Function f(x) = {left( {left{ x right} - frac | vedclass

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(D) The given function is $f(x) = (\{x\} - \frac{1}{2})^2$.$1$. Continuity: The fractional part function $\{x\}$ is discontinuous at all integers $n \

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even

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(D) The given function is $f(x) = (\{x\} - \frac{1}{2})^2$.$1$. Continuity: The fractional part function $\{x\}$ is discontinuous at all integers $n \in \mathbb{Z}$. Since $f(x)$ is a composition of a continuous function (the square function) and a discontinuous function $(\{x\})$,$f(x)$ is discontinuous at all integers $n \in \mathbb{Z}$. Thus,option $A$ is correct.$2$. Differentiability: Since $f(x)$ is discontinuous at integers,it is not differentiable at integers. Thus,option $B$ is incorrect.$3$. Periodicity: The fractional part function $\{x\}$ is periodic with period $1$. Therefore,$f(x) = (\{x\} - \frac{1}{2})^2$ is also periodic with period $1$. Thus,option $C$ is incorrect.$4$. Even/Odd: For a function to be even,$f(-x) = f(x)$. Here,$f(-x) = (\{-x\} - \frac{1}{2})^2$. If $x$ is not an integer,$\{-x\} = 1 - \{x\}$. Then $f(-x) = (1 - \{x\} - \frac{1}{2})^2 = (\frac{1}{2} - \{x\})^2 = (\{x\} - \frac{1}{2})^2 = f(x)$. If $x$ is an integer,$f(-x) = (0 - \frac{1}{2})^2 = \frac{1}{4}$ and $f(x) = (0 - \frac{1}{2})^2 = \frac{1}{4}$. Thus,$f(x)$ is an even function. Option $D$ is also correct. However,in the context of standard multiple-choice questions,the most fundamental property derived from the graph is its periodicity and even nature. Given the options,$A$ and $D$ are both technically correct,but $D$ is a defining characteristic of the graph provided.

Function $f(x) = {\left( {\left\{ x \right\} - \frac{1}{2}} \right)^2}$ is (where $\{.\}$ represents the fractional part function).

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