1
JEE Main 2020 (Online) 6th September Evening Slot
+4
-1
If $$y = \left( {{2 \over \pi }x - 1} \right) cosec\,x$$ is the solution

of the differential equation,
$${{dy} \over {dx}} + p\left( x \right)y = {2 \over \pi } cosec\,x$$,

$$0 < x < {\pi \over 2}$$, then the function p(x) is equal to :
A
cot x
B
sec x
C
tan x
D
cosec x
2
JEE Main 2020 (Online) 6th September Morning Slot
+4
-1
The general solution of the differential equation
$$\sqrt {1 + {x^2} + {y^2} + {x^2}{y^2}}$$ + xy$${{dy} \over {dx}}$$ = 0 is :

(where C is a constant of integration)
A
$$\sqrt {1 + {y^2}} + \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} - 1} \over {\sqrt {1 + {x^2}} + 1}}} \right) + C$$
B
$$\sqrt {1 + {y^2}} - \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} - 1} \over {\sqrt {1 + {x^2}} + 1}}} \right) + C$$
C
$$\sqrt {1 + {y^2}} + \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right) + C$$
D
$$\sqrt {1 + {y^2}} - \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right) + C$$
3
JEE Main 2020 (Online) 5th September Evening Slot
+4
-1
Let y = y(x) be the solution of the differential equation
cosx$${{dy} \over {dx}}$$ + 2ysinx = sin2x, x $$\in$$ $$\left( {0,{\pi \over 2}} \right)$$.
If y$$\left( {{\pi \over 3}} \right)$$ = 0, then y$$\left( {{\pi \over 4}} \right)$$ is equal to :
A
$${1 \over {\sqrt 2 }} - 1$$
B
$${\sqrt 2 - 2}$$
C
$${2 - \sqrt 2 }$$
D
$${2 + \sqrt 2 }$$
4
JEE Main 2020 (Online) 5th September Morning Slot
+4
-1
If y = y(x) is the solution of the differential
equation $${{5 + {e^x}} \over {2 + y}}.{{dy} \over {dx}} + {e^x} = 0$$ satisfying y(0) = 1, then a value of y(loge13) is:
A
-1
B
1
C
0
D
2
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