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. $Noneofthese$

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
$v+x\frac{dv}{dx}=v-co{s}^{2}v$ or $\frac{dv}{co{s}^{2}v}=-\frac{dx}{x}$
On integrating, we get $tanv=-logx+logc$
$\Rightarrow tan\left(\frac{y}{x}\right)=-logx+logC$
This passes through $(1,\frac{\pi }{4})$, therefore 1 =log c
$\Rightarrow tan\left(\frac{y}{x}\right)=-logx+loge\Rightarrow y=xta{n}^{-1}\left[log\left(\frac{e}{x}\right)\right]$.