### Related Formula For a logarithmic term log_b(g(x))$\log_b(g(x))$ to be defined: - g(x) > 0$g(x) > 0$ - b > 0, b neq 1$b > 0, b \neq 1$ ### Core Logic Let's set defining inequalities sequentially: 1. Inside the outer logarithm: 1 - log_4(x^2 - 9x + 18) > 0 implies log_4(x^2 - 9x + 18) < 1$1 - \log_4(x^2 - 9x + 18) > 0 \implies \log_4(x^2 - 9x + 18) < 1$ Since base is 4 > 1$4 > 1$: x^2 - 9x + 18 < 4 implies x^2 - 9x + 14 < 0$x^2 - 9x + 18 < 4 \implies x^2 - 9x + 14 < 0$ (x-2)(x-7) < 0 implies x in (2, 7) quad text--- (1)$(x-2)(x-7) < 0 \implies x \in (2, 7) \quad \text{--- (1)}$ ### Step 1: Finding bounds for inner logarithmic term 2. Inside the inner logarithm: x^2 - 9x + 18 > 0$x^2 - 9x + 18 > 0$ (x-3)(x-6) > 0 implies x in (-infty, 3) cup (6, infty) quad text--- (2)$(x-3)(x-6) > 0 \implies x \in (-\infty, 3) \cup (6, \infty) \quad \text{--- (2)}$ ### Step 2: Intersection of regions Taking the intersection of (1) and (2): x in (2, 3) cup (6, 7)$x \in (2, 3) \cup (6, 7)$ This gives: alpha = 2, quad beta = 3, quad gamma = 6, quad delta = 7$\alpha = 2, \quad \beta = 3, \quad \gamma = 6, \quad \delta = 7$ Calculating the sum: alpha + beta + gamma + delta = 2 + 3 + 6 + 7 = 18$\alpha + \beta + \gamma + \delta = 2 + 3 + 6 + 7 = 18$ ### Pattern Recognition Logarithmic domains must check arguments from the innermost level to the outermost level. Remember that bases >1$>1$ maintain inequality direction upon exponentiation, while bases <1$<1$ reverse it. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Relations and Functions

### Related Formula For a relation R$R$ on set A = \1, 2, 3\$A = \{1, 2, 3\}$: - **Reflexive**: Must contain \(1,1), (2,2), (3,3)\$\{(1,1), (2,2), (3,3)\}$. - **Transitive**: If (a,b) in R$(a,b) \in R$ and (b,c) in R$(b,c) \in R$, then (a,c) in R$(a,c) \in R$. - **Not Symmetric**: Contains at least one element (a,b)$(a,b)$ whose inverse (b,a) notin R$(b,a) \notin R$. ### Core Logic Since R$R$ is reflexive, it must contain exactly 3$3$ initial diagonal elements: R_textbase = \(1,1), \, (2,2), \, (3,3)\$R_{\text{base}} = \{(1,1), \, (2,2), \, (3,3)\}$ We are given that (1,2) in R$(1,2) \in R$. So R$R$ must contain at least these 4$4$ mandatory pairs: R supseteq \(1,1), \, (2,2), \, (3,3), \, (1,2)\$R \supseteq \{(1,1), \, (2,2), \, (3,3), \, (1,2)\}$ Total elements currently = 4. The problem sets a boundary constraint of le 6$\le 6$ total elements. Available remaining elements to selectively append: (2,1), (2,3), (1,3), (3,1), (3,2)$(2,1), (2,3), (1,3), (3,1), (3,2)$. ### Step 1: Analyze Cases based on Element Length - **Case 1**: Exactly 4 elements. R = \(1,1), (2,2), (3,3), (1,2)\$R = \{(1,1), (2,2), (3,3), (1,2)\}$ This is reflexive, transitive, and not symmetric (since (2,1) notin R$(2,1) \notin R$). implies 1 text way$\implies 1 \text{ way}$. ### Step 2: Evaluate 5 and 6 Element Configurations - **Case 2**: Exactly 5 elements. We add one pair from the available pool. To ensure transitivity, we choose pairs like (1,3)$(1,3)$ or (3,2)$(3,2)$. - If we add (1,3)$(1,3)$: R = dots cup \(1,3)\$R = \dots \cup \{(1,3)\}$ implies$\implies$ valid (transitive, non-symmetric). - If we add (3,2)$(3,2)$: R = dots cup \(3,2)\$R = \dots \cup \{(3,2)\}$ implies$\implies$ valid. Adding (2,1)$(2,1)$ or others directly breaks either transitivity or symmetric constraints. implies 2 text ways$\implies 2 \text{ ways}$. - **Case 3**: Exactly 6 elements. Valid configuration groups that satisfy all transitive linkages without triggering full symmetry across the board are: 1. \(2,3), (1,3)\$\{(2,3), (1,3)\}$ added 2. \(1,3), (3,2)\$\{(1,3), (3,2)\}$ added 3. \(3,1), (3,2)\$\{(3,1), (3,2)\}$ added This yields 3 text ways$3 \text{ ways}$. ### Step 3: Calculate the Comprehensive Sum Sum the valid configurations across all operational boundaries: textTotal Relations = 1 + 2 + 3 = 6 quad (textour Analysis)$\text{Total Relations} = 1 + 2 + 3 = 6 \quad (\text{our Analysis})$ *(Note: Official NTA keys accepted 5 due to variant interpretation filters on transitivity bounds).* ### Pattern Recognition When dealing with small set elements counts like n=3$n=3$, building explicit tracking trees of allowed pairs is far safer than calculating raw combinations using generalized formula subsets. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Relations and Functions

### Related Formula max(x,y) = max(y,x)$\max(x,y) = \max(y,x)$ ### Core Logic Enumerate order metrics generated by the max mapping filter to assess population sizes and map properties against equivalence rule standards. ### Step 1: Enumerate Set Components Listing combinations matching the upper caps constraint parameters: R = \(0, 3), (3, 0), (0, 4), (4, 0), (1, 3), (3, 1), (1, 4), (4, 1), (2, 3), (3, 2), (2, 4), (4, 2), (3, 3), (3, 4), (4, 3), (4, 4)\$R = \{(0, 3), (3, 0), (0, 4), (4, 0), (1, 3), (3, 1), (1, 4), (4, 1), (2, 3), (3, 2), (2, 4), (4, 2), (3, 3), (3, 4), (4, 3), (4, 4)\}$ Total element count equals 16 items. Therefore, statement S_1$S_1$ is false. ### Step 2: Analyze Reflexivity and Symmetry Properties * Symmetry: Order switches do not alter peak size values. Since (x,y) in R implies (y,x) in R$(x,y) \in R \implies (y,x) \in R$, symmetry holds. * Reflexivity: Disjoint small pairs like (0,0)$(0,0)$ present peak values below target requirements, breaking reflexivity equations. ### Step 3: Test Transitivity Bounds Pick subset tracking variables showing breakdown trends: (0,3) in R quad textand quad (3,1) in R$(0,3) \in R \quad \text{and} \quad (3,1) \in R$ However, direct boundary tracking combination elements (0,1) notin R$(0,1) \notin R$ because max(0,1) = 1 notin \3,4\$\max(0,1) = 1 \notin \{3,4\}$. Thus, transitivity fails. Only S_2$S_2$ maps correctly. ### Pattern Recognition Max properties natively preserve system balance ordering directions, establishing automatic symmetry maps but struggling with linked cascading elements needed for transitivity rules. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Relations and Functions

### Related Formula -1 le textarg(cos^-1) le 1$-1 \le \text{arg}(\cos^{-1}) \le 1$ textarg(log) > 0$\text{arg}(\log) > 0$ ### Core Logic Isolate boundary inputs on logarithmic filters and inverse cosine boundaries using simple inequality signs to extract set endpoints. ### Step 1: Solve Inverse Cosine Bounds -1 le frac4x+53x-7 le 1 implies frac7x-23x-7 ge 0 quad textand quad fracx+123x-7 le 0$-1 \le \frac{4x+5}{3x-7} \le 1 \implies \frac{7x-2}{3x-7} \ge 0 \quad \text{and} \quad \frac{x+12}{3x-7} \le 0$ {{SOL_IMG_72_1}} {{SOL_IMG_72_2}} Intersecting sets maps out: [-12, 2/7] implies alpha = -12, beta = frac27$[-12, 2/7] \implies \alpha = -12, \beta = \frac{2}{7}$ ### Step 2: Solve Logarithmic Core Domain 2 - 6log_27(2x+5) > 0 implies log_27(2x+5) < frac13$2 - 6\log_{27}(2x+5) > 0 \implies \log_{27}(2x+5) < \frac{1}{3}$ 2x + 5 < 27^1/3 = 3 implies x < -1$2x + 5 < 27^{1/3} = 3 \implies x < -1$ Also structural logging arguments force: 2x+5 > 0 implies x > -5/2$2x+5 > 0 \implies x > -5/2$. Domain is: (-5/2, -1) implies gamma = -frac52, delta = -1$(-5/2, -1) \implies \gamma = -\frac{5}{2}, \delta = -1$ ### Step 3: Combined Metric Equation left| 7(alpha + beta) + 4(gamma + delta) right| = left| 7left(-12 + frac27right) + 4left(-frac52 - 1right) right|$\left| 7(\alpha + \beta) + 4(\gamma + \delta) \right| = \left| 7\left(-12 + \frac{2}{7}\right) + 4\left(-\frac{5}{2} - 1\right) \right|$ = |-82 - 14| = 96$= |-82 - 14| = 96$ ### Pattern Recognition Always align multiple bounds tracks sequentially. Missing internal tracking restrictions like checking if base log variables stay over zero can alter endpoint coordinates. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Relations and Functions

### Related Formula For a logarithmic term log_b(a)$\log_b(a)$ to be valid: a > 0, quad b > 0, quad b neq 1$a > 0, \quad b > 0, \quad b \neq 1$ ### Core Logic Analyzing the first function f_1(x) = log_5(18x - x^2 - 77)$f_1(x) = \log_5(18x - x^2 - 77)$: 18x - x^2 - 77 > 0 implies x^2 - 18x + 77 < 0$18x - x^2 - 77 > 0 \implies x^2 - 18x + 77 < 0$ (x - 7)(x - 11) < 0 implies x in (7, 11)$(x - 7)(x - 11) < 0 \implies x \in (7, 11)$ Hence, alpha = 7, beta = 11$\alpha = 7, \beta = 11$. ### Step 1: Check Second Function Base and Argument Analyzing f_2(x) = log_(x - 1)left(frac2x^2 + 3x - 2x^2 - 3x - 4right)$f_2(x) = \log_{(x - 1)}\left(\frac{2x^2 + 3x - 2}{x^2 - 3x - 4}\right)$: Base constraints: x - 1 > 0 implies x > 1$x - 1 > 0 \implies x > 1$ x - 1 neq 1 implies x neq 2$x - 1 \neq 1 \implies x \neq 2$ Argument constraints: frac2x^2 + 3x - 2x^2 - 3x - 4 > 0 implies frac(2x - 1)(x + 2)(x - 4)(x + 1) > 0$\frac{2x^2 + 3x - 2}{x^2 - 3x - 4} > 0 \implies \frac{(2x - 1)(x + 2)}{(x - 4)(x + 1)} > 0$ ### Step 2: Apply Sign Scheme Using the wave-curve method to determine where the rational fraction is positive:

Combining this with x > 1$x > 1$ and x neq 2$x \neq 2$, the common interval is: x in (4, infty)$x \in (4, \infty)$ Thus, gamma = 4$\gamma = 4$. ### Step 3: Calculate final sum alpha^2 + beta^2 + gamma^2 = 7^2 + 11^2 + 4^2 = 49 + 121 + 16 = 186$\alpha^2 + \beta^2 + \gamma^2 = 7^2 + 11^2 + 4^2 = 49 + 121 + 16 = 186$ ### Pattern Recognition For domain intersections involving variables in both the log base and argument, always list base rules (>0, neq 1$>0, \neq 1$) first to eliminate invalid sign fields early on. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Relations and Functions