Match List-I with List-II
| List-I (Region) | List-II (Area in sq. units) | ||
| (A) | Region bounded by the line y = 2x, the x-axis and the lines x = 0 and x = 3 | (I) | 6 |
| (B) | Region bounded by the line x + 2y = 8 and the coordinate axes | (II) | 16 |
| (C) | Region bounded by the line y = 3x + 6, the x-axis and the lines x = −2 and x = 0 | (III) | \(\displaystyle\frac{13}{2}\) |
| (D) | Region bounded by the line y = 3 − x, the x-axis and the lines x = 0 and x = 5 | (IV) | 9 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(IV), (B)-(II), (C)-(I), (D)-(III)
(A) Region bounded by the line y = 2x, the x-axis and the lines x = 0 and x = 3
This matches (IV) 9.
\( \displaystyle A = \int_{0}^{3}2x\,dx = \left[x^{2}\right]_{0}^{3} = 9 \)
(B) Region bounded by the line x + 2y = 8 and the coordinate axes
This matches (II) 16.
The line meets the axes at (8, 0) and (0, 4), and \(y = \frac{8-x}{2}\):
\( \displaystyle A = \int_{0}^{8}\frac{8-x}{2}\,dx = \frac{1}{2}\left[8x-\frac{x^{2}}{2}\right]_{0}^{8} \)
\( \displaystyle = \frac{1}{2}(64-32) = 16 \)
(C) Region bounded by the line y = 3x + 6, the x-axis and the lines x = −2 and x = 0
This matches (I) 6.
The line is zero at x = −2 and above the x-axis on this interval:
\( \displaystyle A = \int_{-2}^{0}(3x+6)\,dx = \left[\frac{3x^{2}}{2}+6x\right]_{-2}^{0} \)
\( = 0-(6-12) = 6 \)
(D) Region bounded by the line y = 3 − x, the x-axis and the lines x = 0 and x = 5
This matches (III) \(\frac{13}{2}\).
The line crosses the x-axis at x = 3, so the part from 3 to 5 lies below the axis and is taken with its absolute value:
\( \displaystyle \int_{0}^{3}(3-x)\,dx = \left[3x-\frac{x^{2}}{2}\right]_{0}^{3} = \frac{9}{2} \)
\( \displaystyle \int_{3}^{5}(3-x)\,dx = \left(15-\frac{25}{2}\right)-\left(9-\frac{9}{2}\right) = -2 \)
\( \displaystyle A = \frac{9}{2}+|-2| = \frac{13}{2} \)