1
JEE Main 2021 (Online) 20th July Evening Shift
+4
-1
Let y = y(x) satisfies the equation $${{dy} \over {dx}} - |A| = 0$$, for all x > 0, where $$A = \left[ {\matrix{ y & {\sin x} & 1 \cr 0 & { - 1} & 1 \cr 2 & 0 & {{1 \over x}} \cr } } \right]$$. If $$y(\pi ) = \pi + 2$$, then the value of $$y\left( {{\pi \over 2}} \right)$$ is :
A
$${\pi \over 2} + {4 \over \pi }$$
B
$${\pi \over 2} - {1 \over \pi }$$
C
$${{3\pi } \over 2} - {1 \over \pi }$$
D
$${\pi \over 2} - {4 \over \pi }$$
2
JEE Main 2021 (Online) 20th July Morning Shift
+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
$${1 \over 6}(\pi - 1)$$
3
JEE Main 2021 (Online) 20th July Morning Shift
+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
4
JEE Main 2021 (Online) 18th March Evening Shift
+4
-1
Let y = y(x) be the solution of the differential equation

$${{dy} \over {dx}} = (y + 1)\left( {(y + 1){e^{{x^2}/2}} - x} \right)$$, 0 < x < 2.1, with y(2) = 0. Then the value of $${{dy} \over {dx}}$$ at x = 1 is equal to :
A
$${{{e^{5/2}}} \over {{{(1 + {e^2})}^2}}}$$
B
$${{5{e^{1/2}}} \over {{{({e^2} + 1)}^2}}}$$
C
$$- {{2{e^2}} \over {{{(1 + {e^2})}^2}}}$$
D
$${{ - {e^{3/2}}} \over {{{({e^2} + 1)}^2}}}$$
EXAM MAP
Medical
NEET