If y = cos^{-1}(cos(|x| - f(x))),where f(x) = | vedclass

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(B) Given $f(x) = 1$ for $x > 0$. Since $x = \frac{5\pi}{4} > 0$,we have $f(x) = 1$.Thus,$y = \cos^{-1}(\cos(|x| - 1))$.Since $x = \frac{5\pi}{4} > 0$

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$1$

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(B) Given $f(x) = 1$ for $x > 0$. Since $x = \frac{5\pi}{4} > 0$,we have $f(x) = 1$.Thus,$y = \cos^{-1}(\cos(|x| - 1))$.Since $x = \frac{5\pi}{4} > 0$,$|x| = x$,so $y = \cos^{-1}(\cos(x - 1))$.For $x = \frac{5\pi}{4}$,$x - 1 = \frac{5\pi}{4} - 1 \approx 3.927 - 1 = 2.927$.Since $0 \le 2.927 \le \pi$ (where $\pi \approx 3.14159$),the expression simplifies to $y = x - 1$.Therefore,$\frac{dy}{dx} = \frac{d}{dx}(x - 1) = 1$.Evaluating at $x = \frac{5\pi}{4}$,we get $\left. \frac{dy}{dx} \right|_{x = \frac{5\pi}{4}} = 1$.

If $y = \cos^{-1}(\cos(|x| - f(x)))$,where $f(x) = \begin{cases} 1, & \text{if } x > 0 \\ -1, & \text{if } x < 0 \\ 0, & \text{if } x = 0 \end{cases}$,then $\left. \frac{dy}{dx} \right|_{x = \frac{5\pi}{4}}$ is

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