JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

The value of the integral $\int_{1/2}^{2} \frac{t a n ^{- 1} x}{x} d x$ is equal to:

Choose the correct answer:

Explanation

Solution

Maana ki integral $I$ hai:

I = \int_{1/2}^{2} \frac{\tan^{-1} x}{x} dx \quad \dots(1)

Ab hum substitution ka upyog karenge. Let $x = \frac{1}{t}$ , toh $dx = -\frac{1}{t^2} dt$ .

Limits change karte hain:

Jab $x = 1/2$ , toh $t = 2$ .
Jab $x = 2$ , toh $t = 1/2$ .

Integral ban jayega:

I = \int_{2}^{1/2} \frac{\tan^{-1}(1/t)}{1/t} \left( -\frac{1}{t^2} \right) dt

I = \int_{1/2}^{2} \frac{\tan^{-1}(1/t)}{t} dt

Hume pata hai ki $\tan^{-1}(1/t) = \cot^{-1}(t)$ , aur variable badalne se integral ki value nahi badalti ( $t$ ko wapas $x$ likh sakte hain):

I = \int_{1/2}^{2} \frac{\cot^{-1} x}{x} dx \quad \dots(2)

Ab equation (1) aur (2) ko jodte hain:

I + I = \int_{1/2}^{2} \frac{\tan^{-1} x}{x} dx + \int_{1/2}^{2} \frac{\cot^{-1} x}{x} dx

2I = \int_{1/2}^{2} \frac{\tan^{-1} x + \cot^{-1} x}{x} dx

Hume trigonometric identity pata hai: $\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$ .

2I = \int_{1/2}^{2} \frac{\pi/2}{x} dx

2I = \frac{\pi}{2} \int_{1/2}^{2} \frac{1}{x} dx

Integration karte hain:

2I = \frac{\pi}{2} [\log_e x]_{1/2}^{2}

2I = \frac{\pi}{2} [\log_e 2 - \log_e(1/2)]

2I = \frac{\pi}{2} [\log_e 2 - (-\log_e 2)]

2I = \frac{\pi}{2} [2\log_e 2]

2I = \pi \log_e 2

I = \frac{\pi}{2} \log_e 2

Correct Option: (1)