Question

Let the domain of the function $ f(x) = \cos^{-1} \left( \frac{4x + 5}{3x - 7} \right) $ be $ [\alpha, \beta] $ and the domain of $ g(x) = \log_2 \left( 2 - 6 \log_2 \left( 2x + 5 \right) \right) $ be $ (\gamma, \delta) $. Then $ |7(\alpha + \beta) + 4(\gamma + \delta)| $ is equal to:

Hint:

When working with domains of functions involving logarithms or trigonometric functions, ensure that the expressions inside the logarithms or inverse trigonometric functions stay within their respective domains.

Answer 1

Correct Answer: 96

The given function is: \( f(x) = \cos^{-1} \left( \frac{4x + 5}{3x - 7} \right) \)

We are given the following inequalities: 

\( -1 \leq \frac{4x + 5}{3x - 7} \leq 1 \)

First, solve the inequalities:

\( \frac{7x - 2}{3x - 7} \geq 0 \), and \( \frac{x + 12}{3x - 7} \leq 0 \)

Solving for \( x \), we get:

\( x \in \left( -\infty, \frac{2}{7} \right) \cup \left( \frac{7}{3}, \infty \right) \), and \( x \in \left[ -\frac{12}{7}, \frac{7}{3} \right] \)

Thus, we have \( x \in \left[ -12, \frac{2}{7} \right] \). Let \( \alpha = -12 \) and \( \beta = \frac{2}{7} \)

The next function is: \( g(x) = \log_2 \left( 2 - 6 \log_2(2x + 5) \right) \)

To solve for \( g(x) \), we simplify:

\( 2 - 6 \log_2(2x + 5) \geq 0 \), which gives \( 2x + 5 \geq 0 \), so \( x \geq -\frac{5}{2} \)

Next, solving for \( \log_2(2x + 5) \), we get:

\( \log_2(2x + 5) \leq \frac{1}{3} \), which implies \( (2x + 5)^3 \leq 27 \)

Thus, \( 2x + 5 \leq 3 \) gives \( x \leq -\frac{5}{2} - 1 \), so \( \gamma = -\frac{5}{2} \) and \( \delta = -1 \)

The final expression is:

\( |7(\alpha + \beta) + 4(\gamma + \delta)| = 96 \)

Answer 2

1.Domain of \(f(x) = \cos^{-1}\left(\frac{4x+5}{3x-7}\right)\) For the arccosine function to be defined, we require \(-1 \leq \frac{4x+5}{3x-7} \leq 1\). \(\frac{4x+5}{3x-7} \leq 1\):
\[ \frac{4x+5}{3x-7} - 1 \leq 0 \] \[ \frac{4x+5 - (3x-7)}{3x-7} \leq 0 \] \[ \frac{x+12}{3x-7} \leq 0 \] The critical points are \(x = -12\) and \(x = \frac{7}{3}\).
Testing intervals:

\(x<-12\): Both \((x+12)\) and \((3x-7)\) are negative, so the fraction is positive.

\(-12<x<\frac{7}{3}\): \((x+12)\) is positive, \((3x-7)\) is negative, so the fraction is negative.

\(x>\frac{7}{3}\): Both \((x+12)\) and \((3x-7)\) are positive, so the fraction is positive.

Therefore, \(\frac{4x+5}{3x-7} \leq 1\) when \(-12 \leq x<\frac{7}{3}\). \(\frac{4x+5}{3x-7} \geq -1\): \[ \frac{4x+5}{3x-7} + 1 \geq 0 \] \[ \frac{4x+5 +(3x-7)}{3x-7} \geq 0 \] \[ \frac{7x-2}{3x-7} \geq 0 \] 

The critical points are \(x = \frac{2}{7}\) and \(x = \frac{7}{3}\).

Testing intervals:

\(x<\frac{2}{7}\): Both \((7x-2)\) and \((3x-7)\) are negative, so the fraction is positive.

\(\frac{2}{7}<x<\frac{7}{3}\): \((7x-2)\) is positive, \((3x-7)\) is negative, so the fraction is negative.

\(x>\frac{7}{3}\): Both \((7x-2)\) and \((3x-7)\) are positive, so the fraction is positive.

Therefore, \(\frac{4x+5}{3x-7} \geq -1\) when \(x \leq \frac{2}{7}\) or \(x>\frac{7}{3}\).

We need both conditions to hold.

The first condition says \(-12 \leq x<\frac{7}{3}\).

The second says \(x \leq \frac{2}{7}\) or \(x>\frac{7}{3}\).

Taking the intersection of the two solutions, we get \(\left[-12, \frac{2}{7}\right]\).

Therefore, \(\alpha = -12\) and \(\beta = \frac{2}{7}\).

2. Domain of \(g(x) = \log_2\left(2 - 6\log_{27}(2x+5)\right)\) 

For the logarithm to be defined, we need a positive argument.

We have two logarithm expressions.

\(2x+5>0 \implies x>-\frac{5}{2} = -2.5\) \(2 - 6\log_{27}(2x+5)>0 \implies 2>6\log_{27}(2x+5) \implies \frac{1}{3}>\log_{27}(2x+5)\) \(27^{\frac{1}{3}}>2x+5 \implies 3>2x+5 \implies -2>2x \implies x<-1\)

So, we need \(-2.5<x<-1\).

Thus, \(\gamma = -2.5\) and \(\delta = -1\).
3. Compute the Final Expression \[ 7(\alpha + \beta) + 4(\gamma + \delta) = 7\left(-12 + \frac{2}{7}\right) + 4(-2.5 - 1) \] \[ = 7\left(-\frac{84}{7} + \frac{2}{7}\right) + 4(-3.5) \] \[ = 7\left(-\frac{82}{7}\right) - 14 \] \[ = -82 - 14 \] \[ = -96 \] \[ \left|7(\alpha + \beta) + 4(\gamma + \delta)\right| = |-96| = 96.\]