Let `f(x)=(x^(2)+2)/([x]),1 le x le3`, where [.] is the greatest integer function. Then the least value of f(x) is

Options

A. 2
B. 3
C. `3//2`
D. 1

Explanation

To find the least value of the function \( f(x) = \frac{x^2 + 2}{[x]} \) for \( 1 \leq x \leq 3 \), where \([x]\) is the greatest integer function, we will analyze the function on the intervals defined by the greatest integer function.

### Step 1: Identify the intervals
The greatest integer function \([x]\) takes integer values. For \( 1 \leq x < 2 \), \([x] = 1\). For \( 2 \leq x < 3 \), \([x] = 2\). At \( x = 3 \), \([x] = 3\). Therefore, we will evaluate \( f(x) \) in the intervals:
- \( 1 \leq x < 2 \) (where \([x] = 1\))
- \( 2 \leq x < 3 \) (where \([x] = 2\))
- At \( x = 3 \) (where \([x] = 3\))

### Step 2: Calculate \( f(x) \) for \( 1 \leq x < 2 \)
In this interval, we have:
\[ f(x) = \frac{x^2 + 2}{1} = x^2 + 2 \]
Now, we need to find the minimum value of \( f(x) \) in this interval. The function \( x^2 + 2 \) is a parabola opening upwards, and its minimum occurs at the left endpoint \( x = 1 \):
\[ f(1) = 1^2 + 2 = 3 \]

### Step 3: Calculate \( f(x) \) for \( 2 \leq x < 3 \)
In this interval, we have:
\[ f(x) = \frac{x^2 + 2}{2} \]
Now, we need to find the minimum value of \( f(x) \) in this interval. The function \( \frac{x^2 + 2}{2} \) is also a parabola opening upwards. We will evaluate it at the endpoints:
- At \( x = 2 \):
\[ f(2) = \frac{2^2 + 2}{2} = \frac{4 + 2}{2} = \frac{6}{2} = 3 \]
- At \( x = 3 \):
\[ f(3) = \frac{3^2 + 2}{2} = \frac{9 + 2}{2} = \frac{11}{2} = 5.5 \]

### Step 4: Compare the values
Now we compare the minimum values found in each interval:
- For \( 1 \leq x < 2 \), the minimum value is \( 3 \).
- For \( 2 \leq x < 3 \), the minimum value is \( 3 \).
- At \( x = 3 \), the value is \( 5.5 \).

### Conclusion
The least value of \( f(x) \) in the interval \( 1 \leq x \leq 3 \) is:
\[ \boxed{3} \]