Question

The equation of the curve passing through the point $(1,\frac{\pi }{4})$ and having slope of tangent at any point $(x,y)$ as $\frac{y}{x}-{\cos }^2\frac{y}{x}$ is

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

Answer 1

The correct option is A $x=e^{1-\tan \left(y/x\right)}$

Given, $\frac{dy}{dx}=\frac{y}{x}-{\cos }^2\frac{y}{x}$
$\Rightarrow \frac{xdy-ydx}{x}=-\left({\cos }^2\frac{y}{x}\right)dx$
$\Rightarrow {\sec }^2\frac{y}{x}\left(\frac{xdy-ydx}{x^2}\right)=-\frac{dx}{x}$
$\Rightarrow {\sec }^2\frac{y}{x}d\left(\frac{y}{x}\right)=-\frac{dx}{x}$
On integrating both sides, we get
$\tan \frac{y}{x}=-\log x+C$,
where $x=1,y=\frac{\pi }{4},C=1=\log e$
Therefore, $\tan \frac{y}{x}=\log e-\log x$
$\Rightarrow x=e^{1-\tan \frac{y}{x}}$