Question

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

• 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
$v+x\frac{dv}{dx}=v-co{s}^{2}v\Rightarrow \frac{dv}{co{s}^{2}v}=-\frac{dx}{x}\Rightarrow se{c}^{2}vdv=\frac{1}{x}dx$
on integration, we get tan v $=-logx+logC$
$\Rightarrow tan\left(\frac{y}{x}\right)=-logx+logc$
This passes through $\left(1.\frac{\pi }{4}\right)$, therefore 1 = log C.
So tan $\left(\frac{y}{x}\right)=-logx+1$
$\Rightarrow tan\left(\frac{y}{x}\right)=-logx+loge\Rightarrow y=xta{n}^{-1}\left(log\frac{e}{x}\right)$