The greatest vaue of the function
`f(x) = cot^(-1) x + (1//2) log x " on " [1, sqrt3]` is

Options

A. `(pi//6) + 0.25 log 3`
B. `(pi//3) -0.25 log 3`
C. `pi//4`
D. `tan^(-1) e-1//2`

Answer

To find the greatest value of the function \( f(x) = \cot^{-1}(x) + \frac{1}{2} \log(x) \) on the interval \([1, \sqrt{3}]\), we will follow these steps:

### Step 1: Differentiate the function
We need to find the derivative of \( f(x) \):
\[ f'(x) = \frac{d}{dx} \left( \cot^{-1}(x) \right) + \frac{1}{2} \frac{d}{dx} \left( \log(x) \right) \]
Using the derivatives:
\[ \frac{d}{dx} \cot^{-1}(x) = -\frac{1}{1+x^2} \quad \text{and} \quad \frac{d}{dx} \log(x) = \frac{1}{x} \]
Thus,
\[ f'(x) = -\frac{1}{1+x^2} + \frac{1}{2x} \]

### Step 2: Set the derivative to zero
To find critical points, we set \( f'(x) = 0 \):
\[ -\frac{1}{1+x^2} + \frac{1}{2x} = 0 \]
Rearranging gives:
\[ \frac{1}{2x} = \frac{1}{1+x^2} \]
Cross-multiplying:
\[ 1 + x^2 = 2x \]
Rearranging leads to:
\[ x^2 - 2x + 1 = 0 \]
Factoring gives:
\[ (x-1)^2 = 0 \]
Thus, \( x = 1 \) is a critical point.

### Step 3: Evaluate the function at the endpoints and critical points
We need to evaluate \( f(x) \) at the endpoints \( x = 1 \) and \( x = \sqrt{3} \), and at the critical point \( x = 1 \).

1. **At \( x = 1 \)**:
\[ f(1) = \cot^{-1}(1) + \frac{1}{2} \log(1) = \frac{\pi}{4} + 0 = \frac{\pi}{4} \]

2. **At \( x = \sqrt{3} \)**:
\[ f(\sqrt{3}) = \cot^{-1}(\sqrt{3}) + \frac{1}{2} \log(\sqrt{3}) = \frac{\pi}{6} + \frac{1}{2} \cdot \frac{1}{2} \log(3) \]
Simplifying gives:
\[ f(\sqrt{3}) = \frac{\pi}{6} + \frac{1}{4} \log(3) \]

### Step 4: Compare the values
Now we compare \( f(1) \) and \( f(\sqrt{3}) \):
- \( f(1) = \frac{\pi}{4} \)
- \( f(\sqrt{3}) = \frac{\pi}{6} + \frac{1}{4} \log(3) \)

### Step 5: Determine which is greater
To determine which value is greater, we can approximate:
- \( \frac{\pi}{4} \approx 0.785 \)
- \( \frac{\pi}{6} \approx 0.524 \) and \( \frac{1}{4} \log(3) \approx 0.25 \times 1.098 \approx 0.2745 \)

Thus,
\[ f(\sqrt{3}) \approx 0.524 + 0.2745 \approx 0.7985 \]
Since \( 0.7985 > 0.785 \), we conclude that:
\[ f(\sqrt{3}) > f(1) \]

### Conclusion
The greatest value of the function \( f(x) \) on the interval \([1, \sqrt{3}]\) is:
\[ f(\sqrt{3}) = \frac{\pi}{6} + \frac{1}{4} \log(3) \]