The domain of the function
`f(x)=sqrt({(-log_0.3(x-1))/(-x^2+3x+18)})` is
(a)  `[2,6]`  (b)  `]2,6[`
(b)  `[2,6[`  (d)  None of these

Options

A. [2,6]
B. (2,6)
C. [2,6)
D. none of these

Answer

To find the domain of the function \( f(x) = \sqrt{\frac{-\log_{0.3}(x-1)}{-x^2 + 3x + 18}} \), we need to ensure that the expression inside the square root is non-negative. This means we need to solve the inequality:

\[ \frac{-\log_{0.3}(x-1)}{-x^2 + 3x + 18} \geq 0 \]

### Step 1: Analyze the numerator and denominator

1. **Numerator**: \( -\log_{0.3}(x-1) \)
- The logarithm \( \log_{0.3}(x-1) \) is defined for \( x - 1 > 0 \) or \( x > 1 \).
- The logarithm is negative when \( x - 1 < 1 \) or \( x < 2 \) (since \( 0.3 < 1 \)).
- Therefore, \( -\log_{0.3}(x-1) \) is positive when \( 1 < x < 2 \) and negative when \( x > 2 \).

2. **Denominator**: \( -x^2 + 3x + 18 \)
- To find where this expression is positive, we can first find its roots by solving \( -x^2 + 3x + 18 = 0 \).
- Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = -1, b = 3, c = 18 \).
- The discriminant \( D = b^2 - 4ac = 3^2 - 4(-1)(18) = 9 + 72 = 81 \).
- The roots are:
\[ x = \frac{-3 \pm \sqrt{81}}{2(-1)} = \frac{-3 \pm 9}{-2} \]
- This gives us \( x = -3 \) and \( x = 6 \).

- The quadratic opens downwards (since the coefficient of \( x^2 \) is negative), so it is positive between its roots:
- Therefore, \( -x^2 + 3x + 18 > 0 \) for \( -3 < x < 6 \).

### Step 2: Combine the conditions

We need to find the intersection of the intervals where the numerator and denominator are both non-negative:

1. From the numerator, we have:
- \( x > 1 \) and \( x < 2 \) gives \( (1, 2) \) and \( x > 2 \) gives \( (2, \infty) \).

2. From the denominator, we have:
- \( -3 < x < 6 \).

### Step 3: Find the valid intervals

- For \( (1, 2) \) from the numerator and \( (-3, 6) \) from the denominator, the intersection is:
\[ (1, 2) \]

- For \( (2, \infty) \) from the numerator and \( (-3, 6) \) from the denominator, the intersection is:
\[ (2, 6) \]

### Step 4: Combine the intervals

The valid intervals for \( f(x) \) are:
\[ (1, 2) \cup (2, 6) \]

Since the function is not defined at \( x = 2 \) (the logarithm becomes zero), we can write the domain as:
\[ (1, 6) \]

### Conclusion

The domain of the function is \( [2, 6) \).