Question

The function $f\left(x\right)=\frac{|x-1|}{x^2}$ is monotonically decreasing on

• A. $(1,\infty )$
• B. $(0,2)$
• C. $(0,1)$
• D. $(2,\infty )$

Answer 1

Answer: (C), (D)

The correct options are
C $(0,1)$
D $(2,\infty )$

$f\left(x\right)=\{\begin{array}{c}\frac{-(x-1)}{x^2}x<1\\ \frac{(x-1)}{x^2}x\ge 1\end{array}{f}^{\prime }\left(x\right)=\{\begin{array}{c}\frac{(x-2)}{x^3}x<1\\ \frac{-(x-2)}{x^3}x\ge 1\end{array}$

$x<1$, if $f^{\prime }\left(x\right)<0$ (for $f\left(x\right)$ to be monotonically decreasing

$\Rightarrow \frac{(x-2)}{x^3}<0\Rightarrow x\in (0,2)$

But $x<1\Rightarrow x\in (0,1)$

For $x\ge 1$, if $f^{\prime }\left(x\right)<0\Rightarrow \frac{-(x-2)}{x^3}<0\Rightarrow \frac{(x-2)}{x^3}>0\Rightarrow x\in (-\infty ,0)\cup (2,\infty )$

But, $x\ge 1\Rightarrow x\in (2,\infty )$

Hence, $x\in (0,1)\cup (2,\infty )$