Let $$f : (0,1) \to \mathbb{R}$$ be the function defined as $$f(x) = [4x]\left(x - \frac{1}{4}\right)^2\left(x - \frac{1}{2}\right)$$, where $$[x]$$ denotes the greatest integer less than or equal to $$x$$. Then which of the following statements is(are) true?

For $$x \in (0,1)$$ the term $$[4x]$$ (greatest-integer function) is constant on each open sub-interval of length $$\tfrac14$$.
Explicitly

$$[4x]=\begin{cases} 0,& 0\lt x\lt \tfrac14\\ 1,& \tfrac14\le x\lt \tfrac12\\ 2,& \tfrac12\le x\lt \tfrac34\\ 3,& \tfrac34\le x\lt 1 \end{cases}$$

Hence the given function is

$$f(x)=\begin{cases} 0,& 0\lt x\lt \tfrac14\\[4pt] (x-\tfrac14)^2\,(x-\tfrac12),& \tfrac14\le x\lt \tfrac12\\[4pt] 2\,(x-\tfrac14)^2\,(x-\tfrac12),& \tfrac12\le x\lt \tfrac34\\[4pt] 3\,(x-\tfrac14)^2\,(x-\tfrac12),& \tfrac34\le x\lt 1 \end{cases}$$

The only points where discontinuity or non-differentiability can occur are the junctions $$x=\tfrac14,\ \tfrac12,\ \tfrac34$$ because inside each open interval the formula is a polynomial.

Left-hand value: $$\lim_{x\to(\tfrac14)^-}f(x)=0$$ (since $$f=0$$ on the first interval).
Function value: $$f(\tfrac14)=1\cdot0^2(-\tfrac14)=0$$.
Right-hand value: $$\lim_{x\to(\tfrac14)^+}(x-\tfrac14)^2(x-\tfrac12)=0$$.
All three coincide, so the function is continuous.

For $$x\ge\tfrac14$$ write $$g(x)=(x-\tfrac14)^2(x-\tfrac12)$$. Then $$g'(x)=2(x-\tfrac14)(x-\tfrac12)+(x-\tfrac14)^2$$.
Evaluating at $$x=\tfrac14$$ gives $$g'(\tfrac14)=0$$, whereas the left derivative (from the constant zero part) is also $$0$$. Thus $$f$$ is differentiable at $$x=\tfrac14$$.

Left-hand value: $$\lim_{x\to(\tfrac12)^-}(x-\tfrac14)^2(x-\tfrac12)=0$$.
Right-hand value: $$\lim_{x\to(\tfrac12)^+}2(x-\tfrac14)^2(x-\tfrac12)=0$$.
Function value: $$f(\tfrac12)=2\cdot(\tfrac14)^2\cdot0=0$$.
Therefore $$f$$ is continuous at $$x=\tfrac12$$.

Left derivative: $$g'(\tfrac12)=\tfrac1{16}$$.
Right derivative: $$\dfrac{d}{dx}[\,2g(x)\,]\big|_{x=\tfrac12}=2g'(\tfrac12)=\tfrac18$$.
Because $$\tfrac1{16}\ne\tfrac18$$, the derivatives differ and $$f$$ is not differentiable at $$x=\tfrac12$$.

Left-hand value: $$\lim_{x\to(\tfrac34)^-}2(x-\tfrac14)^2(x-\tfrac12)=2\bigl(\tfrac12\bigr)^2\bigl(\tfrac14\bigr)=\tfrac18$$.
Right-hand value: $$\lim_{x\to(\tfrac34)^+}3(x-\tfrac14)^2(x-\tfrac12)=3\bigl(\tfrac12\bigr)^2\bigl(\tfrac14\bigr)=\tfrac{3}{16}$$.
Since the limits are unequal, $$f$$ is discontinuous at $$x=\tfrac34$$.

There are no other points to check, so

• $$f$$ is discontinuous only at $$x=\tfrac34$$ ⇒ exactly one point of discontinuity in $$(0,1)$$.
• Among the remaining junctions, only $$x=\tfrac12$$ is continuous but not differentiable ⇒ exactly one such point.

Hence Options A and B are true.

Concerning Option C: non-differentiability occurs at $$x=\tfrac12$$ (corner) and $$x=\tfrac34$$ (discontinuity), i.e. at two points, not “more than three” ⇒ Option C is false.

Concerning Option D: the function can be negative only on $$(\tfrac14,\tfrac12)$$ where $$h(x)=(x-\tfrac14)^2(x-\tfrac12)$$. Setting $$h'(x)=0$$ yields the critical point $$x=\tfrac5{12}$$. At this point $$h\!\left(\tfrac5{12}\right)=\left(\tfrac1{6}\right)^2\!\left(-\tfrac1{12}\right)=-\tfrac1{432}$$. Since this is less than $$-\tfrac1{512}$$, the claimed minimum is incorrect ⇒ Option D is false.

Final result: Option A and Option B are the only correct statements.