JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $S = \left\{ x \in (0, 1) : 2\tan^{-1}\left(\frac{1-x}{1+x}\right) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \right\}$ . If $n(S)$ is the number of elements in $S$ , then:

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Explanation

Solution

Let $x = \tan \theta$ . Since $x \in (0, 1)$ , $\theta \in (0, \pi/4)$ .

LHS: $2\tan^{-1}\left(\frac{1-\tan \theta}{1+\tan \theta}\right) = 2\tan^{-1}(\tan(\frac{\pi}{4}-\theta)) = 2(\frac{\pi}{4}-\theta) = \frac{\pi}{2} - 2\theta$ .
RHS: $\cos^{-1}\left(\frac{1-\tan^2 \theta}{1+\tan^2 \theta}\right) = \cos^{-1}(\cos 2\theta) = 2\theta$ .
Equating LHS and RHS: $\frac{\pi}{2} - 2\theta = 2\theta \implies 4\theta = \frac{\pi}{2} \implies \theta = \frac{\pi}{8}$ .
Find $x$ : $x = \tan(\frac{\pi}{8}) = \sqrt{2} - 1 \approx 1.414 - 1 = 0.414$ .

Since $0.414 < 0.5$ , $n(S) = 1$ and the element is less than $1/2$ .

Correct Option: (4)

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