Question

The function $\frac{\left|x-1\right|}{x^2}$ is monotonically decreasing at the point

• A. $x=3$
• B. $x=1$
• C. $x=2$
• D. none of these

Answer 1

The correct option is B $x=1$

$f\left(x\right)=\frac{|x-1|}{x^2}$

when $=\frac{1-x}{x^2},\frac{1}{x},-\frac{1}{x},f^{\prime }\left(x\right)=\frac{-2}{x^3}+\frac{1}{x^2}{f}^{\prime }\left(x\right)=\frac{1}{x^2}-\frac{2}{x^3}=\frac{1}{x^2}[1-\frac{2}{x}]f^{\prime }\left(x\right)<0,1-\frac{2}{x}<0=\frac{x-2}{x-0}<0x\epsilon (0,2)$

but $x\le 1$

if $x$ is an $integer$ $x=1$