Question

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

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

Answer 1

Answer: (B)

$x<0$

$f\left(x\right)=e^{ax}+e^{-ax}$
$\Rightarrow f^{\prime }\left(x\right)=a[e^{ax}-e^{-ax}$]
$=2a[ax+\frac{a^3{x}^3}{3}+\frac{a^5{x}^5}{5}+.......]$
$=2a^{2}x[1+\frac{a^2{x}^2}{3}+\frac{a^4{x}^4}{5}+.......]$
Now , $f^{\prime }\left(x\right)<0$ for $x<0$
Hence, $f\left(x\right)$ is decreasing for $x<0$.