JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

If $\int \sqrt{\sec 2x - 1} dx = \alpha \log_e \left| \cos 2x + \beta + \sqrt{\cos 2x (1 + \cos \frac{1}{\beta}x)} \right| + C$ , then $\beta - \alpha$ is equal to ________.

Choose the correct answer:

Explanation

Solution

Step 1: Integral ko simplify karein

I = \int \sqrt{\sec 2x - 1} \, dx = \int \sqrt{\frac{1 - \cos 2x}{\cos 2x}} \, dx

Kyunki $1 - \cos 2x = 2\sin^2 x$ , toh:

I = \int \frac{\sqrt{2} \sin x}{\sqrt{\cos 2x}} \, dx

Step 2: Substitution ka upyog karein

Let $\sqrt{2} \cos x = t$ . Iska derivative hoga $-\sqrt{2} \sin x \, dx = dt$ .

Iske alawa, $\cos 2x = 2\cos^2 x - 1 = t^2 - 1$ .

I = \int \frac{-dt}{\sqrt{t^2 - 1}}

Step 3: Standard Integral formula lagayein

$\int \frac{1}{\sqrt{x^2 - a^2}} \, dx = \log |x + \sqrt{x^2 - a^2}|$ . Toh:

I = -\log_e \left| t + \sqrt{t^2 - 1} \right| + C

$t$ ki value wapas rakhne par:

I = -\log_e \left| \sqrt{2} \cos x + \sqrt{\cos 2x} \right| + C

Step 4: Comparison aur Values nikaalna

Hume is result ko $\cos 2x$ ke terms mein badalna hoga. Calculation aur simplification ke baad values aati hain:

$\alpha = -\frac{1}{2}$
$\beta = \frac{1}{2}$ (Form ko match karne ke liye constant adjustments ke saath)

Step 5: Final Calculation

\beta - \alpha = \frac{1}{2} - \left( -\frac{1}{2} \right)

\beta - \alpha = \frac{1}{2} + \frac{1}{2} = 1

Sahi Option: (D) 1