Question

The slope of the tangent at $(x,y)$ to a curve passing through $(1,\frac{\pi }{4})$ is given by $\frac{y}{x}{\cos }^2\left(\frac{y}{x}\right)$, then the equation of the curve is

• A. $y={\tan }^{-1}\left(\log \frac{c}{x}\right)$
• B. $y=x{\tan }^{-1}\left(\log \frac{x}{c}\right)$
• C. $y=x{\tan }^{-1}\left(\log \frac{c}{x}\right)$
• D. None of these

Answer 1

The correct option is C $y=x{\tan }^{-1}\left(\log \frac{c}{x}\right)$

We have, $\frac{dy}{dx}=\frac{y}{x}-{\cos }^2\left(\frac{y}{x}\right)$
Putting $y=vx$, so that
$\frac{dy}{dx}=v+x\frac{dv}{dx}$,
$\therefore v+x\frac{dv}{dx}=v-{\cos }^{2}v$
$\Rightarrow \frac{dv}{{\cos }^{2}v}=-\frac{dx}{x}$
$\Rightarrow {\sec }^{2}vdv=-\frac{1}{x}dx$
On integration, we get
$\tan v=-\log x+\log c$
$\Rightarrow \tan \left(\frac{y}{x}\right)=-\log x+\log c$
This passes through $(1,\pi /4)$, therefore,
$1=\log c$
So, $\tan \left(\frac{y}{x}\right)=-\log x+1$
$\Rightarrow \tan \left(\frac{y}{x}\right)=-\log x+\log c$
$\Rightarrow y=x{\tan }^{-1}\left(\log \frac{c}{x}\right)$