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 find the value of \( 5\beta - 4\alpha \), we first need to determine the domain of the function \( f(x) \), which is the intersection of the domains of its two component functions.

Concept Used:

The domain of a function is the set of all possible input values for which the function is defined.

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

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

3. Domain of Combined Function: 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 part \( \log_e \left( \frac{2x + 3}{4x^2 + x - 3} \right) \).

The argument of the logarithm must be positive:

\[ \frac{2x + 3}{4x^2 + x - 3} > 0 \]

First, we factor the quadratic denominator: \( 4x^2 + x - 3 = 4x^2 + 4x - 3x - 3 = 4x(x+1) - 3(x+1) = (4x-3)(x+1) \). The inequality becomes:

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

The critical points are \( x = -3/2, x = -1, \) and \( x = 3/4 \). Using the wavy curve method (sign analysis), we find the intervals where the expression is positive.

The solution to this inequality is \( x \in \left(-\frac{3}{2}, -1\right) \cup \left(\frac{3}{4}, \infty\right) \). Let this be Domain \( D_1 \).

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

The argument of the inverse cosine function must satisfy:

\[ -1 \leq \frac{2x - 1}{x + 2} \leq 1 \]

This can be split into two separate inequalities. Also, note that \( x \neq -2 \).

First inequality:

\[ \frac{2x - 1}{x + 2} \geq -1 \implies \frac{2x - 1}{x + 2} + 1 \geq 0 \implies \frac{2x - 1 + x + 2}{x + 2} \geq 0 \implies \frac{3x + 1}{x + 2} \geq 0 \]

The solution for this part is \( x \in (-\infty, -2) \cup [-1/3, \infty) \).

Second inequality:

\[ \frac{2x - 1}{x + 2} \leq 1 \implies \frac{2x - 1}{x + 2} - 1 \leq 0 \implies \frac{2x - 1 - (x + 2)}{x + 2} \leq 0 \implies \frac{x - 3}{x + 2} \leq 0 \]

The solution for this part is \( x \in (-2, 3] \).

The domain \( D_2 \) is the intersection of the solutions to these two inequalities: \( \left( (-\infty, -2) \cup [-1/3, \infty) \right) \cap (-2, 3] \). This gives:

\[ D_2 = \left[-\frac{1}{3}, 3\right] \]

Step 3: Find the overall domain of \( f(x) \) by taking the intersection of \( D_1 \) 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 \( (\frac{3}{4}, 3] \).

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

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

The given domain format is \( (\alpha, \beta] \). Comparing this with our result \( (\frac{3}{4}, 3] \), we get:

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

Final Computation & Result:

Now, we calculate the value of the expression \( 5\beta - 4\alpha \).

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

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

Answer 2

Determine the domain for each part of \( f(x) \): Logarithmic Part: \( \log_{e} \left( \frac{2x + 3}{4x^2 + x - 3} \right) \) requires
\[ \frac{2x + 3}{4x^2 + x - 3} > 0. \]
Critical points are \( x = -\frac{3}{2}, x = -1, \) and \( x = \frac{3}{4} \).
Solution: \( x \in \left( -\frac{3}{2}, -1 \right) \cup \left( \frac{3}{4}, \infty \right) \).

Inverse Cosine Part: \( \cos^{-1} \left( \frac{2x - 1}{x + 2} \right) \) requires
\[ -1 \leq \frac{2x - 1}{x + 2} \leq 1. \]
Solution: \( x \in \left[ -\frac{1}{3}, 3 \right] \).
Intersection of Domains:
The combined domain is \( \left( \frac{3}{4}, 3 \right] \), giving \( (\alpha, \beta] = \left( \frac{3}{4}, 3 \right] \).
Calculate \( 5\beta - 4\alpha \): \( \alpha = \frac{3}{4}, \beta = 3 \).

\[ 5\beta - 4\alpha = 5 \times 3 - 4 \times \frac{3}{4} = 15 - 3 = 12. \]