JEE 2022 — Mathematics PYQ

Q: How do I solve JEE 2022 Mathematics questions?

Practice JEE 2022 Mathematics previous year questions with step-by-step solutions on Mathem Solvex. Each question includes a detailed explanation and video solution to help you master the concept quickly.

Q: What is the difficulty level of JEE Mathematics questions?

JEE Mathematics questions range from moderate to advanced difficulty. Consistent practice with official previous year papers and topic-wise drills on Mathem Solvex is the most effective preparation strategy.

Q: Are JEE previous year papers available for free?

Yes. Mathem Solvex provides free access to JEE previous year question papers and topic-wise PDFs. You can download them or attempt them as timed live mock tests directly on the platform.

Question

JEE 2022 — Mathematics PYQ

JEE | Mathematics | 2022

For real number $a, b (a > b > 0)$ , let: $\text{Area } \{(x, y) : x^2 + y^2 \le a^2 \text{ and } \frac{x^2}{a^2} + \frac{y^2}{b^2} \ge 1\} = 30\pi$ $\text{Area } \{(x, y) : x^2 + y^2 \ge b^2 \text{ and } \frac{x^2}{a^2} + \frac{y^2}{b^2} \le 1\} = 18\pi$ Then the value of $(a - b)^2$ is equal to ________.

Choose the correct answer:

Explanation

Solution

1. Understand the Areas

Total Circle Area: $\pi a^2$
Total Ellipse Area: $\pi ab$
Total Smaller Circle Area: $\pi b^2$

2. Formulate Equations

The first area is the Circle minus the Ellipse (where they overlap): $\pi a^2 - \pi ab = 30\pi \implies a^2 - ab = 30$ .

The second area is the Ellipse minus the small Circle: $\pi ab - \pi b^2 = 18\pi \implies ab - b^2 = 18$ .

3. Solve for $a$ and $b$

Adding the two equations: $a^2 - b^2 = 48 \implies (a-b)(a+b) = 48$ .

Subtracting the second from the first: $a^2 - 2ab + b^2 = 30 - 18 = 12$ .

The expression $a^2 - 2ab + b^2$ is exactly $(a-b)^2$ .

Result: 12.

Choose the correct answer:

Explore More