Question

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

• A. increasing on $[1,\infty )$
• B. decreasing on $[0,\infty )$
• C. increasing on $[0,\frac{\pi }{e}]$ and decreasing on $[\frac{\pi }{e},0]$
• D. decreasing on $[0,\frac{\pi }{e}]$ and increasing on $[\frac{\pi }{e},0]$

Answer 1

The correct option is B decreasing on $[0,\infty )$

$f\left(x\right)=\frac{\log (\pi +x)}{\log (e+x)}$

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

Let $h\left(x\right)=\frac{\log (e+x)}{\pi +x}-\frac{\log (\pi +x)}{e+x}$

Let us consider $g\left(x\right)=x\ln x$

$\Rightarrow g^{\prime }\left(x\right)=x\times \frac{1}{x}+\ln x=1+\ln x$

$g^{\prime }\left(x\right)>0\forall x\in (\frac{1}{e},\infty )$ and $g^{\prime }\left(x\right)<0\forall x\in (0,\frac{1}{e})$

Now $e<\pi \Rightarrow e+x<\pi +x\forall x\in (0,\infty )$

$\Rightarrow g(e+x)<g(\pi +x)$ $[\because g(x)$ is an increasing function for $x>\frac{1}{e}]$

$\Rightarrow (e+x)\ln (e+x)<(\pi +x)\ln (\pi +x)$

$\Rightarrow \frac{\ln (e+x)}{\pi +x}<\frac{\ln (\pi +x)}{e+x}$

$\Rightarrow \frac{\ln (e+x)}{\pi +x}-\frac{\ln (\pi +x)}{e+x}<0$

$\Rightarrow h\left(x\right)<0$

$\therefore f^{\prime }\left(x\right)=\frac{h\left(x\right)}{{\left(\ln (e+x)\right)}^2}<0\forall x\in [0,\infty )$

Hence $f\left(x\right)$ decreases for $[0,\infty )$