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?

$f(x) = [4x]\left(x-\tfrac14\right)^2\left(x-\tfrac12\right),\quad x\in(0,1)$

$[x] = \text{greatest integer } \le x$

$\text{check } (1)\text{--}(4):\ \text{continuity},\ \text{differentiability},\ \min f$

$1.\ [4x] \text{ on each interval}$

$2.\ \text{continuity} + \text{differentiability at } \tfrac14,\tfrac12,\tfrac34$

$3.\ \text{minimum value of } f$

$x\in(0,\tfrac14):\ f=0;\quad x\in[\tfrac14,\tfrac12):\ f=g$

$x\in[\tfrac12,\tfrac34):\ f=2g;\quad x\in[\tfrac34,1):\ f=3g$

$g(x) = \left(x-\tfrac14\right)^2\left(x-\tfrac12\right)$

$\lim_{x\to1/4}f(x) = 0 = f(\tfrac14),\quad \lim_{x\to1/2}f(x) = 0 = f(\tfrac12)$

$\Rightarrow \text{continuous at } \tfrac14,\ \tfrac12$

$g(\tfrac34) = (\tfrac12)^2(\tfrac14) = \tfrac{1}{16}$

$x\to\tfrac34^-:\ f\to 2\cdot\tfrac{1}{16}=\tfrac18;\quad x\to\tfrac34^+:\ f\to 3\cdot\tfrac{1}{16}=\tfrac{3}{16}$

$\tfrac18 \ne \tfrac{3}{16} \Rightarrow (1)\ \text{TRUE}$

$g'(x) = 3x^2-2x+\tfrac{5}{16}$

$f'({\tfrac14}^-) = 0,\quad f'({\tfrac14}^+) = g'(\tfrac14) = 0$

$\Rightarrow \text{differentiable at } \tfrac14$

$f'({\tfrac12}^-) = g'(\tfrac12) = \tfrac{1}{16};\quad f'({\tfrac12}^+) = 2g'(\tfrac12) = \tfrac{1}{8}$

$\tfrac{1}{16} \ne \tfrac18 \Rightarrow (2)\ \text{TRUE};\ \text{only 2 non-diff. points} \Rightarrow (3)\ \text{FALSE}$

$g'(x) = 0 \Rightarrow x = \tfrac14 \text{ or } \tfrac{5}{12}$

$g(\tfrac{5}{12}) = \left(\tfrac16\right)^2\left(-\tfrac{1}{12}\right) = -\dfrac{1}{432}$

$\min f = -\dfrac{1}{432} \ne -\dfrac{1}{512} \Rightarrow (4)\ \text{FALSE}$

$\text{TRUE statements: } (1)\ \text{and}\ (2)$

$\boxed{(1)\ \text{and}\ (2)}$