Domain of Inverse Trigonometric Terms
The function is defined as $f(x) = \cos^{-1}\left( \frac{2x - 5}{11 - 3x} \right) + \sin^{-1}(2x^2 - 3x + 1)$. Its domain is the intersection of the domains of the individual inverse trigonometric terms.
Domain for $\cos^{-1}\left( \frac{2x - 5}{11 - 3x} \right)$
The argument of $\cos^{-1}(u)$ must be in the interval $[-1, 1]$. Thus, we require:
$-1 \le \frac{2x - 5}{11 - 3x} \le 1$
This inequality splits into two parts:
- $\frac{2x - 5}{11 - 3x} \le 1 \implies \frac{5x - 16}{11 - 3x} \le 0$. The solution is $x \in (-\infty, 16/5] \cup (11/3, \infty)$.
- $\frac{2x - 5}{11 - 3x} \ge -1 \implies \frac{6 - x}{11 - 3x} \ge 0$. The solution is $x \in (-\infty, 11/3) \cup [6, \infty)$.
The intersection of these solutions gives the domain for the first term:
$D_1 = (-\infty, 16/5] \cup [6, \infty)$
Domain for $\sin^{-1}(2x^2 - 3x + 1)$
The argument of $\sin^{-1}(v)$ must be in the interval $[-1, 1]$. Thus, we require:
$-1 \le 2x^2 - 3x + 1 \le 1$
This inequality splits into two parts:
- $2x^2 - 3x + 1 \le 1 \implies 2x^2 - 3x \le 0$. Factoring gives $x(2x - 3) \le 0$. The solution is $x \in [0, 3/2]$.
- $2x^2 - 3x + 1 \ge -1 \implies 2x^2 - 3x + 2 \ge 0$. The discriminant is $(-3)^2 - 4(2)(2) = 9 - 16 = -7$. Since the leading coefficient is positive and the discriminant is negative, this quadratic is always positive. The solution is $x \in \mathbb{R}$.
The intersection gives the domain for the second term:
$D_2 = [0, 3/2]$
Overall Domain and Calculation
The domain of the function $f(x)$ is the intersection $D_1 \cap D_2$:
$D = ( (-\infty, 16/5] \cup [6, \infty) ) \cap [0, 3/2] = [0, 3/2]$
The problem states the domain is the interval $[\alpha, \beta]$. Comparing this with our result, we have:
$\alpha = 0$
$\beta = 3/2$
We need to compute $\alpha + 2\beta$:
$\alpha + 2\beta = 0 + 2 \times \frac{3}{2} = 0 + 3 = 3$.
The calculated value is 3. The provided correct answer is Option C (1).
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