Question

If the domain of the function \( f(x) = \dfrac{1}{\sqrt{10 + 3x - x^2}} + \dfrac{1}{\sqrt{x + |x|}} \) is \( (a, b) \), then \((1 + a)^2 + b^2\) 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

Determine the domain of \( f(x) = \frac{1}{\sqrt{10+3x-x^2}} + \frac{1}{\sqrt{x+|x|}} \), which is \( (a, b) \). Calculate \( (1 + a)^2 + b^2 \).

Concept Used:

The domain encompasses all valid input \(x\) values. For \( \frac{1}{\sqrt{g(x)}} \), \( g(x) \) must be greater than 0. The domain of a sum of functions is the intersection of their individual domains.

Step-by-Step Solution:

Step 1: For \( f(x) \) to be defined, the radicands must be strictly positive. Consider each term:

Term 1: \( 10 + 3x - x^2 > 0 \)

Term 2: \( x + |x| > 0 \)

The domain of \( f(x) \) is the intersection of the solutions to these inequalities.

Step 2: Solve \( 10 + 3x - x^2 > 0 \).

Rearranging gives \( x^2 - 3x - 10 < 0 \).

Factoring yields \( (x-5)(x+2) < 0 \).

The solution is \( -2 < x < 5 \), or \( x \in (-2, 5) \).

Step 3: Solve \( x + |x| > 0 \).

Case 1: \( x \ge 0 \). Then \( |x| = x \). The inequality becomes \( x + x > 0 \implies 2x > 0 \implies x > 0 \).

Case 2: \( x < 0 \). Then \( |x| = -x \). The inequality becomes \( x + (-x) > 0 \implies 0 > 0 \), which is false.

The solution for the second inequality is \( x > 0 \), or \( x \in (0, \infty) \).

Step 4: Intersect the domains from Step 2 and Step 3.

Domain \( = (-2, 5) \cap (0, \infty) = (0, 5) \).

Step 5: Compare with the given domain \( (a, b) \).

Thus, \( a = 0 \) and \( b = 5 \).

Final Computation & Result:

Calculate \( (1 + a)^2 + b^2 \):

\[ (1 + a)^2 + b^2 = (1 + 0)^2 + (5)^2 \] \[ = (1)^2 + 25 \] \[ = 1 + 25 = 26 \]

The value is 26.