Match List-I with List-II
| List-I (Region) | List-II (Area in sq. units) | ||
| (A) | \(\{(x, y): 1 \le x \le 3,\ 0 \le y \le x+2\}\) | (I) | \(\pi\) |
| (B) | \(\{(x, y): 0 \le x \le 2,\ 0 \le y \le 3x-x^{2}\}\) | (II) | \(\displaystyle\frac{5}{2}\) |
| (C) | \(\{(x, y): x^{2}+y^{2} \le 4,\ y \ge |x|\}\) | (III) | \(\displaystyle\frac{10}{3}\) |
| (D) | \(\{(x, y): 0 \le x \le 3,\ 0 \le y \le |x-1|\}\) | (IV) | 8 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
(A) \(\{(x, y): 1 \le x \le 3,\ 0 \le y \le x+2\}\)
This matches (IV) 8.
\( \displaystyle A = \int_{1}^{3}(x+2)\,dx = \left[\frac{x^{2}}{2}+2x\right]_{1}^{3} \)
\( \displaystyle = \frac{21}{2}-\frac{5}{2} = 8 \)
(B) \(\{(x, y): 0 \le x \le 2,\ 0 \le y \le 3x-x^{2}\}\)
This matches (III) \(\displaystyle\frac{10}{3}\).
Since \(3x-x^{2} = x(3-x) \ge 0\) for \(0 \le x \le 2\):
\( \displaystyle A = \int_{0}^{2}(3x-x^{2})\,dx = \left[\frac{3x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{2} \)
\( \displaystyle = 6-\frac{8}{3} = \frac{10}{3} \)
(C) \(\{(x, y): x^{2}+y^{2} \le 4,\ y \ge |x|\}\)
This matches (I) \(\pi\).
The condition \(y \ge |x|\) keeps the part above both lines y = x and y = −x. These lines are perpendicular, so inside the circle of radius 2 the region is a quarter of the disc:
\( \displaystyle A = \frac{1}{4}\times\pi\times2^{2} = \pi \)
(D) \(\{(x, y): 0 \le x \le 3,\ 0 \le y \le |x-1|\}\)
This matches (II) \(\displaystyle\frac{5}{2}\).
The modulus changes form at x = 1:
\( \displaystyle A = \int_{0}^{1}(1-x)\,dx+\int_{1}^{3}(x-1)\,dx \)
\( \displaystyle = \frac{1}{2}+2 = \frac{5}{2} \)