Question

If the gradient of the tangent at any point $(x,y)$ of a curve which passes through the point $\left(1,\frac{\pi }{4}\right)$ is $\left\{\frac{y}{x}-si{n}^2\left(\frac{y}{x}\right)\right\}$, then equation of the curve is

• A. $y=co{t}^{-1}\left(lo{g}_{{}_e}x\right)$
• B. $y=co{t}^{-1}\left(lo{g}_{{}_e}\frac{x}{e}\right)$
• C. $y=xco{t}^{-1}\left(lo{g}_{{}_e}ex\right)$
• D. $y=co{t}^{-1}\left(lo{g}_{{}_e}\frac{e}{x}\right)$

Answer 1

Answer: (C)

Given, $\frac{dy}{dx}=\frac{y}{x}-{\sin }^2\left(\frac{y}{x}\right)$
Put $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$v+x\frac{dv}{dx}=v-{\sin }^{2}v$
$\Rightarrow -{\mathrm{cosec}}^{2}vdv=\frac{dx}{x}$
Integrating both sides, we get $-\int {\mathrm{cosec}}^{2}vdv=\int \frac{dx}{x}$
$\Rightarrow \cot v=\log x+C$
$\cot \frac{y}{x}=\log x+C$
This curve passes through the ponit $\left(1,\frac{\pi }{4}\right)$
$\therefore C=1$
$\Rightarrow \cot \frac{y}{x}=\log x+{\log }_{e}e$
$\Rightarrow \cot \frac{y}{x}=\log xe$
$\Rightarrow y=x{\cot }^{-1}(\log xe)$