1
JEE Main 2022 (Online) 24th June Evening Shift
+4
-1

If $$y = {\tan ^{ - 1}}\left( {\sec {x^3} - \tan {x^3}} \right),{\pi \over 2} < {x^3} < {{3\pi } \over 2}$$, then

A
$$xy'' + 2y' = 0$$
B
$${x^2}y'' - 6y + {{3\pi } \over 2} = 0$$
C
$${x^2}y'' - 6y + 3\pi = 0$$
D
$$xy'' - 4y' = 0$$
2
JEE Main 2021 (Online) 27th August Evening Shift
+4
-1
If $$y(x) = {\cot ^{ - 1}}\left( {{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} } \over {\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}} \right),x \in \left( {{\pi \over 2},\pi } \right)$$, then $${{dy} \over {dx}}$$ at $$x = {{5\pi } \over 6}$$ is :
A
$$- {1 \over 2}$$
B
$$-$$1
C
$${1 \over 2}$$
D
0
3
JEE Main 2021 (Online) 26th August Morning Shift
+4
-1
Let $$f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {{{1 - x} \over x}} } \right)} \right)$$, 0 < x < 1. Then :
A
$${(1 - x)^2}f'(x) - 2{(f(x))^2} = 0$$
B
$${(1 + x)^2}f'(x) + 2{(f(x))^2} = 0$$
C
$${(1 - x)^2}f'(x) + 2{(f(x))^2} = 0$$
D
$${(1 + x)^2}f'(x) - 2{(f(x))^2} = 0$$
4
JEE Main 2020 (Online) 5th September Evening Slot
+4
-1
The derivative of
$${\tan ^{ - 1}}\left( {{{\sqrt {1 + {x^2}} - 1} \over x}} \right)$$ with
respect to $${\tan ^{ - 1}}\left( {{{2x\sqrt {1 - {x^2}} } \over {1 - 2{x^2}}}} \right)$$ at x = $${1 \over 2}$$ is :
A
$${{2\sqrt 3 } \over 3}$$
B
$${{2\sqrt 3 } \over 5}$$
C
$${{\sqrt 3 } \over {10}}$$
D
$${{\sqrt 3 } \over {12}}$$
EXAM MAP
Medical
NEET