1
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 + $${x \over {\sec x + \tan x}}$$
2
JEE Main 2016 (Online) 9th April Morning Slot
+4
-1
If   f(x) is a differentiable function in the interval (0, $$\infty$$) such that f (1) = 1 and

$$\mathop {\lim }\limits_{t \to x}$$   $${{{t^2}f\left( x \right) - {x^2}f\left( t \right)} \over {t - x}} = 1,$$ for each x > 0, then $$f\left( {{\raise0.5ex\hbox{\scriptstyle 3} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}} \right)$$ equal to :
A
$${{13} \over 6}$$
B
$${{23} \over 18}$$
C
$${{25} \over 9}$$
D
$${{31} \over 18}$$
3
JEE Main 2016 (Offline)
+4
-1
If a curve $$y=f(x)$$ passes through the point $$(1,-1)$$ and satisfies the differential equation, $$y(1+xy) dx=x$$ $$dy$$, then $$f\left( { - {1 \over 2}} \right)$$ is equal to :
A
$${2 \over 5}$$
B
$${4 \over 5}$$
C
$$-{2 \over 5}$$
D
$$-{4 \over 5}$$
4
JEE Main 2015 (Offline)
+4
-1
Let $$y(x)$$ be the solution of the differential equation

$$\left( {x\,\log x} \right){{dy} \over {dx}} + y = 2x\,\log x,\left( {x \ge 1} \right).$$ Then $$y(e)$$ is equal to :
A
$$2$$
B
$$2e$$
C
$$e$$
D
$$0$$
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