Consider the function $f(x)=\frac{|x-1|}{x^2}$, then $f(x)$ is

• A

  increasing in $(0,1) \cup(2, \infty)$
• B

  increasing in $(-\infty, 0) \cup(0,1)$
• C

  decreasing in $(-\infty, 0) \cup(2, \infty)$
• Correct answer
  D

  decreasing in $(0,1) \cup(2, \infty)$

Solution

Given, $f(x)=\left\{\begin{aligned} \frac{x-1}{x^2} ; & x \geq 1 \\ \frac{-x+1}{x^2} ; & x \lt 1\end{aligned}\right.$ $\begin{aligned} & f(x)= \begin{cases}\frac{1}{x}-\frac{1}{x^2} ; & x \geq 1 \\ -\frac{1}{x}+\frac{1}{x^2} ; & x \lt 1\end{cases} \\ & f^{\prime}(x)= \begin{cases}\frac{-1}{x^2}+\frac{2}{x^3} ; & x \geq 1 \\ \frac{1}{x^2}-\frac{2}{x^3} ; & x \lt 1\end{cases} \\ & f^{\prime}(x)= \begin{cases}\frac{2-x}{x^3} ; & x \geq 1 \\ \frac{x-2}{x^3} ; & x \lt 1\end{cases} \end{aligned}$ By observation, $f^{\prime}(x)$ will be Decreasing in $(0,1) \cup(2, \infty)$ and increasing in $(-\infty, 0) \cup(1,2)$

Hence, the correct answer is option: 'D'.