Chapter 1: Relations and Functions

Q1. Determine whether each of the following relations are reflexive, symmetric and transitive:
R in the set A = {1, 2, 3, ..., 14} defined by R = {(x, y) : 3x − y = 0}

Solution

R = {(x, y) : 3x − y = 0} = {(1,3), (2,6), (3,9), (4,12)}

Reflexive: (1,1) ∉ R since 3(1) − 1 = 2 ≠ 0. ✗ Not reflexive.

Symmetric: (1,3) ∈ R but (3,1) ∉ R since 3(3) − 1 = 8 ≠ 0. ✗ Not symmetric.

Transitive: If (x,y) ∈ R and (y,z) ∈ R, then 3x = y and 3y = z, so z = 9x. For (x,z) in R we need z = 3x, but z = 9x ≠ 3x. ✗ Not transitive.

Q2. Show that the function f : R → R defined by f(x) = 3x + 7 is a one-one function.

Solution

Let f(x₁) = f(x₂). Then:

3x₁ + 7 = 3x₂ + 7 ⇒ 3x₁ = 3x₂ ⇒ x₁ = x₂

Since f(x₁) = f(x₂) implies x₁ = x₂, f is one-one. ✓

Q3. Prove that the function f : N → N defined by f(n) = n² is one-one but not onto.

Solution

One-one: Let f(n₁) = f(n₂), then n₁² = n₂². Since n₁, n₂ ∈ N (positive), n₁ = n₂. ✓

Not onto: Take m = 2 ∈ N. We need n² = 2, so n = √2 ∉ N. Since no natural number maps to 2, f is not onto. ✗

Q4. Show that the Binary operation * on N defined by a * b = LCM(a, b) is commutative and associative. Find the identity element.

Solution

Commutative: a * b = LCM(a, b) = LCM(b, a) = b * a. ✓

Associative: (a * b) * c = LCM(LCM(a,b), c) = LCM(a, LCM(b,c)) = a * (b * c). ✓

Identity: We need a * e = a, i.e., LCM(a, e) = a for all a. This holds when e = 1. Identity element is 1.

Q5. Let f : R → R be defined as f(x) = x³. Show that f is a bijection.

Solution

One-one: Let f(x₁) = f(x₂). Then x₁³ = x₂³. Taking cube root: x₁ = x₂. ✓

Onto: For any y ∈ R, let x = yⅅ(cube root). Then f(x) = (yⅅ)³ = y. So every y has a pre-image. ✓

∴ f is a bijection.

Q6. Find gof and fog if f(x) = 8x³ and g(x) = x^1/3.

Solution

gof(x) = g(f(x)) = g(8x³) = (8x³)^1/3 = 2x
fog(x) = f(g(x)) = f(x^1/3) = 8(x^1/3)³ = 8x

Chapter 2: Inverse Trigonometric Functions →

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