Question

Let $f:\left(-1,\infty \right)\to \mathrm{ℝ}$ be defined by $f\left(0\right)=1$ and $f\left(x\right)=\frac{1}{x}{\log }_e\left(1+x\right),x\ne 0$. Then the function $f$

• A. increases in $\left(-1,\infty \right)$
• B. decreases in $\left(-1,0\right)$ and increases in $\left(0,\infty \right)$
• C. increases in $\left(-1,0\right)$ and decreases in $\left(0,\infty \right)$
• D. decreases in $\left(-1,\infty \right)$

Answer 1

The correct option is D

decreases in $\left(-1,\infty \right)$

Explanation for the correct option:

Determining the range of the given function:

$f\left(x\right)=\frac{1}{x}{\log }_e\left(1+x\right),x\ne 0$

Differentiating the above function

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

The function in the denominator is always positive so we just have to check for the function in the numerator

$h\left(x\right)=\frac{x}{1+x}-{\log }_e(1+x)$

Substituting

$$\begin{gathered} \mathrm{x}=0.5\mathrm{and}\mathrm{x}=-0.5 \\ h(0.5)<0 \\ h(-0.5)<0 \end{gathered}$$

Therefore $f\text{'}\left(x\right)<0$for $x=0.5,x=-0.5$

Therefore $f$ is decreasing in $\left(-1,\infty \right)$ since it eliminates option (C)

Therefore, the correct answer is option (D).