Question:

**Step 1: Analyze the points of discontinuity** The function $f(x) = [4x] g(x)$ where $g(x) = \left(x - \frac{1}{4}\right)^2 \left(x - \frac{1}{2}\right)$. The term $[4x]$ has discontinuities at $x = \frac{1}{4}, \frac{1}{2}, \frac{3}{4}$ in the interval $(0, 1)$. We check if $f(x)$ is continuous at these points.\nAt $x = 1/4$, $g(1/4) = 0$. Since $g(x)$ is continuous, $f(x)$ is continuous at $x = 1/4$.\nAt $x = 1/2$, $g(1/2) = 0$. So $f(x)$ is continuous at $x = 1/2$.\nAt $x = 3/4$, $g(3/4) = (1/2)^2 (1/4) = 1/16 \neq 0$. The left limit has $[4x]=2 \implies f(3/4^-) = 1/8$. The right limit has $[4x]=3 \implies f(3/4^+) = 3/16$. Thus, $f$ is discontinuous at $x = 3/4$.\nSo $f$ is discontinuous exactly at one point. (A) is true. **Step 2: Analyze differentiability** We check the differentiability at $x = 1/4$ and $x = 1/2$.\nAt $x = 1/4$: The right derivative is $\lim_{h \to 0^+} \frac{1 \cdot h^2 (1/4+h-1/2)}{h} = 0$. The left derivative is $\lim_{h \to 0^-} \frac{0 \cdot g(1/4+h)}{h} = 0$. So $f$ is differentiable at $x = 1/4$.\nAt $x = 1/2$: The right derivative is $\lim_{h \to 0^+} \frac{2 \cdot (1/4+h)^2 h}{h} = 2(1/4)^2 = 1/8$. The left derivative is $\lim_{h \to 0^-} \frac{1 \cdot (1/4+h)^2 h}{h} = 1/16$. The left and right derivatives are not equal, so $f$ is not differentiable at $x = 1/2$.\nSince it is discontinuous at $x = 3/4$, it's not differentiable there. So $f$ is not differentiable at exactly two points ($1/2$ and $3/4$). Option (B) is true, Option (C) is false. **Step 3: Find the minimum value of f** For $x \in [1/4, 1/2)$, $f(x) = (x - 1/4)^2 (x - 1/2)$. $f'(x) = 2(x - 1/4)(x - 1/2) + (x - 1/4)^2 = (x - 1/4)(3x - 5/4)$. So $f'(x) = 0$ at $x = 5/12$. The value is $f(5/12) = (1/6)^2 (-1/12) = -1/432$.\nFor $x \in [1/2, 1)$, $f(x) \ge 0$. So the global minimum is $-1/432$. The option (D) states $-1/512$, so it is false.