Let the area of the bounded region (x,y):0≤ 9x≤ y²,y≥ 3x - 6${(x,y):0\leq 9x\leq y^2,y\geq 3x - 6}$ be A. Then 6A is equal to

### Related Formula Area of an ellipse (x²)/(a²) + (y²)/(b²) = 1$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by: Area = π a b$$\text{Area} = \pi a b$$ Area of a rhombus bounded by |x| + |y| = a$|x| + |y| = a$ is 2a²$2a^2$. ### Core Logic The given curves form a bounded geometric area. Ellipse: x² + 4y² = 4 ⇒ (x²)/(4) + (y²)/(1) = 1$x^2 + 4y^2 = 4 \Rightarrow \frac{x^2}{4} + \frac{y^2}{1} = 1$. Here, a = 2$a = 2$, b = 1$b = 1$. The region to be excluded is bounded by y = |x| - 1$y = |x| - 1$ and y = 1 - |x|$y = 1 - |x|$, which rearranges to |x| + |y| = 1$|x| + |y| = 1$. This forms a square/rhombus centered at the origin with vertices at (1, 0), (0, 1), (-1, 0), (0, -1)$(1, 0), (0, 1), (-1, 0), (0, -1)$. ### Step 1: Calculate Total and Excluded Areas Total Area of the Ellipse: Area = π (2)(1) = 2π$$\text{Area} = \pi (2)(1) = 2\pi$$ Excluded Area (Rhombus |x| + |y| = 1$|x| + |y| = 1$): The rhombus consists of 4 identical right-angled triangles in each quadrant. Area of one triangle = (1)/(2) × base × height = (1)/(2) × 1 × 1 = (1)/(2)$\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}$. Total excluded area = 4 × (1)/(2) = 2$4 \times \frac{1}{2} = 2$. ### Step 2: Calculate Required Area

Required Area = Area of ellipse -$-$ Shaded Area = 2π - 2$= 2\pi - 2$ = 2(π - 1)$= 2(\pi - 1)$ ### Pattern Recognition Transform absolute value equations y = ±(|x| - a)$y = \pm(|x| - a)$ into |x| + |y| = a$|x| + |y| = a$ to instantly recognize a standard rhombus, allowing direct geometry formulas instead of integration. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Area Under Curves Class 11 Maths: Conic Sections

### Related Formula Area = ∫x₁x₂ (f(x) - g(x)) dx$$\text{Area} = \int_{x_1}^{x_2} (f(x) - g(x)) dx$$ Area of Parabola piece: ∫₀² (4-x²) dx$$\text{Area of Parabola piece: } \int_{0}^{2} (4-x^2) dx$$ ### Core Logic

The region is bounded above by y = 4 - x²$y = 4 - x^2$, below by y = 1 - 2x$y = 1 - 2x$, and constrained to x ≥ 0, y ≥ 0$x \geq 0, y \geq 0$. The parabola intersects the x-axis at x=2$x=2$ (since 4-x²=0, x≥ 0$4-x^2=0, x\geq 0$). The line intersects the x-axis at x=(1)/(2)$x=\frac{1}{2}$ (since 1-2x=0$1-2x=0$) and y-axis at y=1$y=1$. The required area is the area under the parabola in the first quadrant minus the small triangular region bounded by y = 1-2x, x=0, y=0$y = 1-2x, x=0, y=0$. ### Step 1: Calculate the Area Total area under parabola in 1st quadrant = ∫₀² (4 - x²) dx = [ 4x - (x³)/(3) ]₀² = 8 - (8)/(3) = (16)/(3)$$\text{Total area under parabola in 1st quadrant} = \int_0^2 (4 - x^2) dx = \left[ 4x - \frac{x^3}{3} \right]_0^2 = 8 - \frac{8}{3} = \frac{16}{3}$$ Area of the small right triangle formed by the line y=1-2x$y=1-2x$ in the first quadrant: Vertices are (0,0), (1/2,0), (0,1)$(0,0), (1/2,0), (0,1)$. Area of triangle = (1)/(2) × base × height = (1)/(2) × (1)/(2) × 1 = (1)/(4)$$\text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{1}{2} \times 1 = \frac{1}{4}$$ ### Step 2: Subtraction and Format Match Required Area = (16)/(3) - (1)/(4) = (64 - 3)/(12) = (61)/(12)$$\text{Required Area} = \frac{16}{3} - \frac{1}{4} = \frac{64 - 3}{12} = \frac{61}{12}$$ Here, α = 61, β = 12$\alpha = 61, \beta = 12$. Check (61, 12) = 1$\gcd(61, 12) = 1$. This matches. So, α + β = 61 + 12 = 73$\alpha + \beta = 61 + 12 = 73$. ### Pattern Recognition For areas defined by y ≥ g(x)$y \geq g(x)$ when g(x)$g(x)$ forms a simple geometric shape (like a line), subtract the geometric area directly rather than splitting the integral algebraically. It eliminates integration errors. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Application of Integrals

### Related Formula Area = ∫ₐb (yupper - ylower) dx$$\text{Area} = \int_{a}^{b} (y_{\text{upper}} - y_{\text{lower}}) \,dx$$ ### Core Logic First, find the points of intersection for the curves y = 1 + x²$y = 1 + x^2$ and y = 3 - x$y = 3 - x$: 1 + x² = 3 - x$1 + x^2 = 3 - x$ x² + x - 2 = 0$x^2 + x - 2 = 0$ (x + 2)(x - 1) = 0 x = -2, x = 1$$(x + 2)(x - 1) = 0 \implies x = -2, x = 1$$ So the total region is bounded between x = -2$x = -2$ and x = 1$x = 1$. The line x = -1$x = -1$ divides this region into two parts.

### Step 1: Setting up the Areas Let the area to the left of x = -1$x = -1$ be proportional to m$m$ and the area to the right be proportional to n$n$. Am = ∫₋₂⁻¹ [(3 - x) - (1 + x²)] dx$$A_m = \int_{-2}^{-1} [(3 - x) - (1 + x^2)] \,dx$$ Aₙ = ∫₋₁¹ [(3 - x) - (1 + x²)] dx$$A_n = \int_{-1}^{1} [(3 - x) - (1 + x^2)] \,dx$$ The integrand simplifies to 2 - x - x²$2 - x - x^2$.

### Step 2: Integration ∫ (2 - x - x²) dx = 2x - (x²)/(2) - (x³)/(3)$$\int (2 - x - x^2) \,dx = 2x - \frac{x^2}{2} - \frac{x^3}{3}$$ Evaluate Am$A_m$: Am = [2x - (x²)/(2) - (x³)/(3)]₋₂⁻¹$$A_m = \left[2x - \frac{x^2}{2} - \frac{x^3}{3}\right]_{-2}^{-1}$$ Am = (-2 - (1)/(2) + (1)/(3)) - (-4 - 2 + (8)/(3)) = (-(13)/(6)) - (-(10)/(3)) = (7)/(6)$$A_m = \left(-2 - \frac{1}{2} + \frac{1}{3}\right) - \left(-4 - 2 + \frac{8}{3}\right) = \left(-\frac{13}{6}\right) - \left(-\frac{10}{3}\right) = \frac{7}{6}$$ Evaluate Aₙ$A_n$: Aₙ = [2x - (x²)/(2) - (x³)/(3)]₋₁¹$$A_n = \left[2x - \frac{x^2}{2} - \frac{x^3}{3}\right]_{-1}^{1}$$ Aₙ = (2 - (1)/(2) - (1)/(3)) - (-2 - (1)/(2) + (1)/(3)) = ((7)/(6)) - (-(13)/(6)) = (20)/(6)$$A_n = \left(2 - \frac{1}{2} - \frac{1}{3}\right) - \left(-2 - \frac{1}{2} + \frac{1}{3}\right) = \left(\frac{7}{6}\right) - \left(-\frac{13}{6}\right) = \frac{20}{6}$$ ### Step 3: Finding the Ratio The ratio of the areas m:n$m:n$ is: (m)/(n) = (Aₙ)/(Am) or (Am)/(Aₙ)$$\frac{m}{n} = \frac{A_n}{A_m} \text{ or } \frac{A_m}{A_n}$$ Wait, the solution designates (m)/(n) = ∫₋₁¹∫₋₂⁻¹ = (20/6)/(7/6) = (20)/(7)$\frac{m}{n} = \frac{\int_{-1}^{1}}{\int_{-2}^{-1}} = \frac{20/6}{7/6} = \frac{20}{7}$. (Since (20,7)=1$\gcd(20,7)=1$, m=20$m=20$ and n=7$n=7$). Therefore, m+n = 20 + 7 = 27$m+n = 20 + 7 = 27$. ### Pattern Recognition When a vertical line divides an area into a ratio, calculate the definite integral on both sides of the splitting line independently. Keep fractions with a common denominator until the final ratio step. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Applications of Integrals

### Core Logic To evaluate the area under y = x, x$y = \max \{\sin x, \cos x\}$, we must identify where one function is greater than the other in [0, (3π)/(2)]$[0, \frac{3\pi}{2}]$.

From 0$0$ to (π)/(4)$\frac{\pi}{4}$: x > x ⇒ y = x$\cos x > \sin x \Rightarrow y = \cos x$ From (π)/(4)$\frac{\pi}{4}$ to (5π)/(4)$\frac{5\pi}{4}$: x > x ⇒ y = x$\sin x > \cos x \Rightarrow y = \sin x$ From (5π)/(4)$\frac{5\pi}{4}$ to (3π)/(2)$\frac{3\pi}{2}$: x > x ⇒ y = x$\cos x > \sin x \Rightarrow y = \cos x$ ### Step 1: Formulate the Area Integral Since area is bounded by the x-axis, we must take the absolute value if the function goes below the axis. Wait, the function x$\sin x$ is negative from π$\pi$ to (5π)/(4)$\frac{5\pi}{4}$, and x$\cos x$ is negative from (5π)/(4)$\frac{5\pi}{4}$ to (3π)/(2)$\frac{3\pi}{2}$. Let's integrate carefully: A = ∫₀π/4 x dx + ∫π/4π x dx + ∫π5π/4 | x| dx + ∫5π/43π/2 | x| dx$$A = \int_0^{\pi/4} \cos x \, dx + \int_{\pi/4}^{\pi} \sin x \, dx + \int_{\pi}^{5\pi/4} |\sin x| \, dx + \int_{5\pi/4}^{3\pi/2} |\cos x| \, dx$$ Because area is geometric, absolute values are explicitly integrated: A = ∫₀π/4 x dx + ∫π/4π x dx + ∫π5π/4 (- x) dx + ∫5π/43π/2 (- x) dx$$A = \int_0^{\pi/4} \cos x \, dx + \int_{\pi/4}^{\pi} \sin x \, dx + \int_{\pi}^{5\pi/4} (-\sin x) \, dx + \int_{5\pi/4}^{3\pi/2} (-\cos x) \, dx$$ ### Step 2: Evaluate Integrals ∫₀π/4 x dx = [ x]₀π/4 = 1√(2) - 0 = 1√(2)$$\int_0^{\pi/4} \cos x \, dx = [\sin x]_0^{\pi/4} = \frac{1}{\sqrt{2}} - 0 = \frac{1}{\sqrt{2}}$$ ∫π/4π x dx = [- x]π/4π = -(-1) - (- 1√(2)) = 1 + 1√(2)$$\int_{\pi/4}^{\pi} \sin x \, dx = [-\cos x]_{\pi/4}^{\pi} = -(-1) - \left(-\frac{1}{\sqrt{2}}\right) = 1 + \frac{1}{\sqrt{2}}$$ ∫π5π/4 (- x) dx = [ x]π5π/4 = - 1√(2) - (-1) = 1 - 1√(2)$$\int_{\pi}^{5\pi/4} (-\sin x) \, dx = [\cos x]_{\pi}^{5\pi/4} = -\frac{1}{\sqrt{2}} - (-1) = 1 - \frac{1}{\sqrt{2}}$$ ∫5π/43π/2 (- x) dx = [- x]5π/43π/2 = -(-1) - ( -(- 1√(2)) ) = 1 - 1√(2)$$\int_{5\pi/4}^{3\pi/2} (-\cos x) \, dx = [-\sin x]_{5\pi/4}^{3\pi/2} = -(-1) - \left( -\left(-\frac{1}{\sqrt{2}}\right) \right) = 1 - \frac{1}{\sqrt{2}}$$ ### Step 3: Total Area Sum the areas together: A = 1√(2) + (1 + 1√(2)) + (1 - 1√(2)) + (1 - 1√(2))$$A = \frac{1}{\sqrt{2}} + \left(1 + \frac{1}{\sqrt{2}}\right) + \left(1 - \frac{1}{\sqrt{2}}\right) + \left(1 - \frac{1}{\sqrt{2}}\right)$$ A = 1√(2) + 1 + 1√(2) + 1 - 1√(2) + 1 - 1√(2) = 3$$A = \frac{1}{\sqrt{2}} + 1 + \frac{1}{\sqrt{2}} + 1 - \frac{1}{\sqrt{2}} + 1 - \frac{1}{\sqrt{2}} = 3$$ Now calculate A² + A$A^2 + A$: A² + A = (3)² + 3 = 9 + 3 = 12$$A^2 + A = (3)^2 + 3 = 9 + 3 = 12$$ ### Pattern Recognition The expression ( x, x)$\max(\sin x, \cos x)$ always splits segments exactly at nπ + π/4$n\pi + \pi/4$. When bounded by the x-axis, parts below y=0$y=0$ explicitly require negation. Charting the piecewise transitions is non-negotiable. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Application of Integrals

### Related Formula ∫ √(a² - x²) dx = (x)/(2)√(a² - x²) + (a²)/(2) ⁻¹((x)/(a))$$\int \sqrt{a^2 - x^2} dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$$ ### Core Logic

The intersection points of x² + y² = 4$x^2 + y^2 = 4$ and x² + (y-2)² = 4$x^2 + (y-2)^2 = 4$: Subtracting the equations gives y² - (y-2)² = 0 y² - (y² - 4y + 4) = 0 4y = 4 y = 1$y^2 - (y-2)^2 = 0 \implies y^2 - (y^2 - 4y + 4) = 0 \implies 4y = 4 \implies y = 1$. Substitute y=1$y=1$ back: x² + 1 = 4 x = ± √(3)$x^2 + 1 = 4 \implies x = \pm \sqrt{3}$. The points of intersection are (√(3), 1)$(\sqrt{3}, 1)$ and (-√(3), 1)$(-\sqrt{3}, 1)$. The area is symmetric about the y-axis, so we integrate from x = 0$x = 0$ to x = √(3)$x = \sqrt{3}$ and double the result. Upper curve is the lower arc of the top circle: y = 2 - √(4-x²)$y = 2 - \sqrt{4-x^2}$ -- wait, the area is bounded by the top arc of the bottom circle and the bottom arc of the top circle. Upper curve for area: y = √(4-x²)$y = \sqrt{4-x^2}$ (from x²+y²=4$x^2+y^2=4$) Lower curve for area: y = 2 - √(4-x²)$y = 2 - \sqrt{4-x^2}$ (from x²+(y-2)²=4$x^2+(y-2)^2=4$) A = 2∫₀√(3) [√(4-x²) - (2 - √(4-x²))] dx$$A = 2\int_{0}^{\sqrt{3}} \left[\sqrt{4-x^2} - (2 - \sqrt{4-x^2})\right] dx$$ ### Step 1: Integration A = 2∫₀√(3) (2√(4-x²) - 2) dx = 4∫₀√(3) (√(4-x²) - 1) dx$$A = 2\int_{0}^{\sqrt{3}} (2\sqrt{4-x^2} - 2) dx = 4\int_{0}^{\sqrt{3}} (\sqrt{4-x^2} - 1) dx$$ A = 4[ (1)/(2)(x√(4-x²) + 4 ⁻¹(x)/(2)) - x ]₀√(3)$$A = 4\left[ \frac{1}{2}\left(x\sqrt{4-x^2} + 4\sin^{-1}\frac{x}{2}\right) - x \right]_{0}^{\sqrt{3}}$$ Evaluating the limits: = 4[ (1)/(2)(√(3)(1) + 4 ⁻¹( √(3)2)) - √(3) - (0) ]$$= 4\left[ \frac{1}{2}\left(\sqrt{3}(1) + 4\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)\right) - \sqrt{3} - (0) \right]$$ = 4[ (1)/(2)(√(3) + 4((π)/(3))) - √(3) ]$$= 4\left[ \frac{1}{2}\left(\sqrt{3} + 4\left(\frac{\pi}{3}\right)\right) - \sqrt{3} \right]$$ = 4[ √(3)2 + (2π)/(3) - √(3) ]$$= 4\left[ \frac{\sqrt{3}}{2} + \frac{2\pi}{3} - \sqrt{3} \right]$$ = 4[ (2π)/(3) - √(3)2 ]$$= 4\left[ \frac{2\pi}{3} - \frac{\sqrt{3}}{2} \right]$$ = (8π)/(3) - 2√(3) = (2)/(3)(4π - 3√(3)) (Sq. units)$$= \frac{8\pi}{3} - 2\sqrt{3} = \frac{2}{3}(4\pi - 3\sqrt{3}) \text{ (Sq. units)}$$ ### Pattern Recognition For intersecting identical circles with centres on an axis, symmetry simplifies the integration drastically. Recognize that integrating the circle arc function handles the bulk of the calculation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Area Under Curves