Let f:(0,1) rightarrow mathbb{R} be the funct | vedclass

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(B) The function $f(x) = [4x](x-\frac{1}{4})^2(x-\frac{1}{2})$ can be written as:$f(x) = \begin{cases} 0 & 0 < x < \frac{1}{4} \ (x-\frac{1}{4})^2(x-

Vedclass · Accepted Answer

$A, B$

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(B) The function $f(x) = [4x](x-\frac{1}{4})^2(x-\frac{1}{2})$ can be written as:$f(x) = \begin{cases} 0 & 0 $1$. Continuity: The function is discontinuous at $x = \frac{1}{2}$ and $x = \frac{3}{4}$ because the limits from the left and right do not match at these points. Thus,statement $(A)$ is false.$2$. Differentiability: The function is continuous at $x = \frac{1}{4}$ but not differentiable there. It is discontinuous at $x = \frac{1}{2}$ and $x = \frac{3}{4}$. Thus,there is exactly one point $(x = \frac{1}{4})$ where it is continuous but not differentiable. Statement $(B)$ is true.$3$. Non-differentiability: The function is non-differentiable at $x = \frac{1}{4}, \frac{1}{2}, \frac{3}{4}$. This is exactly three points. Statement $(C)$ is false.$4$. Minimum value: For $x \in [\frac{1}{4}, \frac{1}{2}]$,$f(x) = (x-\frac{1}{4})^2(x-\frac{1}{2})$. Let $t = x-\frac{1}{4}$,then $f(t) = t^2(t-\frac{1}{4}) = t^3 - \frac{1}{4}t^2$. $f'(t) = 3t^2 - \frac{1}{2}t = t(3t - \frac{1}{2})$. Setting $f'(t) = 0$,we get $t = \frac{1}{6}$. The value is $(\frac{1}{6})^2(\frac{1}{6}-\frac{1}{4}) = \frac{1}{36}(-\frac{1}{12}) = -\frac{1}{432}$. Thus,statement $(D)$ is false.

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