Question

The function $f(x) = \frac{\log(\pi + x)}{\log(e + x)}(x \geq 0)$ is

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

Answer 1

Answer: (B)

Decreasing on $\lbrack 0,\infty)$

Answer 2

$f'(x) = \frac{\log(e + x)\text{ x }\frac{1}{\pi + x} - \log(\pi + x)\frac{1}{e + x}}{\left( \log(e + x) \right)^{2}}$= $\frac{\log(e + x)\text{ x }(e + x) - (\pi + x)\log(\pi + x)}{(\pi + x)(e + x)\left( \log(e + x) \right)^{2}}$since log

function is an increasing function and $e < \pi,\log(e + x) < \log(\pi + x)$.

Thus$(e + x)\log(e + x) < (e + x)\log(\pi + x) < (\pi + x)\log(\pi + x)$ for all $x > 0$.

Thus,$f'(x) < 0$ for $\forall x > 0$ ⇒ $f(x)$ decreases on $\lbrack 0,\infty\rbrack$