Let the area of the region (x,y):0leq xleq 3,0leq yleq x2 +2,2x + 2 be A. Then 12A is equal to

Year	2026	2025	2024	Total
Questions	9	11	7	27

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

Area A = \int f(x), g(x) dx

Core Logic

Let us find the point of intersection of the curves $y = x² + 2$ $y = x^2 + 2$ and $y = 2x + 2$ $y = 2x + 2$:

x² + 2 = 2x + 2 x² - 2x = 0 x = 0 or x = 2

Evaluating behaviors over boundaries:

For $0 \leq x \leq 2$ $0 \leq x \leq 2$: $x² + 2 \leq 2x + 2 = x² + 2$
For $2 \leq x \leq 3$ $2 \leq x \leq 3$: $2x + 2 \leq x² + 2 = 2x + 2$

Step 1: Integration Resolution

Setting up continuous area integral steps:

A = \int₀² (x² + 2) dx + \int₂³ (2x + 2) dx

A = [ (x³)/(3) + 2x ]₀² + [ x² + 2x ]₂³

A = ( (8)/(3) + 4 ) + ( (9 + 6) - (4 + 4) )

A = (20)/(3) + (15 - 8) = (20)/(3) + 7 = (41)/(3)

Area Under Curves diagram for Q25 - JEE Main 2024 Evening

Step 2: Scaling the Output Value

We need to compute $12A$ $12A$:

12A = 12 \times (41)/(3) = 4 \times 41 = 164

Pattern Recognition

For min/max boundary sets, always compute intersections first to accurately split integration domains into separate regions.

Chapter Mix

Class 12 Mathematics: Application of Integrals

More Application 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. $(32)/(9)$ $\frac{32}{9}$
B. $4$ $4$
C. $6$ $6$
D. $(14)/(3)$ $\frac{14}{3}$

Solution

Related Formula

Area = \intₐb (yupper - ylower) dx

Core Logic

Area Under Curves diagram for Q11 - JEE Main 2024 Evening

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

3(4x - x²) = x² - 8x + 16

12x - 3x² = x² - 8x + 16

4x² - 20x + 16 = 0 x² - 5x + 4 = 0

Roots are $x = 1, 4$ $x = 1, 4$.

Area integral:

Area = \int₁⁴ [ (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 \leq 4x, x < 4, (xy(x - 1)(x - 2))/((x - 3)(x - 4)) > 0, x \neq 3

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 $y² \leq 4x$ $y^2 \le 4x$ 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 $= 2 \int₀⁴ \sqrt(x) dx = 2 \cdot (2)/(3)[x3/2]₀⁴ = (4)/(3) \cdot 8 = (32)/(3)$ .

Chapter Mix

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

Let the area of the region $(x,y):0\leq x\leq 3,0\leq y\leq x² +2,2x + 2$ be $A$ $A$. Then $12A$ $12A$ is equal to

Solution & Explanation

Related Formula

Core Logic

Step 1: Integration Resolution

Step 2: Scaling the Output Value

Pattern Recognition

Chapter Mix

More Application of Integrals Previous-Year Questions — Page 6

Solution

Related Formula

Core Logic

Chapter Mix

Solution

Core Logic

Step 1: Case I (y > 0)

Step 2: Case II (y < 0)

Step 3: Area Computation

Chapter Mix

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Let the area of the region (x,y):0≤ x≤ 3,0≤ y≤ x² +2,2x + 2$\{(x,y):0\leq x\leq 3,0\leq y\leq \min \{x^2 +2,2x + 2\} \}$ be A$A$. Then 12A$12A$ is equal to

Solution & Explanation

Related Formula

Core Logic

Step 1: Integration Resolution

Step 2: Scaling the Output Value

Pattern Recognition

Chapter Mix

More Application of Integrals Previous-Year Questions — Page 6

Solution

Related Formula

Core Logic

Chapter Mix

Solution

Core Logic

Step 1: Case I (y > 0)

Step 2: Case II (y < 0)

Step 3: Area Computation

Chapter Mix

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Let the area of the region $(x,y):0\leq x\leq 3,0\leq y\leq x² +2,2x + 2$ be $A$ $A$. Then $12A$ $12A$ is equal to