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

Let $f(x) = \int \frac{2x}{(x^2 + 1)(x^2 + 3)} dx$ . If $f(3) = \frac{1}{2}(\log_e 5 - \log_e 6)$ , then $f(4)$ is equal to:

Choose the correct answer:

Explanation

Solution

Integration by Substitution:

Let $x^2 = t \implies 2x \, dx = dt$

f(x) = \int \frac{1}{(t + 1)(t + 3)} dt

Partial Fractions:

$\frac{1}{(t+1)(t+3)} = \frac{1}{2} \left[ \frac{1}{t+1} - \frac{1}{t+3} \right]$

Integrating:

f(x) = \frac{1}{2} \int \left( \frac{1}{t+1} - \frac{1}{t+3} \right) dt

f(x) = \frac{1}{2} [\log_e(t+1) - \log_e(t+3)] + C

f(x) = \frac{1}{2} [\log_e(x^2+1) - \log_e(x^2+3)] + C

Finding Constant $C$ :

Given $f(3) = \frac{1}{2}(\log_e 5 - \log_e 6)$

$\frac{1}{2} [\log_e(3^2+1) - \log_e(3^2+3)] + C = \frac{1}{2}(\log_e 5 - \log_e 6)$

$\frac{1}{2} [\log_e 10 - \log_e 12] + C = \frac{1}{2}(\log_e 5 - \log_e 6)$

$\frac{1}{2} \log_e \left( \frac{10}{12} \right) + C = \frac{1}{2} \log_e \left( \frac{5}{6} \right)$

$\frac{1}{2} \log_e \left( \frac{5}{6} \right) + C = \frac{1}{2} \log_e \left( \frac{5}{6} \right) \implies C = 0$

Calculating $f(4)$ :

f(4) = \frac{1}{2} [\log_e(4^2+1) - \log_e(4^2+3)]

f(4) = \frac{1}{2} (\log_e 17 - \log_e 19)

Correct Option: (4)