Question

When the tangent to the curve y = x log x is parallel to the chord joining the points (1, 0) and (e, e), the value of x is

(a) e^1/1−e

(b) e^{(e−1)(2e−1)}

(c) $e^{\frac{2e-1}{e-1}}$

(d) $\frac{e-1}{e}$

Answer 1

(a) e^1/1−e

Given:
$y=f\left(x\right)=x\log x$

Differentiating the given function with respect to x, we get

$f\text{'}\left(x\right)=1+\log x$

$\Rightarrow$ Slope of the tangent to the curve = $1+\log x$

Also,
Slope of the chord joining the points $\left(1,0\right)\mathrm{and}\left(e,e\right)$, (m) = $\frac{e}{e-1}$

The tangent to the curve is parallel to the chord joining the points $\left(1,0\right)\mathrm{and}\left(e,e\right)$.

$\therefore m=1+\log x$

$\Rightarrow \frac{e}{e-1}=1+\log x$

$$\begin{gathered} \Rightarrow \frac{e}{e-1}-1=\log x \\ \Rightarrow \frac{e-e+1}{e-1}=\log x \\ \Rightarrow \frac{1}{e-1}=\log x \\ \Rightarrow x=e^{\frac{1}{e-1}} \end{gathered}$$