Match List-I with List-II
| List-I (Region) | List-II (Area in sq. units) | ||
| (A) | Region bounded by the line y = x + 2, the x-axis and the lines x = −2 and x = 1 | (I) | 4 |
| (B) | Region bounded by the curve \(y = \sqrt{4-x^{2}}\) and the x-axis | (II) | \(\displaystyle\frac{9}{2}\) |
| (C) | Region bounded by the lines y = x, x + y = 4 and the y-axis | (III) | \(3\pi\) |
| (D) | Region in the second quadrant bounded by the circle \(x^{2}+y^{2}=12\) and the coordinate axes | (IV) | \(2\pi\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
(A) Region bounded by the line y = x + 2, the x-axis and the lines x = −2 and x = 1
This matches (II) \(\displaystyle\frac{9}{2}\).
The line meets the x-axis at x = −2 and is above it up to x = 1:
\( \displaystyle \int_{-2}^{1}(x+2)\,dx = \left[\frac{x^{2}}{2}+2x\right]_{-2}^{1} \)
\( \displaystyle = \frac{5}{2}-(-2) = \frac{9}{2} \)
(B) Region bounded by the curve \(y = \sqrt{4-x^{2}}\) and the x-axis
This matches (IV) \(2\pi\).
The curve is the upper half of the circle \(x^{2}+y^{2}=4\), so the region is a half-disc of radius 2:
\( \displaystyle \frac{1}{2}\pi(2)^{2} = 2\pi \)
(C) Region bounded by the lines y = x, x + y = 4 and the y-axis
This matches (I) 4.
The lines meet at (2, 2). For \(0 \le x \le 2\) the top is y = 4 − x and the bottom is y = x:
\( \displaystyle \int_{0}^{2}\left((4-x)-x\right)dx = \left[4x-x^{2}\right]_{0}^{2} = 4 \)
(D) Region in the second quadrant bounded by the circle \(x^{2}+y^{2}=12\) and the coordinate axes
This matches (III) \(3\pi\).
It is one quarter of the disc with \(r^{2} = 12\):
\( \displaystyle \frac{1}{4}\pi\times12 = 3\pi \)