Question

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

• A. increasing in $(0,1) \cup(2, \infty)$
• B. Decreasing in $(-\infty, 0) \cup(1,2)$
• C. Increasing in $(0,1) \cup(2, \infty)$
• D. decreasing in $(0,1) \cup(2, \infty)$

Answer 1

Answer: (D)

$f(x)=\frac{|x-1|}{x^{2}}= \begin{cases}\frac{1-x}{x^{2}}, & x<1 x \neq 0 \\ \frac{x-1}{x^{2}}, & x>1\end{cases}$
Clearly, $f(x)$ is continuous for all $x$ in $R$ except at $x=0$.
$f^{\prime}(x)= \begin{cases}\frac{x-2}{x^{3}}, & x<1 x \neq 0 \\ \frac{2-x}{x^{3}}, & x>1\end{cases}$
$f^{\prime}(x)>0 \Rightarrow x < 0 $ or $ 1 < x < 2 $
$f^{\prime}(x) < 0 \Rightarrow 0 < x < 1 $ or $ x>2$
Hence, $f(x)$ is increasing in $(-\infty, 0) \cup(1,2)$ and decreasing in $(0,1) \cup(2, \infty)$.