Question

The function \[ f(x)=e^{ax}+e^{-ax}, \quad x\in R \] and \(a<0\), is strictly decreasing for all values of \(x\), where:

• A. \(x<0\)
• B. \(x>0\)
• C. \(x<1\)
• D. \(x>1\)

Hint:

Whenever inequalities are divided by a negative quantity, the inequality sign reverses.

Answer 1

The Correct Option is A

To determine for which values of \(x\) the function \(f(x) = e^{ax} + e^{-ax}\) is strictly decreasing, we need to analyze its derivative. 

1. The given function is \(f(x) = e^{ax} + e^{-ax}\).
2. Calculate the derivative \(f'(x)\) with respect to \(x\): \(f'(x) = \frac{d}{dx}(e^{ax}) + \frac{d}{dx}(e^{-ax}) = ae^{ax} - ae^{-ax} = a(e^{ax} - e^{-ax})\).
3. For the function to be strictly decreasing, the derivative must be less than zero: \(a(e^{ax} - e^{-ax}) < 0\).
4. Given \(a < 0\), the inequality reduces to: \(e^{ax} < e^{-ax}\).
5. The inequality \(e^{ax} < e^{-ax}\) implies, \[ (e^{ax})^2 < 1 \implies e^{2ax} < 1. \]
6. Taking the natural logarithm on both sides, we get: \(-2ax < 0 \implies x < 0\).

Thus, the function \(f(x)\) is strictly decreasing for all values of \(x\) when \(x < 0\).

Therefore, the correct option is \(x < 0\).