JEE MainMathematics

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Integral Calculus: JEE Main Mathematics Question with Solution

Q: Does the rhombus lie strictly inside the ellipse?

Yes, the vertices of the rhombus are and . For the ellipse, the intercepts are and . The straight segments connect points on the axes that lie on or strictly inside the ellipse.

The area of the region, inside the ellipse

x^2 + 4y^2 = 4

and outside the region bounded by the curves

y = |x| - 1

and

y = 1 - |x|

, is :

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2(\pi - 1)

2\pi - \frac{1}{2}

3(\pi - 1)

2\pi - 1

Step-by-step solutionView

Correct answer

The total area is obtained by subtracting the area of the square/rhombus bounded by $y = \pm(1 - |x|)$ (which is $2$ ) from the area of the ellipse $\frac{x^2}{4} + \frac{y^2}{1} = 1$ (which is $2\pi$ ), giving $2\pi - 2 = 2(\pi - 1)$ .

Option analysis

Why each option works or fails

A ·

2(\pi - 1)

None. This is the correct calculation. The area of the ellipse is $\pi ab = \pi(2)(1) = 2\pi$ , and the area bounded by the lines is a square of area $2$ , resulting in $2\pi - 2 = 2(\pi - 1)$ .

B ·

2\pi - \frac{1}{2}

The student computed the area of only one quadrant of the bounded diamond region (area $\frac{1}{2}$ ) and subtracted it from the total ellipse area. Multiply the single-quadrant triangle area $\frac{1}{2}$ by $4$ to get the full rhombus area of $2$ before subtracting.

C ·

3(\pi - 1)

The student misidentified the semi-major axis as $a = 3$ or miscalculated the ellipse area as $3\pi$ . Rewrite $x^2 + 4y^2 = 4$ in standard form $\frac{x^2}{2^2} + \frac{y^2}{1^2} = 1$ , giving semi-axes $a = 2$ and $b = 1$ , so the ellipse area is $2\pi$ .

D ·

2\pi - 1

The student subtracted only half of the inner diamond's area ( $1$ instead of $2$ ) from the ellipse area. The four triangular parts forming the diamond each have area $\frac{1}{2}$ , yielding a total inner area of $4 \times \frac{1}{2} = 2$ .

Reviewed route

Solution

StepWorking

01given

Given the ellipse $x^2 + 4y^2 = 4 \implies \frac{x^2}{4} + \frac{y^2}{1} = 1$ and the inner region bounded by $y = |x| - 1$ and $y = 1 - |x|$ .

02goal

Find the area of the region inside the ellipse and outside the region bounded by $y = |x| - 1$ and $y = 1 - |x|$ .

03approach

Notice that the region bounded by $y = 1 - |x|$ and $y = |x| - 1$ is a square/rhombus with vertices at $(1, 0)$ , $(0, 1)$ , $(-1, 0)$ , and $(0, -1)$ , which lies entirely inside the ellipse (semi-axes $a=2, b=1$ ). The required area is $\text{Area}(\text{ellipse}) - \text{Area}(\text{rhombus})$ .

04execute

Area of ellipse $= \pi a b = \pi (2)(1) = 2\pi$ . The inner shape consists of 4 right triangles of base $1$ and height $1$ , so its area $= 4 \times \left(\frac{1}{2} \times 1 \times 1\right) = 2$ . Thus, required area $= 2\pi - 2 = 2(\pi - 1)$ .

✓verify

Area of ellipse is $2\pi \approx 6.28$ . Inner square has diagonal length 2 along both axes, area $= \frac{1}{2} d_1 d_2 = \frac{1}{2}(2)(2) = 2$ . $2\pi - 2 = 2(\pi - 1) \approx 4.28 > 0$ , perfectly consistent.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 7 September 2026

Quick checks

Students also ask

Does the rhombus lie strictly inside the ellipse?

Yes, the vertices of the rhombus are $(\pm 1, 0)$ and $(0, \pm 1)$ . For the ellipse, the intercepts are $(\pm 2, 0)$ and $(0, \pm 1)$ . The straight segments connect points on the axes that lie on or strictly inside the ellipse.

Answer

The total area is obtained by subtracting the area of the square/rhombus bounded by $y = \pm(1 - |x|)$ (which is $2$ ) from the area of the ellipse $\frac{x^2}{4} + \frac{y^2}{1} = 1$ (which is $2\pi$ ), giving $2\pi - 2 = 2(\pi - 1)$ .

Why each option works or fails

A · correct: $2(\pi - 1)$ - None. This is the correct calculation. The area of the ellipse is $\pi ab = \pi(2)(1) = 2\pi$ , and the area bounded by the lines is a square of area $2$ , resulting in $2\pi - 2 = 2(\pi - 1)$ .
B: $2\pi - \frac{1}{2}$ - The student computed the area of only one quadrant of the bounded diamond region (area $\frac{1}{2}$ ) and subtracted it from the total ellipse area. Multiply the single-quadrant triangle area $\frac{1}{2}$ by $4$ to get the full rhombus area of $2$ before subtracting.
C: $3(\pi - 1)$ - The student misidentified the semi-major axis as $a = 3$ or miscalculated the ellipse area as $3\pi$ . Rewrite $x^2 + 4y^2 = 4$ in standard form $\frac{x^2}{2^2} + \frac{y^2}{1^2} = 1$ , giving semi-axes $a = 2$ and $b = 1$ , so the ellipse area is $2\pi$ .
D: $2\pi - 1$ - The student subtracted only half of the inner diamond's area ( $1$ instead of $2$ ) from the ellipse area. The four triangular parts forming the diamond each have area $\frac{1}{2}$ , yielding a total inner area of $4 \times \frac{1}{2} = 2$ .

Step-by-step solution

given: Given the ellipse $x^2 + 4y^2 = 4 \implies \frac{x^2}{4} + \frac{y^2}{1} = 1$ and the inner region bounded by $y = |x| - 1$ and $y = 1 - |x|$ .
goal: Find the area of the region inside the ellipse and outside the region bounded by $y = |x| - 1$ and $y = 1 - |x|$ .
approach: Notice that the region bounded by $y = 1 - |x|$ and $y = |x| - 1$ is a square/rhombus with vertices at $(1, 0)$ , $(0, 1)$ , $(-1, 0)$ , and $(0, -1)$ , which lies entirely inside the ellipse (semi-axes $a=2, b=1$ ). The required area is $\text{Area}(\text{ellipse}) - \text{Area}(\text{rhombus})$ .
execute: Area of ellipse $= \pi a b = \pi (2)(1) = 2\pi$ . The inner shape consists of 4 right triangles of base $1$ and height $1$ , so its area $= 4 \times \left(\frac{1}{2} \times 1 \times 1\right) = 2$ . Thus, required area $= 2\pi - 2 = 2(\pi - 1)$ .
verify: Area of ellipse is $2\pi \approx 6.28$ . Inner square has diagonal length 2 along both axes, area $= \frac{1}{2} d_1 d_2 = \frac{1}{2}(2)(2) = 2$ . $2\pi - 2 = 2(\pi - 1) \approx 4.28 > 0$ , perfectly consistent.