1
JEE Main 2023 (Online) 30th January Morning Shift
+4
-1 Let the solution curve $$y=y(x)$$ of the differential equation $$\frac{\mathrm{d} y}{\mathrm{~d} x}-\frac{3 x^{5} \tan ^{-1}\left(x^{3}\right)}{\left(1+x^{6}\right)^{3 / 2}} y=2 x \exp \left\{\frac{x^{3}-\tan ^{-1} x^{3}}{\sqrt{\left(1+x^{6}\right)}}\right\} \text { pass through the origin. Then } y(1) \text { is equal to : }$$

A
$$\exp \left(\frac{1-\pi}{4 \sqrt{2}}\right)$$
B
$$\exp \left(\frac{4-\pi}{4 \sqrt{2}}\right)$$
C
$$\exp \left(\frac{4+\pi}{4 \sqrt{2}}\right)$$
D
$$\exp \left(\frac{\pi-4}{4 \sqrt{2}}\right)$$
2
JEE Main 2023 (Online) 30th January Morning Shift
+4
-1 The number of points on the curve $$y=54 x^{5}-135 x^{4}-70 x^{3}+180 x^{2}+210 x$$ at which the normal lines are parallel to $$x+90 y+2=0$$ is :

A
2
B
3
C
4
D
0
3
JEE Main 2023 (Online) 29th January Evening Shift
+4
-1 Let $$y=y(x)$$ be the solution of the differential equation $$x{\log _e}x{{dy} \over {dx}} + y = {x^2}{\log _e}x,(x > 1)$$. If $$y(2) = 2$$, then $$y(e)$$ is equal to

A
$${{1 + {e^2}} \over 2}$$
B
$${{1 + {e^2}} \over 4}$$
C
$${{2 + {e^2}} \over 2}$$
D
$${{4 + {e^2}} \over 4}$$
4
JEE Main 2023 (Online) 29th January Morning Shift
+4
-1

Let $$y=f(x)$$ be the solution of the differential equation $$y(x+1)dx-x^2dy=0,y(1)=e$$. Then $$\mathop {\lim }\limits_{x \to {0^ + }} f(x)$$ is equal to

A
$${e^2}$$
B
0
C
$${1 \over {{e^2}}}$$
D
$${1 \over e}$$
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