MCQ
The function $f(x) = {x^{ - x}},\,(x\, \in \,R)$ attains a maximum value at $x =$
- A$x = 2$
- B$x = 3$
- ✓$x = 1/e $
- D$x = 1 $
✓
Answer
Correct option: C.
$x = 1/e $
c
(c) $f(x) = y = {x^{ - x}}$ ==> $\log y = - \,x\log x$
(c) $f(x) = y = {x^{ - x}}$ ==> $\log y = - \,x\log x$
Differentiating w.r.t. $x$ , $\frac{1}{y}.\frac{{dy}}{{dx}} = - \left[ {x.\frac{1}{x} + \log x} \right]$
==> $\frac{1}{y}.\frac{{dy}}{{dx}} = - [1 + \log x]$ ==> $\frac{{dy}}{{dx}} = - {x^{ - x}}[1 + \log x]$
==> $\frac{{dy}}{{dx}} = {x^{ - x}}\left[ {\log \frac{1}{x} - 1} \right]$
Put $\frac{{dy}}{{dx}} = 0$ ==> ${\log _e}\frac{1}{x} = {\log _e}e$
==> $\frac{1}{x} = e \Rightarrow x = \frac{1}{e}$.
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