If the tangent to the curve y = f(x) = x log_ | vedclass

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(C) The slope of the line segment joining $(1, 0)$ and $(e, e)$ is given by $m = \frac{e - 0}{e - 1} = \frac{e}{e - 1}$.The derivative of the function

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$e^{\left(\frac{1}{e - 1}\right)}$

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(C) The slope of the line segment joining $(1, 0)$ and $(e, e)$ is given by $m = \frac{e - 0}{e - 1} = \frac{e}{e - 1}$.The derivative of the function $f(x) = x \log_{e} x$ is $f'(x) = \frac{d}{dx}(x) \cdot \log_{e} x + x \cdot \frac{d}{dx}(\log_{e} x) = 1 \cdot \log_{e} x + x \cdot \frac{1}{x} = \log_{e} x + 1$.Since the tangent at $(c, f(c))$ is parallel to the line segment,its slope $f'(c)$ must be equal to $m$:$f'(c) = \log_{e} c + 1 = \frac{e}{e - 1}$.Solving for $\log_{e} c$:$\log_{e} c = \frac{e}{e - 1} - 1 = \frac{e - (e - 1)}{e - 1} = \frac{1}{e - 1}$.Therefore,$c = e^{\frac{1}{e - 1}}$.

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 $c$ is equal to:

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