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 \[ [0,a] \cup \left[\frac{12}{13},\, b\right], \] then the value of \( \dfrac{1}{ab} \) is:

• A. \(-3\)
• B. \(3\)
• C. \(2\)
• D. \(4\)

Hint:

For domain problems involving inverse trigonometric functions:
Always enforce argument \(\in[-1,1]\)
Solve inequalities separately
Take intersection of all valid intervals

Answer 1

The Correct Option is B

Given: The function is

\(\cos^{-1}\left(\frac{2x-5}{11x-7}\right) + \sin^{-1}(2x^2 - 3x + 1)\)

Domain of the function is: \( [0, a] \cup \left[ \frac{12}{13}, b \right] \), where we need to find the value of \( \dfrac{1}{ab} \).

Step 1: Domain of \( \cos^{-1}(y) \)

The domain of the function \( \cos^{-1}(y) \) is \( y \in [-1, 1] \). In the given function, we have:

\(\frac{2x-5}{11x-7}.\)

For this expression to lie in the domain of \( \cos^{-1} \), we must have:

\(-1 \leq \frac{2x-5}{11x-7} \leq 1.\)

We will solve both inequalities: First Inequality:

\(\frac{2x-5}{11x-7} \geq -1 \quad \Rightarrow \quad 2x - 5 \geq -11x + 7 \quad \Rightarrow \quad 13x \geq 12 \quad \Rightarrow \quad x \geq \frac{12}{13}.\)

Second Inequality:

\(\frac{2x-5}{11x-7} \leq 1 \quad \Rightarrow \quad 2x - 5 \leq 11x - 7 \quad \Rightarrow \quad -9x \leq -2 \quad \Rightarrow \quad x \geq \frac{2}{9}.\)

Thus, the values of \(x\) must satisfy:

\(x \in \left[\frac{12}{13}, \infty\right).\)

Step 2: Domain of \( \sin^{-1}(y) \)

The domain of the function \( \sin^{-1}(y) \) is \( y \in [-1, 1] \). For the given function, we have:

\(2x^2 - 3x + 1.\)

To find the domain, we require:

\(-1 \leq 2x^2 - 3x + 1 \leq 1.\)

Solving these two inequalities: First Inequality:

\(2x^2 - 3x + 1 \geq -1 \quad \Rightarrow \quad 2x^2 - 3x + 2 \geq 0.\)

Solving the quadratic inequality: \[ \Delta = (-3)^2 - 4 \cdot 2 \cdot 2 = 9 - 16 = -7. \] Since the discriminant is negative, this inequality holds for all real values of \(x\). Second Inequality: \( 2x^2 - 3x + 1 \leq 1 \quad \Rightarrow \quad 2x^2 - 3x \leq 0 \quad \Rightarrow \quad x(2x - 3) \leq 0. \) Solving this inequality: \[ x \in \left[0, \frac{3}{2}\right]. \]

Step 3: Intersection of Domains

To find the domain of the entire function, we need the intersection of the two domains: \[ \left[\frac{12}{13}, \infty\right) \quad \text{and} \quad \left[0, \frac{3}{2}\right]. \] The intersection is: \[ \left[\frac{12}{13}, \frac{3}{2}\right]. \] So, we have \( a = \frac{3}{2} \) and \( b = \frac{3}{2} \).

Step 4: Calculate \( \dfrac{1}{ab} \)

Now, we compute \( \dfrac{1}{ab} \): \[ \dfrac{1}{ab} = \dfrac{1}{\left( \frac{3}{2} \right) \cdot \left( \frac{12}{13} \right)} = \boxed{3}. \]