1
JEE Main 2022 (Online) 27th June Evening Shift
+4
-1

If the solution curve of the differential equation

$$(({\tan ^{ - 1}}y) - x)dy = (1 + {y^2})dx$$ passes through the point (1, 0), then the abscissa of the point on the curve whose ordinate is tan(1), is

A
2e
B
$${2 \over e}$$
C
2
D
$${1 \over e}$$
2
JEE Main 2022 (Online) 27th June Morning Shift
+4
-1

If $${{dy} \over {dx}} + {{{2^{x - y}}({2^y} - 1)} \over {{2^x} - 1}} = 0$$, x, y > 0, y(1) = 1, then y(2) is equal to :

A
$$2 + {\log _2}3$$
B
$$2 + {\log _3}2$$
C
$$2 - {\log _3}2$$
D
$$2 - {\log _2}3$$
3
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
$${4 \over e} - e$$
4
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
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