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 $(1,0)\mathrm{and}(e,e)$, then $c$ is equal to:

• A. $e^{\frac{1}{1-e}}$
• B. $\frac{e-1}{e}$
• C. $\frac{1}{e-1}$
• D. $e^{\frac{1}{e-1}}$

Answer 1

The correct option is D

$e^{\frac{1}{e-1}}$

Explanation for the correct answer.

Step 1: Find the slope of the tangent and slope of the line segment.

Slope of the line-segment joining the points $(1,0)\mathrm{and}(e,e)$ will be

$\frac{e}{e-1}\left[\mathrm{by}\mathrm{slope}\mathrm{of}\mathrm{line}=\frac{y_2-y_1}{x_2-x_1}\right]$

This slope should be equal to the slope of the tangent to the curve, $y=f\left(x\right)=x{\log }_{e}x$, $(x>0)$ at a point $(c,f(c\left)\right)$ as the tangent is parallel to the line segment.

Slope of the tangent will be $f\text{'}\left(x\right)$ and

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{d}{dx}\left(x{\log }_{e}x\right)\\ & =& {\log }_{e}x\left(1\right)+x\left(\frac{1}{x}\right)\\ & =& {\log }_{e}x+1\end{array}$

So, slope at $(c,f(c\left)\right)$ will be $1+\log c$.

Step 2: Find the value of $c$.

Now, we have

$$\begin{gathered} \begin{array}{rcl}& & 1+{\log }_{e}c\end{array}=\frac{e}{e-1} \\ \Rightarrow {\log }_{e}c=\frac{e}{e-1}-1 \\ \Rightarrow {\log }_{e}c=\frac{e-e+1}{e-1} \\ \Rightarrow {\log }_{e}c=\frac{1}{e-1} \\ \Rightarrow c=e^{\frac{1}{e-1}} \end{gathered}$$

Hence, option D is correct.