QJEEM21Ma3deb: JEE Main 2021 Mathematics - Class 12 JEEM21Ma3deb Solution: Let R be such that the function f ( x ) = { cos − 1 ( 1 − { x } …

Question

Verified Solution • 2025–26 Session moderate

Let

\alpha

\in

R be such that the function

f (x) = {\begin{matrix} \frac{\cos^{- 1} (1 - {x}^{2}) \sin^{- 1} (1 - {x})}{{x} - {x}^{3}}, & x \neq 0 \\ α, & x = 0 \end{matrix}

is continuous at x = 0, where {x} = x

-

[ x ] is the greatest integer less than or equal to x. Then :

Accepted Answer

RHL = \mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}(1 - {x^2}){{\sin }^{ - 1}}(1 - x)} \over {x(1 - {x^2})}}

= {\pi \over 2}\mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}(1 - {x^2})} \over x}

= {\pi \over 2}\mathop {\lim }\limits_{x \to {0^ + }} {{ - 1} \over {\sqrt {1 - {{(1 - {x^2})}^2}} }}( - 2x)

(L' Hospital Rule)

= \pi \mathop {\lim }\limits_{x \to {0^ + }} {x \over {\sqrt {2{x^2} - {x^4}} }} = \pi \mathop {\lim }\limits_{x \to {0^ + }} {1 \over {\sqrt {2 - {x^2}} }} = {\pi \over {\sqrt 2 }}

LHL = \mathop {\lim }\limits_{x \to {0^ - }} {{{{\cos }^{ - 1}}(1 - {{(1 + x)}^2}){{\sin }^{ - 1}}( - x)} \over {(1 + x) - {{(1 + x)}^3}}}

= {\pi \over 2}\mathop {\lim }\limits_{x \to {0^ - }} {{{{\sin }^{ - 1}}x} \over {(1 - x)\left[ {{{(1 + x)}^2} - 1} \right]}} = {\pi \over 2}\mathop {\lim }\limits_{x \to {0^ - }} {{{{\sin }^{ - 1}}x} \over {{x^2} + 2x}}

= {\pi \over 2}\left( {{1 \over 2}} \right) = {\pi \over 4}

As LHL

\ne

RHL so f(x) is not continuous at x = 0

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