Question

If the tangent to the curve, $y=f\left(x\right)=x{\log }_{e}x,$ $(x>0)$ at a point $(c,f(c\left)\right)$ is parallel to the line-segment joining the points $(1,0)$ and $(e,e)$, then $c$ is eqaul to

• A. $e^{\left(\frac{1}{1-e}\right)}$
  
• B. $\frac{e-1}{e}$
  
• C. $\frac{1}{e-1}$
  
• D. $e^{\left(\frac{1}{e-1}\right)}$
  

Answer 1

The correct option is D $e^{\left(\frac{1}{e-1}\right)}$


$y=f\left(x\right)=x\ln x$
$m_1={\frac{dy}{dx}|}_{(c,f(c\left)\right)}={\left|\right(\ln x+1\left)\right|}_{(c,f(c\left)\right)}=\ln c+1$
slope of the line joining $(1,0),(e,e)$
$m_2=\frac{e}{e-1}$
$m_2=m_1\Rightarrow \ln c+1=\frac{e}{e-1}$
$\Rightarrow \ln c=\frac{e}{e-1}-1=\frac{1}{e-1}$
$c=e^{\left(\frac{1}{e-1}\right)}$