1
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
$${(1 + x)^2}f'(x) - 2{(f(x))^2} = 0$$
2
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
$${{\sqrt 3 } \over {12}}$$
3
JEE Main 2020 (Online) 4th September Morning Slot
+4
-1
If $$\left( {a + \sqrt 2 b\cos x} \right)\left( {a - \sqrt 2 b\cos y} \right) = {a^2} - {b^2}$$

where a > b > 0, then $${{dx} \over {dy}}\,\,at\left( {{\pi \over 4},{\pi \over 4}} \right)$$ is :
A
$${{a - 2b} \over {a + 2b}}$$
B
$${{a - b} \over {a + b}}$$
C
$${{a + b} \over {a - b}}$$
D
$${{2a + b} \over {2a - b}}$$
4
JEE Main 2020 (Online) 3rd September Morning Slot
+4
-1
If y2 + loge (cos2x) = y,
$$x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$$, then :
A
|y''(0)| = 2
B
|y'(0)| + |y''(0)| = 3
C
y''(0) = 0
D
|y'(0)| + |y"(0)| = 1
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