Question

Let \( f(x) = |1 - 2x| \), then

• A. \( f(x) \) is continuous but not differentiable at \( x=\frac{1}{2} \).
• B. \( f(x) \) is differentiable but not continuous at \( x=\frac{1}{2} \).
• C. \( f(x) \) is both continuous and differentiable at \( x=\frac{1}{2} \).
• D. \( f(x) \) is neither differentiable nor continuous at \( x=\frac{1}{2} \).

Hint:

For \( |g(x)| \): \begin{itemize} \item Continuous everywhere. \item Not differentiable where \( g(x)=0 \) and slope changes. \end{itemize}

Answer 1

The Correct Option is A

Concept: Absolute value functions are continuous everywhere but may fail differentiability at points where inside expression is zero. Step 1: Find critical point. \[ 1 - 2x = 0 \Rightarrow x = \frac{1}{2} \] Step 2: Check continuity. Absolute value is continuous everywhere ⇒ continuous at \( \frac{1}{2} \). Step 3: Check differentiability. Piecewise form: \[ f(x)= \begin{cases} 1-2x, & x \le \frac{1}{2} \\ 2x-1, & x > \frac{1}{2} \end{cases} \] Left derivative: \[ f'_-(x) = -2 \] Right derivative: \[ f'_+(x) = 2 \] Since LHD \( \ne \) RHD, not differentiable.