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1
JEE Main 2021 (Online) 25th July Morning Shift
MCQ (Single Correct Answer)  +4  -1
Let y = y(x) be the solution of the differential equation $${{dy} \over {dx}} = 1 + x{e^{y - x}}, - \sqrt 2 < x < \sqrt 2 ,y(0) = 0$$

then, the minimum value of $$y(x),x \in \left( { - \sqrt 2 ,\sqrt 2 } \right)$$ is equal to :
A
$$\left( {2 - \sqrt 3 } \right) - {\log _e}2$$
B
$$\left( {2 + \sqrt 3 } \right) + {\log _e}2$$
C
$$\left( {1 + \sqrt 3 } \right) - {\log _e}\left( {\sqrt 3 - 1} \right)$$
D
$$\left( {1 - \sqrt 3 } \right) - {\log _e}\left( {\sqrt 3 - 1} \right)$$
2
JEE Main 2021 (Online) 22th July Evening Shift
MCQ (Single Correct Answer)  +4  -1
Let y = y(x) be the solution of the differential equation $$\cos e{c^2}xdy + 2dx = (1 + y\cos 2x)\cos e{c^2}xdx$$, with $$y\left( {{\pi \over 4}} \right) = 0$$. Then, the value of $${(y(0) + 1)^2}$$ is equal to :
A
e1/2
B
e$$-$$1/2
C
e$$-$$1
D
e
3
JEE Main 2021 (Online) 20th July Morning Shift
MCQ (Single Correct Answer)  +4  -1
Let y = y(x) be the solution of the differential equation $$x\tan \left( {{y \over x}} \right)dy = \left( {y\tan \left( {{y \over x}} \right) - x} \right)dx$$, $$ - 1 \le x \le 1$$, $$y\left( {{1 \over 2}} \right) = {\pi \over 6}$$. Then the area of the region bounded by the curves x = 0, $$x = {1 \over {\sqrt 2 }}$$ and y = y(x) in the upper half plane is :
A
$${1 \over 8}(\pi - 1)$$
B
$${1 \over {12}}(\pi - 3)$$
C
$${1 \over 4}(\pi - 2)$$
D
$${1 \over 6}(\pi - 1)$$
4
JEE Main 2021 (Online) 20th July Morning Shift
MCQ (Single Correct Answer)  +4  -1
Let y = y(x) be the solution of the differential equation $${e^x}\sqrt {1 - {y^2}} dx + \left( {{y \over x}} \right)dy = 0$$, y(1) = $$-$$1. Then the value of (y(3))2 is equal to :
A
1 $$-$$ 4e3
B
1 $$-$$ 4e6
C
1 + 4e3
D
1 + 4e6
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