Question

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

• A. $y={\cot }^{-1}\left({\log }_{e}x\right)$
• B. $y={\cot }^{-1}\left({\log }_e\frac{x}{e}\right)$
• C. $y=x{\cot }^{-1}\left({\log }_{e}ex\right)$
• D. $y={\cot }^{-1}\left({\log }_e\frac{e}{x}\right)$

Answer 1

The correct option is C $y=x{\cot }^{-1}\left({\log }_{e}ex\right)$

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

Put $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$

$v+x\frac{dv}{dx}=v-{\sin }^{2}v$
$\Rightarrow -cose{c}^{2}vdv=\frac{dx}{x}$

Integrating both sides, we get

$-\int cose{c}^{2}vdv=\int \frac{dx}{x}$
$\Rightarrow \cot v=\log x+C$
$\cot \frac{y}{x}=\log x+C$

Curve passes through the point $(1,\frac{\pi }{4})$

$\therefore C=1$
$\Rightarrow \cot \frac{y}{x}=\log x+{\log }_{e}e$

$\Rightarrow \cot \frac{y}{x}=\log xe$

$\Rightarrow y=x{\cot }^{-1}\left(\log xe\right)$ is the equation of curve.