1
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
$${1 + {{\log }_e}2}$$
2
JEE Main 2020 (Online) 2nd September Morning Slot
+4
-1
Let y = y(x) be the solution of the differential equation,
$${{2 + \sin x} \over {y + 1}}.{{dy} \over {dx}} = - \cos x$$, y > 0,y(0) = 1.
If y($$\pi$$) = a and $${{dy} \over {dx}}$$ at x = $$\pi$$ is b, then the ordered pair (a, b) is equal to :
A
(2, 1)
B
$$\left( {2,{3 \over 2}} \right)$$
C
(1, -1)
D
(1, 1)
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
Out of Syllabus
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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