Question

Let $f$ be a real-valued function, defined on $R-\left\{-1,1\right\}$ and given by $f\left(x\right)=3{\log }_e\left|\frac{\left(x-1\right)}{\left(x+1\right)}\right|-\frac{2}{\left(x-1\right)}$. Then in which of the following intervals, function $f\left(x\right)$ is increasing?

• A. $\left(-\infty ,-1\right)\cup \left([1/2,\infty )-\left\{1\right\}\right)$
• B. $(-1,1/2]$
• C. $\left(-\infty ,\infty \right)-\left\{-1,1\right\}$
• D. $(-\infty ,1/2]-\left\{-1\right\}$

Answer 1

The correct option is A

$\left(-\infty ,-1\right)\cup \left([1/2,\infty )-\left\{1\right\}\right)$

Explanation for correct option(s)

Option A: $\left(-\infty ,-1\right)\cup \left([1/2,\infty )-\left\{1\right\}\right)$

Consider the given equation as,

$f\left(x\right)=3{\log }_e\left|\frac{\left(x-1\right)}{\left(x+1\right)}\right|-\frac{2}{\left(x-1\right)}......\left(1\right)$

If $f\text{'}\left(x\right)>0$ then $f\left(x\right)$ is increasing function

If $f\text{'}\left(x\right)<0$ then $f\left(x\right)$ is decreasing function

We Know that

$\log \left(\frac{m}{n}\right)=\log m-\log n$

According to the above Equation we can Rewrite the Equation (1)

$$\begin{gathered} f\left(x\right)=3\left\{\log \left(x-1\right)-\log \left(x+1\right)\right\}-\frac{2}{\left(x-1\right)} \\ f\left(x\right)=3\log \left(x-1\right)-3\log \left(x+1\right)-\frac{2}{\left(x-1\right)} \\ f\text{'}\left(x\right)=3\frac{1}{x-1}-3\frac{1}{x+1}+\frac{2}{{\left(x-1\right)}^2} \\ f\text{'}\left(x\right)=\frac{3\left(x-1\right)\left(x+1\right)-3{\left(x-1\right)}^2+2\left(x+1\right)}{{\left(x-1\right)}^{2}x+1} \\ f\text{'}\left(x\right)=\frac{3\left(x^2-1\right)-3\left(x^2+1-2x\right)+2\left(x+1\right)}{{\left(x-1\right)}^{2}x+1} \\ f\text{'}\left(x\right)=\frac{3x^2-3-3x^2-3+6x+2x+2}{{\left(x-1\right)}^{2}x+1} \\ f\text{'}\left(x\right)=\frac{8x-4}{{\left(x-1\right)}^{2}x+1} \\ f\text{'}\left(x\right)=\frac{4\left(2x-1\right)}{{\left(x-1\right)}^{2}x+1} \end{gathered}$$

For increasing $f\text{'}\left(x\right)\ge 0$ then the critical points are

$$\begin{gathered} 2x-1=0 \\ x=\frac{1}{2} \\ x-1=0 \\ x=1 \\ x+1=0 \\ x=-1 \end{gathered}$$

Hence, the following intervals are $\left(-\infty ,-1\right)\cup \left([1/2,\infty )-\left\{1\right\}\right)$

Therefore, The correct answer is Option A.