Medium MCQ +4 / -1 PYQ · JEE Mains 2025

Let $x=x(y)$ 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 $x(1)=\frac{\pi}{2}$. Then $\cos (x(2))$ is equal to :

  1. A $2\left(\log _e 2\right)-1$
  2. B $1-2\left(\log _e 2\right)^2$
  3. C $1-2\left(\log _{\mathrm{e}} 2\right)$
  4. D $2\left(\log _e 2\right)^2-1$ Correct answer

Solution

<p>$$\begin{aligned} \quad y d y & =(x d y-y d x) \sin \left(\frac{x}{y}\right) \\ \frac{d y}{y} & =\left(\frac{x d y-y d x}{y^2}\right) \sin \left(\frac{x}{y}\right) \\ \frac{d y}{y} & =\sin \left(\frac{x}{y}\right) d\left(-\frac{x}{y}\right) \\ \Rightarrow \quad \ell n y & =\cos \frac{x}{y}+C \end{aligned}$$</p> <p>$$\begin{aligned} & x(1)=\frac{\pi}{2} \Rightarrow 0=\cos \frac{\pi}{2}+C \Rightarrow C=0 \\ & \text { थny }=\cos \frac{x}{y} \\ & \text { but } y=2 \Rightarrow \cos \frac{x}{2}=\ln 2 \\ & \qquad \begin{aligned} \cos x & =2 \cos ^2 \frac{x}{2}-1 \\ & =2(\ln 2)^2-1 \end{aligned} \end{aligned}$$</p>

About this question

Subject: Mathematics · Chapter: Differential Equations · Topic: Order and Degree

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