1
JEE Main 2026 (Online) 4th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

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

A

$\sqrt{3} \tan \left(\frac{11 \sqrt{3}}{6}\right)$

B

$\frac{\sqrt{3}}{2} \tan \left(\frac{11 \sqrt{3}}{12}\right)$

C

$\sqrt{3} \tan \left(\frac{11 \sqrt{3}}{12}\right)$

D

$^{\frac{\sqrt{3}}{2}} \tan \left(\frac{11 \sqrt{3}}{6}\right)$

2
JEE Main 2026 (Online) 2nd April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $x = x(y)$ be the solution of the differential equation $2y^2 \frac{dx}{dy} - 2xy + x^2 = 0$, $y > 1$, $x(e) = e$.

Then $x(e^2)$ is equal to:

A

$\frac{3}{2} e^2$

B

$\frac{2}{3} e^2$

C

$e^2$

D

$2e^2$

3
JEE Main 2026 (Online) 2nd April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

If the curve $y = f(x)$ passes through the point $(1, e)$ and satisfies the differential equation $dy = y(2 + \\log_e x) dx$, $x > 0$, then $f(e)$ is equal to :

A

$e^e$

B

$e^{e^2}$

C

$e^{2e}$

D

$e^{2e}$

4
JEE Main 2026 (Online) 2nd April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $y = y(x)$ be the solution curve of the differential equation

$(1 + \sin x)\dfrac{dy}{dx} + (y + 1)\cos x = 0,\ y(0) = 0.$ If the curve $y = y(x)$ passes through the point $\left( \alpha , \dfrac{-1}{2} \right)$,

then a value of $\alpha$ is:

A

$\dfrac{\pi}{6}$

B

$\dfrac{\pi}{4}$

C

$\dfrac{\pi}{3}$

D

$\dfrac{\pi}{2}$

JEE Main Subjects

Browse all chapters by subject