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

Concept: A function is strictly decreasing where: \[ f'(x)<0 \]

Step 1: Differentiating the function.
Given: \[ f(x)=e^{ax}+e^{-ax} \] Differentiate: \[ f'(x)=ae^{ax}-ae^{-ax} \] \[ f'(x)=a(e^{ax}-e^{-ax}) \]

Step 2: Applying decreasing condition.
For decreasing function: \[ a(e^{ax}-e^{-ax})<0 \] Given: \[ a<0 \] Dividing inequality by negative number reverses sign: \[ e^{ax}-e^{-ax}>0 \] \[ e^{ax}>e^{-ax} \] Taking logarithm: \[ ax>-ax \] \[ 2ax>0 \] \[ ax>0 \]

Step 3: Using \(a<0\).
Since \(a\) is negative: \[ x<0 \] Hence function is strictly decreasing for: \[ x<0 \]