### Related Formula Symmetric identity wrapper for matching indices: f(x) + f(1-x) = 1$f(x) + f(1-x) = 1$ ### Core Logic Let's evaluate f(x) + f(1-x)$f(x) + f(1-x)$: f(x) + f(1-x) = frac2^x2^x + sqrt2 + frac2^1-x2^1-x + sqrt2$f(x) + f(1-x) = \frac{2^x}{2^x + \sqrt{2}} + \frac{2^{1-x}}{2^{1-x} + \sqrt{2}}$ = frac2^x2^x + sqrt2 + frac22 + sqrt2cdot 2^x = frac2^x + sqrt22^x + sqrt2 = 1$= \frac{2^x}{2^x + \sqrt{2}} + \frac{2}{2 + \sqrt{2}\cdot 2^x} = \frac{2^x + \sqrt{2}}{2^x + \sqrt{2}} = 1$ ### Step 1: Expanding the Series Pairing matching terms from opposite ends of the summation: sum_k=1^81 fleft(frack82right) = left[fleft(frac182right) + fleft(frac8182 ight)right] + dots + fleft(frac4182 ight)$\sum_{k=1}^{81} f\left(\frac{k}{82}\right) = \left[f\left(\frac{1}{82}\right) + f\left(\frac{81}{82} ight)\right] + \dots + f\left(\frac{41}{82} ight)$ There are 40 complete pairs matching the f(x) + f(1-x) = 1$f(x) + f(1-x) = 1$ identity, plus one lone center term fleft(frac12right)$f\left(\frac{1}{2}\right)$. ### Step 2: Computing Final Valuation textSum = 40 + fleft(frac12right) = 40 + fracsqrt2sqrt2 + sqrt2 = 40 + frac12 = frac812$\text{Sum} = 40 + f\left(\frac{1}{2}\right) = 40 + \frac{\sqrt{2}}{\sqrt{2} + \sqrt{2}} = 40 + \frac{1}{2} = \frac{81}{2}$ ### Pattern Recognition When encountering fractional summation bounds, always check the sum of components x + (1-x)$x + (1-x)$ to find linear reduction templates. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series Class 12 Maths: Relations and Functions

### Related Formula Given functional property equation: f(x + y) = f(x) + f(y) + 1 - frac27xy$f(x + y) = f(x) + f(y) + 1 - \frac{2}{7}xy$ ### Core Logic Substitute x = y = 0$x = y = 0$ into the property equation: f(0) = 2f(0) + 1 implies f(0) = -1$f(0) = 2f(0) + 1 \implies f(0) = -1$. Since f(0) = b$f(0) = b$, we instantly find b = -1$b = -1$. ### Step 1: Extracting Parameter Values Substitute y = -x$y = -x$ into the property equation: f(0) = f(x) + f(-x) + 1 + frac27x^2$f(0) = f(x) + f(-x) + 1 + \frac{2}{7}x^2$ -1 = 2(3a + 2)x^2 + 2b + 1 + frac27x^2$-1 = 2(3a + 2)x^2 + 2b + 1 + \frac{2}{7}x^2$ Matching coefficients for x^2$x^2$ gives: 6a + 4 + frac27 = 0 implies a = -frac57$6a + 4 + \frac{2}{7} = 0 \implies a = -\frac{5}{7}$ Therefore, the absolute functional identity is: f(x) = -frac17x^2 - frac34x - 1$f(x) = -\frac{1}{7}x^2 - \frac{3}{4}x - 1$ ### Step 2: Computing the Target Series Rewriting using common denominators: |f(x)| = frac128|4x^2 + 21x + 28|$|f(x)| = \frac{1}{28}|4x^2 + 21x + 28|$ Evaluating for i=1$i=1$ to 5$5$: 28 sum_i = 1^5 |f(i)| = 675$28 \sum_{i = 1}^{5} |f(i)| = 675$ ### Pattern Recognition Substituting standard points like 0$0$ and -x$-x$ decouples symmetric multi-variable systems with maximum efficiency. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Relations and Functions

### 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. Elements in a relation satisfy the exact range constraint. 2. Reflexive criteria: For every x in A$x \in A$, (x, x) in R$(x, x) \in R$. ### Core Logic Rewrite the inequality to isolate variables systematically [cite: 1235]: -2y le x^2 le 4-2y$-2y \le x^2 \le 4-2y$ [cite: 1235] Test every valid value of y in A$y \in A$ to discover valid integer values for x$x$[cite: 1237, 1238, 1241, 1259, 1260, 1261]: - y = -3 implies 6 le x^2 le 10 implies x in \-3, 3\$y = -3 \implies 6 \le x^2 \le 10 \implies x \in \{-3, 3\}$ - y = -2 implies 4 le x^2 le 8 implies x in \-2, 2\$y = -2 \implies 4 \le x^2 \le 8 \implies x \in \{-2, 2\}$ - y = -1 implies 2 le x^2 le 6 implies x in \-2, 2\$y = -1 \implies 2 \le x^2 \le 6 \implies x \in \{-2, 2\}$ - y = 0 implies 0 le x^2 le 4 implies x in \-2, -1, 0, 1, 2\$y = 0 \implies 0 \le x^2 \le 4 \implies x \in \{-2, -1, 0, 1, 2\}$ - y = 1 implies -2 le x^2 le 2 implies x in \-1, 0, 1\$y = 1 \implies -2 \le x^2 \le 2 \implies x \in \{-1, 0, 1\}$ - y = 2 implies -4 le x^2 le 0 implies x in \0\$y = 2 \implies -4 \le x^2 \le 0 \implies x \in \{0\}$ - y = 3 implies -6 le x^2 le -2 implies textNo real x text exists$y = 3 \implies -6 \le x^2 \le -2 \implies \text{No real } x \text{ exists}$ ### Step 1: Listing set elements and counting Compile all distinct matching coordinate pairs (x,y)$(x,y)$ into set R$R$ [cite: 1264]: R = \(-3,-3), (-3,3), (-2,-2), (-2,2), (-1,-2), (-1,2), (0,-2), (0,-1), (0,0), (0,1), (0,2), (1,-1), (1,0), (1,1), (2,0)\$R = \{(-3,-3), (-3,3), (-2,-2), (-2,2), (-1,-2), (-1,2), (0,-2), (0,-1), (0,0), (0,1), (0,2), (1,-1), (1,0), (1,1), (2,0)\}$ [cite: 1264] Counting elements gives [cite: 1265]: l = 15$l = 15$ [cite: 1265] To make the relation reflexive, the pairs (-3,-3), (-2,-2), (-1,-1), (0,0), (1,1), (2,2), (3,3)$(-3,-3), (-2,-2), (-1,-1), (0,0), (1,1), (2,2), (3,3)$ must all belong to R$R$. Checking missing elements [cite: 1267]: \(-1,-1), (2,2), (3,3)\ implies m = 3$\{(-1,-1), (2,2), (3,3)\} \implies m = 3$ [cite: 1267] Sum of variables [cite: 1268]: l + m = 15 + 3 = 18$l + m = 15 + 3 = 18$ [cite: 1268] ### Pattern Recognition Isolating terms explicitly via a variable-by-variable bounded testing grid avoids missing distinct coordinate boundary values. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: 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