Question

If  \(f(x) = \frac{\log(\pi + x)}{\log(e + x) }\), then the function is:

• A. Increasing in \( [0, \infty) \)
• B. Decreasing in \( [0, \infty) \)
• C. Decreasing in \( [0, \frac{\pi}{e}] \) and increasing in \( [\frac{\pi}{e}, \infty) \)
• D. Increasing in \( [0, \pi] \) and decreasing in \( [\pi, \infty) \)

Hint:

For analyzing the increasing or decreasing nature of a function, compute \( f'(x) \) and determine where it is positive or negative.

Answer 1

The Correct Option is B

To determine the behavior of \( f(x) \), we differentiate it using the quotient rule: \[ f'(x) = \frac{\frac{1}{\pi + x} \log(e + x) - \frac{1}{e + x} \log(\pi + x)}{\{\log(e + x)\}^2} \] Rewriting the numerator: \[ f'(x) = \frac{(e + x) \log(e + x) - (\pi + x) \log(\pi + x)}{(\pi + x)(e + x)\{\log(e + x)\}^2} \] Step 1: Analyze the sign of \( f'(x) \) Since \( e + x<\pi + x \), the numerator is negative for all \( x \geq 0 \). The denominator is always positive. Thus, \( f'(x)<0 \) for all \( x \in [0, \infty) \), implying that \( f(x) \) is monotonically decreasing. Conclusion: The function \( f(x) \) is decreasing in \( [0, \infty) \).