Let f(x) = begin{cases} 1 + 6x - 3x^2, & x le | vedclass

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(A) Given $f(x) = \begin{cases} 1 + 6x - 3x^2, & x \leq 1 \ x + \log_2(b^2 + 7), & x > 1 \end{cases}$.For $f(1)$ to be the maximum value,$f(x)$ must

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$[-1, 1]$

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(A) Given $f(x) = \begin{cases} 1 + 6x - 3x^2, & x \leq 1 \ x + \log_2(b^2 + 7), & x > 1 \end{cases}$.For $f(1)$ to be the maximum value,$f(x)$ must be less than or equal to $f(1)$ for all $x$.First,calculate $f(1) = 1 + 6(1) - 3(1)^2 = 1 + 6 - 3 = 4$.For $x > 1$,we require $f(x) \leq 4$,which means $x + \log_2(b^2 + 7) \leq 4$.Since this must hold for all $x > 1$,we look at the limit as $x 	o 1^+$,which gives $1 + \log_2(b^2 + 7) \leq 4$.$\log_2(b^2 + 7) \leq 3$.$b^2 + 7 \leq 2^3 = 8$.$b^2 \leq 1$.Therefore,$-1 \leq b \leq 1$,so $b \in [-1, 1]$.

Let $f(x) = \begin{cases} 1 + 6x - 3x^2, & x \leq 1 \\ x + \log_2(b^2 + 7), & x > 1 \end{cases}$. Then the set of all possible values of $b$ such that $f(1)$ is the maximum value of $f(x)$ is

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