1
JEE Main 2022 (Online) 26th June Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

If $$\int {{1 \over x}\sqrt {{{1 - x} \over {1 + x}}} dx = g(x) + c} $$, $$g(1) = 0$$, then $$g\left( {{1 \over 2}} \right)$$ is equal to :

A
$${\log _e}\left( {{{\sqrt 3 - 1} \over {\sqrt 3 + 1}}} \right) + {\pi \over 3}$$
B
$${\log _e}\left( {{{\sqrt 3 + 1} \over {\sqrt 3 - 1}}} \right) + {\pi \over 3}$$
C
$${\log _e}\left( {{{\sqrt 3 + 1} \over {\sqrt 3 - 1}}} \right) - {\pi \over 3}$$
D
$${1 \over 2}{\log _e}\left( {{{\sqrt 3 - 1} \over {\sqrt 3 + 1}}} \right) - {\pi \over 6}$$
2
JEE Main 2022 (Online) 26th June Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

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
$${4 \over e} - e$$
3
JEE Main 2022 (Online) 26th June Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

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
4
JEE Main 2022 (Online) 26th June Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The locus of the mid point of the line segment joining the point (4, 3) and the points on the ellipse $${x^2} + 2{y^2} = 4$$ is an ellipse with eccentricity :

A
$${{\sqrt 3 } \over 2}$$
B
$${1 \over {2\sqrt 2 }}$$
C
$${1 \over {\sqrt 2 }}$$
D
$${1 \over 2}$$
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