### Core Logic We have two functional equations to solve before integrating. Equation 1: f(x) + 2fleft(frac1xright) = x^2 + 5$f(x) + 2f\left(\frac{1}{x}\right) = x^2 + 5$ Replace x$x$ with frac1x$\frac{1}{x}$: fleft(frac1xright) + 2f(x) = frac1x^2 + 5$f\left(\frac{1}{x}\right) + 2f(x) = \frac{1}{x^2} + 5$ Multiplying this new equation by 2$2$ and subtracting the original Equation 1 eliminates the fleft(frac1xright)$f\left(\frac{1}{x}\right)$ term: 4f(x) + 2fleft(frac1xright) - left(f(x) + 2fleft(frac1xright)right) = 2left(frac1x^2 + 5right) - (x^2 + 5)$4f(x) + 2f\left(\frac{1}{x}\right) - \left(f(x) + 2f\left(\frac{1}{x}\right)\right) = 2\left(\frac{1}{x^2} + 5\right) - (x^2 + 5)$ 3f(x) = frac2x^2 - x^2 + 5 implies f(x) = frac23x^2 - fracx^23 + frac53$3f(x) = \frac{2}{x^2} - x^2 + 5 \implies f(x) = \frac{2}{3x^2} - \frac{x^2}{3} + \frac{5}{3}$ ### Step 1: Finding alpha Integrate f(x)$f(x)$ from 1$1$ to 2$2$: alpha = int_1^2 left( frac23x^2 - fracx^23 + frac53 right) dx = left[ -frac23x - fracx^39 + frac5x3 right]_1^2$\alpha = \int_{1}^{2} \left( \frac{2}{3x^2} - \frac{x^2}{3} + \frac{5}{3} \right) dx = \left[ -\frac{2}{3x} - \frac{x^3}{9} + \frac{5x}{3} \right]_{1}^{2}$ alpha = left( -frac13 - frac89 + frac103 right) - left( -frac23 - frac19 + frac53 right) = frac199 - frac89 = frac119$\alpha = \left( -\frac{1}{3} - \frac{8}{9} + \frac{10}{3} \right) - \left( -\frac{2}{3} - \frac{1}{9} + \frac{5}{3} \right) = \frac{19}{9} - \frac{8}{9} = \frac{11}{9}$ Thus, 9alpha = 11$9\alpha = 11$. ### Step 2: Solving for g(x) and finding beta We are given 2g(x) - 3gleft(frac12right) = x$2g(x) - 3g\left(\frac{1}{2}\right) = x$. Substitute x = frac12$x = \frac{1}{2}$: 2gleft(frac12right) - 3gleft(frac12right) = frac12 implies -gleft(frac12right) = frac12 implies gleft(frac12right) = -frac12$2g\left(\frac{1}{2}\right) - 3g\left(\frac{1}{2}\right) = \frac{1}{2} \implies -g\left(\frac{1}{2}\right) = \frac{1}{2} \implies g\left(\frac{1}{2}\right) = -\frac{1}{2}$ Substitute this constant value back into the original equation: 2g(x) - 3left(-frac12right) = x implies 2g(x) + frac32 = x implies g(x) = fracx2 - frac34$2g(x) - 3\left(-\frac{1}{2}\right) = x \implies 2g(x) + \frac{3}{2} = x \implies g(x) = \frac{x}{2} - \frac{3}{4}$ Now find beta$\beta$: beta = int_1^2 left( fracx2 - frac34 right) dx = left[ fracx^24 - frac3x4 right]_1^2 = left( 1 - frac32 right) - left( frac14 - frac34 right) = -frac12 - left(-frac12right) = 0$\beta = \int_{1}^{2} \left( \frac{x}{2} - \frac{3}{4} \right) dx = \left[ \frac{x^2}{4} - \frac{3x}{4} \right]_{1}^{2} = \left( 1 - \frac{3}{2} \right) - \left( \frac{1}{4} - \frac{3}{4} \right) = -\frac{1}{2} - \left(-\frac{1}{2}\right) = 0$ ### Step 3: Calculating 9alpha + beta Combining our values: 9alpha + beta = 11 + 0 = 11$9\alpha + \beta = 11 + 0 = 11$ ### Pattern Recognition Functional equations involving x to frac1x$x \to \frac{1}{x}$ are easily solved by treating the swapped forms as a system of linear equations, allowing direct isolation of the underlying function. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integrals Class 12 Mathematics: Functional Equations

### Related Formula King's property of definite integrals: int_a^b f(x)mathrmdx = int_a^b f(a+b-x)mathrmdx$\int_{a}^{b} f(x)\mathrm{d}x = \int_{a}^{b} f(a+b-x)\mathrm{d}x$ ### Core Logic Let the given integral be I$I$. Apply King's property by substituting x to -x$x \to -x$: I = int_-1^1fracleft(1 + sqrt|x| + xright)e^-x + left(sqrt|x| + xright)e^xe^-x + e^x\,mathrmdx$I = \int_{-1}^{1}\frac{\left(1 + \sqrt{|x| + x}\right)e^{-x} + \left(\sqrt{|x| + x}\right)e^{x}}{e^{-x} + e^{x}}\,\mathrm{d}x$ Add both integral expressions 2I = I + I$2I = I + I$: 2I = int_-1^1 frac(e^x + e^-x) + left(sqrt|x| - x + sqrt|x| + xright)(e^x + e^-x)e^x + e^-x\,mathrmdx$2I = \int_{-1}^{1} \frac{(e^x + e^{-x}) + \left(\sqrt{|x| - x} + \sqrt{|x| + x}\right)(e^x + e^{-x})}{e^x + e^{-x}}\,\mathrm{d}x$ 2I = int_-1^1 left(1 + sqrt|x| - x + sqrt|x| + xright)\,mathrmdx$2I = \int_{-1}^{1} \left(1 + \sqrt{|x| - x} + \sqrt{|x| + x}\right)\,\mathrm{d}x$ ### Step 1: Apply Symmetry Properties The integrand is completely even. Hence, convert intervals: 2I = 2int_0^1 left(1 + sqrt|x| - x + sqrt|x| + xright)\,mathrmdx$2I = 2\int_{0}^{1} \left(1 + \sqrt{|x| - x} + \sqrt{|x| + x}\right)\,\mathrm{d}x$ For x in [0,1]$x \in [0,1]$, |x| = x implies sqrt|x| - x = 0$|x| = x \implies \sqrt{|x| - x} = 0$ and sqrt|x| + x = sqrt2x$\sqrt{|x| + x} = \sqrt{2x}$: I = int_0^1 (1 + sqrt2x)\,mathrmdx$I = \int_{0}^{1} (1 + \sqrt{2x})\,\mathrm{d}x$ ### Step 2: Final Integration Execution I = left[ x + sqrt2 cdot fracx^3/23/2 right]_0^1 = left[ x + frac2sqrt23x^3/2 right]_0^1$I = \left[ x + \sqrt{2} \cdot \frac{x^{3/2}}{3/2} \right]_{0}^{1} = \left[ x + \frac{2\sqrt{2}}{3}x^{3/2} \right]_{0}^{1}$ I = 1 + frac2sqrt23$I = 1 + \frac{2\sqrt{2}}{3}$ ### Pattern Recognition When functions involve combinations of exponential components (e^x, e^-x$e^x, e^{-x}$) over symmetric boundaries, adding the variable reflection eliminates exponential fractions instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integration

### Related Formula The beta function integral format satisfies: I(m,n) = int_0^1 x^m-1 (1-x)^n-1 dx$I(m,n) = \int_{0}^{1} x^{m-1} (1-x)^{n-1} dx$ ### Core Logic Let's combine the terms of the requested sum directly by inserting their respective definitions: I(9,14) = int_0^1 x^9-1 (1-x)^14-1 dx = int_0^1 x^8 (1-x)^13 dx$I(9,14) = \int_{0}^{1} x^{9-1} (1-x)^{14-1} dx = \int_{0}^{1} x^8 (1-x)^{13} dx$ I(10,13) = int_0^1 x^10-1 (1-x)^13-1 dx = int_0^1 x^9 (1-x)^12 dx$I(10,13) = \int_{0}^{1} x^{10-1} (1-x)^{13-1} dx = \int_{0}^{1} x^9 (1-x)^{12} dx$ ### Step 1: Factoring out common algebraic terms Summing the two components: I(9,14) + I(10,13) = int_0^1 left[ x^8 (1-x)^13 + x^9 (1-x)^12 right] dx$I(9,14) + I(10,13) = \int_{0}^{1} \left[ x^8 (1-x)^{13} + x^9 (1-x)^{12} \right] dx$ Factor out the common term x^8 (1-x)^12$x^8 (1-x)^{12}$ inside the integrand: = int_0^1 x^8 (1-x)^12 left[ (1-x) + x right] dx$= \int_{0}^{1} x^8 (1-x)^{12} \left[ (1-x) + x \right] dx$ = int_0^1 x^8 (1-x)^12 (1) dx$= \int_{0}^{1} x^8 (1-x)^{12} (1) dx$ = int_0^1 x^9-1 (1-x)^13-1 dx = I(9,13)$= \int_{0}^{1} x^{9-1} (1-x)^{13-1} dx = I(9,13)$ ### Pattern Recognition When dealing with linear combinations of beta functions with shifting parameter indices, directly writing down the definite integral expression often results in immediate algebraic cancellation or simplification via basic factoring. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integrals

### Related Formula The Leibniz Integral Rule template allows direct differentiation of an integral with variable limits: fracddxleft( int_0^x g(t) dt right) = g(x)$\frac{d}{dx}\left( \int_{0}^{x} g(t) dt \right) = g(x)$ ### Core Logic Differentiate both sides of the given functional equation with respect to x$x$ using the product rule: fracddxleft[ 2(x+2)^2 f(x) - 3(x+2)^2 right] = fracddxleft[ 10int_0^x(t+2)f(t)dt right]$\frac{d}{dx}\left[ 2(x+2)^2 f(x) - 3(x+2)^2 \right] = \frac{d}{dx}\left[ 10\int_{0}^{x}(t+2)f(t)dt \right]$ 4(x+2)f(x) + 2(x+2)^2 f'(x) - 6(x+2) = 10(x+2)f(x)$4(x+2)f(x) + 2(x+2)^2 f'(x) - 6(x+2) = 10(x+2)f(x)$ Since x geq 0$x \geq 0$, the factor (x+2)$(x+2)$ is strictly non-zero. Divide the entire equation by 2(x+2)$2(x+2)$: 2f(x) + (x+2)f'(x) - 3 = 5f(x)$2f(x) + (x+2)f'(x) - 3 = 5f(x)$ (x+2)f'(x) - 3f(x) = 3$(x+2)f'(x) - 3f(x) = 3$ ### Step 1: Solve the First-Order Differential Equation Rearrange the expression into standard linear differential equation form where y = f(x)$y = f(x)$: fracdydx - frac3x+2y = frac3x+2$\frac{dy}{dx} - \frac{3}{x+2}y = \frac{3}{x+2}$ Compute the Integrating Factor (I.F.): textI.F. = e^int -frac3x+2 dx = e^-3ln(x+2) = (x+2)^-3$\text{I.F.} = e^{\int -\frac{3}{x+2} dx} = e^{-3\ln(x+2)} = (x+2)^{-3}$ Multiply through by the I.F. and integrate: y cdot (x+2)^-3 = int frac3x+2 cdot (x+2)^-3 dx = int 3(x+2)^-4 dx$y \cdot (x+2)^{-3} = \int \frac{3}{x+2} \cdot (x+2)^{-3} dx = \int 3(x+2)^{-4} dx$ fracf(x)(x+2)^3 = 3 cdot frac(x+2)^-3-3 + C = -(x+2)^-3 + C$\frac{f(x)}{(x+2)^3} = 3 \cdot \frac{(x+2)^{-3}}{-3} + C = -(x+2)^{-3} + C$ f(x) = -1 + C(x+2)^3$f(x) = -1 + C(x+2)^3$ ### Step 2: Apply the Boundary Condition Find the boundary condition by substituting x = 0$x = 0$ into the original integral equation equation: 2(0+2)^2 f(0) - 3(0+2)^2 = 10 int_0^0 (t+2)f(t) dt$2(0+2)^2 f(0) - 3(0+2)^2 = 10 \int_{0}^{0} (t+2)f(t) dt$ 8f(0) - 12 = 0 implies f(0) = frac128 = frac32$8f(0) - 12 = 0 \implies f(0) = \frac{12}{8} = \frac{3}{2}$ Substitute x = 0$x = 0$ into our general solution formula: f(0) = -1 + C(0+2)^3 implies frac32 = -1 + 8C$f(0) = -1 + C(0+2)^3 \implies \frac{3}{2} = -1 + 8C$ frac52 = 8C implies C = frac516$\frac{5}{2} = 8C \implies C = \frac{5}{16}$ Thus, the explicit function is: f(x) = -1 + frac516(x+2)^3$f(x) = -1 + \frac{5}{16}(x+2)^3$ ### Step 3: Evaluate at target point x = 2 Substitute x = 2$x = 2$ into the final function equation: f(2) = -1 + frac516(2+2)^3 = -1 + frac516(64)$f(2) = -1 + \frac{5}{16}(2+2)^3 = -1 + \frac{5}{16}(64)$ f(2) = -1 + 5(4) = -1 + 20 = 19$f(2) = -1 + 5(4) = -1 + 20 = 19$ ### Pattern Recognition When an equation contains a variable integral limit int_0^x$\int_0^x$, differentiating both sides using the Leibniz rule converts it into a standard differential equation. The initial value is found by setting x = 0$x = 0$ directly in the original expression. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integrals Class 12 Mathematics: Differential Equations

### Related Formula Integration by Parts formula: int u cdot v \, dx = u int v \, dx - int left( u' int v \, dx right) dx$\int u \cdot v \, dx = u \int v \, dx - \int \left( u' \int v \, dx \right) dx$ ### Core Logic Given F(x) = xf(x)$F(x) = xf(x)$ and f(2) = 1 implies F(2) = 2f(2) = 2$f(2) = 1 \implies F(2) = 2f(2) = 2$. Let's apply Integration by Parts to the first given integral: int_0^2 x F'(x) dx = 6$\int_{0}^{2} x F'(x) dx = 6$ left[ xF(x) right]_0^2 - int_0^2 F(x) dx = 6$\left[ xF(x) \right]_0^2 - \int_{0}^{2} F(x) dx = 6$ 2F(2) - 0 - int_0^2 F(x) dx = 6$2F(2) - 0 - \int_{0}^{2} F(x) dx = 6$ 2(2) - int_0^2 F(x) dx = 6 implies int_0^2 F(x) dx = 4 - 6 = -2$2(2) - \int_{0}^{2} F(x) dx = 6 \implies \int_{0}^{2} F(x) dx = 4 - 6 = -2$ ### Step 1: Simplify Second Integral via Integration by Parts Now look at the second integral: int_0^2 x^2 F''(x) dx = 40$\int_{0}^{2} x^2 F''(x) dx = 40$ Applying Integration by parts (taking u = x^2$u = x^2$ and v = F''(x)$v = F''(x)$): left[ x^2 F'(x) right]_0^2 - int_0^2 2x F'(x) dx = 40$\left[ x^2 F'(x) \right]_0^2 - \int_{0}^{2} 2x F'(x) dx = 40$ 4 F'(2) - 0 - 2 int_0^2 x F'(x) dx = 40$4 F'(2) - 0 - 2 \int_{0}^{2} x F'(x) dx = 40$ We already know int_0^2 x F'(x) dx = 6$\int_{0}^{2} x F'(x) dx = 6$: 4 F'(2) - 2(6) = 40$4 F'(2) - 2(6) = 40$ 4 F'(2) - 12 = 40 implies 4 F'(2) = 52 implies F'(2) = 13$4 F'(2) - 12 = 40 \implies 4 F'(2) = 52 \implies F'(2) = 13$ ### Step 2: Sum the Values We need to find F'(2) + int_0^2 F(x) dx$F'(2) + \int_{0}^{2} F(x) dx$: 13 + (-2) = 11$13 + (-2) = 11$ ### Pattern Recognition Notice how the definition of f(x)$f(x)$ is mostly a distraction to find F(2)=2$F(2)=2$. The problem is fundamentally testing consecutive applications of integration by parts to reduction structures. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integration