Mathematics - Continuity - Differentiability Question with Solution | TestHub

Given,

$f:(0,1)\to \mathrm{ℝ}$

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

We know that, $\left[4x\right]$ will be discontinuous function at integer points,

So, $\text{critical point }=\frac{1}{4},\frac{1}{2},\frac{3}{4}$ as $x\in \left(0,1\right)$

And function will be discontinuous at $x=\frac{3}{4}$ because for $\frac{1}{2} \& \frac{1}{4}$, $\left(x-\frac{1}{2}\right) \& {\left(x-\frac{1}{4}\right)}^2$ 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,

Now,

$f\left(x\right)=\left\{\begin{array}{c}0, 0<x<\frac{1}{4}\\ 1·{\left(x-\frac{1}{4}\right)}^2\left(x-\frac{1}{2}\right), \frac{1}{4}\le x<\frac{1}{2}\\ 2·{\left(x-\frac{1}{4}\right)}^2\left(x-\frac{1}{2}\right), \frac{1}{2}\le x<\frac{3}{4}\\ 3·{\left(x-\frac{1}{4}\right)}^2\left(x-\frac{1}{2}\right), \frac{3}{4}\le x<1\end{array}\right.$

Now differentiating the above function we get,

$f^{\text{'}}\left(x\right)=\left\{\begin{array}{c}0, 0<x<\frac{1}{4}\\ {\left(x-\frac{1}{4}\right)}^2+2\left(x-\frac{1}{4}\right)\left(x-\frac{1}{2}\right), \frac{1}{4}\le x<\frac{1}{2}\\ 4·\left(x-\frac{1}{4}\right)\left(x-\frac{1}{2}\right)+2{\left(x-\frac{1}{4}\right)}^2, \frac{1}{2}\le x<\frac{3}{4}\\ 3·{\left(x-\frac{1}{4}\right)}^2+6\left(x-\frac{1}{4}\right)\left(x-\frac{1}{2}\right), \frac{3}{4}\le x<1\end{array}\right.$

Now from above function we get,

$f^{\text{'}}\left({\frac{1}{4}}^{-}\right)=f^{\text{'}}\left({\frac{1}{4}}^{+}\right)=0$ so function is differentiable at $x=\frac{1}{4}$

And $f^{\text{'}}\left({\frac{1}{2}}^{-}\right)=\frac{1}{16}\& f^{\text{'}}\left({\frac{1}{2}}^{+}\right)=2\times \frac{1}{16}=\frac{1}{8}$ so function is non-differentiable at $x=\frac{1}{2}$, hence option $\left(B\right)$ is correct,

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

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

So, function non-differentiable at two points so option $\left(C\right)$ is not correct.

Now plotting the graph from the above value we get,

Now $f\left(x\right)$ will be minimum between $\frac{1}{4}\le x<\frac{1}{2}$

So, $f\left(x\right)={\left(x-\frac{1}{4}\right)}^2\left(x-\frac{1}{2}\right), \frac{1}{4}\le x<\frac{1}{2}$

$\Rightarrow f^{\text{'}}\left(x\right)={\left(x-\frac{1}{4}\right)}^2+2\left(x-\frac{1}{4}\right)\left(x-\frac{1}{2}\right)=\left(x-\frac{1}{4}\right)\left(3x-\frac{5}{4}\right)$

Now critical points will be $f^{\text{'}}\left(x\right)=0\Rightarrow x=\frac{1}{4} \text{or} x=\frac{5}{12}$

Now putting the value in first derivative we get,

$f^{\text{'}}\left({\frac{5}{12}}^{-}\right)<0 \& f^{\text{'}}\left({\frac{5}{12}}^{+}\right)>0$

Hence, $x=\frac{5}{12}$ is point of minima,

So, $f{\left(\frac{5}{12}\right)}_{min}={\left(\frac{5}{12}-\frac{1}{4}\right)}^2\left(\frac{5}{12}-\frac{1}{2}\right)=\frac{-1}{432}$

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