JEE Main 2026 · 4 Apr · Shift 1 · Mathematics Q20

← JEE Main 2026 · 4 Apr · Shift 1

MathematicsSingle correct+4 / -1Question 20 of 75

Let $y=y(x)$ be the solution of the differential equation $\dfrac{dy}{dx}=\left(1+x+x^2\right)\left(1-y+y^2\right)$ , $y(0)=\dfrac12$ . Then $(2y(1)-1)$ is equal to

\sqrt3\tan\left(\dfrac{11\sqrt3}6\right)

\dfrac{\sqrt3}2\tan\left(\dfrac{11\sqrt3}{12}\right)

\sqrt3\tan\left(\dfrac{11\sqrt3}{12}\right)

\dfrac{\sqrt3}2\tan\left(\dfrac{11\sqrt3}6\right)

Answer and worked solution

Official answer: C

Separate. $\displaystyle\int\frac{dy}{y^2-y+1}=\frac2{\sqrt3}\tan^{-1}\frac{2y-1}{\sqrt3}$ , which is 0 at $y=\frac12$ .

Integrate the right side from 0 to 1: $1+\frac12+\frac13=\frac{11}6$ .

Solve. $\tan^{-1}\dfrac{2y-1}{\sqrt3}=\dfrac{11\sqrt3}{12}$ , so $2y(1)-1=\sqrt3\tan\dfrac{11\sqrt3}{12}$ .

(C)

Topics:Variable separable Standard integrals

← Previous question Next question →

JEE Main 2026 · 4 Apr · Shift 1, Mathematics question 20