1
JEE Main 2022 (Online) 26th June Evening Shift
+4
-1 If $$y = y(x)$$ is the solution of the differential equation

$$x{{dy} \over {dx}} + 2y = x\,{e^x}$$, $$y(1) = 0$$ then the local maximum value

of the function $$z(x) = {x^2}y(x) - {e^x},\,x \in R$$ is :

A
1 $$-$$ e
B
0
C
$${1 \over 2}$$
D
$${4 \over e} - e$$
2
JEE Main 2022 (Online) 26th June Evening Shift
+4
-1 If the solution of the differential equation

$${{dy} \over {dx}} + {e^x}\left( {{x^2} - 2} \right)y = \left( {{x^2} - 2x} \right)\left( {{x^2} - 2} \right){e^{2x}}$$ satisfies $$y(0) = 0$$, then the value of y(2) is _______________.

A
$$-$$1
B
1
C
0
D
e
3
JEE Main 2022 (Online) 25th June Evening Shift
+4
-1 If $$y = y(x)$$ is the solution of the differential equation

$$2{x^2}{{dy} \over {dx}} - 2xy + 3{y^2} = 0$$ such that $$y(e) = {e \over 3}$$, then y(1) is equal to

A
$${1 \over 3}$$
B
$${2 \over 3}$$
C
$${3 \over 2}$$
D
3
4
JEE Main 2022 (Online) 25th June Morning Shift
+4
-1 Let $$g:(0,\infty ) \to R$$ be a differentiable function such that

$$\int {\left( {{{x(\cos x - \sin x)} \over {{e^x} + 1}} + {{g(x)\left( {{e^x} + 1 - x{e^x}} \right)} \over {{{({e^x} + 1)}^2}}}} \right)dx = {{x\,g(x)} \over {{e^x} + 1}} + c}$$, for all x > 0, where c is an arbitrary constant. Then :

A
g is decreasing in $$\left( {0,{\pi \over 4}} \right)$$
B
g' is increasing in $$\left( {0,{\pi \over 4}} \right)$$
C
g + g' is increasing in $$\left( {0,{\pi \over 2}} \right)$$
D
g $$-$$ g' is increasing in $$\left( {0,{\pi \over 2}} \right)$$
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