When the tangent to the curve y = x log x is parallel

Solution:

(a) $e^{1 / 1-e}$

Given:

$y=f(x)=x \log x$

Differentiating the given function with respect to x,  we get

$f^{\prime}(x)=1+\log x$

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

Also,

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

The tangent to the curve is parallel to the chord joining the points $(1,0)$ and $(e, e)$.

$\therefore m=1+\log x$

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

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

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

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

 

$\Rightarrow x=e^{\frac{1}{e-1}}$