Question

If the domain of the function \[ \cos^{-1} \left( \frac{2x - 5}{11x - 7} \right) + \sin^{-1} \left( 2x^2 - 3x + 1 \right) \] is \[ [0, a] \cup [12/13, b] \] then \( \frac{1}{ab} \) is equal to}

• A. -3
• B. 3
• C. 2
• D. 4

Hint:

For domain problems involving inverse trigonometric functions, always ensure the argument of the function lies within its valid range, and solve the resulting inequalities.

Answer 1

The Correct Option is B

Step 1: Solve for the domain of the first term.
The domain of the inverse cosine function is given by the condition: \[ \frac{2x - 5}{11x - 7} \in [-1, 1] \] Solving this inequality gives the domain for the first term. Step 2: Solve for the domain of the second term.
For the inverse sine function, the domain is given by: \[ -1 \leq 2x^2 - 3x + 1 \leq 1 \] Solving this inequality gives the domain for the second term. Step 3: Determine the values of \( a \) and \( b \).
By solving the inequalities, we find that the domain is \( [0, a] \cup [12/13, b] \), and \( ab = 3 \). Therefore, \( \frac{1}{ab} = 3 \).