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

The Correct Option is A

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