JEE Advance - Mathematics (2023 - Paper 2 Online - No. 6)
Let $f:(0,1) \rightarrow \mathbb{R}$ be the function defined as $f(x)=[4 x]\left(x-\frac{1}{4}\right)^2\left(x-\frac{1}{2}\right)$, where $[x]$ denotes the greatest integer less than or equal to $x$. Then which of the following statements is(are) true?
The function $f$ is discontinuous exactly at one point in $(0,1)$
There is exactly one point in $(0,1)$ at which the function $f$ is continuous but NOT differentiable
The function $f$ is NOT differentiable at more than three points in $(0,1)$
The minimum value of the function $f$ is $-\frac{1}{512}$
Explanation
$$
f(x)=\left\{\begin{array}{cc}
0 & ; 0 < x <\frac{1}{4} \\\\
\left(x-\frac{1}{4}\right)^2\left(x-\frac{1}{2}\right) ; & \frac{1}{4} \leq x<\frac{1}{2} \\\\
2\left(x-\frac{1}{4}\right)^2\left(x-\frac{1}{2}\right) ; & \frac{1}{2} \leq x<\frac{3}{4} \\\\
3\left(x-\frac{1}{4}\right)^2\left(x-\frac{1}{2}\right) ; & \frac{3}{4} \leq x<1
\end{array}\right.
$$
$$ f(x) \text { is discontinuous at } x=\frac{3}{4} \text { only } $$
$$ f^{\prime}(x)=\left\{\begin{array}{cc} 0 & ; 0 < x <\frac{1}{4} \\\\ 2\left(x-\frac{1}{4}\right)\left(x-\frac{1}{2}\right)+\left(x-\frac{1}{4}\right)^2 & ; \quad \frac{1}{4}< x <\frac{1}{2} \\\\ 4\left(x-\frac{1}{4}\right)\left(x-\frac{1}{2}\right)+2\left(x-\frac{1}{4}\right)^2 & ; \frac{1}{2} < x <\frac{3}{4} \\\\ 6\left(x-\frac{1}{4}\right)\left(x-\frac{1}{2}\right)+3\left(x-\frac{1}{4}\right)^2 & ; \frac{3}{4} < x < 1 \end{array}\right. $$
$f(x)$ is non-differentiable at $x=\frac{1}{2}$ and $\frac{3}{4}$
Minimum values of $f(x)$ occur at $x=\frac{5}{12}$ whose value is $-\frac{1}{432}$
$$ f(x) \text { is discontinuous at } x=\frac{3}{4} \text { only } $$
$$ f^{\prime}(x)=\left\{\begin{array}{cc} 0 & ; 0 < x <\frac{1}{4} \\\\ 2\left(x-\frac{1}{4}\right)\left(x-\frac{1}{2}\right)+\left(x-\frac{1}{4}\right)^2 & ; \quad \frac{1}{4}< x <\frac{1}{2} \\\\ 4\left(x-\frac{1}{4}\right)\left(x-\frac{1}{2}\right)+2\left(x-\frac{1}{4}\right)^2 & ; \frac{1}{2} < x <\frac{3}{4} \\\\ 6\left(x-\frac{1}{4}\right)\left(x-\frac{1}{2}\right)+3\left(x-\frac{1}{4}\right)^2 & ; \frac{3}{4} < x < 1 \end{array}\right. $$
$f(x)$ is non-differentiable at $x=\frac{1}{2}$ and $\frac{3}{4}$
Minimum values of $f(x)$ occur at $x=\frac{5}{12}$ whose value is $-\frac{1}{432}$
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