Verified Solution • 2025–26 Session moderate QJEEM25Ma2f38: JEE Main 2025 Mathematics - Class 12 JEEM25Ma2f38 Solution: Let be the solution of the differential equation and . Then is… Let x=x(y)x=x(y)x=x(y) be the solution of the differential equation y=\left(x-y \frac{\mathrm{~d} x}{\mathrm{~d} y}\right) \sin \left(\frac{x}{y}\right), y>0 and x(1)=π2x(1)=\frac{\pi}{2}x(1)=2π . Then cos(x(2))\cos (x(2))cos(x(2)) is equal to : Multiple Choice Options A 2(loge2)−12\left(\log _e 2\right)-12(loge2)−1 B 1−2(loge2)21-2\left(\log _e 2\right)^21−2(loge2)2 C 1−2(loge2)1-2\left(\log _{\mathrm{e}} 2\right)1−2(loge2) D 2(loge2)2−12\left(\log _e 2\right)^2-12(loge2)2−1 Step-by-Step Verified Solution ydyamp;=(xdy−ydx)sin(xy)dyyamp;=(xdy−ydxy2)sin(xy)dyyamp;=sin(xy)d(−xy)⇒ℓnyamp;=cosxy+C\begin{aligned}\quad y d y & =(x d y-y d x) \sin \left(\frac{x}{y}\right) \\\frac{d y}{y} & =\left(\frac{x d y-y d x}{y^2}\right) \sin \left(\frac{x}{y}\right) \\\frac{d y}{y} & =\sin \left(\frac{x}{y}\right) d\left(-\frac{x}{y}\right) \\\Rightarrow \quad \ell n y & =\cos \frac{x}{y}+C\end{aligned}ydyydyydy⇒ℓnyamp;=(xdy−ydx)sin(yx)amp;=(y2xdy−ydx)sin(yx)amp;=sin(yx)d(−yx)amp;=cosyx+C amp;x(1)=π2⇒0=cosπ2+C⇒C=0amp; थny =cosxyamp; but y=2⇒cosx2=ln2amp;cosxamp;=2cos2x2−1amp;=2(ln2)2−1\begin{aligned}& x(1)=\frac{\pi}{2} \Rightarrow 0=\cos \frac{\pi}{2}+C \Rightarrow C=0 \\& \text { थny }=\cos \frac{x}{y} \\& \text { but } y=2 \Rightarrow \cos \frac{x}{2}=\ln 2 \\& \qquad \begin{aligned}\cos x & =2 \cos ^2 \frac{x}{2}-1 \\& =2(\ln 2)^2-1\end{aligned}\end{aligned}amp;x(1)=2π⇒0=cos2π+C⇒C=0amp; थny =cosyxamp; but y=2⇒cos2x=ln2amp;cosxamp;=2cos22x−1amp;=2(ln2)2−1 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring