N

Thus,

R = {(x, y); x + y = 10, x, y ∈ N}

R = {(1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)}

It’s clear (1, 1) ∉ R

So, R is not reflexive.

(x, y) ∈ R ⇒ (y, x) ∈ R

Therefore, R is symmetric.

Now (1, 9) ∈ R, (9, 1) ∈ R, but (1, 1) ∉ R

Therefore, R is not transitive.

(iii) Given, xy is square of an integer x, y ∈ N

R = {(x, y) : xy is a square of an integer x, y ∈ N}

It’s clearly (x, x) ∈ R, ∀ x ∈ N

As x² is square of an integer for any x ∈ N

Thus, R is reflexive.

If (x, y) ∈ R ⇒ (y, x) ∈ R

So, R is symmetric.

Now, if xy is square of an integer and yz is square of an integer.

Then, let xy = m² and yz = n² for some m, n ∈ Z

x = m²/y and z = x²/y

xz = m²n²/ y², which is square of an integer.

Thus, R is transitive.

(iv) x + 4y = 10; x, y ∈ N

R = {(x, y): x + 4y = 10; x, y ∈ N}

R = {(2, 2), (6, 1)}

It’s clearly seen (1, 1) ∉ R

Hence, R is not symmetric.

(x, y) ∈ R ⇒ x + 4y = 10

And (y, z) ∈ R ⇒ y + 4z = 10

⇒ x – 16z = -30

⇒ (x, z) ∉ R

Therefore, R is not transitive.

Given, A = {1, 2, 3, … 9} and (a, b) R (c, d) if a + d = b + c for (a, b), (c, d) ∈ A ×A.

Let (a, b) R(a, b)

So, a + b = b + a, ∀ a, b ∈ A which is true for any a, b ∈ A.

Thus, R is reflexive.

Let (a, b) R(c, d)

Then,

a + d = b + c

c + b = d + a

(c, d) R(a, b)

Thus, R is symmetric.

Let (a, b) R(c, d) and (c, d) R(e, f)

a + d = b + c and c + f = d + e

a + d = b + c and d + e = c + f

(a + d) – (d + e = (b + c) – (c + f)

a – e = b – f

a + f = b + e

(a, b) R(e, f)

So, R is transitive.

Therefore, R is an equivalence relation.

And, [(2,5)=(1,4)(2,5),(3,6),(4,7),(5,8),(6,9)] is the equivalent class under relation R.

Let f: A → B be many-one function.

Let f(a) = p and f(b) = p

So, for inverse function we will have f^-1(p) = a and f^-1(p) = b

Thus, in this case inverse function is not defined as we have two images ‘a and b’ for one pre-image ‘p’.

But for f to be invertible it must be one-one.

Now, let f: A → B is not onto function.

Let B = {p, q, r} and range of f be {p, q}.

Here image ‘r’ has not any pre-image, which will have no image in set A.

And for f to be invertible it must be onto.

Thus, ‘f’ is invertible if and only if ‘f’ is both one-one and onto.

A function f = X → Y is invertible iff f is a bijective function.

Given, f(x) = x² + 3x + 1, g (x) = 2x – 3

(i) fog = f(g(x))

= f(2x – 3)

= (2x – 3)² + 3(2x – 3) + 1

= 4x² + 9 – 12x + 6x – 9 + 1

= 4x² – 6x + 1

(ii) gof = g(f(x))

= g(x² + 3x + 1)

= 2(x² + 3x + 1) – 3

= 2x² + 6x – 1

(iii) fof = f(f(x))

= f(x² + 3x + 1)

= (x² + 3x + 1)² + 3(x² + 3x + 1) + 1

= x⁴ + 9x² + 1 + 6x³ + 6x + 2x² + 3x² + 9x + 3 + 1

= x⁴ + 6x³ + 14x² + 15x + 5

(iv) gog = g(g(x))

= g(2x – 3)

= 2(2x – 3) – 3

= 4x – 6 – 3

= 4x – 9

Given that * is a binary operation defined on Q.

(i) a * b = a – b, ∀ a, b ∈Q and b * a = b – a

So, a * b ≠ b * a

Thus, * is not commutative.

(ii) a * b = a² + b²

b * a = b² + a²

Thus, * is commutative.

(iii) a * b = a + ab

b * a = b + ab

So clearly, a + ab ≠ b + ab

Thus, * is not commutative.

(iv) a * b = (a – b)², ∀ a, b ∈Q

b * a = (b –a)²

Since, (a – b)² = (b – a)²

Thus, * is commutative.

(i) Given that * is a binary operation defined on R by a * b = 1 + ab, ∀ a, b ∈ R

So, we have a * b = ab + 1 = b * a

So, * is a commutative binary operation.

Now, a * (b * c) = a * (1 + bc) = 1 + a (1 + bc) = 1 + a + abc

Also,

(a * b) * c = (1 + ab) * c = 1 + (1 + ab) c = 1 + c + abc

Thus, a * (b * c) ≠ (a * b) * c

Hence, * is not associative.

Therefore, * is commutative but not associative.

Objective Type Questions

Choose the correct answer out of the given four options in each of the Exercises from 28 to 47 (M.C.Q.)

(C) equivalence

Given aRb, if a is congruent to b, ∀ a, b ∈ T.

Then, we have aRa ⇒ a is congruent to a; which is always true.

So, R is reflexive.

Let aRb ⇒ a ~ b

b ~ a

bRa

So, R is symmetric.

Let aRb and bRc

a ~ b and b ~ c

a ~ c

aRc

So, R is transitive.

Therefore, R is equivalence relation.

(B) transitive but not symmetric

aRb ⇒ a is brother of b.

This does not mean b is also a brother of a as b can be a sister of a.

Thus, R is not symmetric.

aRb ⇒ a is brother of b.

and bRc ⇒ b is brother of c.

So, a is brother of c.

Therefore, R is transitive.

(D) 5

Given, set A = {1, 2, 3}

Now, the number of equivalence relations as follows

R₁ = {(1, 1), (2, 2), (3, 3)}

R₂ = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}

R₃ = {(1, 1), (2, 2), (3, 3), (1, 3), (3, 1)}

R₄ = {(1, 1), (2, 2), (3, 3), (2, 3), (3, 2)}

R₅ = {(1, 2, 3) ⇔ A x A = A²}

Thus, maximum number of equivalence relation is ‘5’.