Question

If the domain of the function \(f(x) = \cos^{-1}\left(\frac{2x-5}{11-3x}\right) + \sin^{-1}(2x^2-3x+1)\) is the interval \([\alpha, \beta]\), then \(\alpha + 2\beta\) is equal to:}

• A. 3
• B. 5
• C. 1
• D. 2

Hint:

When solving rational inequalities, the wavy curve method is very efficient. Always bring all terms to one side to get the form \(\frac{P(x)}{Q(x)} \ge 0\) or \(\le 0\), then find the roots of P(x) and Q(x) to determine the intervals on the number line.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The domain of a function that is a sum of two other functions is the intersection of their individual domains. We need to find the domains of \(\cos^{-1}(u)\) and \(\sin^{-1}(v)\) and then find their intersection.
Step 2: Key Formula or Approach:
The domain of both \(\cos^{-1}(u)\) and \(\sin^{-1}(u)\) is defined by the condition \(-1 \le u \le 1\).
So we need to solve the following two inequalities: 1. \(-1 \le \frac{2x-5}{11-3x} \le 1\) 2. \(-1 \le 2x^2-3x+1 \le 1\) The final domain will be the intersection of the solution sets of these two inequalities.
Step 3: Detailed Explanation:
Part 1: Solving \(-1 \le \frac{2x-5}{11-3x} \le 1\)
This inequality splits into two parts: \(\frac{2x-5}{11-3x} \le 1\) and \(\frac{2x-5}{11-3x} \ge -1\).
For the first part: \(\frac{2x-5}{11-3x} - 1 \le 0 \implies \frac{2x-5 - (11-3x)}{11-3x} \le 0 \implies \frac{5x-16}{11-3x} \le 0\).
Using the wavy curve method, critical points are \(x=16/5=3.2\) and \(x=11/3 \approx 3.67\). The expression is \(\le 0\) for \(x \in (-\infty, 16/5] \cup (11/3, \infty)\).
For the second part: \(\frac{2x-5}{11-3x} + 1 \ge 0 \implies \frac{2x-5 + 11-3x}{11-3x} \ge 0 \implies \frac{-x+6}{11-3x} \ge 0 \implies \frac{x-6}{3x-11} \ge 0\).
Critical points are \(x=6\) and \(x=11/3\). The expression is \(\ge 0\) for \(x \in (-\infty, 11/3) \cup [6, \infty)\).
The intersection of these two solution sets gives the domain \(D_1 = (-\infty, 16/5] \cup [6, \infty)\).
Part 2: Solving \(-1 \le 2x^2-3x+1 \le 1\)
This splits into two inequalities:
% Option (i) \(2x^2-3x+1 \ge -1 \implies 2x^2-3x+2 \ge 0\).
The discriminant of this quadratic is \(\Delta = (-3)^2 - 4(2)(2) = 9 - 16 = -7<0\). Since the leading coefficient (2) is positive, the quadratic is always positive. So this inequality holds for all \(x \in \mathbb{R}\).
% Option (ii) \(2x^2-3x+1 \le 1 \implies 2x^2-3x \le 0 \implies x(2x-3) \le 0\).
The roots are \(x=0\) and \(x=3/2\). The parabola opens upwards, so the expression is \(\le 0\) between the roots.
The solution is \(0 \le x \le 3/2\).
The domain for the second term is \(D_2 = [0, 3/2]\).
Part 3: Finding the intersection \(D_1 \cap D_2\)
We need to find the intersection of \(D_1 = (-\infty, 16/5] \cup [6, \infty)\) and \(D_2 = [0, 3/2]\). Since \(16/5 = 3.2\) and \(3/2 = 1.5\). We are intersecting \((-\infty, 3.2] \cup [6, \infty)\) with \([0, 1.5]\). The intersection is clearly \([0, 1.5]\). So the final domain is \([\alpha, \beta] = [0, 3/2]\). This means \(\alpha = 0\) and \(\beta = 3/2\). We need to calculate \(\alpha + 2\beta\): \[ \alpha + 2\beta = 0 + 2\left(\frac{3}{2}\right) = 3 \] Step 4: Final Answer:
The value of \(\alpha + 2\beta\) is 3.