If $a < 0, f(x)=e^{a x}+e^{-a x}$ and S = {x : f(x) is monotonically increasing}, then S equals

{x : x > 0}

We have,

$f(x)=e^{a x}+e^{-a x} \Rightarrow f'(x)=a\left(e^{a x}-e^{-a x}\right)$

For f(x) to be increasing, we must have

$f'(x)>0$

$\Rightarrow a\left(e^{a x}-e^{-a x}\right)>0$

$\Rightarrow e^{a x}-e^{-a x}<0$                     [∵ a < 0]

$\Rightarrow e^{-b x}-e^{b x}<0$ where, a = -b and b > 0

$\Rightarrow e^{b x}-e^{-b x}>0 \Rightarrow x > 0$

Hence, S = {x : x > 0}