If the tangent to the curve,
Question:

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 segement joining the points $(1,0)$ and $(e, e)$, then $c$ is equal to:

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

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

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

  4. (4) $\frac{e}{e-1}$


Correct Option: , 2

Solution:

The given tangent to the curve is,

$y=x \log _{e} x \quad(x>0)$

$\Rightarrow \frac{d y}{d x}=1+\log _{e} x$

$\left.\Rightarrow \frac{d y}{d x}\right]_{x=c}=1+\log _{e} c$ (slope)

$\because$ The tangent is parallel to line joining $(1,0),(e, e)$

$\therefore 1+\log _{e} c=\frac{e-0}{e-1}$

$\Rightarrow \log _{e} c=\frac{e}{e-1}-1 \Rightarrow \log _{e} c=\frac{1}{e-1} \Rightarrow c=e^{\frac{1}{e-1}}$

 

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