Question

If the domain of the function \( f(x) = \log_e \left( \frac{2x + 3}{4x^2 + x - 3} \right) + \cos^{-1} \left( \frac{2x - 1}{x + 2} \right) \) is \( (\alpha, \beta] \), then the value of \( 5\beta - 4\alpha \) is equal to

• A. 12
• B. 10
• C. 11
• D. 9

Answer 1

The Correct Option is A

To determine the value of \( 5\beta - 4\alpha \), we must first ascertain the domain of \( f(x) \). This is achieved by finding the intersection of the domains of its two constituent functions.

Concept Used:

The domain of a function encompasses all permissible input values for which the function yields a defined output.

1. Logarithmic Function Domain: For \( g(x) = \log_e(u(x)) \), the argument \( u(x) \) must be strictly positive (\( u(x)>0 \)).

2. Inverse Cosine Function Domain: For \( h(x) = \cos^{-1}(v(x)) \), the argument \( v(x) \) must lie within the closed interval \( [-1, 1] \) (\( -1 \leq v(x) \leq 1 \)).

3. Combined Function Domain: The domain of \( f(x) = g(x) + h(x) \) is the intersection of the domains of \( g(x) \) and \( h(x) \).

Step-by-Step Solution:

Step 1: Determine the domain of the logarithmic component \( \log_e \left( \frac{2x + 3}{4x^2 + x - 3} \right) \).

The argument of the logarithm requires \( \frac{2x + 3}{4x^2 + x - 3}>0 \).

Factoring the quadratic denominator yields \( 4x^2 + x - 3 = (4x-3)(x+1) \). The inequality becomes:

\[ \frac{2x + 3}{(4x - 3)(x + 1)} > 0 \]

Critical points are \( x = -3/2, x = -1, \) and \( x = 3/4 \). Sign analysis reveals the intervals where the expression is positive.

The solution for this inequality is \( D_1 = \left(-\frac{3}{2}, -1\right) \cup \left(\frac{3}{4}, \infty\right) \).

Step 2: Determine the domain of the inverse cosine component \( \cos^{-1} \left( \frac{2x - 1}{x + 2} \right) \).

The argument of the inverse cosine function must satisfy \( -1 \leq \frac{2x - 1}{x + 2} \leq 1 \), with \( x eq -2 \).

The first inequality is \( \frac{2x - 1}{x + 2} \geq -1 \), which simplifies to \( \frac{3x + 1}{x + 2} \geq 0 \). The solution is \( (-\infty, -2) \cup [-1/3, \infty) \).

The second inequality is \( \frac{2x - 1}{x + 2} \leq 1 \), which simplifies to \( \frac{x - 3}{x + 2} \leq 0 \). The solution is \( (-2, 3] \).

The domain \( D_2 \) is the intersection of these two solutions: \( D_2 = \left( (-\infty, -2) \cup [-1/3, \infty) \right) \cap (-2, 3] = \left[-\frac{1}{3}, 3\right] \).

Step 3: Calculate the overall domain of \( f(x) \) by intersecting \( D_1 \) and \( D_2 \).

\[ \text{Domain}(f) = D_1 \cap D_2 = \left( \left(-\frac{3}{2}, -1\right) \cup \left(\frac{3}{4}, \infty\right) \right) \cap \left[-\frac{1}{3}, 3\right] \]

The intersection of \( (-\frac{3}{2}, -1) \) and \( [-\frac{1}{3}, 3] \) is empty. The intersection of \( (\frac{3}{4}, \infty) \) and \( [-\frac{1}{3}, 3] \) is \( (\frac{3}{4}, 3] \).

Therefore, the domain of \( f(x) \) is \( (\frac{3}{4}, 3] \).

Step 4: Identify \( \alpha \) and \( \beta \) from the determined domain.

Comparing the domain \( (\frac{3}{4}, 3] \) with the format \( (\alpha, \beta] \), we find:

\[ \alpha = \frac{3}{4} \quad \text{and} \quad \beta = 3 \]

Final Computation & Result:

Compute the value of \( 5\beta - 4\alpha \).

\[ 5\beta - 4\alpha = 5(3) - 4\left(\frac{3}{4}\right) \] \[ = 15 - 3 \] \[ = 12 \]

The computed value of \( 5\beta - 4\alpha \) is 12.