JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Area Enclosed Between a Parabola and an Absolute Value Function

Q: Why does substituting X = x+2 not change the value of the area?

A substitution X = x + c is a rigid horizontal translation of the entire region, and rigid translations preserve area (dX = dx).

The area of the region {(x,y):x^2+4x+2≤ y≤ |x+2|} is equal to

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A7B24/5C20/3D5

Step-by-step solutionView

Correct answer

By shifting coordinates with $t = x + 2$ , the region becomes symmetric with bounding curves $y = t^{2} - 2$ and $y = ∣ t ∣$ , yielding an area of $20 / 3$ .

Option analysis

Why each option works or fails

A · 7

Integrating only one half of the symmetric region and incorrectly doubling or miscalculating $\int_{0}^{2} (∣ t ∣ - (t^{2} - 2)) d t$ as $7 / 2$ . Evaluate $\int_{0}^{2} (t - t^{2} + 2) d t = [\frac{t^{2}}{2} - \frac{t^{3}}{3} + 2 t]_{0}^{2} = 2 - \frac{8}{3} + 4 = \frac{10}{3}$ , then double it to get $20 / 3$ .

B · 24/5

Using an incorrect exponent or integration rule for the polynomial terms, leading to a denominator of 5. Integrate powers using $\int t^{n} d t = \frac{t^{n + 1}}{n + 1}$ , which yields denominators of 2 and 3, not 5.

C · 20/3

This is the correct option. The curves intersect at $x + 2 = \pm 2$ , and integrating the upper curve $∣ x + 2 ∣$ minus the lower parabola $(x + 2)^{2} - 2$ across $[- 4, 0]$ gives $\frac{20}{3}$ .

D · 5

Calculating the area without doubling the symmetric half, or dropping the linear term during integration. Ensure both symmetric sides on $t \in [- 2, 2]$ are included, giving $2 \times \frac{10}{3} = \frac{20}{3}$ .

Reviewed route

Solution

StepWorking

01Given

Region bounded by $y \geq x^{2} + 4 x + 2$ and $y \leq ∣ x + 2 ∣$ .

02Approach

Rewrite the parabola as $y = (x + 2)^{2} - 2$ . By shifting the origin horizontally using $X = x + 2$ , the region is bounded by $Y = X^{2} - 2$ and $Y = ∣ X ∣$ , which preserves the area and provides symmetry about the Y-axis.

03Execute

Find the points of intersection: $∣ X ∣^{2} - ∣ X ∣ - 2 = 0 ⟹ (∣ X ∣ - 2) (∣ X ∣ + 1) = 0 ⟹ ∣ X ∣ = 2 ⟹ X = \pm 2$ .

04Execute

Set up the area integral using even symmetry: $Area = 2 \int_{0}^{2} (X - (X^{2} - 2)) d X = 2 {[\frac{X^{2}}{2} - \frac{X^{3}}{3} + 2 X]}_{0}^{2} = 2 (2 - \frac{8}{3} + 4) = 2 (\frac{10}{3}) = \frac{20}{3}$ .

✓Verify

The enclosing triangle with base from $X = - 2$ to $X = 2$ at $Y = 2$ has area $\frac{1}{2} \times 4 \times 2 = 4$ , and the region below $Y = 0$ between the parabola and Y=0 adds $\approx 2 \times \int_{0}^{\sqrt{2}} (2 - X^{2}) d X = \frac{8 \sqrt{2}}{3} \approx 3.77$ . Total area $\approx 4 + 3.77 - overlap = 6.67 = 20 / 3$ , which is consistent.

Hints that build this answer step by step

Using the substitution $t = x + 2$ , how do the bounding inequalities simplify?
$t^{2} - 2 \leq y \leq ∣ t ∣$
What are the points of intersection between the upper boundary $y = ∣ t ∣$ and the lower boundary $y = t^{2} - 2$ ?
$t = - 2$ and $t = 2$
What is the total enclosed area computed by integrating over $t \in [- 2, 2]$ ?
$\frac{20}{3}$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 7 September 2026

Quick checks

Students also ask

Why does substituting

X = x + 2

not change the value of the area?

A substitution $X = x + c$ is a rigid horizontal translation of the entire region, and rigid translations preserve area ( $d X = d x$ ).

Answer

By shifting coordinates with $t = x + 2$ , the region becomes symmetric with bounding curves $y = t^{2} - 2$ and $y = ∣ t ∣$ , yielding an area of $20 / 3$ .

Why each option works or fails

A: 7 - Integrating only one half of the symmetric region and incorrectly doubling or miscalculating $\int_{0}^{2} (∣ t ∣ - (t^{2} - 2)) d t$ as $7 / 2$ . Evaluate $\int_{0}^{2} (t - t^{2} + 2) d t = [\frac{t^{2}}{2} - \frac{t^{3}}{3} + 2 t]_{0}^{2} = 2 - \frac{8}{3} + 4 = \frac{10}{3}$ , then double it to get $20 / 3$ .
B: 24/5 - Using an incorrect exponent or integration rule for the polynomial terms, leading to a denominator of 5. Integrate powers using $\int t^{n} d t = \frac{t^{n + 1}}{n + 1}$ , which yields denominators of 2 and 3, not 5.
C · correct: 20/3 - This is the correct option. The curves intersect at $x + 2 = \pm 2$ , and integrating the upper curve $∣ x + 2 ∣$ minus the lower parabola $(x + 2)^{2} - 2$ across $[- 4, 0]$ gives $\frac{20}{3}$ .
D: 5 - Calculating the area without doubling the symmetric half, or dropping the linear term during integration. Ensure both symmetric sides on $t \in [- 2, 2]$ are included, giving $2 \times \frac{10}{3} = \frac{20}{3}$ .

Step-by-step solution

Given: Region bounded by $y \geq x^{2} + 4 x + 2$ and $y \leq ∣ x + 2 ∣$ .
Approach: Rewrite the parabola as $y = (x + 2)^{2} - 2$ . By shifting the origin horizontally using $X = x + 2$ , the region is bounded by $Y = X^{2} - 2$ and $Y = ∣ X ∣$ , which preserves the area and provides symmetry about the Y-axis.
Execute: Find the points of intersection: $∣ X ∣^{2} - ∣ X ∣ - 2 = 0 ⟹ (∣ X ∣ - 2) (∣ X ∣ + 1) = 0 ⟹ ∣ X ∣ = 2 ⟹ X = \pm 2$ .
Execute: Set up the area integral using even symmetry: $Area = 2 \int_{0}^{2} (X - (X^{2} - 2)) d X = 2 {[\frac{X^{2}}{2} - \frac{X^{3}}{3} + 2 X]}_{0}^{2} = 2 (2 - \frac{8}{3} + 4) = 2 (\frac{10}{3}) = \frac{20}{3}$ .
Verify: The enclosing triangle with base from $X = - 2$ to $X = 2$ at $Y = 2$ has area $\frac{1}{2} \times 4 \times 2 = 4$ , and the region below $Y = 0$ between the parabola and Y=0 adds $\approx 2 \times \int_{0}^{\sqrt{2}} (2 - X^{2}) d X = \frac{8 \sqrt{2}}{3} \approx 3.77$ . Total area $\approx 4 + 3.77 - overlap = 6.67 = 20 / 3$ , which is consistent.

Area Enclosed Between a Parabola and an Absolute Value Function

What feels right?

By shifting coordinates with $t = x + 2$ , the region becomes symmetric with bounding curves $y = t^{2} - 2$ and $y = ∣ t ∣$ , yielding an area of $20 / 3$ .

Why each option works or fails

Solution

01Given

02Approach

03Execute

04Execute

✓Verify

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