1
JEE Main 2021 (Online) 22th July Evening Shift
MCQ (Single Correct Answer)
+4
-1
Let y = y(x) be the solution of the differential equation $$\cos e{c^2}xdy + 2dx = (1 + y\cos 2x)\cos e{c^2}xdx$$, with $$y\left( {{\pi \over 4}} \right) = 0$$. Then, the value of $${(y(0) + 1)^2}$$ is equal to :
A
e1/2
B
e$$-$$1/2
C
e$$-$$1
D
e
2
JEE Main 2021 (Online) 20th July Evening Shift
MCQ (Single Correct Answer)
+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
$${\pi \over 2} - {4 \over \pi }$$
3
JEE Main 2021 (Online) 20th July Morning Shift
MCQ (Single Correct Answer)
+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
$${1 \over 6}(\pi - 1)$$
4
JEE Main 2021 (Online) 20th July Morning Shift
MCQ (Single Correct Answer)
+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
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