Let f: (0, 1) → ℝ be the function defined as f(x) = [4x] (x -...

Given,

$f: ( 0, 1 ) \to \mathbb{R}$

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

We know that, \left[4x\right]$will be discontinuous function at integer points,</p><p>So,$\text{critical point} = \dfrac{1}{4}, \dfrac{1}{2}, \dfrac{3}{4}$as$x \in \left(0, 1\right)$</p><p>And function will be discontinuous at$x = \dfrac{3}{4}$because for$\dfrac{1}{2} \text{ and } \dfrac{1}{4}$, \left(x - \dfrac{1}{2}\right) \text{ and } {\left(x - \dfrac{1}{4}\right)}^{2}$ will make the function continuos

Hence, the function is discontinuous exactly at one point in $\left(0, 1\right)$, so option \left(A\right)$is correct,</p><p>Now,</p><p>$f\left(x\right) = \left\{$$\begin{array}{ll}0, 0 < x < \dfrac{1}{4} \\ 1 \cdot {\left(x - \dfrac{1}{4}\right)}^{2}\left(x - \dfrac{1}{2}\right), \dfrac{1}{4} \le x < \dfrac{1}{2} \\ 2 \cdot {\left(x - \dfrac{1}{4}\right)}^{2}\left(x - \dfrac{1}{2}\right), \dfrac{1}{2} \le x < \dfrac{3}{4} \\ 3 \cdot {\left(x - \dfrac{1}{4}\right)}^{2}\left(x - \dfrac{1}{2}\right), \dfrac{3}{4} \le x < 1\end{array}$$\right)

Now differentiating the above function we get,

{f}^{'}\left(x\right) = \left\{$\begin{array}{ll}0, 0 < x < \dfrac{1}{4} \\ {\left(x - \dfrac{1}{4}\right)}^{2} + 2\left(x - \dfrac{1}{4}\right)\left(x - \dfrac{1}{2}\right), \dfrac{1}{4} \le x < \dfrac{1}{2} \\ 4 \cdot \left(x - \dfrac{1}{4}\right)\left(x - \dfrac{1}{2}\right) + 2{\left(x - \dfrac{1}{4}\right)}^{2}, \dfrac{1}{2} \le x < \dfrac{3}{4} \\ 3 \cdot {\left(x - \dfrac{1}{4}\right)}^{2} + 6\left(x - \dfrac{1}{4}\right)\left(x - \dfrac{1}{2}\right), \dfrac{3}{4} \le x < 1\end{array}$\right)</p><p>Now from above function we get,</p><p>{f}^{'}\left(\dfrac{1}{4} -\right) = {f}^{'}\left(\dfrac{1}{4} +\right) = 0$so function is differentiable at$x = \dfrac{1}{4}$</p><p>And${f}^{'}\left(\dfrac{1}{2} -\right) = \dfrac{1}{16} \text{ and } {f}^{'}\left(\dfrac{1}{2} +\right) = 2 \times \dfrac{1}{16} = \dfrac{1}{8}$so function is non-differentiable at$x = \dfrac{1}{2}$, hence option \left(B\right)$ is correct,

Now ${f}^{'}\left(\dfrac{3}{4} -\right) = 4 \times \dfrac{1}{2} \times \dfrac{1}{4} + 2 \times \dfrac{1}{4} = 1 \text{ and } {f}^{'}\left(\dfrac{3}{4} +\right) = 3 \times \dfrac{1}{4} + 6 \times \dfrac{1}{2} \times \dfrac{1}{4} = \dfrac{3}{2}$

Hence, the function is not differentiable at $x = \dfrac{3}{4}$

So, function non-differentiable at two points so option \left(C\right)$is not correct.</p><p>Now plotting the graph from the above value we get,</p><p><img alt="" src="https://cdn-question-pool.Quantrex.app/pyq/jee_advanced/61a98865-8d84-478f-b0b9-4c2166748046.png" style="width: 304px; height: 172px;" /></p><p>Now$f\left(x\right)$will be minimum between$\dfrac{1}{4} \le x < \dfrac{1}{2}$</p><p>So,$f\left(x\right) = {\left(x - \dfrac{1}{4}\right)}^{2}\left(x - \dfrac{1}{2}\right), \dfrac{1}{4} \le x < \dfrac{1}{2}$</p><p>$⇒ {f}^{'}\left(x\right) = {\left(x - \dfrac{1}{4}\right)}^{2} + 2\left(x - \dfrac{1}{4}\right)\left(x - \dfrac{1}{2}\right) = \left(x - \dfrac{1}{4}\right)\left(3x - \dfrac{5}{4}\right)$</p><p>Now critical points will be${f}^{'}\left(x\right) = 0 ⇒ x = \dfrac{1}{4}\text{or}x = \dfrac{5}{12}$</p><p>Now putting the value in first derivative we get,</p><p>${f}^{'}\left(\dfrac{5}{12} -\right) 0$</p><p>Hence,$x = \dfrac{5}{12}$is point of minima,</p><p>So,$f{\left(\dfrac{5}{12}\right)}_\min = {\left(\dfrac{5}{12} - \dfrac{1}{4}\right)}^{2}\left(\dfrac{5}{12} - \dfrac{1}{2}\right) = \dfrac{- 1}{432}$

Hence, option \left(D\right) is wrong.