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JEE MAIN 2026
08-04-2026_S2
Question
Let $y=y(x)$ be the solution of the differential equation $x \sqrt{1-x^2} d y+\left(y \sqrt{1-x^2}-x \cos ^{-1} x\right) d x=0, x \in(0,1), \lim _{x \rightarrow 1^{-}} y(x)=1$. Then $y\left(\frac{1}{2}\right)$ equals:
Select the correct option:
A
$3-\frac{\pi}{\sqrt{3}}$
B
$4-\sqrt{3} \pi$
C
$4-\frac{2 \pi}{\sqrt{3}}$
D
$3-\frac{\pi}{2 \sqrt{3}}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
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