1
JEE Main 2020 (Online) 7th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Let y = y(x) be a function of x satisfying

$$y\sqrt {1 - {x^2}} = k - x\sqrt {1 - {y^2}}$$ where k is a constant and

$$y\left( {{1 \over 2}} \right) = - {1 \over 4}$$. Then $${{dy} \over {dx}}$$ at x = $${1 \over 2}$$, is equal to :
A
$${2 \over {\sqrt 5 }}$$
B
$$- {{\sqrt 5 } \over 2}$$
C
$${{\sqrt 5 } \over 2}$$
D
$$- {{\sqrt 5 } \over 4}$$
2
JEE Main 2020 (Online) 7th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Let xk + yk = ak, (a, k > 0 ) and $${{dy} \over {dx}} + {\left( {{y \over x}} \right)^{{1 \over 3}}} = 0$$, then k is:
A
$${1 \over 3}$$
B
$${2 \over 3}$$
C
$${4 \over 3}$$
D
$${3 \over 2}$$
3
JEE Main 2020 (Online) 7th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
If $$y\left( \alpha \right) = \sqrt {2\left( {{{\tan \alpha + \cot \alpha } \over {1 + {{\tan }^2}\alpha }}} \right) + {1 \over {{{\sin }^2}\alpha }}} ,\alpha \in \left( {{{3\pi } \over 4},\pi } \right)$$

$${{dy} \over {d\alpha }}\,\,at\,\alpha = {{5\pi } \over 6}is$$ :
A
4
B
-4
C
$${4 \over 3}$$
D
-$${1 \over 4}$$
4
JEE Main 2019 (Online) 12th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
The derivative of $${\tan ^{ - 1}}\left( {{{\sin x - \cos x} \over {\sin x + \cos x}}} \right)$$, with respect to $${x \over 2}$$ , where $$\left( {x \in \left( {0,{\pi \over 2}} \right)} \right)$$ is :
A
1
B
2
C
$${2 \over 3}$$
D
$${1 \over 2}$$
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