### 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 Many finite fractional sum questions involving functional terms rely on identifying an underlying symmetric summation invariant, typically of the form f(x) + f(k-x) = textconstant$f(x) + f(k-x) = \text{constant}$. ### Core Logic First simplify the expression for f(x)$f(x)$ algebraically: f(x) = frac4 cdot 2^x + 162 cdot (2^x)^2 + 16 cdot 2^x + 32$f(x) = \frac{4 \cdot 2^x + 16}{2 \cdot (2^x)^2 + 16 \cdot 2^x + 32}$ Factor out 4$4$ from the numerator and 2$2$ from the denominator: f(x) = frac4(2^x + 4)2[(2^x)^2 + 8 cdot 2^x + 16] = frac2(2^x + 4)(2^x + 4)^2 = frac22^x + 4$f(x) = \frac{4(2^x + 4)}{2[(2^x)^2 + 8 \cdot 2^x + 16]} = \frac{2(2^x + 4)}{(2^x + 4)^2} = \frac{2}{2^x + 4}$ ### Step 1: Establish Symmetry Pairings Let's check the value of f(x) + f(4-x)$f(x) + f(4-x)$: f(4-x) = frac22^4-x + 4 = frac2frac162^x + 4 = frac2 cdot 2^x16 + 4 cdot 2^x = frac2 cdot 2^x4(2^x + 4) = frac2^x2(2^x + 4)$f(4-x) = \frac{2}{2^{4-x} + 4} = \frac{2}{\frac{16}{2^x} + 4} = \frac{2 \cdot 2^x}{16 + 4 \cdot 2^x} = \frac{2 \cdot 2^x}{4(2^x + 4)} = \frac{2^x}{2(2^x + 4)}$ Now compute the sum directly: f(x) + f(4-x) = frac22^x + 4 + frac2^x2(2^x + 4) = frac4 + 2^x2(2^x + 4) = frac12$f(x) + f(4-x) = \frac{2}{2^x + 4} + \frac{2^x}{2(2^x + 4)} = \frac{4 + 2^x}{2(2^x + 4)} = \frac{1}{2}$ Hence, whenever two input arguments sum up to 4$4$, the sum of their functional values is exactly frac12$\frac{1}{2}$. ### Step 2: Group the Finite Series Terms Consider the terms inside the requested sequence: frac115 + frac5915 = frac6015 = 4 implies fleft(frac115 ight) + fleft(frac5915 ight) = frac12$\frac{1}{15} + \frac{59}{15} = \frac{60}{15} = 4 \implies f\left(\frac{1}{15} ight) + f\left(\frac{59}{15} ight) = \frac{1}{2}$ frac215 + frac5815 = frac6015 = 4 implies fleft(frac215 ight) + fleft(frac5815 ight) = frac12$\frac{2}{15} + \frac{58}{15} = \frac{60}{15} = 4 \implies f\left(\frac{2}{15} ight) + f\left(\frac{58}{15} ight) = \frac{1}{2}$ This complementary pairing continues up to: fleft(frac2915 ight) + fleft(frac3115 ight) = frac12$f\left(\frac{29}{15} ight) + f\left(\frac{31}{15} ight) = \frac{1}{2}$ This yields exactly 29$29$ distinct pairs. The single middle term left unpaired corresponds to: textMiddle Term = fleft(frac3015 ight) = f(2) = frac22^2 + 4 = frac28 = frac14$\text{Middle Term} = f\left(\frac{30}{15} ight) = f(2) = \frac{2}{2^2 + 4} = \frac{2}{8} = \frac{1}{4}$ ### Step 3: Evaluate Final Expression Compute the total value by multiplying the grouped sum by 8: textTotal = 8 cdot left[ 29 cdot left(frac12right) + frac14 right]$\text{Total} = 8 \cdot \left[ 29 \cdot \left(\frac{1}{2}\right) + \frac{1}{4} \right]$ textTotal = 8 cdot frac292 + 8 cdot frac14 = 116 + 2 = 118$\text{Total} = 8 \cdot \frac{29}{2} + 8 \cdot \frac{1}{4} = 116 + 2 = 118$ ### Pattern Recognition Whenever a symmetric set of arguments is presented inside a summation matching frackn + fracN-kn = textconstant$\frac{k}{n} + \frac{N-k}{n} = \text{constant}$, look for an algebraic reduction of f(x)$f(x)$ that yields a uniform constant sum for symmetric pairs. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Functions

### Related Formula The domain of sec^-1(y)$\sec^{-1}(y)$ is given by |y| ge 1$|y| \ge 1$, which means: y le -1 quad textor quad y ge 1$y \le -1 \quad \text{or} \quad y \ge 1$ ### Core Logic For f(x) = sec^-1(2[x]+1)$f(x) = \sec^{-1}(2[x]+1)$ to be defined: 2[x] + 1 le -1 quad textor quad 2[x] + 1 ge 1$2[x] + 1 \le -1 \quad \text{or} \quad 2[x] + 1 \ge 1$ ### Step 1: Solve individual inequalities Case 1: 2[x] + 1 le -1 implies 2[x] le -2 implies [x] le -1$2[x] + 1 \le -1 \implies 2[x] \le -2 \implies [x] \le -1$ This holds true for all x < 0$x < 0$, i.e., x in (-infty, 0)$x \in (-\infty, 0)$. Case 2: 2[x] + 1 ge 1 implies 2[x] ge 0 implies [x] ge 0$2[x] + 1 \ge 1 \implies 2[x] \ge 0 \implies [x] \ge 0$ This holds true for all x ge 0$x \ge 0$, i.e., x in [0, infty)$x \in [0, \infty)$. ### Step 2: Take Union of the Solutions textDomain = (-infty, 0) cup [0, infty) = (-infty, infty)$\text{Domain} = (-\infty, 0) \cup [0, \infty) = (-\infty, \infty)$ ### Pattern Recognition Since [x]$[x]$ covers all integer values and 2[x]+1$2[x]+1$ forms all odd integer values, the expression inside sec^-1$\sec^{-1}$ is always a non-zero integer. Non-zero integers always have absolute value ge 1$\ge 1$. Hence, it is valid for all real numbers. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Functions Class 12 Mathematics: Inverse Trigonometric Functions