Let f(x) = operatorname{Max}{cos x, sin x, 0} | vedclass

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(C) Given $f(x) = \max\{\cos x, \sin x, 0\}$.In the interval $[0, 2\pi]$,the function $f(x)$ is defined as:$f(x) = \cos x$ for $x \in [0, \pi/4]$$f(x)

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$3$

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(C) Given $f(x) = \max\{\cos x, \sin x, 0\}$.In the interval $[0, 2\pi]$,the function $f(x)$ is defined as:$f(x) = \cos x$ for $x \in [0, \pi/4]$$f(x) = \sin x$ for $x \in [\pi/4, 5\pi/4]$$f(x) = 0$ for $x \in [5\pi/4, 2\pi]$Checking for non-differentiability at the intersection points:$1$. At $x = \pi/4$,$\cos(\pi/4) = \sin(\pi/4) = 1/\sqrt{2}$. The derivatives are $-\sin(\pi/4) = -1/\sqrt{2}$ and $\cos(\pi/4) = 1/\sqrt{2}$. Since they are not equal,$f(x)$ is not differentiable at $x = \pi/4$.$2$. At $x = 5\pi/4$,$\sin(5\pi/4) = -1/\sqrt{2}$ and $0$. Since $\sin(5\pi/4) $3$. At $x = 2\pi$,$\cos(2\pi) = 1$ and $0$. The transition from $0$ to $\cos x$ occurs at $x = 2\pi - \pi/2 = 3\pi/2$ is not correct,actually $\cos x = 0$ at $x = 3\pi/2$. The function $f(x)$ is $0$ on $[5\pi/4, 2\pi]$ and $\cos x$ starts being positive at $x = 3\pi/2$. At $x = 3\pi/2$,$\cos(3\pi/2) = 0$ and the derivative is $-\sin(3\pi/2) = 1$. Since $1 
eq 0$,it is not differentiable at $x = 3\pi/2$.Thus,there are $3$ points of non-differentiability in every interval of length $2\pi$.In $(0, 2024\pi)$,there are $1012$ such intervals.Total points of non-differentiability $= 3 	imes 1012 = 3036$.Given $1012k = 3036$,we get $k = 3$.

Let $f(x) = \operatorname{Max}\{\cos x, \sin x, 0\}$. If the number of points at which $f(x)$ is not differentiable in $(0, 2024 \pi)$ is $1012 k$,then $k =$

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