Let f(x) = begin{cases} x^3 - x^2 + 10x - 5, | vedclass

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(D) For $f(x)$ to have the greatest value at $x = 1$,we must satisfy $f(x) \le f(1)$ for all $x$ in the domain.First,calculate $f(1)$ using the first

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$[-\sqrt{130}, -\sqrt{2}) \cup (\sqrt{2}, \sqrt{130}]$

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(D) For $f(x)$ to have the greatest value at $x = 1$,we must satisfy $f(x) \le f(1)$ for all $x$ in the domain.First,calculate $f(1)$ using the first part of the function: $f(1) = 1^3 - 1^2 + 10(1) - 5 = 1 - 1 + 10 - 5 = 5$.For $x \le 1$,$f(x) = x^3 - x^2 + 10x - 5$. Let $g(x) = x^3 - x^2 + 10x - 5$. Then $g'(x) = 3x^2 - 2x + 10$. The discriminant of $g'(x)$ is $D = (-2)^2 - 4(3)(10) = 4 - 120 = -116  0$ for all $x$,meaning $f(x)$ is strictly increasing for $x \le 1$. Thus,$f(x) \le f(1)$ is satisfied for all $x \le 1$.For $x > 1$,we require $f(x) \le f(1)$,which means $-2x + \log_2(b^2 - 2) \le 5$. As $x 	o 1^+$,the limit is $-2(1) + \log_2(b^2 - 2) = -2 + \log_2(b^2 - 2)$. For the function to be continuous or have a maximum at $x=1$,we need $-2 + \log_2(b^2 - 2) \le 5$.Also,the argument of the logarithm must be positive: $b^2 - 2 > 0 \implies b^2 > 2 \implies b \in (-\infty, -\sqrt{2}) \cup (\sqrt{2}, \infty)$.Solving $-2 + \log_2(b^2 - 2) \le 5$:$\log_2(b^2 - 2) \le 7$$b^2 - 2 \le 2^7$$b^2 - 2 \le 128$$b^2 \le 130$.Combining $b^2 > 2$ and $b^2 \le 130$,we get $2 Therefore,the correct option is $D$.

Let $f(x) = \begin{cases} x^3 - x^2 + 10x - 5, & x \le 1 \\ -2x + \log_2(b^2 - 2), & x > 1 \end{cases}$. The set of values of $b$ for which $f(x)$ has the greatest value at $x = 1$ is given by:

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