JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

S = \left\{ x \in \mathbb{R} : \sin^{-1} \left( \frac{x + 1}{\sqrt{x^2 + 2x + 2}} \right) - \sin^{-1} \left( \frac{x}{\sqrt{x^2 + 1}} \right) = \frac{\pi}{4} \right\}

Choose the correct answer:

Explanation

Step 1: Substitution (Maan lena)

Equation ke dono terms ko dhyan se dekhiye, ye $\frac{\text{perpendicular}}{\text{hypotenuse}}$ ke form mein hain.

Maan lijiye $x+1 = \tan \theta$ .

Toh $\sqrt{x^2+2x+2} = \sqrt{(x+1)^2+1} = \sqrt{\tan^2 \theta + 1} = \sec \theta$ .

Isliye, $\frac{x+1}{\sqrt{x^2+2x+2}} = \frac{\tan \theta}{\sec \theta} = \sin \theta$ .

Matlab: $\sin^{-1} \left( \frac{x+1}{\sqrt{x^2+2x+2}} \right) = \theta = \tan^{-1}(x+1)$ .
Maan lijiye $x = \tan \phi$ .

Toh $\sqrt{x^2+1} = \sqrt{\tan^2 \phi + 1} = \sec \phi$ .

Isliye, $\frac{x}{\sqrt{x^2+1}} = \frac{\tan \phi}{\sec \phi} = \sin \phi$ .

Matlab: $\sin^{-1} \left( \frac{x}{\sqrt{x^2+1}} \right) = \phi = \tan^{-1}(x)$ .

Step 2: Equation ko simplify karna

Ab hamari equation ban gayi:

\tan^{-1}(x+1) - \tan^{-1}(x) = \frac{\pi}{4}

Yahan formula lagaiye: $\tan^{-1} A - \tan^{-1} B = \tan^{-1} \left( \frac{A-B}{1+AB} \right)$

\tan^{-1} \left( \frac{(x+1) - x}{1 + (x+1)x} \right) = \frac{\pi}{4}

\tan^{-1} \left( \frac{1}{1 + x^2 + x} \right) = \frac{\pi}{4}

Step 3: $x$ ki value nikaalna

Dono taraf tan lene par:

\frac{1}{x^2 + x + 1} = \tan\left(\frac{\pi}{4}\right)

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

Iska matlab:

x^2 + x + 1 = 1

x^2 + x = 0

x(x + 1) = 0

Toh hamare paas $x$ ki do values aati hain:

$x = 0$ ya $x = -1$