JEE Main 2026 · 8 Apr · Shift 2 · Mathematics Q17

← JEE Main 2026 · 8 Apr · Shift 2

MathematicsSingle correct+4 / -1Question 17 of 75

Let $y=y(x)$ be the solution of the differential equation $x\sqrt{1-x^2}\,dy+\left(y\sqrt{1-x^2}-x\cos^{-1}x\right)dx=0$ , $x\in(0,1)$ , $\displaystyle\lim_{x\to1^-}y(x)=1$ . Then $y\left(\dfrac12\right)$ equals :

3-\dfrac\pi{\sqrt3}

4-\sqrt3\,\pi

4-\dfrac{2\pi}{\sqrt3}

3-\dfrac\pi{2\sqrt3}

Answer and worked solution

Official answer: A

Exact form. Dividing by $\sqrt{1-x^2}$ : $d(xy)=\dfrac{x\cos^{-1}x}{\sqrt{1-x^2}}\,dx$ .

Integrate by parts.
$xy=-\sqrt{1-x^2}\cos^{-1}x-x+C.$
As $x\to1^-$ : $1=-1+C$ , so $C=2$ .

At $x=\tfrac12$ .
$\frac y2=-\frac{\sqrt3}2\cdot\frac\pi3+\frac32\ \Rightarrow\ y=3-\frac\pi{\sqrt3}.$

(A)

Topics:Formation and applications Integration by parts

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JEE Main 2026 · 8 Apr · Shift 2, Mathematics question 17