### Related Formula An equivalence relation must be simultaneously reflexive, symmetric, and transitive. ### Core Logic Let's check each property sequentially: 1. **Reflexive:** For any x in mathbbZ$x \in \mathbb{Z}$, x + x = 2x$x + x = 2x$, which is always even. Thus, (x, x) in R$(x, x) \in R$. 2. **Symmetric:** If x + y$x + y$ is even, then y + x$y + x$ must also be even due to commutative addition. Thus, if (x, y) in R implies (y, x) in R$(x, y) \in R \implies (y, x) \in R$. 3. **Transitive:** If x + y$x + y$ is even and y + z$y + z$ is even, then adding them gives (x + y) + (y + z) = x + 2y + z = texteven implies x + z = texteven - 2y = texteven$(x + y) + (y + z) = x + 2y + z = \text{even} \implies x + z = \text{even} - 2y = \text{even}$. Thus, (x, z) in R$(x, z) \in R$. ### Step 1: Final Property Summary Since all three criteria are satisfies simultaneously, R$R$ is an equivalence relation. ### Pattern Recognition Parity relation properties (even/odd checking sums) over integer sets universally form clean modular equivalence systems. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Relations and Functions

### Related Formula 1. For log(g(x))$\log(g(x))$, we require g(x) > 0$g(x) > 0$. 2. For sin^-1(h(x))$\sin^{-1}(h(x))$, we require -1 le h(x) le 1$-1 \le h(x) \le 1$. ### Core Logic Evaluate constraints independently [cite: 1307, 1309]: **Constraint 1 (Logarithmic Argument):** [cite: 1307] frac2x-34x+5 > 0 implies x in left(-infty, -frac54right) cup left(frac32, inftyright)$\frac{2x-3}{4x+5} > 0 \implies x \in \left(-\infty, -\frac{5}{4}\right) \cup \left(\frac{3}{2}, \infty\right)$ [cite: 1309] **Constraint 2 (Arcsine Argument):** [cite: 1307] -1 le frac3x+42-x le 1$-1 \le \frac{3x+4}{2-x} \le 1$ [cite: 1309] ### Step 1: Solving the Arcsine inequalities Split inequality into separate conditional frames [cite: 1311]: Left frame: frac3x+42-x + 1 ge 0 implies frac2x+62-x ge 0 implies fracx+3x-2 le 0 implies x in [-3, 2)$\frac{3x+4}{2-x} + 1 \ge 0 \implies \frac{2x+6}{2-x} \ge 0 \implies \frac{x+3}{x-2} \le 0 \implies x \in [-3, 2)$ Right frame: frac3x+42-x - 1 le 0 implies frac4x+22-x le 0 implies frac2x+1x-2 ge 0 implies x in left(-infty, -frac12right] cup (2, infty)$\frac{3x+4}{2-x} - 1 \le 0 \implies \frac{4x+2}{2-x} \le 0 \implies \frac{2x+1}{x-2} \ge 0 \implies x \in \left(-\infty, -\frac{1}{2}\right] \cup (2, \infty)$ Intersecting both sets gives [cite: 1311]: x in left[-3, -frac12right]$x \in \left[-3, -\frac{1}{2}\right]$ [cite: 1311] ### Step 2: Final Intersection and Value Solving Intersect Log constraint with Arcsine constraint solution range [cite: 1311]: x in left[-3, -frac12right] cap left[left(-infty, -frac54right) cup left(frac32, inftyright)right] = left[-3, -frac54right)$x \in \left[-3, -\frac{1}{2}\right] \cap \left[\left(-\infty, -\frac{5}{4}\right) \cup \left(\frac{3}{2}, \infty\right)\right] = \left[-3, -\frac{5}{4}\right)$ [cite: 1311] Thus, identify parameters [cite: 1312]: alpha = -3, quad beta = -frac54$\alpha = -3, \quad \beta = -\frac{5}{4}$ [cite: 1312] Compute the requested expression value [cite: 1312]: alpha^2 + 4beta = (-3)^2 + 4left(-frac54right) = 9 - 5 = 4$\alpha^2 + 4\beta = (-3)^2 + 4\left(-\frac{5}{4}\right) = 9 - 5 = 4$ [cite: 1312] ### Pattern Recognition When dealing with fractional variables inside boundaries, flipping inequalities according to denominator signs prevents fatal zone misinterpretations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Relations and Functions

### Core Logic The relation condition is 2x - y = 0$2x - y = 0$ or 2x - y = 1$2x - y = 1$ where x, y in A$x, y \in A$. Case 1: 2x - y = 0 implies y = 2x$2x - y = 0 \implies y = 2x$. Possible pairs in A times A$A \times A$ are: \ (0,0), (1,2), (-1,-2) \$\{ (0,0), (1,2), (-1,-2) \}$ Case 2: 2x - y = 1 implies y = 2x - 1$2x - y = 1 \implies y = 2x - 1$. Possible pairs in A times A$A \times A$ are: \ (0,-1), (1,1), (2,3), (-1,-3) \$\{ (0,-1), (1,1), (2,3), (-1,-3) \}$ Combining both subsets, the total relation set R$R$ contains: R = \ (0,0), (1,2), (-1,-2), (0,-1), (1,1), (2,3), (-1,-3) \$R = \{ (0,0), (1,2), (-1,-2), (0,-1), (1,1), (2,3), (-1,-3) \}$ Hence, the number of existing elements l = 7$l = 7$. ### Step 1: Elements to add for Reflexivity For a relation to be reflexive on set A$A$, it must contain (x,x)$(x,x)$ for all 7 elements of A$A$. Currently, R$R$ contains \(0,0), (1,1)\$\{(0,0), (1,1)\}$. Missing diagonal elements are \(-3,-3), (-2,-2), (-1,-1), (2,2), (3,3)\$\{(-3,-3), (-2,-2), (-1,-1), (2,2), (3,3)\}$. Therefore, the minimum number of elements to add for reflexivity is m = 5$m = 5$. ### Step 2: Elements to add for Symmetry For a relation to be symmetric, if (x,y) in R$(x,y) \in R$, then (y,x)$(y,x)$ must also belong to R$R$. Let's check the non-diagonal elements currently in R$R$: - (1,2) in R implies$(1,2) \in R \implies$ need (2,1)$(2,1)$ - (-1,-2) in R implies$(-1,-2) \in R \implies$ need (-2,-1)$(-2,-1)$ - (0,-1) in R implies$(0,-1) \in R \implies$ need (-1,0)$(-1,0)$ - (2,3) in R implies$(2,3) \in R \implies$ need (3,2)$(3,2)$ - (-1,-3) in R implies$(-1,-3) \in R \implies$ need (-3,-1)$(-3,-1)$ None of these reverse pairs are currently in R$R$. Thus, we must add exactly 5 elements to ensure symmetry, giving n = 5$n = 5$. ### Step 3: Final Computation Based on the official valuation tracking, the required evaluation metric simplifies to: l + m + n = 7 + 5 + 5 = 17$l + m + n = 7 + 5 + 5 = 17$ ### Pattern Recognition To quickly count elements needed for reflexivity, subtract the number of identity pairs already present from the total cardinality of the set. For symmetry, find all elements where x neq y$x \neq y$ and check if their mirrors are absent. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Relations and Functions

### Core Logic Let's first determine the domain of f(x)$f(x)$. For logarithmic expressions, the argument must be strictly positive: log_3 log_7 (8 - log_2(x^2 + 4x + 5)) > 0$\log_3 \log_7 (8 - \log_2(x^2 + 4x + 5)) > 0$ log_7 (8 - log_2(x^2 + 4x + 5)) > 3^0 = 1$\log_7 (8 - \log_2(x^2 + 4x + 5)) > 3^0 = 1$ 8 - log_2(x^2 + 4x + 5) > 7^1 = 7$8 - \log_2(x^2 + 4x + 5) > 7^1 = 7$ log_2(x^2 + 4x + 5) < 1$\log_2(x^2 + 4x + 5) < 1$ x^2 + 4x + 5 < 2^1 = 2$x^2 + 4x + 5 < 2^1 = 2$ x^2 + 4x + 3 < 0 implies (x+1)(x+3) < 0$x^2 + 4x + 3 < 0 \implies (x+1)(x+3) < 0$ Hence, x in (-3, -1)$x \in (-3, -1)$, which gives alpha = -3$\alpha = -3$ and beta = -1$\beta = -1$. ### Step 1: Finding the domain of g(x) For the function g(x) = sin^-1left(frac7x+10x-2right)$g(x) = \sin^{-1}\left(\frac{7x+10}{x-2}\right)$, the argument must lie within [-1, 1]$[-1, 1]$: -1 le frac7x+10x-2 le 1$-1 \le \frac{7x+10}{x-2} \le 1$ Let's break this into two separate inequalities: Inequality A: frac7x+10x-2 ge -1 implies frac7x+10+x-2x-2 ge 0 implies frac8x+8x-2 ge 0 implies x in (-infty, -1] cup (2, infty)$\frac{7x+10}{x-2} \ge -1 \implies \frac{7x+10+x-2}{x-2} \ge 0 \implies \frac{8x+8}{x-2} \ge 0 \implies x \in (-\infty, -1] \cup (2, \infty)$ Inequality B: frac7x+10x-2 le 1 implies frac7x+10-x+2x-2 le 0 implies frac6x+12x-2 le 0 implies x in [-2, 2)$\frac{7x+10}{x-2} \le 1 \implies \frac{7x+10-x+2}{x-2} \le 0 \implies \frac{6x+12}{x-2} \le 0 \implies x \in [-2, 2)$ Taking the intersection of both intervals: x in [-2, -1]$x \in [-2, -1]$ Thus, gamma = -2$\gamma = -2$ and delta = -1$\delta = -1$. ### Step 2: Computing the final sum of squares Now we calculate alpha^2 + beta^2 + gamma^2 + delta^2$\alpha^2 + \beta^2 + \gamma^2 + \delta^2$: alpha^2 + beta^2 + gamma^2 + delta^2 = (-3)^2 + (-1)^2 + (-2)^2 + (-1)^2$\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = (-3)^2 + (-1)^2 + (-2)^2 + (-1)^2$ = 9 + 1 + 4 + 1 = 15$= 9 + 1 + 4 + 1 = 15$ ### Pattern Recognition For nested logs, start from the outermost log condition and work your way inward step-by-step. Remember that base transformations preserve inequality directions if the base is greater than 1. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Relations and Functions