Question

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

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

Answer 1

Answer: (C)

We have $\frac{dy}{dx}=\frac{y}{x}-{\cos }^2\left(\frac{y}{x}\right)$
Putting $y=vx$, so that $\frac{dy}{dx}=v+x\frac{dv}{dx}$, we get
$v+x\frac{dv}{dx}=v-{\cos }^{2}v$
$\Rightarrow \frac{dv}{{\cos }^{2}v}=-\frac{dx}{x}$
$\Rightarrow {\sec }^{2}vdv=-\frac{1}{x}dx$
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, $\tan \left(\frac{y}{x}\right)=-\log x+1$
$\Rightarrow \tan \left(\frac{y}{x}\right)=-\log x+\log e$
$\Rightarrow y=x{\tan }^{-1}\left(\log \left(\frac{e}{x}\right)\right)$