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Jee Main 2026
04 - 04 - 2026 S1
Question
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 Options :
Select the correct option:
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
$\frac{\sqrt{3}}{2} \tan \left(\frac{11 \sqrt{3}}{6}\right)$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
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