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 $\frac{d y}{d x}=\frac{y}{x}-\cos ^{2}\left(\frac{y}{x}\right)$
Putting $y=v x$, so that $\frac{d y}{d x}=v+x \frac{d v}{d x}$, we get
$v+x \frac{d v}{d x}=v-\cos ^{2} v$
$\Rightarrow \frac{d v}{\cos ^{2} v}=-\frac{d x}{x}$
$\Rightarrow \sec ^{2} v d v =-\frac{1}{x} d x$
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, $\tan \left(\frac{y}{x}\right)=-\log x+1$
$\Rightarrow \tan \left(\frac{y}{x}\right)=-\log x+\log e$
$\Rightarrow y=x \tan ^{-1}\left(\log \left(\frac{e}{x}\right)\right)$