Match List-I with List-II
| List-I (Region) | List-II (Area in sq. units) | ||
| (A) | Bounded by y = 2x + 1, the x-axis and the lines x = 0 and x = 2 | (I) | 9 |
| (B) | Bounded by y = x − 2, the x-axis and the lines x = 0 and x = 4 | (II) | \(\displaystyle\frac{9}{2}\) |
| (C) | Bounded by the line x + y = 3 and the coordinate axes | (III) | 4 |
| (D) | Bounded by y = |x| and the line y = 3 | (IV) | 6 |
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)-(II), (D)-(I)
(A) Bounded by y = 2x + 1, the x-axis and the lines x = 0 and x = 2
This matches (IV) 6.
The line is above the x-axis for \(0 \le x \le 2\):
\( \displaystyle \int_{0}^{2}(2x+1)\,dx = \left[x^{2}+x\right]_{0}^{2} = 6 \)
(B) Bounded by y = x − 2, the x-axis and the lines x = 0 and x = 4
This matches (III) 4.
The line crosses the x-axis at x = 2, so the two parts are taken separately and the part below the axis with its absolute value:
\( \displaystyle \int_{0}^{2}(x-2)\,dx = \left[\frac{x^{2}}{2}-2x\right]_{0}^{2} = -2 \)
\( \displaystyle \int_{2}^{4}(x-2)\,dx = \left[\frac{x^{2}}{2}-2x\right]_{2}^{4} = 0-(-2) = 2 \)
\(\text{Area} = |-2| + 2 = 4\)
(C) Bounded by the line x + y = 3 and the coordinate axes
This matches (II) \(\frac{9}{2}\).
The line meets the axes at (3, 0) and (0, 3), so the region is a right triangle:
\( \displaystyle \int_{0}^{3}(3-x)\,dx = \left[3x-\frac{x^{2}}{2}\right]_{0}^{3} = \frac{9}{2} \)
(D) Bounded by y = |x| and the line y = 3
This matches (I) 9.
The line y = 3 meets y = |x| at x = −3 and x = 3, and it is above y = |x| between them. Since |x| is even:
\( \displaystyle \int_{-3}^{3}(3-|x|)\,dx = 2\int_{0}^{3}(3-x)\,dx = 2\times\frac{9}{2} = 9 \)