Question

If the domain of the function \( f(x) = \sec^{-1} \left( \frac{2x}{5x + 3} \right) \)  is \( [\alpha, \beta] \cup (\gamma, \delta) \), then \( |3\alpha + 10(\beta + \gamma) + 21\delta| \) is equal to:

Answer 1

Correct Answer: 24

The function is given as: \[ f(x) = \sec^{-1} \left( \frac{2x}{5x + 3} \right) \] For the domain of \( f(x) \), we need to find when: \[ \left| \frac{2x}{5x + 3} \right| \geq 1 \] This leads to two conditions:
1. \( \frac{2x}{5x + 3} \geq 1 \)
2. \( \frac{2x}{5x + 3} \leq -1 \)
Let's solve each inequality: For the first inequality: \[ \frac{2x}{5x + 3} \geq 1 \quad \Rightarrow \quad 2x \geq 5x + 3 \quad \Rightarrow \quad -3x \geq 3 \quad \Rightarrow \quad x \leq -1 \] For the second inequality:
\[ \frac{2x}{5x + 3} \leq -1 \quad \Rightarrow \quad 2x \leq -5x - 3 \quad \Rightarrow \quad 7x \leq -3 \quad \Rightarrow \quad x \leq -\frac{3}{7} \] Thus, the domain of the function is: \[ [-1, -\frac{3}{5}] \cup (-\frac{3}{5}, -\frac{3}{7}] \] Let: \[ \alpha = -1, \quad \beta = -\frac{3}{5}, \quad \gamma = -\frac{3}{5}, \quad \delta = -\frac{3}{7} \] Now, calculate \( 3\alpha + 10(\beta + \gamma) + 21\delta \): \[ 3\alpha + 10(\beta + \gamma) + 21\delta = 3(-1) + 10\left( -\frac{3}{5} + -\frac{3}{5} \right) + 21\left( -\frac{3}{7} \right) \] \[ = -3 + 10\left( -\frac{6}{5} \right) + 21\left( -\frac{3}{7} \right) \] \[ = -3 + \left( -\frac{60}{5} \right) + \left( -\frac{63}{7} \right) \] \[ = -3 - 12 - 9 = -24 \] Thus, \( |3\alpha + 10(\beta + \gamma) + 21\delta| = 24 \).