The Modulus Function |x|: How it Transforms Domain & Range in JEE Problems
Master how |f(x)| affects domain and range - the most frequently tested modulus concept in JEE Main & Advanced.
Why Modulus Function is Crucial for JEE
The modulus function $|x|$ appears in 85% of JEE papers and transforms range in predictable ways. Understanding these patterns can help you solve complex problems in seconds.
🎯 Key Insight:
$|f(x)|$ always makes the range non-negative. If original range was $[a, b]$, then range of $|f(x)|$ becomes $[0, \max(|a|, |b|)]$ when the range spans both positive and negative values.
Basic Modulus Transformation
If $f(x)$ has range $R$, what is the range of $|f(x)|$?
Transformation Rule:
Case 1: If $R \subseteq [0, \infty)$ → Range remains same: $R$
Case 2: If $R \subseteq (-\infty, 0]$ → Range becomes $[0, \max(|x|)]$ where $x \in R$
Case 3: If $R$ spans both positive and negative → Range becomes $[0, M]$ where $M = \max(|min(R)|, |max(R)|)$
Example:
If $f(x) = \sin x$ has range $[-1, 1]$, then $|f(x)| = |\sin x|$ has range $[0, 1]$
Modulus Inside Square Root
Find domain and range of $f(x) = \sqrt{|x| - 2}$
Solution Approach:
Step 1: Domain condition: $|x| - 2 \geq 0$ ⇒ $|x| \geq 2$
Step 2: Domain: $x \leq -2$ or $x \geq 2$
Step 3: For range: $|x| - 2 \geq 0$, so $\sqrt{|x| - 2} \geq 0$
Step 4: As $|x| \to \infty$, $f(x) \to \infty$
Step 5: Minimum at $|x| = 2$: $f(x) = 0$
Step 6: Range: $[0, \infty)$
Composite Modulus Function
Find range of $f(x) = ||x - 2| - 3|$
Solution Approach:
Step 1: Let $g(x) = |x - 2| - 3$
Step 2: Range of $|x - 2|$ is $[0, \infty)$
Step 3: Range of $g(x)$ is $[-3, \infty)$
Step 4: Apply outer modulus: $|g(x)|$ has range $[0, \infty)$
Alternative Graphical Method:
• Graph shifts right by 2 units: $|x-2|$
• Shifts down by 3 units: $|x-2|-3$
• Reflect negative part upward: $||x-2|-3|$
Final Range: $[0, \infty)$
🚀 Modulus Function Quick Strategies
Domain Transformations:
- $|f(x)|$ has same domain as $f(x)$
- $\sqrt{|f(x)|}$ requires $|f(x)| \geq 0$ (always true)
- $\frac{1}{|f(x)|}$ requires $f(x) \neq 0$
- $\log|f(x)|$ requires $|f(x)| > 0$ ⇒ $f(x) \neq 0$
Range Transformations:
- $|f(x)| ≥ 0$ always
- If $f(x)$ range is $[a,b]$, then $|f(x)|$ range is $[0, \max(|a|,|b|)]$
- $|f(x)|$ creates V-shaped graphs
- Piecewise definition helps in complex cases
Modulus in Rational Functions
Find domain and range of $f(x) = \frac{1}{|x| - 1}$
Solution Approach:
Step 1: Domain: Denominator ≠ 0 ⇒ $|x| - 1 \neq 0$ ⇒ $|x| \neq 1$ ⇒ $x \neq \pm1$
Step 2: Domain: $\mathbb{R} - \{-1, 1\}$
Step 3: For range, analyze cases:
• When $|x| > 1$: $|x|-1 > 0$, so $f(x) > 0$
• When $0 \leq |x| < 1$: $|x|-1 < 0$, so $f(x) < 0$
Step 4: As $|x| \to 1^+$, $f(x) \to +\infty$
Step 5: As $|x| \to 1^-$, $f(x) \to -\infty$
Step 6: As $|x| \to \infty$, $f(x) \to 0$
Step 7: Range: $(-\infty, 0) \cup (0, \infty)$
Patterns 5-6 Available in Full Version
Includes Modulus in Trigonometric Functions and Modulus in Inverse Trigonometric Functions with JEE Advanced level problems
📝 Quick Self-Test
Try these modulus function problems to test your understanding:
1. Find range of $f(x) = |x^2 - 4x + 3|$
2. Find domain of $f(x) = \sqrt{4 - |x|}$
3. Find range of $f(x) = \frac{|x|}{x^2 + 1}$
📚 Essential Modulus Formulas
Basic Properties:
- $|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}$
- $|x| \geq 0$ for all real $x$
- $|x| = |-x|$
- $\sqrt{x^2} = |x|$
Inequality Rules:
- $|x| < a ⇔ -a < x < a$
- $|x| > a ⇔ x < -a$ or $x > a$
- $|x ± y| ≤ |x| + |y|$ (Triangle Inequality)
- $|x ± y| ≥ ||x| - |y||$
Ready to Master All 6 Modulus Patterns?
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