MCQ
વિધેય $f(x) = \left( {{{{e^{2x}} - 1} \over {{e^{2x}} + 1}}} \right)$ એ . . . .
- ✓વધતું
- Bઘટતું
- Cયુગ્મ વિધેય
- Dએકપણ નહીં.
==>$f( - x) = \frac{{{e^{ - 2x}} - 1}}{{{e^{ - 2x}} + 1}} = \frac{{1 - {e^{2x}}}}{{1 + {e^{2x}}}}$==>$f(x) = - \frac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}} = - f(x)$
==> $f(x)$ is an odd function.
Again $f(x) = \frac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}} \Rightarrow f'(x) = \frac{{4{e^{2x}}}}{{{{(1 + {e^{2x}})}^2}}} > 0\,\forall \,n \in R$
==> $f(x)$ is an increasing function.
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