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: \[ f(x) = \begin{cases} \frac{x-1}{x^2}; & x \geq 1
\frac{-x+1}{x^2}; & x<1 \end{cases} \] Simplifying \(f(x)\): \[ f(x) = \begin{cases} \frac{1}{x} - \frac{1}{x^2}; & x \geq 1
-\frac{1}{x} + \frac{1}{x^2}; & x<1 \end{cases} \] Differentiating \(f(x)\) with respect to \(x\): \[ f'(x) = \begin{cases} -\frac{1}{x^2} + \frac{2}{x^3}; & x \geq 1
\frac{1}{x^2} - \frac{2}{x^3}; & x<1 \end{cases} \] Further simplification of \(f'(x)\): \[ f'(x) = \begin{cases} \frac{2-x}{x^3}; & x \geq 1
\frac{x-2}{x^3}; & x<1 \end{cases} \] Observation of \(f'(x)\): By observation, \(f'(x)\) will be: \begin{itemize} \item Positive for \(x<0\) and \(1<x<2\) \item Negative for \(0<x<1\) and \(x>2\) \end{itemize} Monotonicity of \(f(x)\): \begin{center} \begin{tabular}{c|c|c|c|c|c} \(-\infty\) & & 0 & & 1 & & 2 & & \(+\infty\)
\hline & + & & - & & + & & - &
& & Undefined & & & & & &
\end{tabular} \end{center} \begin{itemize} \item Decreasing in \((0, 1) \cup (2, \infty)\) \item Increasing in \((-\infty, 0) \cup (1, 2)\) \end{itemize}