Question

If $a<0,$ the function $(e^{ax}+e^{-ax})$is a decreasing function for all values of x, where

• A. $x<0$
• B. $x>0$
• C. $x<1$
• D. $x>1$

Answer 1

The correct option is A

$x<0$

Finding decreasing function:

Let, $f\left(x\right)=(e^{ax}+e^{-ax})$

Now,

$\begin{array}{rcl}f\left(x\right)& =& (e^{ax}+e^{-ax}),a>0\\ & & Differentiatewithrepecttox\\ f\text{'}\left(x\right)& =& (a{e}^{ax}-a{e}^{-ax}),a<0\\ Valueoff\text{'}\left(x\right)& <& 0therefore\\ f\text{'}\left(x\right)& =& (a{e}^{ax}-a{e}^{-ax})<0\\ (a{e}^{ax}-a{e}^{-ax})& <& 0\\ (e^{ax}-e^{-ax})& <& 0\\ (e^{ax}& <& e^{-ax})forx<0\end{array}$

Hence, correct option is $\left(A\right)$