Question

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

• A. $y=ta{n}^{-1}\left(log\left(\frac{e}{x}\right)\right)$
• B. $y=xta{n}^{-1}\left(log\left(\frac{x}{e}\right)\right)$
• C. $y=xta{n}^{-1}\left(log\left(\frac{e}{x}\right)\right)$
• D. None of these

Answer 1

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

we have, $\frac{dy}{dx}=\frac{y}{x}-co{s}^2\left(\frac{y}{x}\right)$
Putting $y=vx$, so that $\frac{dy}{dx}=v+x\frac{dv}{dx}$, we get
$v+x\frac{dv}{dx}=v-co{s}^{2}v$
$\Rightarrow \frac{dv}{co{s}^{2}v}=-\frac{dx}{x}\Rightarrow se{c}^{2}udu=-\frac{1}{x}dx$
On integration, we get
$tanu=-logx+logC$
$\Rightarrow tan\left(\frac{y}{x}\right)=-logx+logC$
This passes through $(1,\frac{\pi }{4})$, therefore 1=log⁡C.
so, $tan\left(\frac{y}{x}\right)=-logx+1$
$\Rightarrow tan\left(\frac{y}{x}\right)=-logx+1\Rightarrow tan\left(\frac{y}{x}\right)=-logx+loge\Rightarrow y=xta{n}^{-1}\left(log\left(\frac{e}{x}\right)\right)$