Question

Consider the function \( f(x) = \frac{|x - 1|}{x^2} \). Then \( f(x) \) is:

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

Hint:

Check the first derivative of the function to determine intervals of increase or decrease.

Answer 1

The Correct Option is D

Step 1: Given function:
\[ f(x) = \begin{cases} \frac{x - 1}{x^2}, & x \geq 1 \\ \frac{-x + 1}{x^2}, & x < 1 \end{cases} \] Simplified \( f(x) \):
\[ f(x) = \begin{cases} \frac{1}{x} - \frac{1}{x^2}, & x \geq 1 \\ -\frac{1}{x} + \frac{1}{x^2}, & x < 1 \end{cases} \] Derivative \( f'(x) \):
\[ f'(x) = \begin{cases} -\frac{1}{x^2} + \frac{2}{x^3}, & x \geq 1 \\ \frac{1}{x^2} - \frac{2}{x^3}, & x < 1 \end{cases} \] Further simplified \( f'(x) \):
\[ f'(x) = \begin{cases} \frac{2 - x}{x^3}, & x \geq 1 \\ \frac{x - 2}{x^3}, & x < 1 \end{cases} \] Sign analysis of \( f'(x) \):

• Positive for \( x < 0 \) and \( 1 < x < 2 \)
• Negative for \( 0 < x < 1 \) and \( x > 2 \)

Monotonicity intervals of \( f(x) \):

\(-\infty\)		0		1		2		\(+\infty\)
	+		-		+		-
		Undefined

• Decreasing in \( (0, 1) \cup (2, \infty) \)
• Increasing in \( (-\infty, 0) \cup (1, 2) \)