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}
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}
Show 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.
Updated On: Feb 5, 2026
- -3
- 3
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The Correct Option is B
Solution and Explanation
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, \( \frac{1}{ab} = 3 \).
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, \( \frac{1}{ab} = 3 \).
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