The domain of the function f(x) = exp ( `sqrt(5x - 3 - 2x^(2)))` is

Options

A. `[3//2 , infty)`
B. `[ 1, 3//2]`
C. `( - infty, 1]`
D. `(1, 3//2)`

Answer

To find the domain of the function \( f(x) = \sqrt{5x - 3 - 2x^2} \), we need to ensure that the expression inside the square root is non-negative, as the square root function is only defined for non-negative values.

### Step-by-step Solution:

1. **Set up the inequality**:
We need to solve the inequality:
\[ 5x - 3 - 2x^2 \geq 0 \]

2. **Rearrange the inequality**:
Rearranging gives us:
\[ -2x^2 + 5x - 3 \geq 0 \]
We can multiply the entire inequality by -1 (remember to reverse the inequality sign):
\[ 2x^2 - 5x + 3 \leq 0 \]

3. **Find the roots of the quadratic equation**:
We can find the roots of the quadratic equation \( 2x^2 - 5x + 3 = 0 \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 2 \), \( b = -5 \), and \( c = 3 \):
\[ x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 3}}{2 \cdot 2} \]
\[ x = \frac{5 \pm \sqrt{25 - 24}}{4} \]
\[ x = \frac{5 \pm 1}{4} \]

4. **Calculate the roots**:
This gives us two roots:
\[ x_1 = \frac{6}{4} = \frac{3}{2}, \quad x_2 = \frac{4}{4} = 1 \]

5. **Determine the intervals**:
The roots divide the number line into intervals. We need to test the intervals to determine where the quadratic is less than or equal to zero. The intervals are:
- \( (-\infty, 1) \)
- \( (1, \frac{3}{2}) \)
- \( (\frac{3}{2}, \infty) \)

6. **Test the intervals**:
- For \( x < 1 \) (e.g., \( x = 0 \)):
\[ 2(0)^2 - 5(0) + 3 = 3 \quad (\text{positive}) \]
- For \( 1 < x < \frac{3}{2} \) (e.g., \( x = 1.2 \)):
\[ 2(1.2)^2 - 5(1.2) + 3 = 2.88 - 6 + 3 = -0.12 \quad (\text{negative}) \]
- For \( x > \frac{3}{2} \) (e.g., \( x = 2 \)):
\[ 2(2)^2 - 5(2) + 3 = 8 - 10 + 3 = 1 \quad (\text{positive}) \]

7. **Conclusion on intervals**:
The quadratic \( 2x^2 - 5x + 3 \) is less than or equal to zero in the interval:
\[ [1, \frac{3}{2}] \]

Thus, the domain of the function \( f(x) \) is:
\[ \boxed{[1, \frac{3}{2}]} \]