`"Let "f(x) ={underset(-2x + log_(2) (b^(2)-2), x gt1)(x^(3) -x^(2) +10 x-5 , x le 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), -sqrt(2)) uu(sqrt(2), sqrt(130)]`

Answer

To solve the problem, we need to analyze the function \( f(x) \) defined as follows:

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

We want to find the set of values of \( b \) for which \( f(x) \) has its greatest value at \( x = 1 \).

### Step 1: Analyze \( f(x) \) for \( x \leq 1 \)

First, we differentiate \( f(x) \) for \( x \leq 1 \):

\[ f'(x) = \frac{d}{dx}(x^3 - x^2 + 10x - 5) = 3x^2 - 2x + 10 \]

### Step 2: Determine the nature of \( f'(x) \)

To find out if \( f(x) \) is increasing or decreasing for \( x \leq 1 \), we need to check the discriminant of the quadratic \( 3x^2 - 2x + 10 \):

\[ D = b^2 - 4ac = (-2)^2 - 4 \cdot 3 \cdot 10 = 4 - 120 = -116 \]

Since the discriminant \( D < 0 \), the quadratic \( f'(x) \) has no real roots and is always positive. Therefore, \( f(x) \) is an increasing function for \( x \leq 1 \).

### Step 3: Analyze \( f(x) \) for \( x > 1 \)

Now, we differentiate \( f(x) \) for \( x > 1 \):

\[ f'(x) = \frac{d}{dx}(-2x + \log_2(b^2 - 2)) = -2 \]

This derivative is negative, indicating that \( f(x) \) is a decreasing function for \( x > 1 \).

### Step 4: Establish conditions for maximum at \( x = 1 \)

Since \( f(x) \) is increasing for \( x \leq 1 \) and decreasing for \( x > 1 \), it will have a maximum at \( x = 1 \). Thus, we need to ensure that:

\[ \lim_{x \to 1^-} f(x) \leq f(1) \leq \lim_{x \to 1^+} f(x) \]

### Step 5: Calculate \( f(1) \)

Calculating \( f(1) \):

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

### Step 6: Calculate \( \lim_{x \to 1^+} f(x) \)

Now, we calculate \( \lim_{x \to 1^+} f(x) \):

\[ \lim_{x \to 1^+} f(x) = -2(1) + \log_2(b^2 - 2) = -2 + \log_2(b^2 - 2) \]

### Step 7: Set up the inequality

We need:

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

This simplifies to:

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

Exponentiating both sides gives:

\[ b^2 - 2 \leq 2^7 \implies b^2 \leq 130 \]

### Step 8: Establish the positivity condition

Additionally, since \( b^2 - 2 \) must be positive (as it is inside a logarithm), we have:

\[ b^2 - 2 > 0 \implies b^2 > 2 \implies |b| > \sqrt{2} \]

### Step 9: Combine the conditions

Now we have two conditions:

1. \( b^2 \leq 130 \) implies \( -\sqrt{130} \leq b \leq \sqrt{130} \)
2. \( |b| > \sqrt{2} \) implies \( b < -\sqrt{2} \) or \( b > \sqrt{2} \)

### Step 10: Final solution

Combining these conditions gives us:

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