JEE MAIN - Mathematics (2021 - 16th March Evening Shift - No. 12)
Let $$\alpha$$ $$\in$$ R be such that the function $$f(x) = \left\{ {\matrix{
{{{{{\cos }^{ - 1}}(1 - {{\{ x\} }^2}){{\sin }^{ - 1}}(1 - \{ x\} )} \over {\{ x\} - {{\{ x\} }^3}}},} & {x \ne 0} \cr
{\alpha ,} & {x = 0} \cr
} } \right.$$ is continuous at x = 0, where {x} = x $$-$$ [ x ] is the greatest integer less than or equal to x. Then :
no such $$\alpha$$ exists
$$\alpha$$ = 0
$$\alpha$$ = $${\pi \over 4}$$
$$\alpha$$ = $${\pi \over {\sqrt 2 }}$$
Explanation
$$RHL = \mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}(1 - {x^2}){{\sin }^{ - 1}}(1 - x)} \over {x(1 - {x^2})}} $$
$$= {\pi \over 2}\mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}(1 - {x^2})} \over x}$$
$$ = {\pi \over 2}\mathop {\lim }\limits_{x \to {0^ + }} {{ - 1} \over {\sqrt {1 - {{(1 - {x^2})}^2}} }}( - 2x)$$ (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
$$= {\pi \over 2}\mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}(1 - {x^2})} \over x}$$
$$ = {\pi \over 2}\mathop {\lim }\limits_{x \to {0^ + }} {{ - 1} \over {\sqrt {1 - {{(1 - {x^2})}^2}} }}( - 2x)$$ (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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