Math Notes For Class 12 Chapter 1 Ex 1.2 Solved & Quiz
What are the Core Formulas and Definitions for this Exercise?
In Class 12 FSc Pre-Engineering Mathematics (Chapter 1: Functions and Limits), Exercise 1.2 focuses on two fundamental algebraic concepts: the Composition of Functions and the Inverse of a Function.
- Composition of Functions: Let $f: X \to Y$ and $g: Y \to Z$ be two functions. The composition of $f$ and $g$, denoted by $g \circ f$ or $g(f(x))$, is a function from $X$ to $Z$ defined by $(g \circ f)(x) = g(f(x))$ for all $x \in X$. Note that function composition is generally non-commutative, i.e., $f(g(x)) \neq g(f(x))$ in general.
- Inverse of a Function: Let $f: X \to Y$ be a one-to-one (injective) and onto (surjective) function (i.e., a bijective function). The inverse function $f^{-1}: Y \to X$ is defined such that $f^{-1}(y) = x$ if and only if $f(x) = y$.
- Fundamental Identity: For any bijective function $f$ and its inverse $f^{-1}$, $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$.
- Domain and Range Inversion: $\text{Domain}(f^{-1}) = \text{Range}(f)$ and $\text{Range}(f^{-1}) = \text{Domain}(f)$.
| Concept / Operation | Mathematical Notation | Key Domain / Structural Condition |
|---|---|---|
| Composite Function $f \circ g$ | $(f \circ g)(x) = f(g(x))$ | $\text{Range}(g) \subseteq \text{Domain}(f)$ |
| Composite Function $g \circ f$ | $(g \circ f)(x) = g(f(x))$ | $\text{Range}(f) \subseteq \text{Domain}(g)$ |
| Self-Composition | $(f \circ f)(x) = f(f(x))$ | Evaluates $f$ at the output of $f(x)$ |
| Inverse Function Rules | $y = f(x) \iff x = f^{-1}(y)$ | $f$ must be a bijective (1-1 and onto) function |
| Domain/Range Relation | $\text{Dom}(f^{-1}) = \text{Ran}(f)$ $\text{Ran}(f^{-1}) = \text{Dom}(f)$ |
Allows finding inverse domain/range without explicit inversion |
How to Solve All Exercise Questions Step-by-Step?
Question 1 (Part i): The real valued functions $f$ and $g$ are defined below. Find (a) $f(g(x))$, (b) $g(f(x))$, (c) $f(f(x))$, and (d) $g(g(x))$ for:
$$f(x) = 2x + 1, \quad g(x) = \frac{3}{x-1}, \quad x \neq 1$$
Solution:
(a) Finding $f(g(x))$:
$$f(g(x)) = f\left(\frac{3}{x-1}\right) = 2\left(\frac{3}{x-1}\right) + 1 = \frac{6}{x-1} + 1 = \frac{6 + (x - 1)}{x - 1} = \frac{x + 5}{x - 1}$$
(b) Finding $g(f(x))$:
$$g(f(x)) = g(2x + 1) = \frac{3}{(2x + 1) - 1} = \frac{3}{2x}$$
(c) Finding $f(f(x))$:
$$f(f(x)) = f(2x + 1) = 2(2x + 1) + 1 = 4x + 2 + 1 = 4x + 3$$
(d) Finding $g(g(x))$:
$$g(g(x)) = g\left(\frac{3}{x-1}\right) = \frac{3}{\frac{3}{x-1} - 1} = \frac{3}{\frac{3 - (x - 1)}{x - 1}} = \frac{3(x - 1)}{4 - x}$$
Answer:
(a) $$f(g(x)) = \frac{x+5}{x-1}$$, (b) $$g(f(x)) = \frac{3}{2x}$$, (c) $$f(f(x)) = 4x+3$$, (d) $$g(g(x)) = \frac{3(x-1)}{4-x}$$
Question 1 (Part ii): Find (a) $f(g(x))$, (b) $g(f(x))$, (c) $f(f(x))$, and (d) $g(g(x))$ for:
$$f(x) = \sqrt{x+1}, \quad g(x) = \frac{1}{x^2}, \quad x \neq 0$$
Solution:
(a) Finding $f(g(x))$:
$$f(g(x)) = f\left(\frac{1}{x^2}\right) = \sqrt{\frac{1}{x^2} + 1} = \sqrt{\frac{1 + x^2}{x^2}} = \frac{\sqrt{x^2 + 1}}{x}$$
(b) Finding $g(f(x))$:
$$g(f(x)) = g(\sqrt{x+1}) = \frac{1}{(\sqrt{x+1})^2} = \frac{1}{x + 1}$$
(c) Finding $f(f(x))$:
$$f(f(x)) = f(\sqrt{x+1}) = \sqrt{\sqrt{x+1} + 1}$$
(d) Finding $g(g(x))$:
$$g(g(x)) = g\left(\frac{1}{x^2}\right) = \frac{1}{\left(\frac{1}{x^2}\right)^2} = \frac{1}{\frac{1}{x^4}} = x^4$$
Answer:
(a) $$f(g(x)) = \frac{\sqrt{x^2+1}}{x}$$, (b) $$g(f(x)) = \frac{1}{x+1}$$, (c) $$f(f(x)) = \sqrt{\sqrt{x+1}+1}$$, (d) $$g(g(x)) = x^4$$
Question 1 (Part iii): Find (a) $f(g(x))$, (b) $g(f(x))$, (c) $f(f(x))$, and (d) $g(g(x))$ for:
$$f(x) = \frac{1}{\sqrt{x-1}}, \quad x > 1, \quad g(x) = (x^2 + 1)^2$$
Solution:
(a) Finding $f(g(x))$:
$$f(g(x)) = f((x^2 + 1)^2) = \frac{1}{\sqrt{(x^2 + 1)^2 - 1}} = \frac{1}{\sqrt{x^4 + 2x^2 + 1 - 1}} = \frac{1}{\sqrt{x^4 + 2x^2}} = \frac{1}{x\sqrt{x^2 + 2}}$$
(b) Finding $g(f(x))$:
$$g(f(x)) = g\left(\frac{1}{\sqrt{x-1}}\right) = \left(\left(\frac{1}{\sqrt{x-1}}\right)^2 + 1\right)^2 = \left(\frac{1}{x - 1} + 1\right)^2 = \left(\frac{1 + x - 1}{x - 1}\right)^2 = \frac{x^2}{(x - 1)^2}$$
(c) Finding $f(f(x))$:
$$f(f(x)) = f\left(\frac{1}{\sqrt{x-1}}\right) = \frac{1}{\sqrt{\frac{1}{\sqrt{x-1}} - 1}} = \frac{1}{\sqrt{\frac{1 - \sqrt{x-1}}{\sqrt{x-1}}}} = \sqrt{\frac{\sqrt{x-1}}{1 - \sqrt{x-1}}}$$
(d) Finding $g(g(x))$:
$$g(g(x)) = g((x^2 + 1)^2) = \left(((x^2 + 1)^2)^2 + 1\right)^2 = \left((x^2 + 1)^4 + 1\right)^2$$
Answer:
(a) $$f(g(x)) = \frac{1}{x\sqrt{x^2+2}}$$, (b) $$g(f(x)) = \frac{x^2}{(x-1)^2}$$, (c) $$f(f(x)) = \sqrt{\frac{\sqrt{x-1}}{1-\sqrt{x-1}}}$$, (d) $$g(g(x)) = ((x^2+1)^4+1)^2$$
Question 1 (Part iv): Find (a) $f(g(x))$, (b) $g(f(x))$, (c) $f(f(x))$, and (d) $g(g(x))$ for:
$$f(x) = 3x^3 - 2, \quad g(x) = \frac{x+1}{2}$$
Solution:
(a) Finding $f(g(x))$:
$$f(g(x)) = f\left(\frac{x+1}{2}\right) = 3\left(\frac{x+1}{2}\right)^3 - 2 = 3\left(\frac{(x+1)^3}{8}\right) - 2 = \frac{3(x+1)^3 - 16}{8}$$
(b) Finding $g(f(x))$:
$$g(f(x)) = g(3x^3 - 2) = \frac{(3x^3 - 2) + 1}{2} = \frac{3x^3 - 1}{2}$$
(c) Finding $f(f(x))$:
$$f(f(x)) = f(3x^3 - 2) = 3(3x^3 - 2)^3 - 2$$
(d) Finding $g(g(x))$:
$$g(g(x)) = g\left(\frac{x+1}{2}\right) = \frac{\left(\frac{x+1}{2}\right) + 1}{2} = \frac{\frac{x+1+2}{2}}{2} = \frac{x+3}{4}$$
Answer:
(a) $$f(g(x)) = \frac{3(x+1)^3-16}{8}$$, (b) $$g(f(x)) = \frac{3x^3-1}{2}$$, (c) $$f(f(x)) = 3(3x^3-2)^3-2$$, (d) $$g(g(x)) = \frac{x+3}{4}$$
Question 2 (Part i): For the real valued function $f(x) = -2x + 8$, find (a) $f^{-1}(x)$, (b) $f^{-1}(-1)$, and verify that $f(f^{-1}(x)) = f^{-1}(f(x)) = x$.
Solution:
(a) Finding $f^{-1}(x)$:
Let $y = f(x) \implies y = -2x + 8$
Solving for $x$:
$$2x = 8 - y \implies x = \frac{8 - y}{2}$$
Since $x = f^{-1}(y)$, we have $f^{-1}(y) = \frac{8 - y}{2}$. Replacing $y$ with $x$:
$$f^{-1}(x) = \frac{8 - x}{2}$$
(b) Finding $f^{-1}(-1)$:
$$f^{-1}(-1) = \frac{8 - (-1)}{2} = \frac{8 + 1}{2} = \frac{9}{2}$$
Verification:
1. $f(f^{-1}(x)) = f\left(\frac{8 - x}{2}\right) = -2\left(\frac{8 - x}{2}\right) + 8 = -(8 - x) + 8 = -8 + x + 8 = x$.
2. $f^{-1}(f(x)) = f^{-1}(-2x + 8) = \frac{8 - (-2x + 8)}{2} = \frac{8 + 2x - 8}{2} = \frac{2x}{2} = x$.
Hence, verified.
Answer:
(a) $$f^{-1}(x) = \frac{8-x}{2}$$, (b) $$f^{-1}(-1) = \frac{9}{2}$$, Verification proved $$f(f^{-1}(x)) = f^{-1}(f(x)) = x$$.
Question 2 (Part ii): For the real valued function $f(x) = 3x^3 + 7$, find (a) $f^{-1}(x)$, (b) $f^{-1}(-1)$, and verify that $f(f^{-1}(x)) = f^{-1}(f(x)) = x$.
Solution:
(a) Finding $f^{-1}(x)$:
Let $y = f(x) \implies y = 3x^3 + 7$
$$3x^3 = y - 7 \implies x^3 = \frac{y - 7}{3} \implies x = \left(\frac{y - 7}{3}\right)^{1/3}$$
Therefore, $$f^{-1}(x) = \left(\frac{x - 7}{3}\right)^{1/3}$$
(b) Finding $f^{-1}(-1)$:
$$f^{-1}(-1) = \left(\frac{-1 - 7}{3}\right)^{1/3} = \left(\frac{-8}{3}\right)^{1/3} = -\frac{2}{3^{1/3}}$$
Verification:
1. $f(f^{-1}(x)) = f\left(\left(\frac{x - 7}{3}\right)^{1/3}\right) = 3\left(\left(\frac{x - 7}{3}\right)^{1/3}\right)^3 + 7 = 3\left(\frac{x - 7}{3}\right) + 7 = x - 7 + 7 = x$.
2. $f^{-1}(f(x)) = f^{-1}(3x^3 + 7) = \left(\frac{(3x^3 + 7) - 7}{3}\right)^{1/3} = \left(\frac{3x^3}{3}\right)^{1/3} = (x^3)^{1/3} = x$.
Hence, verified.
Answer:
(a) $$f^{-1}(x) = \left(\frac{x-7}{3}\right)^{1/3}$$, (b) $$f^{-1}(-1) = -\frac{2}{3^{1/3}}$$, Verification proved.
Question 2 (Part iii): For the real valued function $f(x) = (-x + 9)^3$, find (a) $f^{-1}(x)$, (b) $f^{-1}(-1)$, and verify that $f(f^{-1}(x)) = f^{-1}(f(x)) = x$.
Solution:
(a) Finding $f^{-1}(x)$:
Let $y = (-x + 9)^3 \implies y^{1/3} = -x + 9 \implies x = 9 - y^{1/3}$
Therefore, $$f^{-1}(x) = 9 - x^{1/3
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