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(D) Given the function $f(x) = x - [x] - \cos x$. In the neighborhood of $x = \frac{\pi}{2}$,the value of $[x]$ is constant because $[x] = [1.57...] =

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$2$

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(D) Given the function $f(x) = x - [x] - \cos x$. In the neighborhood of $x = \frac{\pi}{2}$,the value of $[x]$ is constant because $[x] = [1.57...] = 1$. Therefore,for $x$ in an interval around $\frac{\pi}{2}$,we can write $f(x) = x - 1 - \cos x$. Differentiating with respect to $x$,we get $f^{\prime}(x) = \frac{d}{dx}(x - 1 - \cos x) = 1 - 0 - (-\sin x) = 1 + \sin x$. Now,substituting $x = \frac{\pi}{2}$ into the derivative: $f^{\prime}\left(\frac{\pi}{2}\right) = 1 + \sin\left(\frac{\pi}{2}\right) = 1 + 1 = 2$.

If $[x]$ represents the greatest integer function and $f(x) = x - [x] - \cos x$,then $f^{\prime}\left(\frac{\pi}{2}\right) = $

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