JEE MAIN - Mathematics Bengali (2022 - 26th June Morning Shift - No. 5)
$$\mathop {\lim }\limits_{x \to {1 \over {\sqrt 2 }}} \,{{\sin ({{\cos }^{ - 1}}x) - x} \over {1 - \tan ({{\cos }^{ - 1}}x)}}$$ সমান :
$$\sqrt 2 $$
$$ - \sqrt 2 $$
$${1 \over {\sqrt 2 }}$$
$$ - {1 \over {\sqrt 2 }}$$
Comments (0)
