1
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
$$2 + {\pi \over {\sqrt 2 }}$$
2
JEE Main 2022 (Online) 28th June Morning Shift
+4
-1

Let the solution curve $$y = y(x)$$ of the differential equation

$$\left[ {{x \over {\sqrt {{x^2} - {y^2}} }} + {e^{{y \over x}}}} \right]x{{dy} \over {dx}} = x + \left[ {{x \over {\sqrt {{x^2} - {y^2}} }} + {e^{{y \over x}}}} \right]y$$

pass through the points (1, 0) and (2$$\alpha$$, $$\alpha$$), $$\alpha$$ > 0. Then $$\alpha$$ is equal to

A
$${1 \over 2}\exp \left( {{\pi \over 6} + \sqrt e - 1} \right)$$
B
$${1 \over 2}\exp \left( {{\pi \over 6} + e - 1} \right)$$
C
$$\exp \left( {{\pi \over 6} + \sqrt e + 1} \right)$$
D
$$2\exp \left( {{\pi \over 3} + \sqrt e - 1} \right)$$
3
JEE Main 2022 (Online) 28th June Morning Shift
+4
-1

Let y = y(x) be the solution of the differential equation $$x(1 - {x^2}){{dy} \over {dx}} + (3{x^2}y - y - 4{x^3}) = 0$$, $$x > 1$$, with $$y(2) = - 2$$. Then y(3) is equal to :

A
$$-$$18
B
$$-$$12
C
$$-$$6
D
$$-$$3
4
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
$${1 \over e}$$
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