When the tangent to the curve `y=xlogx` is parallel to the chord joining the points (1, 0) and `(e ,\ e)` , the value of `x` is `e^(1//1-e)` (b) `e^((e-1)(2e-1))` (c) `e^((2e-1)/(e-1))` (d) `(e-1)/e`

Answer

Option A `e^(1/(1-e))`

Explanation

Given equation is,

$$ y=x \log x $$
Differentiating the function with respect to ' `x` ',
$$ \frac{d y}{d x}=1+\log \mathrm{x} $$
Slope of tangent to the curve `=1+\log x`

And, slope of the chord joining the points, `(1,0)` & `(e, e)`

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