JEE 2023 — Mathematics PYQ

Q: How do I solve JEE 2023 Mathematics questions?

Practice JEE 2023 Mathematics previous year questions with step-by-step solutions on Mathem Solvex. Each question includes a detailed explanation and video solution to help you master the concept quickly.

Q: What is the difficulty level of JEE Mathematics questions?

JEE Mathematics questions range from moderate to advanced difficulty. Consistent practice with official previous year papers and topic-wise drills on Mathem Solvex is the most effective preparation strategy.

Q: Are JEE previous year papers available for free?

Yes. Mathem Solvex provides free access to JEE previous year question papers and topic-wise PDFs. You can download them or attempt them as timed live mock tests directly on the platform.

Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

If $\phi(x) = \frac{1}{\sqrt{x}} \int_{\frac{\pi}{4}}^{x} (4\sqrt{2} \sin t - 3\phi'(t)) \, dt, x > 0$ , then $ϕ^{'} (\frac{π}{4})$ is equal to :

Choose the correct answer:

Explanation

Solution Steps:

Equation ko rearrange karein: $\sqrt{x}\phi(x) = \int_{\frac{\pi}{4}}^{x} (4\sqrt{2} \sin t - 3\phi'(t)) \, dt$ .

Dono side ko $x$ ke respect mein differentiate karein (Leibniz Rule ka use karke):

\frac{1}{2\sqrt{x}}\phi(x) + \sqrt{x}\phi'(x) = 4\sqrt{2}\sin x - 3\phi'(x)

Ab $x = \frac{\pi}{4}$ rakhein. Hume pata hai ki $\phi(\frac{\pi}{4}) = 0$ (kyuki integral ki limits same ho jayengi).

Value rakhne par: $0 + \sqrt{\frac{\pi}{4}}\phi'(\frac{\pi}{4}) = 4\sqrt{2}\sin(\frac{\pi}{4}) - 3\phi'(\frac{\pi}{4})$

Simplify karein: $\frac{\sqrt{\pi}}{2}\phi'(\frac{\pi}{4}) + 3\phi'(\frac{\pi}{4}) = 4\sqrt{2}(\frac{1}{\sqrt{2}}) = 4$ .

$\phi'(\frac{\pi}{4}) \left[\frac{\sqrt{\pi} + 6}{2}\right] = 4 \implies \phi'(\frac{\pi}{4}) = \frac{8}{6 + \sqrt{\pi}}$ .

Sahi Option: (1)

Choose the correct answer:

Explore More