Let the line x = -1$x = -1$ divide the area of the region left\(x,y):1+x^2leq yleq3-xright\$\left\{(x,y):1+x^{2}\leq y\leq3-x\right\}$ in the ratio m:n$m:n$, gcd (m,n)=1$(m,n)=1$. Then m+n$m+n$ is equal to

### Related Formula Area of an ellipse fracx^2a^2 + fracy^2b^2 = 1$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by: textArea = pi a b$$\text{Area} = \pi a b$$ Area of a rhombus bounded by |x| + |y| = a$|x| + |y| = a$ is 2a^2$2a^2$. ### Core Logic The given curves form a bounded geometric area. Ellipse: x^2 + 4y^2 = 4 Rightarrow fracx^24 + fracy^21 = 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: textArea = pi (2)(1) = 2pi$$\text{Area} = \pi (2)(1) = 2\pi$$ 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 = frac12 times textbase times textheight = frac12 times 1 times 1 = frac12$\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}$. Total excluded area = 4 times frac12 = 2$4 \times \frac{1}{2} = 2$. ### Step 2: Calculate Required Area

Required Area = Area of ellipse -$-$ Shaded Area = 2pi - 2$= 2\pi - 2$ = 2(pi - 1)$= 2(\pi - 1)$ ### Pattern Recognition Transform absolute value equations y = pm(|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 textArea = int_x_1^x_2 (f(x) - g(x)) dx$$\text{Area} = \int_{x_1}^{x_2} (f(x) - g(x)) dx$$ textArea of Parabola piece: int_0^2 (4-x^2) dx$$\text{Area of Parabola piece: } \int_{0}^{2} (4-x^2) dx$$ ### Core Logic

The region is bounded above by y = 4 - x^2$y = 4 - x^2$, below by y = 1 - 2x$y = 1 - 2x$, and constrained to x geq 0, y geq 0$x \geq 0, y \geq 0$. The parabola intersects the x-axis at x=2$x=2$ (since 4-x^2=0, xgeq 0$4-x^2=0, x\geq 0$). The line intersects the x-axis at x=frac12$x=\frac{1}{2}$ (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 textTotal area under parabola in 1st quadrant = int_0^2 (4 - x^2) dx = left[ 4x - fracx^33 right]_0^2 = 8 - frac83 = frac163$$\text{Total area under parabola in 1st quadrant} = \int_0^2 (4 - x^2) dx = \left[ 4x - \frac{x^3}{3} \right]_0^2 = 8 - \frac{8}{3} = \frac{16}{3}$$ 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)$(0,0), (1/2,0), (0,1)$. textArea of triangle = frac12 times textbase times textheight = frac12 times frac12 times 1 = frac14$$\text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{1}{2} \times 1 = \frac{1}{4}$$ ### Step 2: Subtraction and Format Match textRequired Area = frac163 - frac14 = frac64 - 312 = frac6112$$\text{Required Area} = \frac{16}{3} - \frac{1}{4} = \frac{64 - 3}{12} = \frac{61}{12}$$ Here, alpha = 61, beta = 12$\alpha = 61, \beta = 12$. Check gcd(61, 12) = 1$\gcd(61, 12) = 1$. This matches. So, alpha + beta = 61 + 12 = 73$\alpha + \beta = 61 + 12 = 73$. ### Pattern Recognition For areas defined by y geq 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 bounded by functions integrated with respect to y$y$ axis: textArea = int_c^d (x_textright - x_textleft) \, mathrmdy$$\text{Area} = \int_{c}^{d} (x_{\text{right}} - x_{\text{left}}) \, \mathrm{d}y$$ ### Core Logic Identify the bounding graphs and intersection coordinates in the first quadrant, then construct standard definite integrals along the vertical axis.

### Step 1: Unpack Bounding Curves The condition |4-x^2| le y$|4-x^2| \le y$ splits into two sections at x=2$x=2$: 1. For 0 le x le 2 implies 4 - x^2 le y implies x^2 ge 4 - y implies x = sqrt4-y$0 \le x \le 2 \implies 4 - x^2 \le y \implies x^2 \ge 4 - y \implies x = \sqrt{4-y}$ 2. For x ge 2 implies x^2 - 4 le y implies x^2 le 4 + y implies x = sqrt4+y$x \ge 2 \implies x^2 - 4 \le y \implies x^2 \le 4 + y \implies x = \sqrt{4+y}$ Also bounded by y le x^2 implies x ge sqrty$y \le x^2 \implies x \ge \sqrt{y}$, and the outer cap constraint y le 4$y \le 4$. ### Step 2: Construct the Integral Area Formula Integrating with respect to y$y$ covers the region bounded on the left by sqrty$\sqrt{y}$ and sqrt4-y$\sqrt{4-y}$, and on the right by sqrt4+y$\sqrt{4+y}$: A = int_0^4 sqrt4+y \, mathrmdy - int_0^2 sqrt4-y \, mathrmdy - int_2^4 sqrty \, mathrmdy$$A = \int_{0}^{4} \sqrt{4+y} \, \mathrm{d}y - \int_{0}^{2} \sqrt{4-y} \, \mathrm{d}y - \int_{2}^{4} \sqrt{y} \, \mathrm{d}y$$ ### Step 3: Evaluate the Definite Integrals A = left[ frac(4+y)^3/23/2 right]_0^4 + left[ frac(4-y)^3/23/2 right]_0^2 - left[ fracy^3/23/2 right]_2^4$$A = \left[ \frac{(4+y)^{3/2}}{3/2} \right]_0^4 + \left[ \frac{(4-y)^{3/2}}{3/2} \right]_0^2 - \left[ \frac{y^{3/2}}{3/2} \right]_2^4$$ Evaluating these values precisely: A = frac23left(8^3/2 - 4^3/2right) + frac23left(2^3/2 - 4^3/2right) - frac23left(4^3/2 - 2^3/2right)$$A = \frac{2}{3}\left(8^{3/2} - 4^{3/2}\right) + \frac{2}{3}\left(2^{3/2} - 4^{3/2}\right) - \frac{2}{3}\left(4^{3/2} - 2^{3/2}\right)$$ A = frac80sqrt23 - 16$$A = \frac{80\sqrt{2}}{3} - 16$$ ### Step 4: Solve for Constants Compare the final value expression to left(frac80sqrt2alpha - betaright)$\left(\frac{80\sqrt{2}}{\alpha} - \beta\right)$: alpha = 3, quad beta = 16 implies alpha + beta = 3 + 16 = 22$$\alpha = 3, \quad \beta = 16 \implies \alpha + \beta = 3 + 16 = 22$$ ### Pattern Recognition Integrating along the vertical axis (y$y$-direction) is significantly faster here because it avoids splitting the domain across multiple vertical segments on the horizontal x$x$-axis. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Application of Integrals

### Related Formula The area between two continuous curves y_textupper(x)$y_{\text{upper}}(x)$ and y_textlower(x)$y_{\text{lower}}(x)$ from x=a$x=a$ to x=b$x=b$ is given by: textArea = int_a^b left( y_textupper - y_textlower right) \, dx$$\text{Area} = \int_a^b \left( y_{\text{upper}} - y_{\text{lower}} \right) \, dx$$ ### Core Logic Analyze the given inequality condition |x - y| le y le 4sqrtx$|x - y| \le y \le 4\sqrt{x}$: 1. **Left Inequality:** |x - y| le y iff -y le x - y le y$$|x - y| \le y \iff -y \le x - y \le y$$ Adding y$y$ across all terms: 0 le x le 2y$0 \le x \le 2y$ This yields two conditions: * x ge 0$x \ge 0$ * y ge fracx2$y \ge \frac{x}{2}$ (lower bounding line) 2. **Right Inequality:** y le 4sqrtx implies y^2 le 16x quad (y ge 0)$$y \le 4\sqrt{x} \implies y^2 \le 16x \quad (y \ge 0)$$ This gives the upper bounding parabolic curve y = 4sqrtx$y = 4\sqrt{x}$. Thus, the enclosed region lies in the first quadrant bounded between the upper curve y = 4sqrtx$y = 4\sqrt{x}$ and the lower line y = fracx2$y = \frac{x}{2}$. ### Step 1: Finding Points of Intersection Equating the two boundary curves: 4sqrtx = fracx2 implies 8sqrtx = x implies x^2 - 64x = 0$$4\sqrt{x} = \frac{x}{2} \implies 8\sqrt{x} = x \implies x^2 - 64x = 0$$ implies x = 0 quad textor quad x = 64$$\implies x = 0 \quad \text{or} \quad x = 64$$ Corresponding y$y$-values: * At x = 0 implies y = 0$x = 0 \implies y = 0$ * At x = 64 implies y = 32$x = 64 \implies y = 32$ The curves intersect at (0, 0)$(0, 0)$ and (64, 32)$(64, 32)$.

### Step 2: Area Integration Integrating with respect to x$x$ from x = 0$x = 0$ to x = 64$x = 64$: beginaligned textArea &= int_0^64 left( 4sqrtx - fracx2 right) \, dx \\ &= left[ frac83x^3/2 - fracx^24 right]_0^64 \\ &= frac83(64)^3/2 - frac(64)^24 \\ &= frac83(512) - frac40964 \\ &= frac40963 - 1024 \\ &= frac4096 - 30723 \\ &= frac10243 endaligned $$ \begin{aligned} \text{Area} &= \int_0^{64} \left( 4\sqrt{x} - \frac{x}{2} \right) \, dx \\ &= \left[ \frac{8}{3}x^{3/2} - \frac{x^2}{4} \right]_0^{64} \\ &= \frac{8}{3}(64)^{3/2} - \frac{(64)^2}{4} \\ &= \frac{8}{3}(512) - \frac{4096}{4} \\ &= \frac{4096}{3} - 1024 \\ &= \frac{4096 - 3072}{3} \\ &= \frac{1024}{3} \end{aligned} $$ ### Pattern Recognition (Alternative Verification) For the area enclosed between a parabola y^2 = 4ax$y^2 = 4ax$ and a line y = mx$y = mx$: textArea = frac8a^23m^3$$\text{Area} = \frac{8a^2}{3m^3}$$ Given y^2 = 16x implies 4a = 16 implies a = 4$y^2 = 16x \implies 4a = 16 \implies a = 4$, and y = frac12x implies m = frac12$y = \frac{1}{2}x \implies m = \frac{1}{2}$: textArea = frac8(4)^23left(frac12right)^3 = frac8 times 163 times frac18 = frac128 times 83 = frac10243$$\text{Area} = \frac{8(4)^2}{3\left(\frac{1}{2}\right)^3} = \frac{8 \times 16}{3 \times \frac{1}{8}} = \frac{128 \times 8}{3} = \frac{1024}{3}$$ ### Evaluation Rubric / Model Answer **Correct Answer:** Option B (frac10243$\frac{1024}{3}$) ### Chapter Mix * Class 12 Mathematics: Applications of Integrals * Class 11 Mathematics: Conic Sections