Let f (x) = `{{:( x^(3) - x^(2) + 10 x- 5 , "," , x le 1), ( -2x + log _(2) (b^(2) - 2) , "," , x gt 1):}` the set of values of b for which f(x) has greatest value at x = 1 is given by

Options

A. `1 le b le 2`
B. `b = {1 , 2}`
C. `b in (-oo , -1)`
D. `[-sqrt(130) , - sqrt2) cup (sqrt2 , sqrt(130)]`

Answer

To solve the problem, we need to determine the set of values of \( b \) for which the function \( f(x) \) has its greatest value at \( x = 1 \). The function is defined piecewise:

\[ f(x) = \begin{cases} x^3 - x^2 + 10x - 5 & \text{if } x \leq 1 \\ -2x + \log_2(b^2 - 2) & \text{if } x > 1 \end{cases} \]

### Step 1: Evaluate \( f(1) \)

First, we calculate \( f(1) \) using the first piece of the function:

\[ f(1) = 1^3 - 1^2 + 10 \cdot 1 - 5 = 1 - 1 + 10 - 5 = 5 \]

### Step 2: Find the right-hand limit as \( x \) approaches 1 from the right

Next, we evaluate the right-hand limit of \( f(x) \) as \( x \) approaches 1 from the right:

\[ f(1^+) = -2(1) + \log_2(b^2 - 2) = -2 + \log_2(b^2 - 2) \]

### Step 3: Set the right-hand limit less than or equal to \( f(1) \)

For \( f(x) \) to have its greatest value at \( x = 1 \), we need:

\[ -2 + \log_2(b^2 - 2) \leq 5 \]

### Step 4: Solve the inequality

Rearranging the inequality:

\[ \log_2(b^2 - 2) \leq 7 \]

Exponentiating both sides gives:

\[ b^2 - 2 \leq 2^7 \]

\[ b^2 - 2 \leq 128 \]

Adding 2 to both sides:

\[ b^2 \leq 130 \]

### Step 5: Determine the range for \( b \)

Taking the square root of both sides, we find:

\[ -\sqrt{130} \leq b \leq \sqrt{130} \]

### Step 6: Consider the logarithmic condition

Since \( \log_2(b^2 - 2) \) is defined, we need \( b^2 - 2 > 0 \):

\[ b^2 > 2 \implies |b| > \sqrt{2} \]

### Step 7: Combine the inequalities

Combining the two conditions:

1. \( -\sqrt{130} \leq b \leq \sqrt{130} \)
2. \( |b| > \sqrt{2} \)

This means:

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

### Conclusion

The set of values of \( b \) for which \( f(x) \) has its greatest value at \( x = 1 \) is:

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