5 Powerful Methods to Find the Range of a Function for JEE Advanced
Master advanced techniques like Graphical Method, Calculus Approach, and Function Transformations with detailed examples.
Why These Methods Matter for JEE Advanced
Finding the range of functions is a crucial skill for JEE Advanced, appearing in both multiple-choice and numerical answer type questions. These 5 methods cover 95% of range-finding problems in JEE Advanced:
- Graphical Method - Visual approach for complex functions
- Calculus Approach - Using derivatives to find extrema
- Function Transformations - Systematic manipulation
- Quadratic Method - For rational functions
- Inverse Function Method - Finding domain of inverse
Graphical Method
Sketch the graph and observe the y-values covered by the function.
When to Use:
Functions with known graphs: polynomials, trigonometric, exponential, logarithmic functions.
Example: Find range of $f(x) = x^2 - 4x + 3$
Step 1: Complete the square: $f(x) = (x-2)^2 - 1$
Step 2: The graph is a parabola opening upwards with vertex at $(2, -1)$
Step 3: Minimum value is $-1$ at $x=2$
Step 4: As $x \to \pm\infty$, $f(x) \to \infty$
Step 5: Range: $[-1, \infty)$
Example: Find range of $f(x) = \sin x + \cos x$
Step 1: Express as single trigonometric function: $f(x) = \sqrt{2}\sin(x + \frac{\pi}{4})$
Step 2: Range of $\sin$ function is $[-1, 1]$
Step 3: Multiply by $\sqrt{2}$: Range is $[-\sqrt{2}, \sqrt{2}]$
Calculus Approach
Use derivatives to find critical points and determine function behavior.
When to Use:
Differentiable functions where finding extrema is challenging by algebraic methods.
Example: Find range of $f(x) = \frac{x}{x^2 + 1}$
Step 1: Find derivative: $f'(x) = \frac{1-x^2}{(x^2+1)^2}$
Step 2: Set $f'(x) = 0$: $1-x^2=0 \Rightarrow x=\pm1$
Step 3: Evaluate at critical points:
• $f(1) = \frac{1}{2}$
• $f(-1) = -\frac{1}{2}$
Step 4: Check limits: $\lim_{x\to\pm\infty} f(x) = 0$
Step 5: Range: $[-\frac{1}{2}, \frac{1}{2}]$
Example: Find range of $f(x) = x^3 - 3x$ on $[-2, 2]$
Step 1: Find derivative: $f'(x) = 3x^2 - 3 = 3(x^2-1)$
Step 2: Critical points: $x=\pm1$
Step 3: Evaluate:
• $f(-2) = -2$, $f(-1) = 2$, $f(1) = -2$, $f(2) = 2$
Step 4: Range: $[-2, 2]$
Function Transformations
Systematically transform a function with known range to find range of complex functions.
When to Use:
Composite functions, functions with shifts, stretches, or reflections.
Example: Find range of $f(x) = 2\sin(3x - \frac{\pi}{4}) + 1$
Step 1: Start with range of $\sin\theta$: $[-1, 1]$
Step 2: Multiply by 2: Range becomes $[-2, 2]$
Step 3: Add 1: Range becomes $[-1, 3]$
Step 4: Final range: $[-1, 3]$
Example: Find range of $f(x) = \sqrt{4 - x^2}$
Step 1: Domain: $4-x^2 \geq 0 \Rightarrow x \in [-2, 2]$
Step 2: Let $y = \sqrt{4-x^2} \Rightarrow y^2 = 4-x^2 \Rightarrow x^2 = 4-y^2$
Step 3: Since $x^2 \geq 0$, we have $4-y^2 \geq 0 \Rightarrow y^2 \leq 4$
Step 4: Also $y \geq 0$ (square root), so range: $[0, 2]$
🚀 Advanced Problem-Solving Strategies
For Composite Functions:
- Work from inside out
- Find range of inner function first
- Use that as domain for outer function
- Watch for restrictions at each step
For Piecewise Functions:
- Find range for each piece separately
- Take union of all ranges
- Check continuity at boundaries
- Verify if all values are actually attained
Methods 4-5 Available in Full Version
Includes Quadratic Method and Inverse Function Method with detailed examples and practice problems
📝 Quick Self-Test
Try these JEE-level problems to test your understanding:
1. Find range of $f(x) = \frac{x^2 + 1}{x^2 - 1}$
2. Find range of $f(x) = \log_2(x^2 - 4x + 5)$
3. Find range of $f(x) = \sqrt{9 - x^2} + \sqrt{x^2 - 4}$
Ready to Master All 5 Methods?
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