The Square Root Trick for Range | JEE Secret Weapon

Why This Trick is a Game-Changer

Traditional range-finding methods can be time-consuming and messy. The discriminant method gives you a systematic, foolproof approach that works for a wide variety of functions commonly asked in JEE.

🎯 The Core Idea

For a function $y = f(x)$, if we can rearrange it into a quadratic in x, then:

Real x exists ⇔ Discriminant $D \geq 0$

Solving $D \geq 0$ gives us the range of y-values!

The 5-Step Discriminant Method

Step 1: Set y = f(x)

Start with the equation $y = f(x)$

Step 2: Rearrange to Quadratic Form

Express the equation as $Ax^2 + Bx + C = 0$, where A, B, C may contain y

Step 3: Apply Discriminant Condition

For real x to exist: $D = B^2 - 4AC \geq 0$

Step 4: Solve the Inequality

Solve $B^2 - 4AC \geq 0$ to find possible y-values

Step 5: Verify Boundary Values

Check if the boundary y-values are actually attainable

Example 1: Classic Rational Function

Find the range of $f(x) = \frac{x}{x^2 + 1}$

Step 1: Set y = f(x)

$y = \frac{x}{x^2 + 1}$

Step 2: Rearrange to Quadratic

$y(x^2 + 1) = x$

$yx^2 + y - x = 0$

$yx^2 - x + y = 0$

This is quadratic in x with:

$A = y$, $B = -1$, $C = y$

Step 3: Apply Discriminant Condition

$D = B^2 - 4AC \geq 0$

$(-1)^2 - 4(y)(y) \geq 0$

$1 - 4y^2 \geq 0$

Step 4: Solve the Inequality

$1 - 4y^2 \geq 0$

$4y^2 \leq 1$

$y^2 \leq \frac{1}{4}$

$-\frac{1}{2} \leq y \leq \frac{1}{2}$

Step 5: Verify Boundary Values

When $y = \frac{1}{2}$: $\frac{1}{2} = \frac{x}{x^2 + 1}$ gives $x = 1$ ✓

When $y = -\frac{1}{2}$: $-\frac{1}{2} = \frac{x}{x^2 + 1}$ gives $x = -1$ ✓

✅ Final Answer

Range = $\left[-\frac{1}{2}, \frac{1}{2}\right]$

🎯 When to Use the Discriminant Method

Perfect Candidates:

Rational functions: $\frac{P(x)}{Q(x)}$
Functions with $x$ and $x^2$ terms
Expressions where you can isolate $x$
Square root functions (after squaring)
Trigonometric functions (using identities)

Not Suitable For:

Functions that don't form quadratics
When domain is severely restricted
Piecewise defined functions
Functions with vertical asymptotes only

Example 2: Square Root Function

Find the range of $f(x) = \sqrt{x^2 + x + 1}$

Step 1: Set y = f(x)

$y = \sqrt{x^2 + x + 1}$, where $y \geq 0$

Step 2: Rearrange to Quadratic

$y^2 = x^2 + x + 1$ (squaring both sides)

$x^2 + x + 1 - y^2 = 0$

Quadratic in x with:

$A = 1$, $B = 1$, $C = 1 - y^2$

Step 3: Apply Discriminant Condition

$D = B^2 - 4AC \geq 0$

$1^2 - 4(1)(1 - y^2) \geq 0$

$1 - 4 + 4y^2 \geq 0$

$4y^2 - 3 \geq 0$

Step 4: Solve the Inequality

$4y^2 - 3 \geq 0$

$y^2 \geq \frac{3}{4}$

$y \leq -\frac{\sqrt{3}}{2}$ or $y \geq \frac{\sqrt{3}}{2}$

Step 5: Combine with Initial Condition

From Step 1: $y \geq 0$

From Step 4: $y \geq \frac{\sqrt{3}}{2}$ (since $y \geq 0$)

Check $y = \frac{\sqrt{3}}{2}$: $x^2 + x + 1 = \frac{3}{4}$ gives real x ✓

✅ Final Answer

Range = $\left[\frac{\sqrt{3}}{2}, \infty\right)$

⚠️ Important Caveats & Precautions

1. Domain Restrictions Still Apply

The discriminant method finds possible y-values. You must still consider the original domain of the function.

2. Watch for Extraneous Solutions

When squaring (like in Example 2), verify that solutions satisfy the original equation.

3. A ≠ 0 Condition

If A = 0, the equation becomes linear. Check this case separately!

4. Boundary Value Verification

Always check if the boundary y-values are actually attained by some x in the domain.

Special Case: When A = 0

Find the range of $f(x) = \frac{x+1}{x^2 + x + 1}$

Step 1: Set y = f(x)

$y = \frac{x+1}{x^2 + x + 1}$

Step 2: Rearrange

$y(x^2 + x + 1) = x + 1$

$yx^2 + yx + y - x - 1 = 0$

$yx^2 + (y-1)x + (y-1) = 0$

Step 3: Check A = 0 Case

When $y = 0$: Equation becomes $-x - 1 = 0$ ⇒ $x = -1$ ✓

So $y = 0$ is in range

Step 4: Apply Discriminant for y ≠ 0

$D = (y-1)^2 - 4y(y-1) \geq 0$

$(y-1)[(y-1) - 4y] \geq 0$

$(y-1)(-3y-1) \geq 0$

Step 5: Solve Inequality

Critical points: $y = 1$, $y = -\frac{1}{3}$

Solution: $-\frac{1}{3} \leq y \leq 1$

Step 6: Combine Both Cases

From A = 0 case: $y = 0$

From discriminant: $-\frac{1}{3} \leq y \leq 1$

Since $0 \in [-\frac{1}{3}, 1]$, range is $[-\frac{1}{3}, 1]$

✅ Final Answer

Range = $\left[-\frac{1}{3}, 1\right]$

📐 The Discriminant Method Formula

For $y = f(x)$ rearranged as $Ax^2 + Bx + C = 0$

Range = $\{y \in \mathbb{R} : B^2 - 4AC \geq 0\}$

(Plus any additional constraints from the original function)

🔥 Practice These JEE-Level Problems

1. Find range of $f(x) = \frac{x^2 + 1}{x^2 + x + 1}$

Hint: Multiply both sides and form quadratic in x

2. Find range of $f(x) = \frac{x}{x^2 + 4}$

Hint: Watch for the A = 0 case

3. Find range of $f(x) = \sqrt{\frac{x-1}{x+2}}$

Hint: Square and rearrange carefully

4. Find range of $f(x) = \frac{2x}{x^2 + 3}$

Hint: Classic discriminant application

Time Yourself: Try to solve each problem in under 2 minutes using the discriminant method!

💡 Pro Tips for JEE Success

Speed Optimization:

Practice mental rearrangement to quadratic form
Memorize common discriminant patterns
Use quick inequality solving techniques
Verify only suspicious boundary values

Common Pitfalls:

Forgetting the A = 0 special case
Missing domain restrictions
Not verifying boundary values
Algebra errors in rearrangement

Master This Time-Saving Technique!

The discriminant method can save you 3-5 minutes on range problems in JEE

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