Question

The number of non-differentiable points for the function $f(x) = \min\left\{x - \lfloor x \rfloor, 1 - x + \lfloor x \rfloor\right\}$ in $(-2, 2)$ is ($\lfloor x \rfloor$ represents integral part of $x$)

• A. 0
• B. 3
• C. 5
• D. 7

Hint:

For piecewise-defined functions with floor/ceiling/min/max, check for breakpoints to count non-differentiable points.

Answer 1

The Correct Option is D

To determine the number of non-differentiable points for the given function \( f(x) = \min\{x - \lfloor x \rfloor, 1 - x + \lfloor x \rfloor\} \) in the interval \((-2, 2)\), we first analyze the components. The function is defined as the minimum of two expressions:

• \(x - \lfloor x \rfloor = \{ x \}\), which represents the fractional part of \(x\).
• \(1 - x + \lfloor x \rfloor = 1 - \{ x \}\).

We express \(f(x)\) as:

\(f(x) = \min(\{x\}, 1 - \{x\})\)

The function alternates between these two values based on \(\{x\}\). The critical points occur when \(\{x\} = 0.5\) because:

• For \(\{x\} < 0.5\), \(f(x) = \{x\}\).
• For \(\{x\} > 0.5\), \(f(x) = 1 - \{x\}\).
• For \(\{x\} = 0.5\), both are equal, so function is continuous.

Differentiability can be compromised at transitions occurring at half-integers within the range. Therefore, we identify such points:

1. Half-integers: \(-1.5, -0.5, 0.5, 1.5\).
2. Integers (since floor function changes): \(-2, -1, 0, 1\).

Examining \((-2, 2)\), the function is non-differentiable at points where fractional or integral parts change, as differentiability is affected at crucial transitions: Point list: \(-1\), \(-0.5\), \(-1.5\), \(0\), \(0.5\), \(1\), \(1.5\). Thus, there are 7 non-differentiable points in \((-2, 2)\).