Let f : (0, 1) → ℝ be the function defined as f(x) = [4x] (x − 1/4)² (x − 1/2), where [x] denotes the greatest integer less than or equal to x. Which of the following statements is(are) true?

Correct Statements:
1. The function f is discontinuous exactly at one point in (0, 1).
2. There is exactly one point in (0, 1) at which the function f is continuous but NOT differentiable.

Step-by-Step Explanation:

The given function is defined on the domain $x\in (0,1)$ as:

$f\left(x\right)=\left[4x\right]{(x-\frac{1}{4})}^2(x-\frac{1}{2})$

where $[·]$ denotes the greatest integer function.

To analyze the continuity and differentiability of $f\left(x\right)$, let us break the domain $(0,1)$ into sub-intervals based on the values of the step function $\left[4x\right]$:

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

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

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

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

Analysis at potential points of discontinuity/non-differentiability:

1. At $x=\frac{1}{4}$:
- Left-Hand Limit (LHL): $\underset{x\to {\frac{1}{4}}^{-}}{lim}f\left(x\right)=0$
- Right-Hand Limit (RHL): $\underset{x\to {\frac{1}{4}}^{+}}{lim}f\left(x\right)=1·{\left(0\right)}^2(\frac{1}{4}-\frac{1}{2})=0$
- Value at point: $f\left(\frac{1}{4}\right)=0$
Since $\text{LHL}=\text{RHL}=f\left(\frac{1}{4}\right)$, $f\left(x\right)$ is continuous at $x=\frac{1}{4}$.

Now checking derivatives at $x=\frac{1}{4}$:
- Left-Hand Derivative (LHD): $f^{\prime }\left({\frac{1}{4}}^{-}\right)=0$
- Right-Hand Derivative (RHD): $f^{\prime }\left({\frac{1}{4}}^{+}\right)=\frac{d}{dx}{\left[{(x-\frac{1}{4})}^2\right(x-\frac{1}{2}\left)\right]}_{x=\frac{1}{4}}^{}=2(x-\frac{1}{4}){(x-\frac{1}{2})+{(x-\frac{1}{4})}^2|}_{x=\frac{1}{4}}=0$
Since $\text{LHD}=\text{RHD}=0$, $f\left(x\right)$ is differentiable at $x=\frac{1}{4}$.

2. At $x=\frac{1}{2}$:
- LHL: $\underset{x\to {\frac{1}{2}}^{-}}{lim}f\left(x\right)=1·{\left(\frac{\mathrm{1window}}{4}\right)}^2\left(0\right)=0$
- RHL: $\underset{x\to {\frac{1}{2}}^{+}}{lim}f\left(x\right)=2·{\left(\frac{1}{4}\right)}^2\left(0\right)=0$
- Value at point: $f\left(\frac{1}{2}\right)=0$
Thus, $f\left(x\right)$ is continuous at $x=\frac{1}{2}$.

Now checking derivatives at $x=\frac{1}{2}$:
- LHD: $f^{\prime }\left({\frac{1}{2}}^{-}\right)=1·{\left[2\right(x-\frac{1}{4}\left)\right(x-\frac{1}{2})+{(x-\frac{1}{4})}^2]}_{x=\frac{1}{2}}^{}=1·{\left(\frac{1}{4}\right)}^2=\frac{1}{16}$
- RHD: $f^{\prime }\left({\frac{1}{2}}^{+}\right)=2·{\left[2\right(x-\frac{1}{4}\left)\right(x-\frac{1}{2})+{(x-\frac{1}{4})}^2]}_{x=\frac{1}{2}}^{}=2·{\left(\frac{1}{4}\right)}^2=\frac{1}{8}$
Since $\text{LHD}\ne \text{RHD}$, $f\left(x\right)$ is NOT differentiable at $x=\frac{1}{2}$.

3. At $x=\frac{3}{4}$:
- LHL: $\underset{x\to {\frac{3}{4}}^{-}}{lim}f\left(x\right)=2{(\frac{3}{4}-\frac{1}{4})}^2(\frac{3}{4}-\frac{1}{2})=2{\left(\frac{1}{2}\right)}^2\left(\frac{1}{4}\right)=\frac{1}{8}$
- RHL: $\underset{\mathrm{x<mo>\to </mo><msup><mfrac><mn>3</mn><mn>4</mn></mfrac><mo>+</mo></msup>}}{lim}f\left(x\right)=3{(\frac{3}{4}-\frac{1}{4})}^2(\frac{3}{4}-\frac{1}{2})=3{\left(\frac{1}{2}\right)}^2\left(\frac{1}{4}\right)=\frac{3}{16}$
Since $\text{LHL}\ne \text{RHL}$, $f\left(x\right)$ is discontinuous at $x=\frac{3}{4}$ (and consequently non-differentiable at this point as well).

Conclusion:
- Points of discontinuity in $(0,1)$: Exactly one point, $x=\frac{3}{4}$.
- Points where $f$ is continuous but NOT differentiable in $(0,1)$: Exactly one point, $x=\frac{1}{2}$.