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

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

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

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

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

answer is B.

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Detailed Solution

$\begin{array}{l} y = x \log x \\ \frac{d y}{d x} = 1 + \log x \\ 1 + \log C = \frac{e - 0}{e - 1} \\ m = \frac{d y}{d x} a t (c, f (c)) \\ ∴ \log_{e} C = \frac{e}{e - 1} - 1 = \frac{1}{e - 1} \\ \Rightarrow C = e^{\frac{1}{e - 1}} \end{array}$

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