Continuity and Differentiability of Piecewise Function with Greatest Integer Function

To determine the correct options, we analyze the function $f: (0,1) \rightarrow \mathbb{R}$ given by: $f(x) = [4x] \left(x - \frac{1}{4}\right)^2 \left(x - \frac{1}{2}\right)$

We first partition the domain $(0,1)$ into intervals based on the values taken by the greatest integer function $[4x]$:

1. For $x \in \left(0, \frac{1}{4}\right)$, $4x \in (0,1) \implies [4x] = 0$: $f(x) = 0$

2. For $x \in \left[\frac{1}{4}, \frac{1}{2}\right)$, $4x \in [1,2) \implies [4x] = 1$: $f(x) = \left(x - \frac{1}{4}\right)^2 \left(x - \frac{1}{2}\right)$

3. For $x \in \left[\frac{1}{2}, \frac{3}{4}\right)$, $4x \in [2,3) \implies [4x] = 2$: $f(x) = 2\left(x - \frac{1}{4}\right)^2 \left(x - \frac{1}{2}\right)$

4. For $x \in \left[\frac{3}{4}, 1\right)$, $4x \in [3,4) \implies [4x] = 3$: $f(x) = 3\left(x - \frac{1}{4}\right)^2 \left(x - \frac{1}{2}\right)$

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1. Continuity and Differentiability Analysis

The potential points of discontinuity and non-differentiability in $(0,1)$ are $x = \frac{1}{4}, \frac{1}{2}, \frac{3}{4}$.

• At $x = \frac{1}{4}$: $\lim_{x \to \frac{1}{4}^-} f(x) = 0$ $f\left(\frac{1}{4}\right) = 1 \cdot \left(\frac{1}{4} - \frac{1}{4}\right)^2 \left(\frac{1}{4} - \frac{1}{2}\right) = 0$ $\lim_{x \to \frac{1}{4}^+} f(x) = 0$ Thus, $f(x)$ is continuous at $x = \frac{1}{4}$.

  Now checking derivatives:

  • Left-hand derivative: $f' \left(\frac{1}{4}^-\right) = 0$
  • Right-hand derivative: $f'(x) = 2\left(x - \frac{1}{4}\right)\left(x - \frac{1}{2}\right) + \left(x - \frac{1}{4}\right)^2 \implies f'\left(\frac{1}{4}^+\right) = 0$ Since $f'\left(\frac{1}{4}^-\right) = f'\left(\frac{1}{4}^+\right) = 0$, $f(x)$ is differentiable at $x = \frac{1}{4}$.

• At $x = \frac{1}{2}$: $\lim_{x \to \frac{1}{2}^-} f(x) = 1 \cdot \left(\frac{1}{2} - \frac{1}{4}\right)^2 \left(\frac{1}{2} - \frac{1}{2}\right) = 0$ $f\left(\frac{1}{2}\right) = 2 \cdot \left(\frac{1}{2} - \frac{1}{4}\right)^2 \left(\frac{1}{2} - \frac{1}{2}\right) = 0$ $\lim_{x \to \frac{1}{2}^+} f(x) = 0$ Thus, $f(x)$ is continuous at $x = \frac{1}{2}$.

  Now checking derivatives:

  • Left-hand derivative: $f'\left(\frac{1}{2}^-\right) = 2\left(\frac{1}{2} - \frac{1}{4}\right)\left(\frac{1}{2} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{4}\right)^2 = \left(\frac{1}{4}\right)^2 = \frac{1}{16}$
  • Right-hand derivative: $f'\left(\frac{1}{2}^+\right) = 2 \left[ 2\left(\frac{1}{2} - \frac{1}{4}\right)\left(\frac{1}{2} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{4}\right)^2 \right] = 2 \left(\frac{1}{16}\right) = \frac{1}{8}$ Since $f'\left(\frac{1}{2}^-\right) \neq f'\left(\frac{1}{2}^+\right)$, $f(x)$ is continuous but NOT differentiable at $x = \frac{1}{2}$.

• At $x = \frac{3}{4}$: $\lim_{x \to \frac{3}{4}^-} f(x) = 2 \left(\frac{3}{4} - \frac{1}{4}\right)^2 \left(\frac{3}{4} - \frac{1}{2}\right) = 2 \left(\frac{1}{2}\right)^2 \left(\frac{1}{4}\right) = \frac{1}{8}$ $f\left(\frac{3}{4}\right) = 3 \left(\frac{3}{4} - \frac{1}{4}\right)^2 \left(\frac{3}{4} - \frac{1}{2}\right) = 3 \left(\frac{1}{2}\right)^2 \left(\frac{1}{4}\right) = \frac{3}{8}$ Since $\lim_{x \to \frac{3}{4}^-} f(x) \neq f\left(\frac{3}{4}\right)$, $f(x)$ is discontinuous at $x = \frac{3}{4}$. Consequently, it is also NOT differentiable at $x = \frac{3}{4}$.

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2. Evaluation of Options A, B, and C

• Option (A): The function $f$ is discontinuous only at $x = \frac{3}{4}$ in $(0,1)$. Thus, it is discontinuous at exactly one point. (Correct)
• Option (B): The function $f$ is continuous but NOT differentiable only at $x = \frac{1}{2}$ in $(0,1)$. Thus, there is exactly one point satisfying this condition. (Correct)
• Option (C): The function $f$ is not differentiable at $x = \frac{1}{2}$ and $x = \frac{3}{4}$ (total 2 points). It is not non-differentiable at more than three points. (Incorrect)

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3. Minimum Value of $f(x)$ (Option D)

• For $x \in \left(0, \frac{1}{4}\right)$, $f(x) = 0$.
• For $x \in \left[\frac{1}{2}, 1\right)$, $\left(x - \frac{1}{2}\right) \ge 0$, so $f(x) \ge 0$.
• For $x \in \left[\frac{1}{4}, \frac{1}{2}\right)$, $f(x) = \left(x - \frac{1}{4}\right)^2 \left(x - \frac{1}{2}\right) \le 0$.

To find the minimum value of $g(x) = \left(x - \frac{1}{4}\right)^2 \left(x - \frac{1}{2}\right)$ on $\left[\frac{1}{4}, \frac{1}{2}\right]$: $g'(x) = \left(x - \frac{1}{4}\right) \left[ 2\left(x - \frac{1}{2}\right) + \left(x - \frac{1}{4}\right) \right] = \left(x - \frac{1}{4}\right) \left(3x - \frac{5}{4}\right)$

Setting $g'(x) = 0$ yields critical points at $x = \frac{1}{4}$ and $x = \frac{5}{12}$.

Evaluating $g\left(\frac{5}{12}\right)$: $g\left(\frac{5}{12}\right) = \left(\frac{5}{12} - \frac{1}{4}\right)^2 \left(\frac{5}{12} - \frac{1}{2}\right) = \left(\frac{1}{6}\right)^2 \left(-\frac{1}{12}\right) = -\frac{1}{432}$

Hence, the minimum value of $f(x)$ is $-\frac{1}{432}$, not $-\frac{1}{512}$. Option (D) is Incorrect.

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Conclusion

The correct options are A and B.