Question

If $a<0$, the function $\left(e^{a x}+e^{-a x}\right)$ is a decreasing function for all values of $x$, where
A. $x<0$
B. $x>0$
C. $x<1$
D. $x>1$

Answer 1

Hint: In this question, we have given a function and it is decreasing for all values of x. By using the condition of decreasing function i.e, the first order derivative of a function is less than 0, we can get the condition of x for which all the values of x are decreasing for a given function.

Formula used:
 $f'x<0$
$\dfrac{\text{d}e^{x}}{\text{d}x}=e^{x}$

Complete Step-by-step Solution:
The given function is $\left(e^{a x}+e^{-a x}\right)$ and $a<0$
$\mathrm{f}(\mathrm{x})=\mathrm{e}^{\mathrm{ax}}+\mathrm{e}^{-\mathrm{ax}}$
Now we take the first derivative of the function
$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{a}\left[\mathrm{e}^{\mathrm{ax}}-\mathrm{e}^{-\mathrm{ax}}\right]$
The instantaneous rate of change of a function at a certain position is known as the derivative of a function. The derivative provides the precise slope of the curve at a given location.
The function's derivative, or the derivative of $y$ with respect to the variable $x$, is denoted by the symbol $dy/dx$.
We know for f(x) to be decreasing
 $f'x<0$
Then ${{e}^{ax}}-{{e}^{-ax}}>0$ ( as $a<0$)
Hence ${{e}^{ax}}>{{e}^{-ax}}$
Then $\dfrac{{{e}^{ax}}}{{{e}^{-ax}}}>1$
Or ${{e}^{2ax}}>1$
Now taking logs on both sides, we get
$2ax>{{\log }_{e}}1$
$2ax>0$
$ax>0$
As $a<0$
Then $x<0$( for $ax>0$)
Here, $\mathrm{f}^{\prime}(\mathrm{x})<0$ for $\mathrm{x}<0$. We know that, if the first-order derivative of a function $f(x)$ is less than 0 implies that $x<0$, then it is a decreasing function for all values of x, where $x<0$.
Hence, $\mathrm{f}(\mathrm{x})$ is decreasing for $x<0$.

So the correct answer is option A.

Note: It is to be noted that the first-order derivative of a function is always zero in the case of decreasing function whereas in the case of an increasing function it is always greater than zero. So, don’t make mistakes while using the condition.