Let the line x = -1 divide the area of the region (x,y):1+x²≤ y≤3-x in the ratio m:n, gcd (m,n)=1. Then m+n is equal to

Solution & Explanation

Related Formula
Area = ∫ₐb (yupper - ylower) dx
Core Logic

First, find the points of intersection for the curves y = 1 + x² and y = 3 - x:

1 + x² = 3 - x

x² + x - 2 = 0

(x + 2)(x - 1) = 0 x = -2, x = 1

So the total region is bounded between x = -2 and x = 1. The line x = -1 divides this region into two parts.

Area Between Two Curves diagram for Q3 - JEE Main 2026 Morning
Area Between Two Curves diagram for Q3 - JEE Main 2026 Morning

Step 1: Setting up the Areas

Let the area to the left of x = -1 be proportional to m and the area to the right be proportional to n.

Am = ∫₋₂⁻¹ [(3 - x) - (1 + x²)] dx Aₙ = ∫₋₁¹ [(3 - x) - (1 + x²)] dx

The integrand simplifies to 2 - x - x².

Area Between Two Curves diagram for Q3 - JEE Main 2026 Morning
Area Between Two Curves diagram for Q3 - JEE Main 2026 Morning

Step 2: Integration
∫ (2 - x - x²) dx = 2x - (x²)/(2) - (x³)/(3)

Evaluate Am:

Am = [2x - (x²)/(2) - (x³)/(3)]₋₂⁻¹ Am = (-2 - (1)/(2) + (1)/(3)) - (-4 - 2 + (8)/(3)) = (-(13)/(6)) - (-(10)/(3)) = (7)/(6)

Evaluate Aₙ:

Aₙ = [2x - (x²)/(2) - (x³)/(3)]₋₁¹ Aₙ = (2 - (1)/(2) - (1)/(3)) - (-2 - (1)/(2) + (1)/(3)) = ((7)/(6)) - (-(13)/(6)) = (20)/(6)
Step 3: Finding the Ratio

The ratio of the areas m:n is:

(m)/(n) = (Aₙ)/(Am) or (Am)/(Aₙ)

Wait, the solution designates (m)/(n) = ∫₋₁¹∫₋₂⁻¹ = (20/6)/(7/6) = (20)/(7). (Since (20,7)=1, m=20 and n=7).

Therefore, 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.

Chapter Mix

Class 12 Maths: Applications of Integrals

Reference Study Guides

More Applications of Integrals Previous-Year Questions — Page 6

Q11 jee_main_2024_31_jan_evening Area Under Curves
The area of the region enclosed by the parabola y = 4x - x² and 3y = (x - 4)² is equal to
  • A. (32)/(9)
  • B. 4
  • C. 6
  • D. (14)/(3)

Solution

Related Formula
Area = ∫ₐb (yupper - ylower) dx
Core Logic

Area Under Curves diagram for Q11 - JEE Main 2024 Evening
Area Under Curves diagram for Q11 - JEE Main 2024 Evening

Find intersection points of y = 4x - x² and 3y = (x - 4)²:

3(4x - x²) = x² - 8x + 16 12x - 3x² = x² - 8x + 16 4x² - 20x + 16 = 0 x² - 5x + 4 = 0

Roots are x = 1, 4.

Area integral:

Area = ∫₁⁴ [ (4x - x²) - ((x - 4)²)/(3) ] dx = [ (4x²)/(2) - (x³)/(3) - ((x - 4)³)/(9) ]₁⁴ = [ 2(16) - (64)/(3) - 0 ] - [ 2(1) - (1)/(3) - ((-3)³)/(9) ] = ( 32 - (64)/(3) ) - ( 2 - (1)/(3) + 3 ) = (32)/(3) - ( 5 - (1)/(3) ) = (32)/(3) - (14)/(3) = (18)/(3) = 6
Chapter Mix

Class 12 Maths: Applications of the Integrals

Q5 jee_main_2024_31_jan_morning Area 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 is
  • A. (16)/(3)
  • B. (64)/(3)
  • C. (8)/(3)
  • D. (32)/(3)

Solution

Core Logic

Given y² ≤ 4x and x < 4. Analyze the inequality (xy(x-1)(x-2))/((x-3)(x-4)) > 0 considering y > 0 and y < 0 separately.

Area bounded by Parabolas and Inequalities diagram for Q5 - JEE Main 2024 Morning
Area bounded by Parabolas and Inequalities diagram for Q5 - JEE Main 2024 Morning

Step 1: Case I (y > 0)

If y > 0, the inequality reduces to (x(x-1)(x-2))/((x-3)(x-4)) > 0. Using wavy curve method and given x in (0, 4): x in (0, 1) (2, 3).

Step 2: Case II (y < 0)

If y < 0, the inequality reduces to (x(x-1)(x-2))/((x-3)(x-4)) < 0. Using wavy curve method and given x in (0, 4): x in (1, 2) (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).

Chapter Mix

Class 12 Maths: Area Under Curves Class 11 Maths: Linear Inequalities

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JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)