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(D) The function is $f(x) = [x] \cos \left( \frac{2x - 1}{2} \pi \right) = [x] \cos \left( x\pi - \frac{\pi}{2} \right) = [x] \sin(x\pi)$.We check for

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continuous for every real $x$

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(D) The function is $f(x) = [x] \cos \left( \frac{2x - 1}{2} \pi ight) = [x] \cos \left( x\pi - \frac{\pi}{2} ight) = [x] \sin(x\pi)$.We check for continuity at any integer $x = n$,where $n \in \mathbb{Z}$.The value of the function at $x = n$ is $f(n) = [n] \sin(n\pi) = n \cdot 0 = 0$.The Left Hand Limit $(LHL)$ as $x 	o n^-$ is $\lim_{x 	o n^-} [x] \sin(x\pi) = (n - 1) \sin(n\pi) = (n - 1) \cdot 0 = 0$.The Right Hand Limit $(RHL)$ as $x 	o n^+$ is $\lim_{x 	o n^+} [x] \sin(x\pi) = n \sin(n\pi) = n \cdot 0 = 0$.Since $LHL = RHL = f(n) = 0$ for all $n \in \mathbb{Z}$,the function is continuous at all integers.Since $[x]$ is continuous everywhere except at integers and $\sin(x\pi)$ is continuous everywhere,the product is continuous everywhere.Thus,$f(x)$ is continuous for every real $x$.

Let $f: R \to R$ be a function defined by $f(x) = [x] \cos \left( \frac{2x - 1}{2} \pi \right)$,where $[x]$ denotes the greatest integer function. Then $f$ is:

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