Question

Let D be the domain of the function f(x)=sin⁻¹(\( log_{3 x} (\frac{6+2log_3x}{−5x}\))) If the range of the function g : D → R defined by g ( x ) = x − [ x ] is the greatest integer function), is ( α , β ) , then α ² +\(\frac{ 5}{ β}\) is equal to 

• A. 45
• B. 46
• C. 135
• D. 136

Hint:

Domain and range problems for composite functions require careful consideration of the domain and range of each individual function. Review the properties of logarithms, trigonometric functions, and the greatest integer function.

Answer 1

The Correct Option is C

The given inequality is: \[ 6 + 2 \log_3 x - 5x > 0 \quad \text{and} \quad x > 0 \quad \text{and} \quad x = \frac{1}{3}. \] 
From this, we obtain the range: \[ x \in \left(0, \frac{1}{27}\right) \quad \text{... (1)}. \]

Additionally, we consider: \[ -1 \leq \log_3 (6 + 2 \log_3 x - 5x). \] 
Breaking this into cases, we solve: \[ 3x \leq 6 + 2 \log_3 x - 5x \leq 1 \quad \text{and} \quad 3x. \] 
This gives: \[ \leq 1, \, 15x^2 + 6 + 2 \log_3 x \geq 0 \quad \implies x \in \left(0, \frac{1}{27}\right) \quad \text{... (2)}. \] And: \[ 6 + 2 \log_3 x + \frac{5}{3} \geq 0 \quad \implies x \geq 3 - \frac{23}{6} \quad \text{... (3)}. \]

Combining all conditions from (1), (2), and (3): \[ x \in \left(3 - \frac{23}{6}, \frac{1}{27}\right). \]

From this range: Finally, we conclude: \(\alpha\) is a small positive quantity and \(\beta = \frac{1}{27}\). \[ \alpha^2 + \frac{5}{\beta} \, \text{is just greater than 135}. \]