Question

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

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

Answer 1

The correct option is B

Decreasing on $(0,\infty )$

Explanation for correct option

Step 1 Determine the interval of the function

The given function is, $\mathrm{f}\left(\mathrm{x}\right)=\frac{\ln \left(\mathrm{\pi }+\mathrm{x}\right)}{\ln \left(\mathrm{e}+\mathrm{x}\right)}$

Differentiate the given function.

$\Rightarrow$ $\mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\ln \left(\mathrm{\pi }+\mathrm{x}\right)}{\ln \left(\mathrm{e}+\mathrm{x}\right)}\right)$

$\Rightarrow$ $\mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{{\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}}\left(\ln \left(\mathrm{\pi }+\mathrm{x}\right)\right)·\ln \left(\mathrm{e}+\mathrm{x}\right)-{\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}}\left(\ln \left(\mathrm{e}+\mathrm{x}\right)\right)·\ln \left(\mathrm{\pi }+\mathrm{x}\right)}{{\left(\ln \left(\mathrm{e}+\mathrm{x}\right)\right)}^2}$ $\left[{\left(\frac{\mathbf{f}}{\mathbf{g}}\right)}^{,}=\frac{\mathbf{f}\mathbf{\text{'}}\mathbf{·}\mathbf{g}\mathbf{-}\mathbf{g}\mathbf{\text{'}}\mathbf{·}\mathbf{f}}{{\mathbf{g}}^{\mathbf{2}}}\right]$

We will simplify further by using the Chain rule.

$\Rightarrow$ $\mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{\left({\displaystyle \frac{1}{\mathrm{\pi }+\mathrm{x}}}\times {\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}}\left(\mathrm{\pi }+\mathrm{x}\right)\right)·\ln \left(\mathrm{e}+\mathrm{x}\right)-\left({\displaystyle \frac{1}{\mathrm{e}+\mathrm{x}}}\times {\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}}\left(\mathrm{e}+\mathrm{x}\right)\right)\times \ln \left(\mathrm{\pi }+\mathrm{x}\right)}{{\left(\ln \left(\mathrm{e}+\mathrm{x}\right)\right)}^2}$

Step 2: Apply sum/difference rule

We will simplify the function by using the sum/difference rule is, $\left(f\pm g\right)\text{'}\mathbf{=}\mathbf{f}\mathbf{\text{'}}\mathbf{\pm }\mathbf{g}\mathbf{\text{'}}$.

$\Rightarrow$ $\mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{\left({\displaystyle \frac{1}{\mathrm{\pi }+\mathrm{x}}}\times \left(\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{\pi }\right)+{\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}}\left(\mathrm{x}\right)\right)\right)·\ln \left(\mathrm{e}+\mathrm{x}\right)-\left({\displaystyle \frac{1}{\mathrm{e}+\mathrm{x}}}\times \left(\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{e}\right)+{\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}}\left(\mathrm{x}\right)\right)\right)\times \ln \left(\mathrm{\pi }+\mathrm{x}\right)}{{\left(\ln \left(\mathrm{e}+\mathrm{x}\right)\right)}^2}$

$\Rightarrow$ $\mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{\left({\displaystyle \frac{1}{\mathrm{\pi }+\mathrm{x}}}\right)·\ln \left(\mathrm{e}+\mathrm{x}\right)-\left({\displaystyle \frac{1}{\mathrm{e}+\mathrm{x}}}\right)\times \ln \left(\mathrm{\pi }+\mathrm{x}\right)}{{\ln }^2\left(\mathrm{e}+\mathrm{x}\right)}$

Step 3: Use Multiply fraction method.

$\Rightarrow$ $\mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{{\displaystyle \frac{\ln \left(\mathrm{e}+\mathrm{x}\right)}{\mathrm{\pi }+\mathrm{x}}}-{\displaystyle \frac{\ln \left(\mathrm{x}+\mathrm{\pi }\right)}{\mathrm{e}+\mathrm{x}}}}{{\ln }^2\left(\mathrm{e}+\mathrm{x}\right)}$

Combine the fractions

$\Rightarrow$ $\mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{{\displaystyle \ln \left(\mathrm{e}+\mathrm{x}\right)\left(\mathrm{x}+\mathrm{e}\right)-\ln \left(\mathrm{\pi }+\mathrm{x}\right)\left(\mathrm{x}+\mathrm{\pi }\right)}}{{\ln }^2\left(\mathrm{e}+\mathrm{x}\right)\left(\mathrm{x}+\mathrm{\pi }\right)\left(\mathrm{x}+\mathrm{e}\right)}$

Since, $\left(\mathrm{\pi }+\mathrm{x}\right)\ln \left(\mathrm{\pi }+\mathrm{x}\right)>\left(\mathrm{e}+\mathrm{x}\right)\ln \left(\mathrm{\pi }+\mathrm{x}\right)>\left(\mathrm{e}+\mathrm{x}\right)\ln \left(\mathrm{e}+\mathrm{x}\right)$, $\forall ·\mathrm{x}>0$.

So, $\mathrm{f}\text{'}\left(\mathrm{x}\right)<0$

Hence, $\mathrm{f}\left(\mathrm{x}\right)$ is decreasing for $\mathrm{x}\in \left(0,\infty \right)$.

Therefore, option $\mathrm{B}$ is correct answer.