JEE Advanced: The Toughest Domain & Range Problems Solved

Step 1: Innermost constraint

For $\sqrt{1 - x^2}$ to be defined:

Step 2: Outer square root constraint

For the main square root to be defined:

Multiply both sides by 2 (positive):

Step 3: Analyze the inequality

Since $\sqrt{1 - x^2}$ is always non-negative and for $x \in [-1, 1]$, we have $0 \leq \sqrt{1 - x^2} \leq 1$.

The inequality $\sqrt{1 - x^2} \leq 1$ is always true for $x \in [-1, 1]$.

Step 4: Check for equality edge case

When $\sqrt{1 - x^2} = 1$, we get $x = 0$, and the function becomes:

So $x = 0$ is included in the domain.

Step 1: Innermost log constraint

For $\log_{1/2}(x^2 - 5x + 6)$ to be defined:

Factor: $(x - 2)(x - 3) > 0$

Solution: $x < 2$ or $x > 3$

Step 2: Middle log constraint

For $\log_3(\log_{1/2}(x^2 - 5x + 6))$ to be defined:

Since base $1/2 < 1$, the inequality reverses:

So: $x^2 - 5x + 6 < 1 \Rightarrow x^2 - 5x + 5 < 0$

Step 3: Solve quadratic inequality

Roots of $x^2 - 5x + 5 = 0$:

So $x^2 - 5x + 5 < 0$ when:

Step 4: Combine all constraints

We need:

1. $x < 2$ or $x > 3$ (from Step 1)
2. $\frac{5 - \sqrt{5}}{2} < x < \frac{5 + \sqrt{5}}{2}$ (from Step 3)

Numerically: $\frac{5 - \sqrt{5}}{2} \approx 1.38$, $\frac{5 + \sqrt{5}}{2} \approx 3.62$

Intersection: $(1.38, 2) \cup (3, 3.62)$

Step 1: First square root constraint

For $\sqrt{\sin x - \cos x}$ to be defined:

Divide by $\sqrt{2}$:

This is: $\sin\left(x - \frac{\pi}{4}\right) \geq 0$

Solution: $2n\pi \leq x - \frac{\pi}{4} \leq (2n+1)\pi$

So: $2n\pi + \frac{\pi}{4} \leq x \leq (2n+1)\pi + \frac{\pi}{4}$

Step 2: Second square root constraint

For $\sqrt{\frac{1}{\tan x} - 1}$ to be defined:

Simplify: $\frac{1 - \tan x}{\tan x} \geq 0$

This is a rational inequality. Critical points: $\tan x = 0$ and $\tan x = 1$

Step 3: Solve the rational inequality

Sign chart for $\frac{1 - \tan x}{\tan x}$:

• When $0 < \tan x < 1$: numerator positive, denominator positive → positive
• When $\tan x > 1$: numerator negative, denominator positive → negative
• When $\tan x < 0$: numerator positive, denominator negative → negative

So $\frac{1 - \tan x}{\tan x} \geq 0$ when $0 < \tan x \leq 1$

Step 4: Combine both constraints

We need both:

1. $2n\pi + \frac{\pi}{4} \leq x \leq (2n+1)\pi + \frac{\pi}{4}$
2. $0 < \tan x \leq 1$

The intersection occurs in intervals where both conditions are satisfied simultaneously.

Final domain: $x \in \left[2n\pi + \frac{\pi}{4}, 2n\pi + \frac{\pi}{2}\right) \cup \left(2n\pi + \frac{\pi}{2}, 2n\pi + \frac{5\pi}{4}\right]$

Step 1: Domain of inverse sine

For $\sin^{-1}(y)$ to be defined, we need $-1 \leq y \leq 1$

So: $-1 \leq \frac{1 + x^2}{2x} \leq 1$

Step 2: Solve the inequality

We need to solve two inequalities:

1. $\frac{1 + x^2}{2x} \geq -1$
2. $\frac{1 + x^2}{2x} \leq 1$

Step 3: First inequality

$\frac{1 + x^2}{2x} \geq -1$

Multiply by $2x$ (but sign matters):

• If $x > 0$: $1 + x^2 \geq -2x \Rightarrow x^2 + 2x + 1 \geq 0 \Rightarrow (x+1)^2 \geq 0$ (always true)
• If $x < 0$: $1 + x^2 \leq -2x \Rightarrow x^2 + 2x + 1 \leq 0 \Rightarrow (x+1)^2 \leq 0 \Rightarrow x = -1$

So from first inequality: $x > 0$ or $x = -1$

Step 4: Second inequality

$\frac{1 + x^2}{2x} \leq 1$

Multiply by $2x$ (sign matters):

• If $x > 0$: $1 + x^2 \leq 2x \Rightarrow x^2 - 2x + 1 \leq 0 \Rightarrow (x-1)^2 \leq 0 \Rightarrow x = 1$
• If $x < 0$: $1 + x^2 \geq 2x \Rightarrow x^2 - 2x + 1 \geq 0 \Rightarrow (x-1)^2 \geq 0$ (always true)

So from second inequality: $x = 1$ or $x < 0$

Step 5: Combine results

From Step 3: $x > 0$ or $x = -1$

From Step 4: $x = 1$ or $x < 0$

Intersection: $x = 1$ or $x = -1$

Step 1: Square root constraint

For $\sqrt{|x| - x}$ to be defined:

Step 2: Denominator constraint

Since it's in denominator:

Step 3: Analyze by cases

Case 1: $x \geq 0$

Then $|x| = x$, so $|x| - x = 0$

But this violates denominator constraint. So no $x \geq 0$.

Case 2: $x < 0$

Then $|x| = -x$, so $|x| - x = -x - x = -2x$

Since $x < 0$, $-2x > 0$, so $\sqrt{|x| - x}$ is defined and non-zero.

Step 1: Innermost function constraint

For $\sin^{-1}\left(\frac{x^2 - 1}{x^2 + 1}\right)$ to be defined:

Since denominator $x^2 + 1 > 0$ for all $x$, we can analyze directly.

Step 2: Analyze the bounds

Note that $\frac{x^2 - 1}{x^2 + 1} = 1 - \frac{2}{x^2 + 1}$

Since $x^2 + 1 \geq 1$, we have $\frac{2}{x^2 + 1} \leq 2$

So: $1 - \frac{2}{x^2 + 1} \geq 1 - 2 = -1$

Also: $1 - \frac{2}{x^2 + 1} \leq 1 - 0 = 1$

So automatically: $-1 \leq \frac{x^2 - 1}{x^2 + 1} \leq 1$ for all real $x$.

Step 3: Logarithm constraint

For $\log_2\left(\sin^{-1}\left(\frac{x^2 - 1}{x^2 + 1}\right)\right)$ to be defined:

Since $\sin^{-1}(y) > 0$ when $y > 0$:

Step 4: Solve the inequality

$\frac{x^2 - 1}{x^2 + 1} > 0$

Since denominator $x^2 + 1 > 0$ always, this simplifies to:

Step 5: Find the range

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

When $x \to \pm\infty$, $y \to 1^-$

When $x \to 1^+$ or $x \to -1^-$, $y \to 0^+$

So $y \in (0, 1)$ for $x \in (-\infty, -1) \cup (1, \infty)$

Then $\sin^{-1}(y) \in (0, \frac{\pi}{2})$

Finally, $\log_2(\sin^{-1}(y)) \in (-\infty, \log_2(\frac{\pi}{2}))$