JEE MAIN - Mathematics Hindi (2022 - 29th June Evening Shift - No. 3)
$$\mathop {\lim }\limits_{x \to 1} {{({x^2} - 1){{\sin }^2}(\pi x)} \over {{x^4} - 2{x^3} + 2x - 1}}$$ का मान बराबर है -
$${{{\pi ^2}} \over 6}$$
$${{{\pi ^2}} \over 3}$$
$${{{\pi ^2}} \over 2}$$
$$\pi$$2
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