MCQ
$\,\,{f}{\text{ (x) = }}{{\text{e}}^{{\text{ax}}}}{\text{ + }}{{\text{e}}^{{\text{ - ax }}}}{\text{ }}, a > 0$ એ $x$ ની કઇ કિંમત માટે વધતુ વિધેય છે ?
- A$x > 0$
- B$x < 0$
- C$x > 1$
- D$x < 1$
$\therefore \,\,{f}'\,\,\left( x \right)\,\, = \,\,a{e^{ax}}\, - \,a{e^{ - ax}}\,\,$
$ = \,\,a{e^{ - ax}}\,\,\left( {{e^{2ax\,}} - 1} \right)\,\,$
$ = \,\,\frac{a}{{{e^{ax}}}}\,\left( {{e^{2ax}}\, - \,1} \right)\,\, $
$= \,\,\frac{a}{{{e^{ax}}}}\,\,\left( {{e^{2ax}}\,\, - \,\,1} \right)$
તેથી જો $ x > 0 $ તો $ f (x) > 0 [∵ a > 0]$
$ x > 0 $ માટે $f$ વધતુ વિધેય છે.
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