1
JEE Main 2022 (Online) 25th June Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

If the solution curve $$y = y(x)$$ of the differential equation $${y^2}dx + ({x^2} - xy + {y^2})dy = 0$$, which passes through the point (1, 1) and intersects the line $$y = \sqrt 3 x$$ at the point $$(\alpha ,\sqrt 3 \alpha )$$, then value of $${\log _e}(\sqrt 3 \alpha )$$ is equal to :

A
$${\pi \over 3}$$
B
$${\pi \over 2}$$
C
$${\pi \over 12}$$
D
$${\pi \over 6}$$
2
JEE Main 2022 (Online) 24th June Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

If x = x(y) is the solution of the differential equation

$$y{{dx} \over {dy}} = 2x + {y^3}(y + 1){e^y},\,x(1) = 0$$; then x(e) is equal to :

A
$${e^3}({e^e} - 1)$$
B
$${e^e}({e^3} - 1)$$
C
$${e^2}({e^e} + 1)$$
D
$${e^e}({e^2} - 1)$$
3
JEE Main 2021 (Online) 1st September Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
If y = y(x) is the solution curve of the differential equation $${x^2}dy + \left( {y - {1 \over x}} \right)dx = 0$$ ; x > 0 and y(1) = 1, then $$y\left( {{1 \over 2}} \right)$$ is equal to :
A
$${3 \over 2} - {1 \over {\sqrt e }}$$
B
$$3 + {1 \over {\sqrt e }}$$
C
3 + e
D
3 $$-$$ e
4
JEE Main 2021 (Online) 31st August Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
If $${{dy} \over {dx}} = {{{2^x}y + {2^y}{{.2}^x}} \over {{2^x} + {2^{x + y}}{{\log }_e}2}}$$, y(0) = 0, then for y = 1, the value of x lies in the interval :
A
(1, 2)
B
$$\left( {{1 \over 2},1} \right]$$
C
(2, 3)
D
$$\left( {0,{1 \over 2}} \right]$$
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