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)

$f(x)=\frac{|x-1|}{x^2}=\{\begin{array}{ll}\frac{1-x}{x^2},& x<1x\ne 0\\ \frac{x-1}{x^2},& x>1\end{array}$
Clearly, $f(x)$ is continuous for all $x$ in $R$ except at $x=0$.
$f^{\prime }(x)=\{\begin{array}{ll}\frac{x-2}{x^3},& x<1x\ne 0\\ \frac{2-x}{x^3},& x>1\end{array}$
$f^{\prime }(x)>0\Rightarrow x<0$ 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 $(0,1)\cup (2,\infty )$.