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

Answer 1

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

Given that, equation of tangent is $\frac{y}{x}-{\cos }^2\left(\frac{y}{x}\right)$

According to the given condition

$\frac{dy}{dx}=\frac{y}{x}-{\cos }^2\left(\frac{y}{x}\right)$

On putting $y=vx$

$\Rightarrow$ $\frac{dy}{dx}=v+x\frac{dv}{dx}$. 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

$\tan v=-\log x+\log c$

$\Rightarrow \tan \left(\frac{y}{x}\right)=-\log x+\log c$

Since, this curve is passing through $(1,\pi /4)$

$\therefore \tan \left(\frac{\pi }{4}\right)=-\log 1+\log c$

$\Rightarrow \log c=1$

$\therefore \tan \left(\frac{y}{x}\right)=-\log x+1$

$\Rightarrow \tan \left(\frac{y}{x}\right)=-\log x+\log e$

$\Rightarrow y=x{\tan }^{-1}\left[\log \left(\frac{e}{x}\right)\right]$