JEE Main 2022 · 26 Jul · Shift 2 · Mathematics Q10

← JEE Main 2022 · 26 Jul · Shift 2

MathematicsSingle correct+4 / -1Question 10 of 90

Let the solution curve $y=f(x)$ of the differential equation $\dfrac{dy}{dx}+\dfrac{xy}{x^2-1}=\dfrac{x^4+2x}{\sqrt{1-x^2}}$ , $x\in(-1,1)$ pass through the origin. Then $\displaystyle\int_{-\frac{\sqrt3}2}^{\frac{\sqrt3}2}f(x)\,dx$ is equal to

\dfrac\pi3-\dfrac14

\dfrac\pi3-\dfrac{\sqrt3}4

\dfrac\pi6-\dfrac{\sqrt3}4

\dfrac\pi6-\dfrac{\sqrt3}2

Answer and worked solution

Official answer: B

Integrating factor $e^{\int\frac x{x^2-1}dx}=\sqrt{1-x^2}$ on $(-1,1)$ .

Solve. $y\sqrt{1-x^2}=\frac{x^5}5+x^2+c$ and $y(0)=0$ gives $c=0$ , so $f(x)=\frac{x^5/5+x^2}{\sqrt{1-x^2}}$ .

Integrate. The odd part vanishes; with $x=\sin\theta$ , $2\int_0^{\pi/3}\sin^2\theta\,d\theta=\frac\pi3-\frac{\sqrt3}4$ .

(B)

Topics:Linear first-order equations Properties of definite integrals

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JEE Main 2022 · 26 Jul · Shift 2, Mathematics question 10