Question

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

• A. $y={\tan }^{-1}\left(\ln \left|\frac{x}{e}\right|\right)$
• B. $y=x{\tan }^{-1}\left(\ln \left|\frac{x}{e}\right|\right)$
• C. $y={\tan }^{-1}\left(\ln \left|\frac{e}{x}\right|\right)$
• D. $y=x{\tan }^{-1}\left(\ln \left|\frac{e}{x}\right|\right)$

Answer 1

The correct option is D $y=x{\tan }^{-1}\left(\ln \left|\frac{e}{x}\right|\right)$

We have, $\frac{dy}{dx}=\frac{y}{x}-{\cos }^2\left(\frac{y}{x}\right)$
Putting $y=vx,$ so that $\frac{dy}{dx}=v+x\frac{dv}{dx},$ 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 integrating both sides, we get
$\tan v=-\ln |x|+\ln |C|$
$\Rightarrow \tan \left(\frac{y}{x}\right)=-\ln |x|+\ln |C|$
This curve passes through $(1,\frac{\pi }{4}),$ so we have $1=\ln |C|.$
$\therefore \tan \left(\frac{y}{x}\right)=-\ln |x|+1$
$\Rightarrow \tan \left(\frac{y}{x}\right)=-\ln |x|+\ln |e|$
$\Rightarrow y=x{\tan }^{-1}\left(\ln \left|\frac{e}{x}\right|\right)$