The function f(x) = [x] cos left( frac{2x - 1 | vedclass

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(B) Given the function $f(x) = [x] \cos \left( \frac{2x - 1}{2} \pi \right)$.Let $g(x) = [x]$ and $h(x) = \cos \left( \frac{2x - 1}{2} \pi \right)$.Th

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(B) Given the function $f(x) = [x] \cos \left( \frac{2x - 1}{2} \pi ight)$.Let $g(x) = [x]$ and $h(x) = \cos \left( \frac{2x - 1}{2} \pi ight)$.The function $g(x) = [x]$ is the greatest integer function,which is known to be discontinuous at all integer values of $x$.For any integer $n$,we have $h(n) = \cos \left( \frac{2n - 1}{2} \pi ight) = \cos \left( n\pi - \frac{\pi}{2} ight) = 0$.Since $h(n) = 0$ for all integers $n$,the product $f(x) = g(x) \cdot h(x)$ remains continuous at all integers because the jump discontinuity of $[x]$ is multiplied by $0$ at these points.However,the question asks for the points of discontinuity. Upon re-evaluating,$[x]$ is discontinuous at all $x \in \mathbb{Z}$. At $x = n \in \mathbb{Z}$,$\lim_{x 	o n^-} f(x) = (n-1) \cdot 0 = 0$ and $\lim_{x 	o n^+} f(x) = n \cdot 0 = 0$. Since $f(n) = n \cdot 0 = 0$,the function is actually continuous at all integers.Therefore,the function is continuous for all $x$. Thus,there is no $x$ where the function is discontinuous.

The function $f(x) = [x] \cos \left( \frac{2x - 1}{2} \pi \right)$,where $[.]$ denotes the greatest integer function,is discontinuous at

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