Mathematics · 2026
JEE Main · 8 April 2026, Shift 2 · Q17
Let y=y(x) be the solution of the differential equation x √(1-x^2) d y+(y √(1-x^2)-x cos^-1 x) d x=0, x ∈(0,1), lim_x → 1^- y(x)=1. Then y(1/2)…
Let $\displaystyle y=y(x)$ be the solution of the differential equation
$\displaystyle x \sqrt{1-x^2} d y+\left(y \sqrt{1-x^2}-x \cos ^{-1} x\right) d x=0, x \in(0,1), \lim _{x \rightarrow 1^{-}} y(x)=1$. Then $\displaystyle y\left(\frac{1}{2}\right)$ equals :
Official answer
From NTA’s final answer key for this paper.
(1)
$\displaystyle 3-\frac{\pi}{\sqrt{3}}$
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JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.