NCERT Class 12 Maths Chapter 1: Exercise 1.3 (Relations and Functions)

Understanding Composition of Functions

To understand the composition of functions, imagine an industrial assembly line. Let function \( f: A \to B \) be the first machine that takes an input \( x \) from set \( A \) and outputs \( f(x) \) in set \( B \). Now, let function \( g: B \to C \) be the second machine that takes \( f(x) \) as its input and outputs \( g(f(x)) \) in set \( C \). This combined operation is called the composition of \( f \) and \( g \), mathematically represented as \( g \circ f \) (read as 'g circle f' or 'g of f'). Crucially, the composition \( g \circ f \) is only defined if the range of \( f \) is a subset of the domain of \( g \). Note that function composition is generally not commutative, meaning \( g \circ f \neq f \circ g \) in most scenarios.

How to Prove a Function is Invertible and Find its Inverse

1. Step 1: Check if the Function is One-to-One (Injective) — Assume \( f(x_1) = f(x_2) \) for any arbitrary elements \( x_1, x_2 \) in the domain. Solve the algebraic equation. If you get \( x_1 = x_2 \) as the only real solution, the function is injective.
2. Step 2: Check if the Function is Onto (Surjective) — Let \( y \) be an arbitrary element in the codomain. Set \( y = f(x) \) and express \( x \) explicitly in terms of \( y \). Check if this expression of \( x \) exists and belongs to the domain for every \( y \) in the codomain. If yes, the function is surjective.
3. Step 3: Define the Inverse Function — Since the function is both injective and surjective (bijective), it is invertible. The inverse function \( f^{-1}(y) \) is simply the expression for \( x \) in terms of \( y \) that you derived in Step 2.

Detailed Step-by-Step Examples

• Example 1: Let \( f: \mathbb{R} \to \mathbb{R} \) and \( g: \mathbb{R} \to \mathbb{R} \) be defined as \( f(x) = x^2 \) and \( g(x) = x + 5 \). Find \( g \circ f \) and \( f \circ g \). Step 1: Find \( g \circ f(x) \). By definition, \( g \circ f(x) = g(f(x)) \). Substitute \( f(x) = x^2 \) into \( g \), which yields \( g(x^2) = x^2 + 5 \). Step 2: Find \( f \circ g(x) \). By definition, \( f \circ g(x) = f(g(x)) \). Substitute \( g(x) = x + 5 \) into \( f \), which yields \( f(x + 5) = (x + 5)^2 = x^2 + 10x + 25 \). Conclusion: Since \( x^2 + 5 \neq x^2 + 10x + 25 \), we have proved that \( g \circ f \neq f \circ g \).
• Example 2: Let \( f: \mathbb{R} - \{3\} \to \mathbb{R} - \{1\} \) be defined by \( f(x) = \frac{x-2}{x-3} \). Show that \( f \) is invertible and find its inverse. Step 1: Injective check. Let \( f(x_1) = f(x_2) \). Thus, \( \frac{x_1-2}{x_1-3} = \frac{x_2-2}{x_2-3} \). Cross-multiplying: \( (x_1-2)(x_2-3) = (x_2-2)(x_1-3) \) which expands to \( x_1 x_2 - 3x_1 - 2x_2 + 6 = x_1 x_2 - 3x_2 - 2x_1 + 6 \). Simplifying gives \( -3x_1 - 2x_2 = -3x_2 - 2x_1 \), leading directly to \( x_1 = x_2 \). Hence, \( f \) is injective. Step 2: Surjective check. Let \( y = \frac{x-2}{x-3} \). Solving for \( x \): \( y(x-3) = x-2 \Rightarrow xy - 3y = x - 2 \Rightarrow x(y-1) = 3y-2 \Rightarrow x = \frac{3y-2}{y-1} \). For any \( y \in \mathbb{R} - \{1\} \), \( x \) is defined and belongs to \( \mathbb{R} - \{3\} \). Hence, \( f \) is surjective. Step 3: Finding inverse. Since \( f \) is bijective, it is invertible. The inverse function is \( f^{-1}(y) = \frac{3y-2}{y-1} \).

CBSE Board Exam Tips & Common Pitfalls

1. Order of Operations: Students often confuse \( g \circ f(x) \) with \( f(x) \cdot g(x) \). Remember that composition is a function of a function, not simple algebraic multiplication.
2. Check the Domain: Always check if the output set of the first function matches the input domain of the second function. If they do not align, the composite function does not exist!
3. Proving Invertibility: CBSE examiners expect you to formally prove BOTH one-to-one (injectivity) and onto (surjectivity) properties before writing down the inverse formula. Simply rearranging the equation for \( x \) without these proofs will lose you marks.

Practice Questions with Solutions

• Q: Let \( f(x) = |x| \) and \( g(x) = |5x - 2| \). Find \( g \circ f \) and \( f \circ g \). A: Step 1: Write down the expressions for both functions. We have \( f(x) = |x| \) and \( g(x) = |5x - 2| \). Step 2: Find \( g \circ f(x) = g(f(x)) \). Substitute \( f(x) \) into the expression of \( g \): \( g(|x|) = |5|x| - 2| \). Step 3: Find \( f \circ g(x) = f(g(x)) \). Substitute \( g(x) \) into the expression of \( f \): \( f(|5x - 2|) = ||5x - 2|| = |5x - 2| \) (since double absolute value is redundant). Final answer: \( g \circ f(x) = |5|x| - 2| \) and \( f \circ g(x) = |5x - 2| \).
• Q: Show that \( f: [-1, 1] \to \mathbb{R} \), given by \( f(x) = \frac{x}{x+2} \) is one-to-one. Find the inverse of the function \( f: [-1, 1] \to \text{Range } f \). A: Step 1: Prove injectivity. Let \( f(x_1) = f(x_2) \) for \( x_1, x_2 \in [-1, 1] \). \( \frac{x_1}{x_1+2} = \frac{x_2}{x_2+2} \Rightarrow x_1(x_2+2) = x_2(x_1+2) \Rightarrow x_1 x_2 + 2x_1 = x_1 x_2 + 2x_2 \Rightarrow 2x_1 = 2x_2 \Rightarrow x_1 = x_2 \). Hence, \( f \) is one-to-one. Step 2: Since the codomain is restricted to \( \text{Range } f \), the function is onto by definition. Thus, it is invertible. Step 3: Let \( y = \frac{x}{x+2} \). Solve for \( x \) in terms of \( y \): \( y(x+2) = x \Rightarrow xy + 2y = x \Rightarrow x(1-y) = 2y \Rightarrow x = \frac{2y}{1-y} \). Final answer: The function is one-to-one and its inverse function is \( f^{-1}(y) = \frac{2y}{1-y} \) for \( y \neq 1 \).
• Q: If \( f: \mathbb{R} \to \mathbb{R} \) is given by \( f(x) = (3 - x^3)^{1/3} \), then find \( f \circ f(x) \). A: Step 1: Write the expression for \( f \circ f(x) = f(f(x)) \). Step 2: Substitute \( f(x) \) in place of \( x \) in the formula for \( f(x) \): \( f(f(x)) = (3 - (f(x))^3)^{1/3} \). Step 3: Replace \( f(x) \) with its definition \( (3 - x^3)^{1/3} \): \( f(f(x)) = (3 - ((3 - x^3)^{1/3})^3)^{1/3} \). Step 4: Simplify the expression: \( (3 - x^3)^{1/3 \times 3} = 3 - x^3 \). Now plug this back: \( (3 - (3 - x^3))^{1/3} = (3 - 3 + x^3)^{1/3} = (x^3)^{1/3} = x \). Final answer: \( f \circ f(x) = x \).
• Q: State with reason whether the function \( f: \{1, 2, 3, 4\} \to \{10\} \) with \( f = \{(1, 10), (2, 10), (3, 10), (4, 10)\} \) has an inverse. A: Step 1: Check if the function is one-to-one. Here, \( f(1) = 10 \), \( f(2) = 10 \), etc. Since distinct elements in the domain have the same image in the codomain, \( f \) is many-to-one. Step 2: A function is invertible if and only if it is bijective (both one-to-one and onto). Step 3: Since \( f \) is not one-to-one, it cannot be bijective. Final answer: The function \( f \) does not have an inverse because it is not a one-to-one function.

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