1
JEE Main 2017 (Online) 9th April Morning Slot
+4
-1
If 2x = y$${^{{1 \over 5}}}$$ + y$${^{ - {1 \over 5}}}$$ and

(x2 $$-$$ 1) $${{{d^2}y} \over {d{x^2}}}$$ + $$\lambda$$x $${{dy} \over {dx}}$$ + ky = 0,

then $$\lambda$$ + k is equal to :
A
$$-$$ 23
B
$$-$$ 24
C
26
D
$$-$$ 26
2
JEE Main 2017 (Online) 8th April Morning Slot
+4
-1
The curve satisfying the differential equation, ydx $$-$$(x + 3y2)dy = 0 and passing through the point (1, 1), also passes through the point :
A
$$\left( {{1 \over 4}, - {1 \over 2}} \right)$$
B
$$\left( { - {1 \over 3},{1 \over 3}} \right)$$
C
$$\left( {{1 \over 3}, - {1 \over 3}} \right)$$
D
$$\left( {{1 \over 4}, {1 \over 2}} \right)$$
3
JEE Main 2017 (Offline)
+4
-1
If $$\left( {2 + \sin x} \right){{dy} \over {dx}} + \left( {y + 1} \right)\cos x = 0$$ and y(0) = 1,

then $$y\left( {{\pi \over 2}} \right)$$ is equal to :
A
$$- {2 \over 3}$$
B
$$- {1 \over 3}$$
C
$${4 \over 3}$$
D
$${1 \over 3}$$
4
JEE Main 2016 (Online) 10th April Morning Slot
+4
-1
The solution of the differential equation

$${{dy} \over {dx}}\, + \,{y \over 2}\,\sec x = {{\tan x} \over {2y}},\,\,$$

where 0 $$\le$$ x < $${\pi \over 2}$$, and y (0) = 1, is given by :
A
y = 1 $$-$$ $${x \over {\sec x + \tan x}}$$
B
y2 = 1 + $${x \over {\sec x + \tan x}}$$
C
y2 = 1 $$-$$ $${x \over {\sec x + \tan x}}$$
D
y = 1 + $${x \over {\sec x + \tan x}}$$
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