JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

If the sum of all the solutions of $\tan^{-1} \left( \frac{2x}{1-x^2} \right) + \cot^{-1} \left( \frac{1-x^2}{2x} \right) = \frac{\pi}{3}, \quad -1 < x < 1, \quad x \neq 0$ is $\alpha - \frac{4}{\sqrt{3}}$ , then $\alpha$ is equal to:

Choose the correct answer:

Explanation

Solution

Step 1: Equation ko simplify karna

Hume di gayi equation hai:

\tan^{-1} \left( \frac{2x}{1-x^2} \right) + \cot^{-1} \left( \frac{1-x^2}{2x} \right) = \frac{\pi}{3}

Hume pata hai ki $\cot^{-1}(\theta) = \tan^{-1}(\frac{1}{\theta})$ jab $\theta > 0$ .

Yahan $\frac{1-x^2}{2x}$ ki value check karte hain. Kyonki $-1 < x < 1$ aur $x \neq 0$ :

Agar $0 < x < 1$ , toh $\frac{1-x^2}{2x} > 0$ .
Agar $-1 < x < 0$ , toh $\frac{1-x^2}{2x} < 0$ .

Case 1: $0 < x < 1$

Equation ban jayegi:

\tan^{-1} \left( \frac{2x}{1-x^2} \right) + \tan^{-1} \left( \frac{2x}{1-x^2} \right) = \frac{\pi}{3}

2 \tan^{-1} \left( \frac{2x}{1-x^2} \right) = \frac{\pi}{3}

\tan^{-1} \left( \frac{2x}{1-x^2} \right) = \frac{\pi}{6}

Hume pata hai $\tan^{-1} \left( \frac{2x}{1-x^2} \right) = 2\tan^{-1}(x)$ (for $|x| < 1$ ):

2\tan^{-1}(x) = \frac{\pi}{6} \implies \tan^{-1}(x) = \frac{\pi}{12}

x_1 = \tan(15^\circ) = 2 - \sqrt{3}

Case 2: $-1 < x < 0$

Is case mein $\cot^{-1}(\theta) = \pi + \tan^{-1}(\frac{1}{\theta})$ hota hai:

\tan^{-1} \left( \frac{2x}{1-x^2} \right) + \pi + \tan^{-1} \left( \frac{2x}{1-x^2} \right) = \frac{\pi}{3}

2 \tan^{-1} \left( \frac{2x}{1-x^2} \right) = \frac{\pi}{3} - \pi = -\frac{2\pi}{3}

\tan^{-1} \left( \frac{2x}{1-x^2} \right) = -\frac{\pi}{3}

2\tan^{-1}(x) = -\frac{\pi}{3} \implies \tan^{-1}(x) = -\frac{\pi}{6}

x_2 = \tan(-30^\circ) = -\frac{1}{\sqrt{3}}

Step 2: Sum of solutions aur $\alpha$ nikalna

Solutions ka sum:

\text{Sum} = x_1 + x_2 = (2 - \sqrt{3}) + \left( -\frac{1}{\sqrt{3}} \right)

\text{Sum} = 2 - \sqrt{3} - \frac{1}{\sqrt{3}} = 2 - \frac{3+1}{\sqrt{3}} = 2 - \frac{4}{\sqrt{3}}

Question ke anusar, sum $\alpha - \frac{4}{\sqrt{3}}$ hai:

\alpha - \frac{4}{\sqrt{3}} = 2 - \frac{4}{\sqrt{3}}

\alpha = 2