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Golden Rules Reading Time: 10 min

The 4 Golden Rules for Finding Domain of Any Function

Master these non-negotiable rules to solve any domain problem in JEE Main & Advanced with confidence.

In this guide:

Why These Rules Are Golden

These 4 rules form the foundation of domain finding in calculus. Master them, and you can tackle any domain problem that appears in JEE. Remember: Violating any of these rules makes a function undefined at that point.

JEE Importance:

  • Direct questions appear every year in both JEE Main and Advanced
  • Essential for solving limits, continuity, and differentiability problems
  • Forms the basis for understanding function behavior
  • Common source of errors if not applied correctly
1

Denominator Cannot Be Zero

For $f(x) = \frac{g(x)}{h(x)}$

We require $h(x) \neq 0$

This is the most fundamental rule. Division by zero is undefined in mathematics, so we must exclude any x-values that make the denominator zero.

Example 1: Simple Rational Function

Find domain of $f(x) = \frac{1}{x-2}$

Step 1: Identify denominator: $x - 2$

Step 2: Set denominator ≠ 0: $x - 2 \neq 0$

Step 3: Solve: $x \neq 2$

Solution: Domain = $\mathbb{R} - \{2\}$ or $(-\infty, 2) \cup (2, \infty)$

Example 2: Complex Denominator

Find domain of $f(x) = \frac{x+1}{x^2 - 4}$

Step 1: Identify denominator: $x^2 - 4$

Step 2: Set denominator ≠ 0: $x^2 - 4 \neq 0$

Step 3: Solve: $x^2 \neq 4 \Rightarrow x \neq \pm 2$

Solution: Domain = $\mathbb{R} - \{-2, 2\}$

2

Expression Inside Square Root (Even Root) Must Be ≥ 0

For $f(x) = \sqrt{g(x)}$ or $\sqrt[n]{g(x)}$ where n is even

We require $g(x) \geq 0$

Square roots (and other even roots) of negative numbers are not real numbers. Therefore, the expression inside must be non-negative.

Example 1: Simple Square Root

Find domain of $f(x) = \sqrt{x-3}$

Step 1: Identify expression under root: $x - 3$

Step 2: Set expression ≥ 0: $x - 3 \geq 0$

Step 3: Solve: $x \geq 3$

Solution: Domain = $[3, \infty)$

Example 2: Quadratic Under Root

Find domain of $f(x) = \sqrt{x^2 - 4}$

Step 1: Identify expression: $x^2 - 4$

Step 2: Set ≥ 0: $x^2 - 4 \geq 0$

Step 3: Solve inequality: $(x-2)(x+2) \geq 0$

Step 4: Critical points: $x = -2, 2$

Solution: Domain = $(-\infty, -2] \cup [2, \infty)$

3

Expression Inside Logarithm Must Be > 0

For $f(x) = \log(g(x))$ or $\ln(g(x))$

We require $g(x) > 0$

Logarithms are only defined for positive real numbers. The argument must be strictly greater than zero.

Example 1: Simple Logarithm

Find domain of $f(x) = \ln(x+4)$

Step 1: Identify argument: $x + 4$

Step 2: Set argument > 0: $x + 4 > 0$

Step 3: Solve: $x > -4$

Solution: Domain = $(-4, \infty)$

Example 2: Logarithm with Quadratic

Find domain of $f(x) = \log(x^2 - 1)$

Step 1: Identify argument: $x^2 - 1$

Step 2: Set > 0: $x^2 - 1 > 0$

Step 3: Solve inequality: $(x-1)(x+1) > 0$

Step 4: Critical points: $x = -1, 1$

Solution: Domain = $(-\infty, -1) \cup (1, \infty)$

4

Special Restrictions for Inverse Trigonometric Functions

$\sin^{-1}x, \cos^{-1}x$

Domain: $[-1, 1]$

Input must be between -1 and 1 inclusive

$\tan^{-1}x, \cot^{-1}x$

Domain: $\mathbb{R}$

All real numbers are allowed

$\sec^{-1}x, \cosec^{-1}x$

Domain: $(-\infty, -1] \cup [1, \infty)$

Input must satisfy $|x| \geq 1$

Example: Inverse Trigonometric Function

Find domain of $f(x) = \sin^{-1}(2x-1)$

Step 1: For $\sin^{-1}$, argument must be in $[-1, 1]$

Step 2: Set up inequality: $-1 \leq 2x-1 \leq 1$

Step 3: Solve: $0 \leq 2x \leq 2$

Step 4: Final: $0 \leq x \leq 1$

Solution: Domain = $[0, 1]$

Applying Multiple Rules Together

In JEE problems, you'll often need to apply multiple rules simultaneously. The final domain is the intersection of all individual restrictions.

JEE Main 2022 Problem

Find the domain of $f(x) = \frac{\sqrt{x-2}}{\log(x^2-1)}$

Solution Approach:

Rule 2: Square Root Condition

$x - 2 \geq 0 \Rightarrow x \geq 2$

Rule 3: Logarithm Condition

$x^2 - 1 > 0 \Rightarrow (x-1)(x+1) > 0$

This gives $x < -1$ or $x > 1$

Rule 1: Denominator Condition

$\log(x^2-1) \neq 0 \Rightarrow x^2-1 \neq 1$

$x^2 \neq 2 \Rightarrow x \neq \pm\sqrt{2}$

Combine All Conditions:

From above: $x \geq 2$ AND $(x < -1$ or $x > 1)$ AND $x \neq \pm\sqrt{2}$

Final Domain: $(2, \infty)$

Golden Rules Quick Reference

The 4 Golden Rules

  • 1 Denominator ≠ 0
    For $\frac{g(x)}{h(x)}$, exclude roots of $h(x)$
  • 2 Even Root ≥ 0
    For $\sqrt{g(x)}$, solve $g(x) \geq 0$
  • 3 Logarithm > 0
    For $\log(g(x))$, solve $g(x) > 0$
  • 4 Inverse Trig Restrictions
    Remember specific domain ranges

Problem Solving Strategy

  1. Identify all rules that apply to the function
  2. Solve each condition separately
  3. Take intersection of all solutions
  4. Express final domain in interval notation
  5. Always check edge cases and boundaries

Ready to Apply These Rules?

Test your understanding with domain finding practice problems:

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