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STD 12 - 6. Application of derivatives — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 6. Application of derivatives1 Mark
MCQ
The function $\frac{{({e^{2x}} - 1)}}{{({e^{2x}} + 1)}}$ is
  • A
    Increasing
  • B
    Odd
  • C
    Even
  • ✓
    $A$ or $B$ both 
✓

Answer

Correct option: D.
$A$ or $B$ both 
d
(d) Since $f( - x) = \frac{{({e^{ - 2x}} - 1)}}{{({e^{ - 2x}} + 1)}} = \frac{{1 - {e^{2x}}}}{{1 + {e^{2x}}}} = - {\rm{ }}\left[ {\frac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}} \right] = - f(x)$
$\therefore f(x)$ is an odd function.
Also $f'(x) = \frac{{({e^{2x}} + 1).2{e^{2x}} - 2{e^{2x}}({e^{2x}} - 1)}}{{{{({e^{2x}} + 1)}^2}}}$
$ = \frac{{2{e^{2x}}({e^{2x}} + 1 - {e^{2x}} + 1)}}{{{{({e^{2x}} + 1)}^2}}} = \frac{{4{e^{2x}}}}{{{{({e^{2x}} + 1)}^2}}} > 0$
$ \Rightarrow f'(x)$ is $ + ve$, $\therefore f(x)$ is an increasing function.

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