Match List-I with List-II
| List-I (Region) | List-II (Area in sq. units) | ||
| (A) | Region bounded by the curve \(y = 2x-x^{2}\) and the x-axis | (I) | \(\displaystyle\frac{9\pi}{4}\) |
| (B) | Region bounded by the curve \(y = 2\sqrt{x}\), the x-axis and the lines x = 0 and x = 9 | (II) | 8 |
| (C) | Region in the first quadrant bounded by the circle \(x^{2}+y^{2}=9\) and the coordinate axes | (III) | \(\displaystyle\frac{4}{3}\) |
| (D) | Region bounded by the line y = 4x, the x-axis and the lines x = 0 and x = 2 | (IV) | 36 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
(A) Region bounded by the curve \(y = 2x-x^{2}\) and the x-axis
This matches (III) \(\frac{4}{3}\).
Since \(2x-x^{2} = x(2-x)\), the curve meets the x-axis at x = 0 and x = 2 and is above it in between:
\( \displaystyle A = \int_{0}^{2}(2x-x^{2})\,dx = \left[x^{2}-\frac{x^{3}}{3}\right]_{0}^{2} \)
\( \displaystyle = 4-\frac{8}{3} = \frac{4}{3} \)
(B) Region bounded by the curve \(y = 2\sqrt{x}\), the x-axis and the lines x = 0 and x = 9
This matches (IV) 36.
\( \displaystyle A = \int_{0}^{9}2x^{1/2}\,dx = \left[\frac{4}{3}x^{3/2}\right]_{0}^{9} \)
\( \displaystyle = \frac{4}{3}\times27 = 36 \)
(C) Region in the first quadrant bounded by the circle \(x^{2}+y^{2}=9\) and the coordinate axes
This matches (I) \(\frac{9\pi}{4}\).
In the first quadrant \(y = \sqrt{9-x^{2}}\), and the region is a quarter of the circle:
\( \displaystyle A = \int_{0}^{3}\sqrt{9-x^{2}}\,dx = \left[\frac{x}{2}\sqrt{9-x^{2}}+\frac{9}{2}\sin^{-1}\frac{x}{3}\right]_{0}^{3} \)
\( \displaystyle = \frac{9}{2}\times\frac{\pi}{2} = \frac{9\pi}{4} \)
(D) Region bounded by the line y = 4x, the x-axis and the lines x = 0 and x = 2
This matches (II) 8.
\( \displaystyle A = \int_{0}^{2}4x\,dx = \left[2x^{2}\right]_{0}^{2} = 8 \)