If the area of the region(x, y) : 1 - 2x ≤ y ≤ 4 - x², x ≥ 0, y ≥ 0$\{(x, y) : 1 - 2x \leq y \leq 4 - x^{2}, x \geq 0, y \geq 0\}$ is (α)/(β), α, β, in N, (α, β) = 1$\frac{\alpha}{\beta}, \alpha, \beta, \in N, \gcd(\alpha, \beta) = 1$, then the value of (α + β)$(\alpha + \beta)$ is :
A.73$73$
B.85$85$
C.91$91$
D.67$67$
Solution & Explanation
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
Area under curve diagram for Q11 - JEE Main 2026 Evening
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)$$\text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{1}{2} \times 1 = \frac{1}{4}$$
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.
Q5jee_main_2024_31_jan_morningArea bounded by Parabolas and Inequalities
The area of the region (x,y): y² ≤ 4x, x < 4, (xy(x - 1)(x - 2))/((x - 3)(x - 4)) > 0, x ≠ 3$\left\{(x,y): y^2 \le 4x, x < 4, \frac{xy(x - 1)(x - 2)}{(x - 3)(x - 4)} > 0, x \neq 3\right\}$ is
A.(16)/(3)$$\frac{16}{3}$$
B.(64)/(3)$$\frac{64}{3}$$
C.(8)/(3)$$\frac{8}{3}$$
D.(32)/(3)$$\frac{32}{3}$$
Solution
Core Logic
Given y² ≤ 4x$y^2 \le 4x$ and x < 4$x < 4$.
Analyze the inequality (xy(x-1)(x-2))/((x-3)(x-4)) > 0$\frac{xy(x-1)(x-2)}{(x-3)(x-4)} > 0$ considering y > 0$y > 0$ and y < 0$y < 0$ separately.
Area bounded by Parabolas and Inequalities diagram for Q5 - JEE Main 2024 Morning
Step 1: Case I (y > 0)
If y > 0$y > 0$, the inequality reduces to (x(x-1)(x-2))/((x-3)(x-4)) > 0$\frac{x(x-1)(x-2)}{(x-3)(x-4)} > 0$.
Using wavy curve method and given x in (0, 4)$x \in (0, 4)$:
x in (0, 1) (2, 3)$x \in (0, 1) \cup (2, 3)$.
Step 2: Case II (y < 0)
If y < 0$y < 0$, the inequality reduces to (x(x-1)(x-2))/((x-3)(x-4)) < 0$\frac{x(x-1)(x-2)}{(x-3)(x-4)} < 0$.
Using wavy curve method and given x in (0, 4)$x \in (0, 4)$:
x in (1, 2) (3, 4)$x \in (1, 2) \cup (3, 4)$.
Step 3: Area Computation
Because the regions map perfectly without overlap in opposite quadrants relative to the x-axis, they form complete parabolic strips when combined:
Area = 2 ∫₀⁴ √(x) dx = 2 · (2)/(3)[x3/2]₀⁴ = (4)/(3) · 8 = (32)/(3)$= 2 \int_{0}^{4} \sqrt{x} dx = 2 \cdot \frac{2}{3}[x^{3/2}]_{0}^{4} = \frac{4}{3} \cdot 8 = \frac{32}{3}$.
Chapter Mix
Class 12 Maths: Area Under Curves
Class 11 Maths: Linear Inequalities
More Application of Integrals Questions — jee_main_2026_21_jan_evening
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
