Question

Let
\(A=\{(x,y) ∈R^2:y≥0,2x≤y≤\sqrt{4-(x-1)^2} \}\)
and
\(B=\{(x,y) ∈R\times R:0≤y≤min \{2x,\sqrt{4-(x-1)^2}\}\}\)
Then the ratio of the area of A to the area of B is

• A. π/π-1
• B. π+1/π-1
• C. π-1/π+1
• D. π/π+1

Answer 1

The Correct Option is C

Step 1: Analyzing the regions.
We are given two areas \( A \) and \( B \). The region \( A \) is bounded by the curve \( y = \sqrt{4 - (x - 1)^2} \), which represents a semicircle with center at \( (1, 0) \) and radius 2. The region \( B \) is bounded by the line \( y = 2x \) and the same semicircle.
The area \( A \) corresponds to the area of the semicircle, and the area \( B \) consists of the area of a sector of the circle minus the area of the triangle formed by the points \( A(1, 0) \), \( B(1, 2) \), and the origin \( O(0, 0) \).
Step 2: Area of A.
The area of the semicircle is: \[ A = \frac{\pi r^2}{2} = \frac{\pi \cdot 4}{2} = 2\pi. \] Thus, the area of \( A \) is \( 2\pi - 1 \), as the area of the triangular portion is subtracted, where \( 1 \) represents the area of the right-angled triangle. \[ A = \pi - 1. \] Step 3: Area of B. The area of \( B \) consists of the area of the sector of the circle minus the area of the triangle: \[ \text{Area of B} = \text{Area of sector OAB} + \text{Area of arc OBC}. \] The area of the triangle is: \[ \text{Area of triangle OAB} = \frac{1}{2} \times 1 \times 2 = 1. \] The area of the sector is: \[ \text{Area of sector OAB} = \frac{\pi}{4} \times (2)^2 = \frac{\pi}{2}. \] Thus, the area of \( B \) is: \[ B = 1 + \frac{\pi}{2} = \pi + 1. \] Step 4: Ratio of the areas. The ratio of the area of \( A \) to the area of \( B \) is: \[ \frac{A}{B} = \frac{\pi - 1}{\pi + 1}. \] Thus, the correct answer is option (3).