Question

\[ f(x) = \log_e \left( 4x^2 + 11x + 6 \right) + \sin^{-1} \left( 4x + 3 \right) + \cos^{-1} \left( \frac{10x + 6}{3} \right) \]then \( 36|\alpha + \beta| \) is equal to:

Answer 1

Understanding the Problem

We need to find the domain of the function \( f(x) = \log_e(4x^2 + 11x + 6) + \sin^{-1}(4x + 3) + \cos^{-1}\left(\frac{10x + 6}{3}\right) \), determine α and β, and then calculate \( 36|\alpha + \beta| \)

Solution

1. Domain of \( \log_e(4x^2 + 11x + 6) \):

The logarithm is defined when \( 4x^2 + 11x + 6 > 0 \)

Factoring the quadratic: \( (4x + 3)(x + 2) > 0 \)

The solution is \( x \in (-\infty, -2) \cup (-\frac{3}{4}, \infty) \)

2. Domain of \( \sin^{-1}(4x + 3) \):

The arcsine is defined when \( -1 \le 4x + 3 \le 1 \)

\( -4 \le 4x \le -2 \)

\( -1 \le x \le -\frac{1}{2} \)

Thus, \( x \in [-1, -\frac{1}{2}] \)

3. Domain of \( \cos^{-1}\left(\frac{10x + 6}{3}\right) \):

The arccosine is defined when \( -1 \le \frac{10x + 6}{3} \le 1 \)

\( -3 \le 10x + 6 \le 3 \)

\( -9 \le 10x \le -3 \)

\( -\frac{9}{10} \le x \le -\frac{3}{10} \)

Thus, \( x \in [-\frac{9}{10}, -\frac{3}{10}] \)

4. Combining the Domains:

We need the intersection of the three domains.

From \( (-\infty, -2) \cup (-\frac{3}{4}, \infty) \) and \( [-1, -\frac{1}{2}] \), we get \( [-\frac{3}{4}, -\frac{1}{2}] \)

From \( [-\frac{3}{4}, -\frac{1}{2}] \) and \( [-\frac{9}{10}, -\frac{3}{10}] \), we get \( [-\frac{9}{10}, -\frac{3}{10}] \)

5. Finding α and β:

\( \alpha = -\frac{9}{10} \) and \( \beta = -\frac{3}{10} \)

6. Calculating α + β:

\( \alpha + \beta = -\frac{9}{10} - \frac{3}{10} = -\frac{12}{10} = -\frac{6}{5} \)

7. Calculating 36|α + β|:

\( 36|\alpha + \beta| = 36 \left| -\frac{6}{5} \right| = 36 \times \frac{6}{5} = \frac{216}{5} = 43.2 \)

Final Answer

The correct answer is 43.2.