1
JEE Main 2025 (Online) 24th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $\mathrm{y}=\mathrm{y}(\mathrm{x})$ be the solution of the differential equation $\left(x y-5 x^2 \sqrt{1+x^2}\right) d x+\left(1+x^2\right) d y=0, y(0)=0$. Then $y(\sqrt{3})$ is equal to

A
$\frac{5 \sqrt{3}}{2}$
B
$\sqrt{\frac{15}{2}}$
C
$\sqrt{\frac{14}{3}}$
D
$2 \sqrt{2}$
2
JEE Main 2025 (Online) 23rd January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $x=x(y)$ be the solution of the differential equation $y=\left(x-y \frac{\mathrm{~d} x}{\mathrm{~d} y}\right) \sin \left(\frac{x}{y}\right), y>0$ and $x(1)=\frac{\pi}{2}$. Then $\cos (x(2))$ is equal to :

A
$2\left(\log _e 2\right)-1$
B
$1-2\left(\log _e 2\right)^2$
C
$1-2\left(\log _{\mathrm{e}} 2\right)$
D
$2\left(\log _e 2\right)^2-1$
3
JEE Main 2025 (Online) 23rd January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let a curve $y=f(x)$ pass through the points $(0,5)$ and $\left(\log _e 2, k\right)$. If the curve satisfies the differential equation $2(3+y) e^{2 x} d x-\left(7+e^{2 x}\right) d y=0$, then $k$ is equal to

A
32
B
8
C
4
D
16
4
JEE Main 2025 (Online) 22nd January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

If $x=f(y)$ is the solution of the differential equation $\left(1+y^2\right)+\left(x-2 \mathrm{e}^{\tan ^{-1} y}\right) \frac{\mathrm{d} y}{\mathrm{~d} x}=0, y \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ with $f(0)=1$, then $f\left(\frac{1}{\sqrt{3}}\right)$ is equal to :

A
$\mathrm{e}^{\pi / 4}$
B
$e^{\pi / 12}$
C
$\mathrm{e}^{\pi / 6}$
D
$e^{\pi / 3}$
JEE Main Subjects
EXAM MAP