JEE Main 2022 · 29 Jun · Shift 2 · Mathematics Q5

MathematicsSingle correct+4 / -1Question 5 of 90

The value of $\displaystyle\lim_{x\to1}\frac{\left(x^2-1\right)\sin^2(\pi x)}{x^4-2x^3+2x-1}$ is equal to:

\dfrac{\pi^2}6

\dfrac{\pi^2}3

\dfrac{\pi^2}2

\pi^2

Answer and worked solution

Official answer: D

Factor. $x^4-2x^3+2x-1=(x-1)^3(x+1)$ and $\sin^2\pi x\sim\pi^2(x-1)^2$ .

Limit. $\dfrac{(x-1)(x+1)\pi^2(x-1)^2}{(x-1)^3(x+1)}\to\pi^2$ .

(D)

Topics:Limits

JEE Main 2022 · 29 Jun · Shift 2, Mathematics question 5