1
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'
2
JEE Main 2020 (Online) 8th January Morning Slot
+4
-1
Let y = y(x) be a solution of the differential equation,

$$\sqrt {1 - {x^2}} {{dy} \over {dx}} + \sqrt {1 - {y^2}} = 0$$, |x| < 1.

If $$y\left( {{1 \over 2}} \right) = {{\sqrt 3 } \over 2}$$, then $$y\left( { - {1 \over {\sqrt 2 }}} \right)$$ is equal to :
A
$$- {{\sqrt 3 } \over 2}$$
B
None of those
C
$${{1 \over {\sqrt 2 }}}$$
D
$$-{{1 \over {\sqrt 2 }}}$$
3
JEE Main 2020 (Online) 7th January Evening Slot
+4
-1
Let y = y(x) be the solution curve of the differential equation,

$$\left( {{y^2} - x} \right){{dy} \over {dx}} = 1$$, satisfying y(0) = 1. This curve intersects the x-axis at a point whose abscissa is :
A
2 + e
B
-e
C
2
D
2 - e
4
JEE Main 2020 (Online) 7th January Morning Slot
+4
-1
If y = y(x) is the solution of the differential equation, $${e^y}\left( {{{dy} \over {dx}} - 1} \right) = {e^x}$$ such that y(0) = 0, then y(1) is equal to:
A
2 + loge2
B
loge2
C
1 + loge2
D
2e
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