JEE Main 2026 — 8 April, Evening Shift — Question 34

$x \sqrt{1-x^{2}} d y+\left(y \sqrt{1-x^{2}}-x \cos ^{-1} x\right) d x=0$ $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}+\frac{\mathrm{y}}{\mathrm{x}}=\frac{\cos ^{-1} \mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}}$ it is L.D.E. ∴ I.F. $=\mathrm{e}^{\int \frac{1}{\mathrm{x}} \mathrm{dx}}=\mathrm{e}^{\ln \mathrm{x}}=\mathrm{x}$ ∴ solution is $\mathrm{y} \cdot \mathrm{x}=\int \frac{\mathrm{x} \cos ^{-1} \mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}+\mathrm{C}$

\mathrm{y} \cdot \mathrm{x}=\mathrm{I}_{1}+\mathrm{C} \end{gathered}$$ $I_{1}=\int \frac{x \cos ^{-1} x}{\sqrt{1-x^{2}}} d x$ Let $\cos ^{-1} \mathrm{x}=\mathrm{t} \Rightarrow \mathrm{x}=\mathrm{cost}$ $-\frac{1}{\sqrt{1-x^{2}}} d x=d t$ $\mathrm{I}_{1}=\int-\mathrm{t} \cdot \cot \mathrm{dt}$ $=-(\mathrm{t} \sin \mathrm{t}+\cos \mathrm{t})$ $$\begin{gathered} I_{1}=-\left(\sqrt{1-x^{2}} \cdot \cos ^{-1} x+x\right) \end{gathered}$$ $\therefore y \cdot x=-\left(\sqrt{1-x^{2}} \cdot \cos ^{-1} x+x\right)+C$ $\because \lim _{\mathrm{x} \rightarrow 1^{-}} \mathrm{y}(\mathrm{x})=1$ $\therefore 1=-(0+1)+C$ $\mathrm{C}=2$ $\therefore \mathrm{y} \cdot \mathrm{x}=-\sqrt{1-\mathrm{x}^{2}} \cdot \cos ^{-1} \mathrm{x}-\mathrm{x}+2$ $\left(\right.$ put $\left.\mathrm{x}=\frac{1}{2}\right)$ $y\left(\frac{1}{2}\right) \times \frac{1}{2}=-\frac{\sqrt{3}}{2} \times \frac{\pi}{3}-\frac{1}{2}+2$ $\Rightarrow \mathrm{y}\left(\frac{1}{2}\right)=3-\frac{\pi}{\sqrt{3}}$