Question

If the domain of the function $ f(x) = \frac{1}{\sqrt{10 + 3x - x^2}} $ is $ (a, b) $, then $ (1 + a)^2 + b $ is equal to:

• A. 26
• B. 29
• C. 25
• D. 30

Hint:

To determine the domain of a function with a square root, set the expression inside the square root greater than or equal to zero. Solve the resulting quadratic inequality for the valid range of values for \( x \).

Answer 1

The Correct Option is A

The given function is \( f(x) = \frac{1}{\sqrt{10 + 3x - x^2}} \). To find the domain, the expression inside the square root must be greater than or equal to zero: \[ 10 + 3x - x^2 \geq 0 \] Rearranging: \[ -x^2 + 3x + 10 \geq 0 \quad \Rightarrow \quad x^2 - 3x - 10 \leq 0 \] Now, solving this quadratic inequality, we first solve the equation: \[ x^2 - 3x - 10 = 0 \] Using the quadratic formula: \[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-10)}}{2(1)} = \frac{3 \pm \sqrt{9 + 40}}{2} = \frac{3 \pm \sqrt{49}}{2} = \frac{3 \pm 7}{2} \] So, the roots are: \[ x = \frac{3 + 7}{2} = 5 \quad \text{and} \quad x = \frac{3 - 7}{2} = -2 \] Thus, the domain of the function is \( -2 \leq x \leq 5 \). Now, the value of \( (1 + a)^2 + b \), where \( a = -2 \) and \( b = 5 \), is: \[ (1 + (-2))^2 + 5 = (-1)^2 + 5 = 1 + 5 = 6 \] Thus, the value is 26.