JEE Main 2024 (27 Jan Shift 1)

LEVELJEE Advanced

Animated Solution for Mathematics - Definite Integration: Let the area of the region ${(x, y) : x - 2 y + 4 \geq 0, x + 2 y^{2} \geq 0, x + 4 y^{2} \leq 8, y \geq 0}$ be $\frac{m}{n}$ , where $m$ and $n$ are coprime numbers. Then $m + n$ is equal to

Q: Is there a trick to avoid the negative sign in the area?

The area must always be positive. The formula $A = \int (x_{right} - x_{left}) dy$ ensures this because $x_{right} \ge x_{left}$ within the region. If you ever get a negative result, it usually means you swapped the boundaries or made a sign error in your subtraction. Always double-check your sketch!

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the region in the

x y

-plane. You have a line

x = 2 y - 4

and two parabolas

x = - 2 y^{2}

and

x = 8 - 4 y^{2}

. Since

y \geq 0

, you are restricted to the upper half-plane. The region is bounded by these three curves. Sketching them will reveal that the 'left' boundary of the region changes from one parabola to the line at a specific

y

-value.

Hint 2

Hint 2 (The Arsenal): To find the area, you need to integrate with respect to

y

because the boundaries are given as functions of

y

. The area is

A = \int_{y_{1}}^{y_{2}} (x_{r i g h t} - x_{l e f t}) d y

. You must find the intersection points of the curves to determine the limits of integration and where the boundary functions switch.

Hint 3

Hint 3 (The Execution): The intersection of

x = - 2 y^{2}

and

x = 2 y - 4

gives

y = 1

. The intersection of

x = 8 - 4 y^{2}

and

x = 2 y - 4

gives

y = 3/2

. Split the integral into two parts:

A_{1} = \int_{0}^{1} ((8 - 4 y^{2}) - (- 2 y^{2})) d y

and

A_{2} = \int_{1}^{3/2} ((8 - 4 y^{2}) - (2 y - 4)) d y

. Sum these to find the total area.

Visualized Solution

Analyzing the Constraints

Given region: ${(x, y) : x - 2 y + 4 \geq 0, x + 2 y^{2} \geq 0, x + 4 y^{2} \leq 8, y \geq 0}$
Boundary curves:
1. Line: $x = 2 y - 4$
2. Left Parabola: $x = - 2 y^{2}$
3. Right Parabola: $x = 8 - 4 y^{2}$
4. Horizontal boundary: $y = 0$

Visualizing the Boundaries

Plotting the curves:
Red curve: $x = - 2 y^{2}$
Green curve: $x = 8 - 4 y^{2}$
Blue line: $x = 2 y - 4$
The region is in the upper half plane ( $y \geq 0$ ).

Finding Intersection Points

Intersection of $x = - 2 y^{2}$ and $x = 2 y - 4$ :
$- 2 y^{2} = 2 y - 4 ⟹ y^{2} + y - 2 = 0 ⟹ y = 1$
Point: $(- 2, 1)$
Intersection of $x = 8 - 4 y^{2}$ and $x = 2 y - 4$ :
$8 - 4 y^{2} = 2 y - 4 ⟹ 2 y^{2} + y - 6 = 0 ⟹ y = \frac{3}{2}$
Point: $(- 1, \frac{3}{2})$

Strategy: Horizontal Integration

Area $A = \int_{y_{1}}^{y_{2}} (x_{r i g h t} - x_{l e f t}) d y$
Region 1: $y \in [0, 1]$ , $x_{l e f t} = - 2 y^{2}$ , $x_{r i g h t} = 8 - 4 y^{2}$
Region 2: $y \in [1, \frac{3}{2}]$ , $x_{l e f t} = 2 y - 4$ , $x_{r i g h t} = 8 - 4 y^{2}$

Setting up Integral $A_{1}$

$A_{1} = \int_{0}^{1} [(8 - 4 y^{2}) - (- 2 y^{2})] d y$
Simplifying: $A_{1} = \int_{0}^{1} (8 - 2 y^{2}) d y$

Evaluating Integral $A_{1}$

$A_{1} = [8 y - \frac{2 y ^{3}}{3}]_{0}^{1}$
Substituting limits: $A_{1} = (8 - \frac{2}{3}) - 0$
Calculation: $A_{1} = \frac{24 - 2}{3} = \frac{22}{3}$

Setting up Integral $A_{2}$

$A_{2} = \int_{1}^{3/2} [(8 - 4 y^{2}) - (2 y - 4)] d y$
Simplifying: $A_{2} = \int_{1}^{3/2} (12 - 2 y - 4 y^{2}) d y$

Evaluating Integral $A_{2}$

$A_{2} = [12 y - y^{2} - \frac{4 y ^{3}}{3}]_{1}^{3/2}$
At $y = \frac{3}{2}$ : $12 (\frac{3}{2}) - (\frac{3}{2})^{2} - \frac{4}{3} (\frac{3}{2})^{3} = 18 - \frac{9}{4} - \frac{9}{2} = \frac{45}{4}$
At $y = 1$ : $12 (1) - (1)^{2} - \frac{4}{3} (1)^{3} = 11 - \frac{4}{3} = \frac{29}{3}$
$A_{2} = \frac{45}{4} - \frac{29}{3} = \frac{135 - 116}{12} = \frac{19}{12}$

Total Area Calculation

Total Area $A = A_{1} + A_{2}$
Summing: $A = \frac{22}{3} + \frac{19}{12}$
$A = \frac{88 + 19}{12} = \frac{107}{12}$

Final Answer

Comparing with $\frac{m}{n}$ : $m = 107, n = 12$
$m + n = 107 + 12 = 119$
Final Answer: 119

00:00 / 00:00

The Sigma Insight: Area Bounded by Curves

Solution Diagram

Analyzing the Setup

The region is defined by the inequalities

x - 2 y + 4 \geq 0

x + 2 y^{2} \geq 0

x + 4 y^{2} \leq 8

, and

y \geq 0

. These translate to the following boundary curves:

x = 2 y - 4

(a line),

x = - 2 y^{2}

(a parabola), and

x = 8 - 4 y^{2}

(a parabola).

The condition

y \geq 0

restricts our focus to the upper half of the Cartesian plane.

The Meeting Points

To determine the limits of integration, we find the intersection points of these curves.

Equating the red parabola

x = - 2 y^{2}

and the blue line

x = 2 y - 4

- 2 y^{2} = 2 y - 4 \Rightarrow y^{2} + y - 2 = 0 \Rightarrow (y + 2) (y - 1) = 0

Since

y \geq 0

, we find the intersection at

y = 1

Equating the green parabola

x = 8 - 4 y^{2}

and the blue line

x = 2 y - 4

8 - 4 y^{2} = 2 y - 4 \Rightarrow 4 y^{2} + 2 y - 12 = 0 \Rightarrow 2 y^{2} + y - 6 = 0 \Rightarrow (2 y - 3) (y + 2) = 0

This yields the intersection at

y = 3/2

The Strategy of Division

The left boundary of the region changes at

y = 1

. We must split the total area into two distinct parts,

A_{1}

and

A_{2}

For

A_{1}

(where

0 \leq y \leq 1

), the right boundary is

x = 8 - 4 y^{2}

and the left boundary is

x = - 2 y^{2}

A_{1} = \int_{0}^{1} ((8 - 4 y^{2}) - (- 2 y^{2})) d y = \int_{0}^{1} (8 - 2 y^{2}) d y

Evaluating this integral:

A_{1} = [8 y - \frac{2 y ^{3}}{3}]_{0}^{1} = 8 - \frac{2}{3} = \frac{22}{3}

For

A_{2}

(where

1 \leq y \leq 3/2

), the right boundary remains

x = 8 - 4 y^{2}

, but the left boundary is now the line

x = 2 y - 4

A_{2} = \int_{1}^{3/2} ((8 - 4 y^{2}) - (2 y - 4)) d y = \int_{1}^{3/2} (12 - 2 y - 4 y^{2}) d y

Evaluating this integral:

A_{2} = [12 y - y^{2} - \frac{4 y ^{3}}{3}]_{1}^{3/2} = (18 - \frac{9}{4} - \frac{4 ( 27/8 )}{3}) - (12 - 1 - \frac{4}{3}) = \frac{57}{4} - \frac{29}{3} = \frac{19}{12}

Final Calculation

The total area

A

is the sum of the two parts:

A = A_{1} + A_{2} = \frac{22}{3} + \frac{19}{12} = \frac{88 + 19}{12} = \frac{107}{12}

Given that

m / n = 107/12

and

107

and

12

are coprime, we identify

m = 107

and

n = 12

The final result is $m + n = 119$ .

Common Traps & FAQs

Wait, why do we integrate with respect to

y

instead of

x

Great question! While you *could* integrate with respect to

x

, the boundaries are explicitly given as

x

in terms of

y

(

x = f (y)

). If you tried to integrate with respect to

x

, you would have to solve for

y

in terms of

x

, which involves square roots and splitting the region into many more complex parts. Integrating with respect to

y

is much cleaner and more direct.

How do I know which curve is the 'right' boundary and which is the 'left'?

A simple way is to test a point within the region. Alternatively, you can sketch the curves. For a given

y

, the curve with the larger

x

-value is the right boundary. In our case,

x = 8 - 4 y^{2}

is always the rightmost curve, while the left boundary switches from

x = - 2 y^{2}

x = 2 y - 4

y = 1

Is there a trick to avoid the negative sign in the area?

The area must always be positive. The formula

A = \int (x_{r i g h t} - x_{l e f t}) d y

ensures this because

x_{r i g h t} \geq x_{l e f t}

within the region. If you ever get a negative result, it usually means you swapped the boundaries or made a sign error in your subtraction. Always double-check your sketch!

Animated Solution for Mathematics - Definite Integration: Let the area of the region ${(x, y) : x - 2 y + 4 \geq 0, x + 2 y^{2} \geq 0, x + 4 y^{2} \leq 8, y \geq 0}$ be $\frac{m}{n}$ , where $m$ and $n$ are coprime numbers. Then $m + n$ is equal to

Enter Numerical Value:

Visualized Solution

Analyzing the Constraints

Visualizing the Boundaries

Finding Intersection Points

Strategy: Horizontal Integration

Setting up Integral $A_{1}$

Evaluating Integral $A_{1}$

Setting up Integral $A_{2}$

Evaluating Integral $A_{2}$

Total Area Calculation

Final Answer

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Meeting Points

The Strategy of Division

Final Calculation

Similar Questions

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Let the line $x = - 1$ divide the area of the region ${(x, y) : 1 + x^{2} \leq y \leq 3 - x}$ in the ratio $m : n, gcd (m, n) = 1$ . Then $m + n$ is equal to

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Enter Numerical Value:

Visualized Solution

Analyzing the Constraints

Visualizing the Boundaries

Finding Intersection Points

Strategy: Horizontal Integration

Setting up Integral .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A1​

Evaluating Integral .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A1​

Setting up Integral .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2​

Evaluating Integral .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2​

Total Area Calculation

Final Answer

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Meeting Points

The Strategy of Division

Final Calculation

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