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

We are asked to find the domain of the function \( f(x) = \frac{1}{\sqrt{10+3x-x^2}} + \frac{1}{\sqrt{x+|x|}} \), which is given as the interval \( (a, b) \). After determining \(a\) and \(b\), we need to compute the value of \( (1 + a)^2 + b^2 \).

Concept Used:

The domain of a function is the set of all possible input values (\(x\)) for which the function is defined. For a function of the form \( \frac{1}{\sqrt{g(x)}} \), the expression inside the square root, \( g(x) \), must be strictly positive, i.e., \( g(x) > 0 \). If a function is a sum of two or more functions, its domain is the intersection of the domains of the individual functions.

Step-by-Step Solution:

Step 1: For the function \( f(x) \) to be defined, the expressions under the square roots in the denominators must be strictly positive. Let's analyze each term separately.

For the first term, \( \frac{1}{\sqrt{10+3x-x^2}} \), we must have:

\[ 10 + 3x - x^2 > 0 \]

For the second term, \( \frac{1}{\sqrt{x+|x|}} \), we must have:

\[ x + |x| > 0 \]

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

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

Multiply by -1 and reverse the inequality sign:

\[ x^2 - 3x - 10 < 0 \]

Factor the quadratic expression:

\[ (x-5)(x+2) < 0 \]

The roots are \( x = 5 \) and \( x = -2 \). The inequality holds for values of \( x \) between the roots. So, the solution is:

\[ -2 < x < 5 \quad \text{or} \quad x \in (-2, 5) \]

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

We consider two cases based on the definition of \( |x| \).

Case 1: If \( x \ge 0 \), then \( |x| = x \). The inequality becomes:

\[ x + x > 0 \implies 2x > 0 \implies x > 0 \]

Case 2: If \( x < 0 \), then \( |x| = -x \). The inequality becomes:

\[ x + (-x) > 0 \implies 0 > 0 \]

This is a false statement, so there are no solutions for \( x < 0 \).

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

Step 4: Find the intersection of the domains from Step 2 and Step 3.

The domain of \( f(x) \) is the intersection of \( (-2, 5) \) and \( (0, \infty) \).

\[ \text{Domain} = (-2, 5) \cap (0, \infty) = (0, 5) \]

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

We have found the domain to be \( (0, 5) \). Therefore:

\[ a = 0 \quad \text{and} \quad b = 5 \]

Final Computation & Result:

Now, we compute the value of the expression \( (1 + a)^2 + b^2 \) using the values of \(a\) and \(b\).

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

The value of the expression is 26.