Question

The function $f(x) = \frac{\ln(\pi + x)}{\ln(e + x)}$ is

• A. Increasing on [0, ∞)
• B. Decreasing on [0, ∞)
• C. Decreasing on $\left\lbrack 0,\frac{\pi}{e} \right)$ and increasing on $\left\lbrack \frac{\pi}{e},\infty \right)$
• D. Increasing on $\left\lbrack 0,\frac{\pi}{e} \right)$ and decreasing on $\left\lbrack \frac{\pi}{e},\infty \right)$

Answer 1

Answer: (B)

Decreasing on [0, ∞)

Answer 2

Let $f(x) = \frac{\ln(\pi + x)}{\ln(e + x)}$

$\therefore$ $f^{'}(x) = \frac{\ln(e + x) \times \frac{1}{\pi + x} - \ln(\pi + x)\frac{1}{e + x}}{\ln^{2}(e + x)}$

= $\frac{(e + x)\ln(e + x) - (\pi + x)\ln(\pi + x)}{\{\ln(e + x)\}^{2} \times (e + x)(\pi + x)}$

⇒ $f^{'}(x) < 0$ for all $x \geq 0$ $\{\because\pi > e\}$

Hence, $f(x)$ is decreasing in $\lbrack 0,\infty)$.