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

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(B) The function is given by $f(x) = [x] \cdot \cos \left( \frac{2x - 1}{2} \pi \right)$.We know that $[x]$ is the greatest integer function,which is

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no $x$.

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(B) The function is given by $f(x) = [x] \cdot \cos \left( \frac{2x - 1}{2} \pi ight)$.We know that $[x]$ is the greatest integer function,which is discontinuous at all integer values of $x$.Let $x = n$,where $n \in \mathbb{Z}$.At $x = n$,the term $\cos \left( \frac{2n - 1}{2} \pi ight) = \cos \left( n\pi - \frac{\pi}{2} ight) = \cos \left( n\pi ight) \cos \left( \frac{\pi}{2} ight) + \sin \left( n\pi ight) \sin \left( \frac{\pi}{2} ight) = 0 + 0 = 0$.Since the product of a discontinuous function and a function that is zero at those points can be continuous,we check the limit at $x = n$.For $x = n + h$ (where $h 	o 0^+$),$f(n+h) = [n+h] \cdot \cos \left( \frac{2(n+h) - 1}{2} \pi ight) = n \cdot \cos \left( n\pi - \frac{\pi}{2} + h\pi ight) = n \cdot \sin(h\pi) \approx n \cdot h\pi 	o 0$.For $x = n - h$ (where $h 	o 0^+$),$f(n-h) = [n-h] \cdot \cos \left( \frac{2(n-h) - 1}{2} \pi ight) = (n-1) \cdot \cos \left( n\pi - \frac{\pi}{2} - h\pi ight) = (n-1) \cdot \sin(-h\pi) \approx -(n-1) \cdot h\pi 	o 0$.Since the left-hand limit and right-hand limit are both $0$,and $f(n) = n \cdot 0 = 0$,the function is continuous at all integers.However,the question implies a standard property check. Re-evaluating: $[x]$ is discontinuous at all integers,but $f(x)$ is continuous everywhere because the product with the zero-valued cosine term compensates for the jump discontinuity of $[x]$ at every integer $n$.

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

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