If the domain of the function f(x) = cos^-1left(frac2x - 511 - 3xright) + sin^-1(2x^2 - 3x + 1)$f(x) = \cos^{-1}\left(\frac{2x - 5}{11 - 3x}\right) + \sin^{-1}(2x^2 - 3x + 1)$ is the interval [alpha, beta]$[\alpha, \beta]$ , then alpha + 2beta$\alpha + 2\beta$ is equal to :

### Related Formula For inverse trigonometric functions sin^-1(g(x))$\sin^{-1}(g(x))$ and cos^-1(h(x))$\cos^{-1}(h(x))$, the arguments must satisfy: -1 leq g(x) leq 1$$-1 \leq g(x) \leq 1$$ -1 leq h(x) leq 1$$-1 \leq h(x) \leq 1$$ ### Core Logic Given f(x) = cos^-1left(frac2x - 511 - 3xright) + sin^-1(2x^2 - 3x + 1)$f(x) = \cos^{-1}\left(\frac{2x - 5}{11 - 3x}\right) + \sin^{-1}(2x^2 - 3x + 1)$ We establish two simultaneous inequalities for the domain: 1) -1 leq frac2x - 511 - 3x leq 1$-1 \leq \frac{2x - 5}{11 - 3x} \leq 1$ 2) -1 leq 2x^2 - 3x + 1 leq 1$-1 \leq 2x^2 - 3x + 1 \leq 1$ ### Step 1: Solve the Quadratic Inequality From -1 leq 2x^2 - 3x + 1 leq 1$-1 \leq 2x^2 - 3x + 1 \leq 1$: Split into two parts: 2x^2 - 3x + 2 geq 0$2x^2 - 3x + 2 \geq 0$ (This is always true as discriminant D < 0, a > 0$D < 0, a > 0$) 2x^2 - 3x leq 0 Rightarrow x(2x - 3) leq 0$2x^2 - 3x \leq 0 \Rightarrow x(2x - 3) \leq 0$ x in left[0, frac32right] dots(i)$$x \in \left[0, \frac{3}{2}\right] \dots(i)$$ ### Step 2: Solve the Rational Inequality From -1 leq frac2x - 511 - 3x leq 1$-1 \leq \frac{2x - 5}{11 - 3x} \leq 1$: Part A: frac2x - 511 - 3x + 1 geq 0 Rightarrow frac2x - 5 + 11 - 3x11 - 3x geq 0 Rightarrow frac6 - x11 - 3x geq 0$\frac{2x - 5}{11 - 3x} + 1 \geq 0 \Rightarrow \frac{2x - 5 + 11 - 3x}{11 - 3x} \geq 0 \Rightarrow \frac{6 - x}{11 - 3x} \geq 0$

x in left(-infty, frac113right) cup [6, infty)$$x \in \left(-\infty, \frac{11}{3}\right) \cup [6, \infty)$$ Part B: frac2x - 511 - 3x - 1 leq 0 Rightarrow frac5x - 1611 - 3x leq 0 Rightarrow x in left(-infty, frac165right] cup left(frac113, inftyright)$\frac{2x - 5}{11 - 3x} - 1 \leq 0 \Rightarrow \frac{5x - 16}{11 - 3x} \leq 0 \Rightarrow x \in \left(-\infty, \frac{16}{5}\right] \cup \left(\frac{11}{3}, \infty\right)$ Intersection of Part A and Part B: $ Intersection of Part A and Part B: $x in left(-infty, frac165right] cup [6, infty) dots(ii)$x \in \left(-\infty, \frac{16}{5}\right] \cup [6, \infty) \dots(ii)$ ### Step 3: Final Intersection Taking the intersection of $ ### Step 3: Final Intersection Taking the intersection of $(i) and $ and $(ii): $: $x in left[0, frac32right]$x \in \left[0, \frac{3}{2}\right]$ Comparing this with $ Comparing this with $[\alpha, \beta], we have $, we have $\alpha = 0, \beta = \frac{3}{2}. Therefore, $. Therefore, $\alpha + 2\beta = 0 + 2\left(\frac{3}{2}\right) = 3 ### Pattern Recognition Whenever dealing with dual inverse trig terms, strictly isolate the bounding intervals $ ### Pattern Recognition Whenever dealing with dual inverse trig terms, strictly isolate the bounding intervals $[-1, 1]$ for each argument separately and use a number line intersection to find the strictest common region. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Functions Class 11 Maths: Linear Inequalities