If the domain of the function f(x) = frac1sqrt10 + 3x - x^2 + frac1sqrtx + |x|$$f(x) = \frac{1}{\sqrt{10 + 3x - x^2}} + \frac{1}{\sqrt{x + |x|}}$$ is (a, b)$(a, b)$, then (1 + a)^2 + b^2$(1 + a)^2 + b^2$ is equal to:

### 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

### 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