Question

Find the intervals of monotonicity of the function

$y=\frac{|x-1|}{x^2}$

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

Answer 1

Answer: (A), (B)

The correct options are
A $(-\infty ,0)\cup (1,2)$ for increasing
B $(0,1)\cup (2,\infty )$ for decreasing

$y=\frac{|x-1|}{x^2}$
Let
$x>1$, then
$y=\frac{x-1}{x^2}$
$=\frac{1}{x}-\frac{1}{x^2}$
$y^{\prime }=\frac{2}{x^3}-\frac{1}{x^2}$
For an increasing function
$y^{\prime }>0$ Or
$\frac{2}{x^3}-\frac{1}{x^2}>0$
Or
$\frac{2-x}{x^3}>0$
Since $x>1$, Hence $x^3>1$
Thus
$2-x>0$ Or $x<2$
Hence it is increasing for $xϵ(1,2)$ and decreasing for $(2,\infty )$...(i)
Hence $f\left(x\right)$ is increasing for $(-\infty ,0)\cup (1,2)$ and decreasing for $(0,1)\cup (2,\infty )$