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(C) Let $x = n$,where $n \in Z$.First,consider the value of the function at $x = n$:$f(n) = [n-1] \cos \left(\frac{2n-1}{2}\right) \pi = (n-1) \cos \l

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continuous for every real $x$

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(C) Let $x = n$,where $n \in Z$.First,consider the value of the function at $x = n$:$f(n) = [n-1] \cos \left(\frac{2n-1}{2}\right) \pi = (n-1) \cos \left(n\pi - \frac{\pi}{2}\right) = (n-1) \cdot 0 = 0$.Now,calculate the Left Hand Limit $(LHL)$:$LHL = \lim_{x \rightarrow n^-} [x-1] \cos \left(\frac{2x-1}{2}\right) \pi$.As $x \rightarrow n^-$,$[x-1] = n-2$.$LHL = (n-2) \cos \left(n\pi - \frac{\pi}{2}\right) = (n-2) \cdot 0 = 0$.Now,calculate the Right Hand Limit $(RHL)$:$RHL = \lim_{x \rightarrow n^+} [x-1] \cos \left(\frac{2x-1}{2}\right) \pi$.As $x \rightarrow n^+$,$[x-1] = n-1$.$RHL = (n-1) \cos \left(n\pi - \frac{\pi}{2}\right) = (n-1) \cdot 0 = 0$.Since $LHL = RHL = f(n) = 0$ for all $n \in Z$,the function is continuous at all integral values of $x$.For non-integral values of $x$,the function is a product of a constant (locally) and a continuous trigonometric function,hence it is continuous.Therefore,$f(x)$ is continuous for every real $x$.

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

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