10 Maths samacheer kalvi maths || Ex 1.5 ||

Exercise 1.5 Solutions

1. Using the functions \( f \) and \( g \) given below, find \( f \circ g \) and \( g \circ f \). Check whether \( f \circ g = g \circ f \).

(i) \( f(x) = x - 6 \), \( g(x) = x^2 \)

\( (f \circ g)(x) = f(g(x)) = f(x^2) = x^2 - 6 \)

\( (g \circ f)(x) = g(f(x)) = g(x - 6) = (x - 6)^2 = x^2 - 12x + 36 \)

Result: \( f \circ g \neq g \circ f \)

(ii) \( f(x) = \frac{2}{x} \), \( g(x) = 2x^2 - 1 \)

\( (f \circ g)(x) = f(g(x)) = f(2x^2 - 1) = \frac{2}{2x^2 - 1} \)

\( (g \circ f)(x) = g(f(x)) = g\left(\frac{2}{x}\right) = 2\left(\frac{2}{x}\right)^2 - 1 = \frac{8}{x^2} - 1 \)

Result: \( f \circ g \neq g \circ f \)

(iii) \( f(x) = \frac{x + 6}{3} \), \( g(x) = 3 - x \)

\( (f \circ g)(x) = f(g(x)) = f(3 - x) = \frac{(3 - x) + 6}{3} = \frac{9 - x}{3} \)

\( (g \circ f)(x) = g(f(x)) = g\left(\frac{x + 6}{3}\right) = 3 - \frac{x + 6}{3} = \frac{9 - (x + 6)}{3} = \frac{3 - x}{3} \)

Result: \( f \circ g \neq g \circ f \)

(iv) \( f(x) = 3 + x \), \( g(x) = x - 4 \)

\( (f \circ g)(x) = f(g(x)) = f(x - 4) = 3 + (x - 4) = x - 1 \)

\( (g \circ f)(x) = g(f(x)) = g(3 + x) = (3 + x) - 4 = x - 1 \)

Result: \( f \circ g = g \circ f \)

(v) \( f(x) = 4x^2 - 1 \), \( g(x) = 1 + x \)

\( (f \circ g)(x) = f(g(x)) = f(1 + x) = 4(1 + x)^2 - 1 = 4(1 + 2x + x^2) - 1 = 4x^2 + 8x + 3 \)

\( (g \circ f)(x) = g(f(x)) = g(4x^2 - 1) = 1 + (4x^2 - 1) = 4x^2 \)

Result: \( f \circ g \neq g \circ f \)

2. Find the value of \( k \), such that \( f \circ g = g \circ f \).

(i) \( f(x) = 3x + 2 \), \( g(x) = 6x - k \)

\( (f \circ g)(x) = f(6x - k) = 3(6x - k) + 2 = 18x - 3k + 2 \)

\( (g \circ f)(x) = g(3x + 2) = 6(3x + 2) - k = 18x + 12 - k \)

Given \( f \circ g = g \circ f \):

\( 18x - 3k + 2 = 18x + 12 - k \)

\( -3k + k = 12 - 2 \implies -2k = 10 \implies \mathbf{k = -5} \)

(ii) \( f(x) = 2x - k \), \( g(x) = 4x + 5 \)

\( (f \circ g)(x) = f(4x + 5) = 2(4x + 5) - k = 8x + 10 - k \)

\( (g \circ f)(x) = g(2x - k) = 4(2x - k) + 5 = 8x - 4k + 5 \)

Given \( f \circ g = g \circ f \):

\( 8x + 10 - k = 8x - 4k + 5 \)

\( -k + 4k = 5 - 10 \implies 3k = -5 \implies \mathbf{k = -\frac{5}{3}} \)

3. If \( f(x) = 2x - 1 \), \( g(x) = \frac{x + 1}{2} \), show that \( f \circ g = g \circ f = x \).

1. Find \( (f \circ g)(x) \):

\( (f \circ g)(x) = f(g(x)) = f\left(\frac{x + 1}{2}\right) = 2\left(\frac{x + 1}{2}\right) - 1 = x + 1 - 1 = x \)

---

2. Find \( (g \circ f)(x) \):

\( (g \circ f)(x) = g(f(x)) = g(2x - 1) = \frac{(2x - 1) + 1}{2} = \frac{2x}{2} = x \)

Hence proved: \( f \circ g = g \circ f = x \)

4. If \( f(x) = x^2 - 1 \), \( g(x) = x - 2 \), find \( a \) if \( (g \circ f)(a) = 1 \).

\( (g \circ f)(a) = g(f(a)) = g(a^2 - 1) \)

\( g(a^2 - 1) = (a^2 - 1) - 2 = a^2 - 3 \)

Given \( (g \circ f)(a) = 1 \):

\( a^2 - 3 = 1 \implies a^2 = 4 \implies \mathbf{a = \pm 2} \)

5. Let \( A, B, C \subseteq \mathbb{N} \) and a function \( f: A \to B \) be defined by \( f(x) = 2x + 1 \) and \( g: B \to C \) be defined by \( g(x) = x^2 \). Find the range of \( f \circ g \) and \( g \circ f \).

1. Range of \( f \circ g \):

\( (f \circ g)(x) = f(g(x)) = f(x^2) = 2x^2 + 1 \)

Since domain \( x \in \mathbb{N} \):

Range of \( f \circ g \): \( \{y \mid y = 2x^2 + 1, x \in \mathbb{N}\} = \{3, 9, 19, 33, \dots\} \)

---

2. Range of \( g \circ f \):

\( (g \circ f)(x) = g(f(x)) = g(2x + 1) = (2x + 1)^2 \)

Since domain \( x \in \mathbb{N} \):

Range of \( g \circ f \): \( \{y \mid y = (2x + 1)^2, x \in \mathbb{N}\} = \{9, 25, 49, 81, \dots\} \)

6. Let \( f(x) = x^2 - 1 \). Find (i) \( f \circ f \), (ii) \( f \circ f \circ f \).

(i) \( f \circ f \)

\( (f \circ f)(x) = f(f(x)) = f(x^2 - 1) = (x^2 - 1)^2 - 1 \)

\( = x^4 - 2x^2 + 1 - 1 = \mathbf{x^4 - 2x^2} \)

(ii) \( f \circ f \circ f \)

\( (f \circ f \circ f)(x) = f((f \circ f)(x)) = f(x^4 - 2x^2) \)

\( = \mathbf{(x^4 - 2x^2)^2 - 1} \)

7. If \( f: \mathbb{R} \to \mathbb{R} \) and \( g: \mathbb{R} \to \mathbb{R} \) are defined by \( f(x) = x^5 \) and \( g(x) = x^4 \), then check if \( f, g \) are one-one and \( f \circ g \) is one-one?

1. Check \( f(x) = x^5 \):

If \( f(x_1) = f(x_2) \implies x_1^5 = x_2^5 \implies x_1 = x_2 \).

So, \( f \) is one-one.

---

2. Check \( g(x) = x^4 \):

\( g(1) = 1^4 = 1 \), and \( g(-1) = (-1)^4 = 1 \).

Distinct elements \( 1 \neq -1 \) have the same image \( 1 \).

So, \( g \) is NOT one-one.

---

3. Check \( (f \circ g)(x) \):

\( (f \circ g)(x) = f(x^4) = (x^4)^5 = x^{20} \)

\( (f \circ g)(1) = 1^{20} = 1 \), and \( (f \circ g)(-1) = (-1)^{20} = 1 \).

So, \( f \circ g \) is NOT one-one.

8. Consider the functions \( f(x), g(x), h(x) \) as given below. Show that \( (f \circ g) \circ h = f \circ (g \circ h) \) in each case.

(i) \( f(x) = x - 1 \), \( g(x) = 3x + 1 \), \( h(x) = x^2 \)

\( (f \circ g)(x) = f(3x + 1) = (3x + 1) - 1 = 3x \)

\( \mathbf{LHS = ((f \circ g) \circ h)(x)} = (f \circ g)(x^2) = 3x^2 \)

\( (g \circ h)(x) = g(x^2) = 3x^2 + 1 \)

\( \mathbf{RHS = (f \circ (g \circ h))(x)} = f(3x^2 + 1) = (3x^2 + 1) - 1 = 3x^2 \)

LHS = RHS

(ii) \( f(x) = x^2 \), \( g(x) = 2x \), \( h(x) = x + 4 \)

\( (f \circ g)(x) = f(2x) = (2x)^2 = 4x^2 \)

\( \mathbf{LHS = ((f \circ g) \circ h)(x)} = 4(x + 4)^2 \)

\( (g \circ h)(x) = g(x + 4) = 2(x + 4) \)

\( \mathbf{RHS = (f \circ (g \circ h))(x)} = [2(x + 4)]^2 = 4(x + 4)^2 \)

LHS = RHS

(iii) \( f(x) = x - 4 \), \( g(x) = x^2 \), \( h(x) = 3x - 5 \)

\( (f \circ g)(x) = f(x^2) = x^2 - 4 \)

\( \mathbf{LHS = ((f \circ g) \circ h)(x)} = (3x - 5)^2 - 4 = 9x^2 - 30x + 25 - 4 = 9x^2 - 30x + 21 \)

\( (g \circ h)(x) = g(3x - 5) = (3x - 5)^2 \)

\( \mathbf{RHS = (f \circ (g \circ h))(x)} = (3x - 5)^2 - 4 = 9x^2 - 30x + 21 \)

LHS = RHS

9. Let \( f = \{(-1, 3), (0, -1), (2, -9)\} \) be a linear function from \( \mathbb{Z} \) into \( \mathbb{Z} \). Find \( f(x) \).

Since \( f(x) \) is a linear function, let \( f(x) = ax + b \).

• Given \( (0, -1) \in f \implies f(0) = -1 \):

\( a(0) + b = -1 \implies \mathbf{b = -1} \)

• Given \( (-1, 3) \in f \implies f(-1) = 3 \):

\( a(-1) + b = 3 \implies -a - 1 = 3 \implies -a = 4 \implies \mathbf{a = -4} \)

Substitute \( a = -4 \) and \( b = -1 \):

Answer: \( f(x) = -4x - 1 \)

10. In electrical circuit theory, a circuit \( C(t) \) is called a linear circuit if it satisfies the superposition principle given by \( C(at_1 + bt_2) = aC(t_1) + bC(t_2) \), where \( a, b \) are constants. Show that the circuit \( C(t) = 3t \) is linear.

Given function: \( C(t) = 3t \)

LHS:

\( C(at_1 + bt_2) = 3(at_1 + bt_2) = 3at_1 + 3bt_2 \)

---

RHS:

\( aC(t_1) + bC(t_2) = a(3t_1) + b(3t_2) = 3at_1 + 3bt_2 \)

---

Since \( \text{LHS} = \text{RHS} \), \( C(t) = 3t \) satisfies the superposition principle.

Hence, the circuit \( C(t) = 3t \) is a linear circuit.