JEE MAIN - Mathematics (2021 - 17th March Morning Shift - No. 2)
The value of
$$\mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}(x - {{[x]}^2}).{{\sin }^{ - 1}}(x - {{[x]}^2})} \over {x - {x^3}}}$$, where [ x ] denotes the greatest integer $$ \le $$ x is :
$$\mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}(x - {{[x]}^2}).{{\sin }^{ - 1}}(x - {{[x]}^2})} \over {x - {x^3}}}$$, where [ x ] denotes the greatest integer $$ \le $$ x is :
$$\pi$$
$${\pi \over 4}$$
$${\pi \over 2}$$
0
Explicació
$$\mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}\left( {x - {{[x]}^2}} \right).{{\sin }^{ - 1}}\left( {x - {{[x]}^2}} \right)} \over {x - {x^3}}}$$
$$ = \mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}x} \over {1 - {x^2}}}.{{{{\sin }^{ - 1}}x} \over x}$$
$$ = {\cos ^{ - 1}}0 = {\pi \over 2}$$
$$ = \mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}x} \over {1 - {x^2}}}.{{{{\sin }^{ - 1}}x} \over x}$$
$$ = {\cos ^{ - 1}}0 = {\pi \over 2}$$