1
JEE Main 2022 (Online) 29th June Evening Shift
+4
-1

If y = y(x) is the solution of the differential equation $$\left( {1 + {e^{2x}}} \right){{dy} \over {dx}} + 2\left( {1 + {y^2}} \right){e^x} = 0$$ and y (0) = 0, then $$6\left( {y'(0) + {{\left( {y\left( {{{\log }_e}\sqrt 3 } \right)} \right)}^2}} \right)$$ is equal to

A
2
B
$$-$$2
C
$$-$$4
D
$$-$$1
2
JEE Main 2022 (Online) 29th June Morning Shift
+4
-1

Let the solution curve of the differential equation

$$x{{dy} \over {dx}} - y = \sqrt {{y^2} + 16{x^2}}$$, $$y(1) = 3$$ be $$y = y(x)$$. Then y(2) is equal to:

A
15
B
11
C
13
D
17
3
JEE Main 2022 (Online) 28th June Evening Shift
+4
-1

Let x = x(y) be the solution of the differential equation

$$2y\,{e^{x/{y^2}}}dx + \left( {{y^2} - 4x{e^{x/{y^2}}}} \right)dy = 0$$ such that x(1) = 0. Then, x(e) is equal to :

A
$$e{\log _e}(2)$$
B
$$- e{\log _e}(2)$$
C
$${e^2}{\log _e}(2)$$
D
$$- {e^2}{\log _e}(2)$$
4
JEE Main 2022 (Online) 28th June Evening Shift
+4
-1

Let the slope of the tangent to a curve y = f(x) at (x, y) be given by 2 $$\tan x(\cos x - y)$$. If the curve passes through the point $$\left( {{\pi \over 4},0} \right)$$, then the value of $$\int\limits_0^{\pi /2} {y\,dx}$$ is equal to :

A
$$(2 - \sqrt 2 ) + {\pi \over {\sqrt 2 }}$$
B
$$2 - {\pi \over {\sqrt 2 }}$$
C
$$(2 + \sqrt 2 ) + {\pi \over {\sqrt 2 }}$$
D
$$2 + {\pi \over {\sqrt 2 }}$$
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