### Related Formula For functional equations with inversion, substituting x to frac1x$x \to \frac{1}{x}$ establishes a solvable system of algebraic equations to isolate f(x)$f(x)$ directly. ### Core Logic The given equation is: f(x) - 6fleft(frac1xright) = frac353x - frac52 quad dots (1)$f(x) - 6f\left(\frac{1}{x}\right) = \frac{35}{3x} - \frac{5}{2} \quad \dots (1)$ Substitute x to frac1x$x \to \frac{1}{x}$ in equation (1): fleft(frac1xright) - 6f(x) = frac35x3 - frac52 quad dots (2)$f\left(\frac{1}{x}\right) - 6f(x) = \frac{35x}{3} - \frac{5}{2} \quad \dots (2)$ ### Step 1: Eliminate f(1/x)$f(1/x)$ Multiply equation (2) by 6$6$ and add it to equation (1): left[ f(x) - 6fleft(frac1xright) right] + 6 left[ fleft(frac1xright) - 6f(x) right] = left( frac353x - frac52 right) + 6 left( frac35x3 - frac52 right)$\left[ f(x) - 6f\left(\frac{1}{x}\right) \right] + 6 \left[ f\left(\frac{1}{x}\right) - 6f(x) \right] = \left( \frac{35}{3x} - \frac{5}{2} \right) + 6 \left( \frac{35x}{3} - \frac{5}{2} \right)$ f(x) - 36f(x) = frac353x - frac52 + 70x - 15$f(x) - 36f(x) = \frac{35}{3x} - \frac{5}{2} + 70x - 15$ -35f(x) = 70x + frac353x - frac352$-35f(x) = 70x + \frac{35}{3x} - \frac{35}{2}$ Divide across by -35$-35$: f(x) = -2x - frac13x + frac12$f(x) = -2x - \frac{1}{3x} + \frac{1}{2}$ ### Step 2: Evaluate the Limit We are given that the following limit evaluates to a finite constant beta$\beta$: lim_x to 0 left( frac1alpha x + f(x) right) = beta$\lim_{x \to 0} \left( \frac{1}{\alpha x} + f(x) \right) = \beta$ lim_x to 0 left( frac1alpha x - 2x - frac13x + frac12 right) = beta$\lim_{x \to 0} \left( \frac{1}{\alpha x} - 2x - \frac{1}{3x} + \frac{1}{2} \right) = \beta$ lim_x to 0 left( left[ frac1alpha - frac13 right] frac1x - 2x + frac12 right) = beta$\lim_{x \to 0} \left( \left[ \frac{1}{\alpha} - \frac{1}{3} \right] \frac{1}{x} - 2x + \frac{1}{2} \right) = \beta$ For the limit to be a finite value, the coefficient of frac1x$\frac{1}{x}$ must vanish completely: frac1alpha - frac13 = 0 implies alpha = 3$\frac{1}{\alpha} - \frac{1}{3} = 0 \implies \alpha = 3$ When alpha = 3$\alpha = 3$, the limit simplifies directly to the constant term: beta = lim_x to 0 left( -2x + frac12 right) = frac12$\beta = \lim_{x \to 0} \left( -2x + \frac{1}{2} \right) = \frac{1}{2}$ ### Step 3: Calculate Final Value Substitute the determined parameters alpha$\alpha$ and beta$\beta$: alpha + 2beta = 3 + 2left(frac12right) = 3 + 1 = 4$\alpha + 2\beta = 3 + 2\left(\frac{1}{2}\right) = 3 + 1 = 4$ ### Pattern Recognition In limit problems involving fractional components where x to 0$x \to 0$, any term like frac1x$\frac{1}{x}$ or higher negative powers must have a net coefficient of zero to guarantee existence of a finite limit value. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Functions Class 11 Mathematics: Limits and Derivatives

### 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 For a function to be onto, its Codomain must equal its Range. ### Core Logic First, find range A$A$ for f(x) = 2x^3 - 15x^2 + 36x + 7$f(x) = 2x^3 - 15x^2 + 36x + 7$ over [0, 3]$[0, 3]$. Differentiating f(x)$f(x)$: f'(x) = 6x^2 - 30x + 36 = 6(x^2 - 5x + 6) = 6(x-2)(x-3)$f'(x) = 6x^2 - 30x + 36 = 6(x^2 - 5x + 6) = 6(x-2)(x-3)$ Critical points are x=2$x=2$ and x=3$x=3$. Evaluate f(x)$f(x)$ at boundary and critical points: - f(0) = 7$f(0) = 7$ - f(2) = 2(8) - 15(4) + 36(2) + 7 = 16 - 60 + 72 + 7 = 35$f(2) = 2(8) - 15(4) + 36(2) + 7 = 16 - 60 + 72 + 7 = 35$ - f(3) = 2(27) - 15(9) + 36(3) + 7 = 54 - 135 + 108 + 7 = 34$f(3) = 2(27) - 15(9) + 36(3) + 7 = 54 - 135 + 108 + 7 = 34$ Thus, Range A = [7, 35]$A = [7, 35]$. ### Step 1: Find Range B for g(x) Now look at g(x) = fracx^2025x^2025+1 = 1 - frac1x^2025+1$g(x) = \frac{x^{2025}}{x^{2025}+1} = 1 - \frac{1}{x^{2025}+1}$ over [0, infty)$[0, \infty)$. - At x = 0$x = 0$, g(0) = 0$g(0) = 0$. - As x to infty$x \to \infty$, g(x) to 1$g(x) \to 1$. Since g(x)$g(x)$ is continuous and monotonically strictly increasing, Range B = [0, 1)$B = [0, 1)$. ### Step 2: Find the Integer Count of Union Set S S = \x in mathbbZ : x in A text or x in B\ = mathbbZ cap (A cup B)$S = \{x \in \mathbb{Z} : x \in A \text{ or } x \in B\} = \mathbb{Z} \cap (A \cup B)$ A cup B = [0, 1) cup [7, 35]$A \cup B = [0, 1) \cup [7, 35]$ The integers in this set are: - From [0, 1)$[0, 1)$: x = 0$x = 0$ - From [7, 35]$[7, 35]$: x = 7, 8, 9, dots, 35$x = 7, 8, 9, \dots, 35$ Total number of integers n(S)$n(S)$: n(S) = 1 + (35 - 7 + 1) = 1 + 29 = 30$n(S) = 1 + (35 - 7 + 1) = 1 + 29 = 30$ ### Pattern Recognition The condition 'or' means union. Be careful not to include integers between 1$1$ and 6$6$ since they are not present in either continuous range segment. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Functions Class 12 Mathematics: Application of Derivatives

### 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

### Related Formula Standard result for functional equation of a polynomial satisfying f(x)f(1/x) = f(x) + f(1/x)$f(x)f(1/x) = f(x) + f(1/x)$: f(x) = 1 pm x^n$f(x) = 1 \pm x^n$ ### Core Logic Given that f(x)$f(x)$ is a polynomial of degree 2, the identity implies: f(x) = 1 + x^2 quad textor quad f(x) = 1 - x^2$f(x) = 1 + x^2 \quad \text{or} \quad f(x) = 1 - x^2$ We are given the range is bounded above: (-infty, 1]$(-\infty, 1]$. - For 1 + x^2$1 + x^2$, the range is [1, infty)$[1, \infty)$. - For 1 - x^2$1 - x^2$, the range is (-infty, 1]$(-\infty, 1]$. Therefore, the correct functional form is f(x) = 1 - x^2$f(x) = 1 - x^2$. ### Step 1: Solve for K Given condition: f(K) = -2K$f(K) = -2K$ 1 - K^2 = -2K implies K^2 - 2K - 1 = 0$1 - K^2 = -2K \implies K^2 - 2K - 1 = 0$ Let the roots of this equation be K_1$K_1$ and K_2$K_2$. From quadratic properties (Vieta's formulas): K_1 + K_2 = 2$K_1 + K_2 = 2$ K_1 cdot K_2 = -1$K_1 \cdot K_2 = -1$ ### Step 2: Calculate Sum of Squares We need the sum of squares of the values of K: K_1^2 + K_2^2 = (K_1 + K_2)^2 - 2K_1K_2$K_1^2 + K_2^2 = (K_1 + K_2)^2 - 2K_1K_2$ K_1^2 + K_2^2 = (2)^2 - 2(-1) = 4 + 2 = 6$K_1^2 + K_2^2 = (2)^2 - 2(-1) = 4 + 2 = 6$ ### Pattern Recognition The functional equation f(x)f(1/x)=f(x)+f(1/x)$f(x)f(1/x)=f(x)+f(1/x)$ uniquely forces polynomials to be 1 pm x^n$1 \pm x^n$. Remembering this shortcut saves valuable time required to derive the template from general coefficients. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Complex Numbers and Quadratic Equations Class 12 Mathematics: Functions