The domain of the function f(x) = sqrt{cos^{- | vedclass

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(B) For the function $f(x) = \sqrt{\cos^{-1}\left(\frac{1-|x|}{2}\right)}$ to be defined,the expression inside the square root must be non-negative,an

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$[-3, 3]$

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(B) For the function $f(x) = \sqrt{\cos^{-1}\left(\frac{1-|x|}{2}\right)}$ to be defined,the expression inside the square root must be non-negative,and the argument of $\cos^{-1}$ must be in the interval $[-1, 1]$.First,we require $\cos^{-1}\left(\frac{1-|x|}{2}\right) \geq 0$. Since the range of $\cos^{-1}(u)$ is $[0, \pi]$,this is always true for any $u \in [-1, 1]$.Next,we solve for the domain of $\cos^{-1}\left(\frac{1-|x|}{2}\right)$ by setting $-1 \leq \frac{1-|x|}{2} \leq 1$.Multiplying by $2$: $-2 \leq 1 - |x| \leq 2$.Subtracting $1$: $-3 \leq -|x| \leq 1$.Multiplying by $-1$ (and reversing the inequalities): $-1 \leq |x| \leq 3$.Since $|x|$ is always non-negative,$|x| \leq 3$ implies $-3 \leq x \leq 3$.Thus,the domain is $x \in [-3, 3]$.

The domain of the function $f(x) = \sqrt{\cos^{-1}\left(\frac{1-|x|}{2}\right)}$ is

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