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(B) For a function $f(x)$ to have a local minimum at $x = 1$,the value of the function at $x = 1$ must be less than or equal to the limit of the funct

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$(\cos 3, 1]$

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(B) For a function $f(x)$ to have a local minimum at $x = 1$,the value of the function at $x = 1$ must be less than or equal to the limit of the function as $x$ approaches $1$ from the left.$f(1) = 4(1) = 4$.$\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^-} (\cos^{-1}(\mu) + x^2) = \cos^{-1}(\mu) + 1$.For a local minimum at $x = 1$,we require $f(1) \leq \lim_{x 	o 1^-} f(x)$.$4 \leq \cos^{-1}(\mu) + 1$.$\cos^{-1}(\mu) \geq 3$.Since the range of $\cos^{-1}(\mu)$ is $[0, \pi]$,we have $3 \leq \cos^{-1}(\mu) \leq \pi$.Taking the cosine of all parts (noting that $\cos$ is a decreasing function in $[0, \pi]$),the inequality signs reverse:$\cos(3) \leq \mu \leq \cos(0)$.Since $\cos(0) = 1$,we have $\cos(3) \leq \mu \leq 1$.However,for the function to be defined,$\mu$ must be in $[-1, 1]$. Thus,the interval is $[\cos 3, 1]$.Given the options provided,the correct interval is $[\cos 3, 1]$. Note: The provided option $A$ is $[-1, \cos 3]$,which is incorrect. The correct interval is $[\cos 3, 1]$. Since the question asks for the interval,and based on standard analysis,the correct range is $[\cos 3, 1]$.

Let a function $f(x)$ be defined as $f(x) = \begin{cases} \cos^{-1}(\mu) + x^2, & 0 < x < 1 \\ 4x, & x \geqslant 1 \end{cases}$. The function $f(x)$ can have a local minimum at $x = 1$ if the value of $\mu$ lies in the interval:

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