Question

If the domain of the function $\sin^{-1} \left( \frac{3x - 22}{2x - 19} \right) + \log_e \left( \frac{3x^2 - 8x + 5}{x^2 - 3x - 10} \right)$ is $(\alpha, \beta]$, then $3\alpha + 10\beta$ is equal to:

• A. 97
• B. 100
• C. 95
• D. 98

Answer 1

Answer: (A)

97

Answer 2

Domain of the $\sin^{-1}$ Function:
For $\sin^{-1}\left(\frac{3x - 22}{2x - 19}\right)$ to be defined, the argument $\frac{3x - 22}{2x - 19}$ must satisfy:
$-1 \leq \frac{3x - 22}{2x - 19} \leq 1$
Solving this inequality involves two cases:

Case 1: $\frac{3x - 22}{2x - 19} \leq 1$
$3x - 22 \leq 2x - 19 \implies x \leq 3$
Case 2: $\frac{3x - 22}{2x - 19} \geq -1$
$3x - 22 \geq -2x + 19 \implies 5x \geq 41 \implies x \geq \frac{41}{5}$
Therefore, the solution for the $\sin^{-1}$ function domain is:
$x \in \left[\frac{41}{5}, 3\right]$

Domain of the $\log_e$ Function:
For $\log_e\left(\frac{3x^2 - 8x + 5}{x^2 - 3x - 10}\right)$ to be defined, the argument $\frac{3x^2 - 8x + 5}{x^2 - 3x - 10}$ must be positive:
$\frac{3x^2 - 8x + 5}{x^2 - 3x - 10} > 0$
Factorize both the numerator and denominator: $\frac{(3x - 5)(x - 1)}{(x - 5)(x + 2)} > 0$
Determine the intervals where this inequality holds by testing values between the critical points $x = -2, 1, 5, \frac{5}{3}$.

The valid intervals are: $x \in \left(\frac{5}{3}, 1\right) \cup (5, \infty)$

Intersection of the Domains:
The domain of the combined function is the intersection of the two domains:
$x \in \left[\frac{41}{5}, 3\right] \cap \left(\frac{5}{3}, 1\right) \cup (5, \infty)$ This simplifies to: $x \in \left[\frac{41}{5}, 3\right]$

Calculate $3\alpha + 10\beta$:
Here, $\alpha = \frac{41}{5}$ and $\beta = 3$.

Then: $3\alpha + 10\beta = 3 \times \frac{41}{5} + 10 \times 3 = \frac{123}{5} + 30 = 97$