### Related Formula sqrt1 - sin 2x = sqrt(cos x - sin x)^2 = |cos x - sin x|$\sqrt{1 - \sin 2x} = \sqrt{(\cos x - \sin x)^2} = |\cos x - \sin x|$ ### Core Logic The absolute function |cos x - sin x|$|\cos x - \sin x|$ flips sign at x = fracpi4$x = \frac{\pi}{4}$ within the boundary range: * For x in left[fracpi6, fracpi4right]$x \in \left[\frac{\pi}{6}, \frac{\pi}{4}\right]$: cos x geq sin x implies |cos x - sin x| = cos x - sin x$\cos x \geq \sin x \implies |\cos x - \sin x| = \cos x - \sin x$ * For x in left[fracpi4, fracpi3right]$x \in \left[\frac{\pi}{4}, \frac{\pi}{3}\right]$: sin x geq cos x implies |cos x - sin x| = sin x - cos x$\sin x \geq \cos x \implies |\cos x - \sin x| = \sin x - \cos x$ ### Step 1: Boundary Splitting Integration I = int_fracpi6^fracpi4 (cos x - sin x) \, dx + int_fracpi4^fracpi3 (sin x - cos x) \, dx$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} (\cos x - \sin x) \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} (\sin x - \cos x) \, dx$ I = Big[ sin x + cos x Big]_fracpi6^fracpi4 + Big[ -cos x - sin x Big]_fracpi4^fracpi3$I = \Big[ \sin x + \cos x \Big]_{\frac{\pi}{6}}^{\frac{\pi}{4}} + \Big[ -\cos x - \sin x \Big]_{\frac{\pi}{4}}^{\frac{\pi}{3}}$ I = left( frac1sqrt2 + frac1sqrt2 - left(frac12 + fracsqrt32right) right) + left( -frac12 - fracsqrt32 - left(-frac1sqrt2 - frac1sqrt2right) right)$I = \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} - \left(\frac{1}{2} + \frac{\sqrt{3}}{2}\right) \right) + \left( -\frac{1}{2} - \frac{\sqrt{3}}{2} - \left(-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right) \right)$ I = left( sqrt2 - frac1 + sqrt32 right) + left( sqrt2 - frac1 + sqrt32 right) = 2sqrt2 - 1 - sqrt3 = -1 + 2sqrt2 - sqrt3$I = \left( \sqrt{2} - \frac{1 + \sqrt{3}}{2} \right) + \left( \sqrt{2} - \frac{1 + \sqrt{3}}{2} \right) = 2\sqrt{2} - 1 - \sqrt{3} = -1 + 2\sqrt{2} - \sqrt{3}$ ### Step 2: Coefficient Matrix Matching Comparing with the given baseline layout: alpha = -1, quad beta = 2, quad gamma = -1$\alpha = -1, \quad \beta = 2, \quad \gamma = -1$ Evaluating the required target metrics: 3alpha + 4beta - gamma = 3(-1) + 4(2) - (-1) = -3 + 8 + 1 = 6$3\alpha + 4\beta - \gamma = 3(-1) + 4(2) - (-1) = -3 + 8 + 1 = 6$ ### Pattern Recognition Never forget that sqrtf(x)^2 = |f(x)|$\sqrt{f(x)^2} = |f(x)|$. Skipping modulus checks inside definite root boundaries leads to incorrect answers. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Integrals

### Related Formula int frac1u^2 + a^2 du = frac1a tan^-1left(fracuaright) + C$\int \frac{1}{u^2 + a^2} du = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C$ ### Core Logic Simplify the given integrand by converting all trigonometric ratios into sin x$\sin x$ and cos x$\cos x$: I = fracoperatornamecosec x + sin xoperatornamecosec x sec x + tan x sin^2 x$I = \frac{\operatorname{cosec} x + \sin x}{\operatorname{cosec} x \sec x + \tan x \sin^2 x}$ Numerator: frac1sin x + sin x = frac1 + sin^2 xsin x$\frac{1}{\sin x} + \sin x = \frac{1 + \sin^2 x}{\sin x}$ Denominator: frac1sin x cos x + fracsin xcos x cdot sin^2 x = frac1 + sin^4 xsin x cos x$\frac{1}{\sin x \cos x} + \frac{\sin x}{\cos x} \cdot \sin^2 x = \frac{1 + \sin^4 x}{\sin x \cos x}$ Divide Numerator by Denominator: I = fracfrac1 + sin^2 xsin xfrac1 + sin^4 xsin x cos x = frac(1 + sin^2 x)cos x1 + sin^4 x$I = \frac{\frac{1 + \sin^2 x}{\sin x}}{\frac{1 + \sin^4 x}{\sin x \cos x}} = \frac{(1 + \sin^2 x)\cos x}{1 + \sin^4 x}$ ### Step 1: Apply Substitution Now rewrite the integral: y(x) = int frac(1 + sin^2 x)cos x1 + sin^4 x dx$y(x) = \int \frac{(1 + \sin^2 x)\cos x}{1 + \sin^4 x} dx$ Let sin x = t$\sin x = t$, then cos x \, dx = dt$\cos x \, dx = dt$. The integral becomes: y(x) = int frac1 + t^21 + t^4 dt$y(x) = \int \frac{1 + t^2}{1 + t^4} dt$ Divide numerator and denominator by t^2$t^2$: y(x) = int frac1 + frac1t^2t^2 + frac1t^2 dt$y(x) = \int \frac{1 + \frac{1}{t^2}}{t^2 + \frac{1}{t^2}} dt$ We know that t^2 + frac1t^2 = left(t - frac1tright)^2 + 2$t^2 + \frac{1}{t^2} = \left(t - \frac{1}{t}\right)^2 + 2$. Let u = t - frac1t$u = t - \frac{1}{t}$, then du = left(1 + frac1t^2right) dt$du = \left(1 + \frac{1}{t^2}\right) dt$. y(x) = int fracduu^2 + (sqrt2)^2 = frac1sqrt2tan^-1left(fracusqrt2right) + C$y(x) = \int \frac{du}{u^2 + (\sqrt{2})^2} = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{u}{\sqrt{2}}\right) + C$ y(x) = frac1sqrt2tan^-1left(fract - 1/tsqrt2right) + C = frac1sqrt2tan^-1left(fracsin x - operatornamecosec xsqrt2right) + C$y(x) = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t - 1/t}{\sqrt{2}}\right) + C = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sin x - \operatorname{cosec} x}{\sqrt{2}}\right) + C$ ### Step 2: Solve for Constant C We are given that lim_xrightarrow(pi/2)^- y(x) = 0$\lim_{x\rightarrow(\pi/2)^-} y(x) = 0$. As x to fracpi2$x \to \frac{\pi}{2}$, sin x to 1$\sin x \to 1$ and operatornamecosec x to 1$\operatorname{cosec} x \to 1$. Thus, sin x - operatornamecosec x to 0$\sin x - \operatorname{cosec} x \to 0$. 0 = frac1sqrt2tan^-1left(frac1 - 1sqrt2right) + C$0 = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{1 - 1}{\sqrt{2}}\right) + C$ 0 = frac1sqrt2tan^-1(0) + C Rightarrow C = 0$0 = \frac{1}{\sqrt{2}}\tan^{-1}(0) + C \Rightarrow C = 0$ ### Step 3: Find y(pi/4) Substitute x = fracpi4$x = \frac{\pi}{4}$ into the exact solution y(x) = frac1sqrt2tan^-1left(fracsin x - operatornamecosec xsqrt2right)$y(x) = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sin x - \operatorname{cosec} x}{\sqrt{2}}\right)$: At x = fracpi4$x = \frac{\pi}{4}$, sinleft(fracpi4right) = frac1sqrt2$\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$ and operatornamecosecleft(fracpi4right) = sqrt2$\operatorname{cosec}\left(\frac{\pi}{4}\right) = \sqrt{2}$. yleft(fracpi4right) = frac1sqrt2tan^-1left(fracfrac1sqrt2 - sqrt2sqrt2right)$y\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\frac{1}{\sqrt{2}} - \sqrt{2}}{\sqrt{2}}\right)$ = frac1sqrt2tan^-1left(fracfrac1 - 2sqrt2sqrt2right) = frac1sqrt2tan^-1left(frac-1/sqrt2sqrt2right)$= \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\frac{1 - 2}{\sqrt{2}}}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{-1/\sqrt{2}}{\sqrt{2}}\right)$ = frac1sqrt2tan^-1left(-frac12right)$= \frac{1}{\sqrt{2}}\tan^{-1}\left(-\frac{1}{2}\right)$ ### Pattern Recognition A high-degree polynomial of sin x$\sin x$ or cos x$\cos x$ in the denominator matched with their derivative elements on top is the hallmark of the t+1/t$t+1/t$ or t-1/t$t-1/t$ structural substitution form. Dividing out the middle power (t^2$t^2$) aligns the differential instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Integral Calculus

### Related Formula textKing's Rule Variant: int_-a^a f(x) dx = int_0^a (f(x) + f(-x)) dx$\text{King's Rule Variant: } \int_{-a}^{a} f(x) dx = \int_{0}^{a} (f(x) + f(-x)) dx$ ### Core Logic Let the integral be I$I$. We apply the property int_-a^a f(x) dx = int_0^a (f(x)+f(-x)) dx$\int_{-a}^a f(x) dx = \int_0^a (f(x)+f(-x)) dx$ to each fractional term independently. For the first term f_1(x) = fracx^2cos x1+pi^x$f_1(x) = \frac{x^2\cos x}{1+\pi^x}$: f_1(x) + f_1(-x) = fracx^2cos x1+pi^x + frac(-x)^2cos(-x)1+pi^-x = x^2cos x left( frac11+pi^x + fracpi^xpi^x+1 right) = x^2cos x$f_1(x) + f_1(-x) = \frac{x^2\cos x}{1+\pi^x} + \frac{(-x)^2\cos(-x)}{1+\pi^{-x}} = x^2\cos x \left( \frac{1}{1+\pi^x} + \frac{\pi^x}{\pi^x+1} \right) = x^2\cos x$ For the second term f_2(x) = frac1+sin^2x1+e^sin x^2023$f_2(x) = \frac{1+\sin^2x}{1+e^{\sin x^{2023}}}$: Observe the exponent: sin(-x)^2023 = sin(-x^2023) = -sin x^2023$\sin(-x)^{2023} = \sin(-x^{2023}) = -\sin x^{2023}$. f_2(x) + f_2(-x) = frac1+sin^2x1+e^p + frac1+sin^2(-x)1+e^-p = (1+sin^2x) left( frac11+e^p + frace^pe^p+1 right) = 1+sin^2x$f_2(x) + f_2(-x) = \frac{1+\sin^2x}{1+e^{p}} + \frac{1+\sin^2(-x)}{1+e^{-p}} = (1+\sin^2x) \left( \frac{1}{1+e^p} + \frac{e^p}{e^p+1} \right) = 1+\sin^2x$ Thus, the integral radically simplifies to: I = int_0^pi/2 (x^2cos x + 1 + sin^2 x) dx$I = \int_0^{\pi/2} (x^2\cos x + 1 + \sin^2 x) dx$ ### Step 1: Execute Integration by Parts Evaluate int_0^pi/2 x^2cos x \, dx$\int_0^{\pi/2} x^2\cos x \, dx$ using integration by parts: u = x^2 Rightarrow du = 2x \, dx$u = x^2 \Rightarrow du = 2x \, dx$ dv = cos x \, dx Rightarrow v = sin x$dv = \cos x \, dx \Rightarrow v = \sin x$ int x^2cos x dx = x^2sin x - int 2xsin x dx$\int x^2\cos x dx = x^2\sin x - \int 2x\sin x dx$ Applying parts again to int 2xsin x dx$\int 2x\sin x dx$: int 2xsin x dx = 2x(-cos x) - int 2(-cos x)dx = -2xcos x + 2sin x$\int 2x\sin x dx = 2x(-\cos x) - \int 2(-\cos x)dx = -2x\cos x + 2\sin x$ Substituting back: left[ x^2sin x + 2xcos x - 2sin x right]_0^pi/2$\left[ x^2\sin x + 2x\cos x - 2\sin x \right]_0^{\pi/2}$ Evaluate at upper limit pi/2$\pi/2$: (pi/2)^2(1) + 0 - 2(1) = fracpi^24 - 2$(\pi/2)^2(1) + 0 - 2(1) = \frac{\pi^2}{4} - 2$ Evaluate at lower limit 0: 0 + 0 - 0 = 0$0 + 0 - 0 = 0$ So, int_0^pi/2 x^2cos x \, dx = fracpi^24 - 2$\int_0^{\pi/2} x^2\cos x \, dx = \frac{\pi^2}{4} - 2$. ### Step 2: Execute Standard Integrals Evaluate int_0^pi/2 (1 + sin^2 x) dx$\int_0^{\pi/2} (1 + \sin^2 x) dx$: = int_0^pi/2 dx + int_0^pi/2 sin^2 x dx$= \int_0^{\pi/2} dx + \int_0^{\pi/2} \sin^2 x dx$ = fracpi2 + left[ frac12 cdot fracpi2 right] = fracpi2 + fracpi4 = frac3pi4$= \frac{\pi}{2} + \left[ \frac{1}{2} \cdot \frac{\pi}{2} \right] = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}$ ### Step 3: Combine and Compare Sum the components to get the total integral I$I$: I = left(fracpi^24 - 2right) + frac3pi4 = fracpi^24 + frac3pi4 - 2$I = \left(\frac{\pi^2}{4} - 2\right) + \frac{3\pi}{4} = \frac{\pi^2}{4} + \frac{3\pi}{4} - 2$ Factor out fracpi4$\frac{\pi}{4}$: I = fracpi4(pi + 3) - 2$I = \frac{\pi}{4}(\pi + 3) - 2$ The problem states I = fracpi4(pi + a) - 2$I = \frac{\pi}{4}(\pi + a) - 2$. Comparing the two expressions gives: a = 3$a = 3$ ### Pattern Recognition Frightening denominators like 1+a^f(x)$1+a^{f(x)}$ in an integral from -L$-L$ to L$L$ where f(x)$f(x)$ is an odd function are almost exclusively designed to cancel out and leave 1 via the f(x)+f(-x)$f(x)+f(-x)$ expansion. Strip the ugly denominators immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Integral Calculus

### Related Formula textSlope of tangent at x=a text is f'(a) = tan theta$\text{Slope of tangent at } x=a \text{ is } f'(a) = \tan \theta$ int u^n du = fracu^n+1n+1$\int u^n du = \frac{u^{n+1}}{n+1}$ ### Core Logic Given y=f(x)$y=f(x)$, the derivatives at the given points correspond to the slope of tangents: left. fracdydx right|_x=1 = f'(1) = tanleft(fracpi6right) = frac1sqrt3$\left. \frac{dy}{dx} \right|_{x=1} = f'(1) = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$ left. fracdydx right|_x=3 = f'(3) = tanleft(fracpi4right) = 1$\left. \frac{dy}{dx} \right|_{x=3} = f'(3) = \tan\left(\frac{\pi}{4}\right) = 1$ We must evaluate the integral: I = int_1^3 left( (f'(t))^2 + 1 right) f''(t) dt$I = \int_{1}^{3} \left( (f'(t))^2 + 1 \right) f''(t) dt$ ### Step 1: Integration by Substitution Let z = f'(t)$z = f'(t)$. Then dz = f''(t)dt$dz = f''(t)dt$. The limits of integration change accordingly: When t = 1$t = 1$, z = f'(1) = frac1sqrt3$z = f'(1) = \frac{1}{\sqrt{3}}$ When t = 3$t = 3$, z = f'(3) = 1$z = f'(3) = 1$ The integral becomes: I = int_1/sqrt3^1 (z^2 + 1) dz$I = \int_{1/\sqrt{3}}^{1} (z^2 + 1) dz$ I = left[ fracz^33 + z right]_1/sqrt3^1$I = \left[ \frac{z^3}{3} + z \right]_{1/\sqrt{3}}^{1}$ ### Step 2: Evaluating the Definite Integral Plug in the limits: I = left( frac1^33 + 1 right) - left( frac13 cdot frac13sqrt3 + frac1sqrt3 right)$I = \left( \frac{1^3}{3} + 1 \right) - \left( \frac{1}{3} \cdot \frac{1}{3\sqrt{3}} + \frac{1}{\sqrt{3}} \right)$ I = frac43 - left( frac19sqrt3 + frac99sqrt3 right)$I = \frac{4}{3} - \left( \frac{1}{9\sqrt{3}} + \frac{9}{9\sqrt{3}} \right)$ I = frac43 - frac109sqrt3 = frac43 - frac10sqrt327$I = \frac{4}{3} - \frac{10}{9\sqrt{3}} = \frac{4}{3} - \frac{10\sqrt{3}}{27}$ ### Step 3: Finding alpha and beta We are given that 27 cdot I = alpha + betasqrt3$27 \cdot I = \alpha + \beta\sqrt{3}$. 27 left( frac43 - frac10sqrt327 right) = 36 - 10sqrt3$27 \left( \frac{4}{3} - \frac{10\sqrt{3}}{27} \right) = 36 - 10\sqrt{3}$ Comparing with alpha + betasqrt3$\alpha + \beta\sqrt{3}$, we get: alpha = 36$\alpha = 36$ beta = -10$\beta = -10$ Therefore, alpha + beta = 36 - 10 = 26$\alpha + \beta = 36 - 10 = 26$. ### Pattern Recognition Integrals containing a function derivative alongside its second derivative are classic substitution traps. Set u = f'(x)$u = f'(x)$ directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Integral Calculus Class 12 Maths: Application of Derivatives

### Related Formula int e^kx dx = frace^kxk$\int e^{kx} dx = \frac{e^{kx}}{k}$ e^log_e x = x$e^{\log_e x} = x$ ### Core Logic Given f(x) = ae^2x + be^x + cx$f(x) = ae^{2x} + be^x + cx$. Condition 1: f(0) = -1$f(0) = -1$ ae^0 + be^0 + c(0) = -1 Rightarrow a + b = -1 quad dots(i)$ae^0 + be^0 + c(0) = -1 \Rightarrow a + b = -1 \quad \dots(i)$ Condition 2: f'(log_e 2) = 21$f'(\log_e 2) = 21$ Find the derivative: f'(x) = 2ae^2x + be^x + c$f'(x) = 2ae^{2x} + be^x + c$ Substitute x = ln 2$x = \ln 2$: f'(ln 2) = 2a(e^ln 2)^2 + b(e^ln 2) + c$f'(\ln 2) = 2a(e^{\ln 2})^2 + b(e^{\ln 2}) + c$ 21 = 2a(2)^2 + b(2) + c$21 = 2a(2)^2 + b(2) + c$ 8a + 2b + c = 21 quad dots(ii)$8a + 2b + c = 21 \quad \dots(ii)$ ### Step 1: Applying the Integral Condition Condition 3: int_0^ln 4 (f(x) - cx) dx = frac392$\int_0^{\ln 4} (f(x) - cx) dx = \frac{39}{2}$ Substitute f(x) - cx = ae^2x + be^x$f(x) - cx = ae^{2x} + be^x$: int_0^ln 4 (ae^2x + be^x) dx = frac392$\int_0^{\ln 4} (ae^{2x} + be^x) dx = \frac{39}{2}$ left[ fracae^2x2 + be^x right]_0^ln 4 = frac392$\left[ \frac{ae^{2x}}{2} + be^x \right]_0^{\ln 4} = \frac{39}{2}$ Substitute the limits: left( fraca2e^2ln 4 + be^ln 4 right) - left( fraca2e^0 + be^0 right) = frac392$\left( \frac{a}{2}e^{2\ln 4} + be^{\ln 4} \right) - \left( \frac{a}{2}e^0 + be^0 \right) = \frac{39}{2}$ left( fraca2(16) + 4b right) - left( fraca2 + b right) = frac392$\left( \frac{a}{2}(16) + 4b \right) - \left( \frac{a}{2} + b \right) = \frac{39}{2}$ 8a + 4b - fraca2 - b = frac392$8a + 4b - \frac{a}{2} - b = \frac{39}{2}$ frac15a2 + 3b = frac392$\frac{15a}{2} + 3b = \frac{39}{2}$ Multiply by 2: 15a + 6b = 39 quad dots(iii)$15a + 6b = 39 \quad \dots(iii)$ ### Step 2: Solving the System of Equations From (i), b = -1 - a$b = -1 - a$. Substitute b$b$ into (iii): 15a + 6(-1 - a) = 39$15a + 6(-1 - a) = 39$ 15a - 6 - 6a = 39$15a - 6 - 6a = 39$ 9a = 45 Rightarrow a = 5$9a = 45 \Rightarrow a = 5$ Thus, b = -1 - 5 = -6$b = -1 - 5 = -6$. Now, substitute a$a$ and b$b$ into (ii) to find c$c$: 8(5) + 2(-6) + c = 21$8(5) + 2(-6) + c = 21$ 40 - 12 + c = 21$40 - 12 + c = 21$ 28 + c = 21 Rightarrow c = -7$28 + c = 21 \Rightarrow c = -7$ ### Step 3: Finding Final Absolute Sum Calculate |a + b + c|$|a + b + c|$: |5 + (-6) + (-7)| = |5 - 13| = |-8| = 8$|5 + (-6) + (-7)| = |5 - 13| = |-8| = 8$ ### Pattern Recognition Translating multiple constraints (f(x)$f(x)$, f'(x)$f'(x)$, int f(x)$\int f(x)$) into a system of linear equations for coefficients a, b, c$a, b, c$ exposes a direct sequential solution without circular dependency. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Integral Calculus Class 12 Maths: Application of Derivatives