### Related Formula textDomain of log_e(A): A gt 0$\text{Domain of } \log_e(A): A \gt 0$ textDomain of cos^-1(B): -1 le B le 1$\text{Domain of } \cos^{-1}(B): -1 \le B \le 1$ ### Core Logic Condition 1: Logarithm domain frac2x + 34x^2 + x - 3 gt 0$\frac{2x + 3}{4x^2 + x - 3} \gt 0$ Factorizing the denominator: 4x^2 + 4x - 3x - 3 = 4x(x+1) - 3(x+1) = (4x-3)(x+1)$4x^2 + 4x - 3x - 3 = 4x(x+1) - 3(x+1) = (4x-3)(x+1)$. frac2x + 3(4x - 3)(x + 1) gt 0$\frac{2x + 3}{(4x - 3)(x + 1)} \gt 0$ Critical points are x = -frac32, -1, frac34$x = -\frac{3}{2}, -1, \frac{3}{4}$. Using the wavy curve method: The intervals where it is positive are left(-frac32, -1right) cup left(frac34, inftyright)$\left(-\frac{3}{2}, -1\right) \cup \left(\frac{3}{4}, \infty\right)$. Let this be set D_1$D_1$. ### Step 1: Inverse Trigonometric Domain Condition 2: Arc cosine domain -1 le frac2x - 1x + 2 le 1$-1 \le \frac{2x - 1}{x + 2} \le 1$ This breaks into two inequalities: (i) frac2x - 1x + 2 ge -1 Rightarrow frac2x - 1x + 2 + 1 ge 0 Rightarrow frac3x + 1x + 2 ge 0$\frac{2x - 1}{x + 2} \ge -1 \Rightarrow \frac{2x - 1}{x + 2} + 1 \ge 0 \Rightarrow \frac{3x + 1}{x + 2} \ge 0$ Critical points are x = -2, -frac13$x = -2, -\frac{1}{3}$. The valid set is (-infty, -2) cup left[-frac13, inftyright)$(-\infty, -2) \cup \left[-\frac{1}{3}, \infty\right)$. (ii) frac2x - 1x + 2 le 1 Rightarrow frac2x - 1x + 2 - 1 le 0 Rightarrow fracx - 3x + 2 le 0$\frac{2x - 1}{x + 2} \le 1 \Rightarrow \frac{2x - 1}{x + 2} - 1 \le 0 \Rightarrow \frac{x - 3}{x + 2} \le 0$ Critical points are x = -2, 3$x = -2, 3$. The valid set is (-2, 3]$(-2, 3]$. Intersection for the cosine domain (D_2$D_2$): (-2, 3] cap left( (-infty, -2) cup left[-frac13, inftyright) right) = left[-frac13, 3right]$(-2, 3] \cap \left( (-\infty, -2) \cup \left[-\frac{1}{3}, \infty\right) \right) = \left[-\frac{1}{3}, 3\right]$. ### Step 2: Finding Total Domain Total domain D = D_1 cap D_2$D = D_1 \cap D_2$: D = left( left(-frac32, -1right) cup left(frac34, inftyright) right) cap left[-frac13, 3right]$D = \left( \left(-\frac{3}{2}, -1\right) \cup \left(\frac{3}{4}, \infty\right) \right) \cap \left[-\frac{1}{3}, 3\right]$ Since left[-frac13, 3right]$\left[-\frac{1}{3}, 3\right]$ starts strictly greater than -1$-1$, the overlap with left(-frac32, -1right)$\left(-\frac{3}{2}, -1\right)$ is empty. Overlap with left(frac34, inftyright)$\left(\frac{3}{4}, \infty\right)$ occurs from frac34$\frac{3}{4}$ to 3$3$. Thus, D = left(frac34, 3right]$D = \left(\frac{3}{4}, 3\right]$. ### Step 3: Calculating Final Expression Comparing with (alpha, beta]$(\alpha, \beta]$: alpha = frac34$\alpha = \frac{3}{4}$ and beta = 3$\beta = 3$. We need to find 5beta - 4alpha$5\beta - 4\alpha$: 5(3) - 4left(frac34right) = 15 - 3 = 12$5(3) - 4\left(\frac{3}{4}\right) = 15 - 3 = 12$ ### Pattern Recognition Compound domains always require individual set resolution (using Wavy Curve) followed by taking the strictest absolute intersection. Identifying disjoint segments (like x lt -1$x \lt -1$ missing the cosine bound) prunes manual checking. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Functions