MCQ
The function $f(x) = {x^{1/x}}$ is
  • A
    Increasing in $(1,\,\,\infty )$
  • B
    Decreasing in $(1,\,\,\infty )$
  • Increasing in $(1,\,e),$ decreasing in $(e,\infty )$
  • D
    Decreasing in $(1,\,e),$ increasing in$(e,\infty )$

Answer

Correct option: C.
Increasing in $(1,\,e),$ decreasing in $(e,\infty )$
c
(c) Let $y = {x^{1/x}}$==> $\log y = \frac{1}{x}\log x$

==>$\frac{1}{y}\frac{{dy}}{{dx}} = \frac{1}{{{x^2}}} - \frac{{\log x}}{{{x^2}}} = \frac{{1 - \log x}}{{{x^2}}}$

==>$\frac{{dy}}{{dx}} = {x^{1/x}}\left( {\frac{{1 - \log x}}{{{x^2}}}} \right)$

Now, ${x^{1/x}} > 0$for all x and $\frac{{1 - \log x}}{{{x^2}}} > 0$ in $ (1, e) $ and 

$\frac{{1 - \log x}}{{{x^2}}} < 0$ in $(e,\infty )$

$\therefore$ $f(x)$ is increasing in $ (1, e) $ and decreasing in $(e,\,\infty ).$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free