Question:

Step 1: Understand the definitions of regions A and B and their geometric interpretation. The region for $y \leq \sqrt{4-(x-1)^2}$ with $y \geq 0$ describes the upper semi-disk of a circle centered at $(1,0)$ with radius $R=2$. Let this semi-disk be $S_{upper}$. Its area is $\frac{1}{2}\pi R^2 = \frac{1}{2}\pi(2^2) = 2\pi$. The line $y=2x$ passes through the origin $(0,0)$. The intersection points of $y=2x$ and the circle $(x-1)^2+y^2=4$ are found by substituting $y=2x$: $$(x-1)^2 + (2x)^2 = 4$$ $$x^2 - 2x + 1 + 4x^2 = 4$$ $$5x^2 - 2x - 3 = 0$$ $$(5x+3)(x-1) = 0$$ This gives $x=1$ or $x=-3/5$. The corresponding $y$ values are $y=2(1)=2$ and $y=2(-3/5)=-6/5$. So the intersection points are $(1,2)$ and $(-3/5, -6/5)$. Region A is defined as $A=\{(x,y)\in\mathbb{R}^2:y\geq0,\,2x\leq y\leq\sqrt{4-(x-1)^2}\}$. This means: 1. $y \geq 0$ (above or on the x-axis). 2. $y \geq 2x$ (above or on the line $y=2x$). 3. $y \leq \sqrt{4-(x-1)^2}$ (below or on the upper semi-circle $(x-1)^2+y^2=4$). Thus, A is the part of the upper semi-disk $S_{upper}$ that lies above or on the line $y=2x$. Region B is defined as $B=\{(x,y)\in\mathbb{R}\times\mathbb{R}:0\leq y\leq\min\{2x,\sqrt{4-(x-1)^2}\}\}$. This means: 1. $y \geq 0$ (above or on the x-axis). 2. $y \leq 2x$ (below or on the line $y=2x$). 3. $y \leq \sqrt{4-(x-1)^2}$ (below or on the upper semi-circle $(x-1)^2+y^2=4$). Thus, B is the part of the upper semi-disk $S_{upper}$ that lies below or on the line $y=2x$. From these interpretations, it is clear that regions A and B partition the entire upper semi-disk $S_{upper}$, with their intersection being the line segment from $(0,0)$ to $(1,2)$, which has zero area. Therefore, Area(A) + Area(B) = Area($S_{upper}$) $= 2\pi$. Step 2: Calculate Area(B). Region B is defined by $0\leq y\leq\min\{2x,\sqrt{4-(x-1)^2}\}$. For the region to be non-empty, we must have $y \geq 0$. Since $y \leq 2x$, this implies $x \geq 0$. The semi-circle arc $y=\sqrt{4-(x-1)^2}$ extends from $x=-1$ to $x=3$. Combining with $x \geq 0$, the $x$-range for region B is $[0,3]$. We need to evaluate $\min\{2x,\sqrt{4-(x-1)^2}\}$. The two functions $y=2x$ and $y=\sqrt{4-(x-1)^2}$ intersect at $x=1$ (for $y \geq 0$). For $x \in [0,1]$: At $x=0$, $2x=0$ and $\sqrt{4-(x-1)^2}=\sqrt{3}$. So $2x \leq \sqrt{4-(x-1)^2}$. Thus $\min\{2x,\sqrt{4-(x-1)^2}\} = 2x$. For $x \in [1,3]$: At $x=2$, $2x=4$ and $\sqrt{4-(x-1)^2}=\sqrt{4-1}=\sqrt{3}$. So $\sqrt{4-(x-1)^2} \leq 2x$. Thus $\min\{2x,\sqrt{4-(x-1)^2}\} = \sqrt{4-(x-1)^2}$. Therefore, Area(B) can be calculated as two separate integrals: $$ \text{Area(B)} = \int_0^1 2x \,dx + \int_1^3 \sqrt{4-(x-1)^2} \,dx $$ Calculate the first integral: $$ \int_0^1 2x \,dx = \left[x^2\right]_0^1 = 1^2 - 0^2 = 1 $$ Calculate the second integral. Let $u=x-1$, so $du=dx$. When $x=1, u=0$. When $x=3, u=2$. $$ \int_1^3 \sqrt{4-(x-1)^2} \,dx = \int_0^2 \sqrt{4-u^2} \,du $$ This integral represents the area of a quarter circle of radius 2. Using the formula $\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$: $$ \int_0^2 \sqrt{4-u^2} \,du = \left[\frac{u}{2}\sqrt{4-u^2} + \frac{4}{2}\sin^{-1}\left(\frac{u}{2}\right)\right]_0^2 $$ $$ = \left(\frac{2}{2}\sqrt{4-2^2} + 2\sin^{-1}\left(\frac{2}{2}\right)\right) - \left(\frac{0}{2}\sqrt{4-0^2} + 2\sin^{-1}\left(\frac{0}{2}\right)\right) $$ $$ = (0 + 2\sin^{-1}(1)) - (0 + 2\sin^{-1}(0)) = 2\left(\frac{\pi}{2}\right) - 0 = \pi $$ Combining the two parts for Area(B): $$ \text{Area(B)} = 1 + \pi $$ Step 3: Calculate Area(A) using the relationship Area(A) + Area(B) = Area($S_{upper}$). As established in Step 1, Area(A) + Area(B) $= 2\pi$. Using Area(B) $= 1+\pi$ from Step 2: $$ \text{Area(A)} + (1+\pi) = 2\pi $$ $$ \text{Area(A)} = 2\pi - (1+\pi) = \pi - 1 $$ Step 4: Calculate the ratio of Area(A) to Area(B). $$ \text{Ratio} = \frac{\text{Area(A)}}{\text{Area(B)}} = \frac{\pi-1}{\pi+1} $$ The final answer is $\boxed{\dfrac{\pi-1}{\pi+1}}$. <div class="key-concept"><strong>Key Concept:</strong> Circle: $y^2+(x-1)^2=4$, centre $(1,0)$, radius 2. Line $y=2x$ intersects circle at $(1,2)$. Area A = quarter circle sector minus triangle $=\pi-1$. Area B = triangle + arc $=1+\pi=\pi+1$.</div> <div class="trap-box"><strong>Trap:</strong> A/B $=\frac{\pi-1}{\pi+1}$.</div>