JEE 2022 — Mathematics PYQ

Q: How do I solve JEE 2022 Mathematics questions?

Practice JEE 2022 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 2022 — Mathematics PYQ

JEE | Mathematics | 2022

The value of $\lim_{x \to 1} \frac{(x^2 - 1)\sin^2(\pi x)}{x^4 - 2x^3 + 2x - 1}$ is equal to:

Choose the correct answer:

Explanation

Solution (for $x \to 1$ )

Factorize Denominator: $x^4 - 2x^3 + 2x - 1 = (x^2 - 1)(x^2 + 1) - 2x(x^2 - 1) = (x^2 - 1)(x^2 - 2x + 1) = (x^2 - 1)(x - 1)^2$ .
Simplify the limit expression:

$\lim_{x \to 1} \frac{(x^2 - 1)\sin^2(\pi x)}{(x^2 - 1)(x - 1)^2} = \lim_{x \to 1} \frac{\sin^2(\pi x)}{(x - 1)^2}$
Substitution: Let $h = x - 1$ . As $x \to 1, h \to 0$ .

$\lim_{h \to 0} \frac{\sin^2(\pi(1 + h))}{h^2} = \lim_{h \to 0} \frac{(-\sin(\pi h))^2}{h^2} = \lim_{h \to 0} \frac{\sin^2(\pi h)}{h^2}$
Standard Limit: $\lim_{h \to 0} (\frac{\sin(\pi h)}{\pi h})^2 \cdot \pi^2 = 1 \cdot \pi^2 = \pi^2$ .

Correct Option: (D) $\pi^2$

Choose the correct answer:

Explore More