1
JEE Main 2020 (Online) 3rd September Morning Slot
+4
-1
The solution curve of the differential equation,

(1 + e-x)(1 + y2)$${{dy} \over {dx}}$$ = y2,

which passes through the point (0, 1), is :
A
y2 + 1 = y$$\left( {{{\log }_e}\left( {{{1 + {e^{ - x}}} \over 2}} \right) + 2} \right)$$
B
y2 + 1 = y$$\left( {{{\log }_e}\left( {{{1 + {e^{ x}}} \over 2}} \right) + 2} \right)$$
C
y2 = 1 + $${y{{\log }_e}\left( {{{1 + {e^{ - x}}} \over 2}} \right)}$$
D
y2 = 1 + $${y{{\log }_e}\left( {{{1 + {e^{ x}}} \over 2}} \right)}$$
2
JEE Main 2020 (Online) 2nd September Evening Slot
+4
-1
If a curve y = f(x), passing through the point (1, 2), is the solution of the differential equation,
2x2dy= (2xy + y2)dx, then $$f\left( {{1 \over 2}} \right)$$ is equal to :
A
$${1 \over {1 - {{\log }_e}2}}$$
B
$${1 \over {1 + {{\log }_e}2}}$$
C
$${{ - 1} \over {1 + {{\log }_e}2}}$$
D
$${1 + {{\log }_e}2}$$
3
JEE Main 2020 (Online) 9th January Evening Slot
+4
-1
If $${{dy} \over {dx}} = {{xy} \over {{x^2} + {y^2}}}$$; y(1) = 1; then a value of x satisfying y(x) = e is :
A
$$\sqrt 2 e$$
B
$${1 \over 2}\sqrt 3 e$$
C
$${e \over {\sqrt 2 }}$$
D
$$\sqrt 3 e$$
4
JEE Main 2020 (Online) 8th January Evening Slot
+4
-1
The differential equation of the family of curves, x2 = 4b(y + b), b $$\in$$ R, is
A
x(y')2 = x – 2yy'
B
x(y')2 = 2yy' – x
C
xy" = y'
D
x(y')2 = x + 2yy'
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