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6. Application of derivatives — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths6. Application of derivatives1 Mark
MCQ
${x^x}$ has a stationary point at
  • A
    $x = e$
  • ✓
    $x = {1 \over e}$
  • C
    $x = 1$
  • D
    $x = \sqrt e $
✓

Answer

Correct option: B.
$x = {1 \over e}$
b
(b) Let $y = {x^x}$==> $\log y = x.\log x,\;\;\;\;(x > 0)$

Differentiating $\frac{{dy}}{{dx}} = {x^x}(1 + \log x)$;

$\therefore \frac{{dy}}{{dx}} = 0$

==> $\log x = - 1$==>$x = {e^{ - 1}} = \frac{1}{e}$

$\therefore $ Stationary point is $x = \frac{1}{e}$

$\frac{{{d^2}y}}{{d{x^2}}} = {x^x}{(1 + \log x)^2} + {x^x}.\frac{1}{x}$

When $x = \frac{1}{e},\;\;\frac{{{d^2}y}}{{d{x^2}}} = {\left( {\frac{1}{e}} \right)^{(1/e) - 1}} > 0$

Therefore $y$ is minimum at $x = \frac{1}{e}$ and

minimum value $ = {\left( {\frac{1}{e}} \right)^{1/e}} = {e^{ - 1/e}}$.

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