The area bounded by the curves \(|x|+|y| \geq 2\) and \(x^{2}+y^{2} \leq 4\) is...... (a) \(4 \pi-4\) (b) \(4 \pi-2\) (c) \(4(\pi-2)\) (d) \(4(\pi-1)\)

Let's rewrite the absolute value functions without the absolute value signs: 1. \(|x|+|y| \geq 2 \implies x+y\geq2, -x+y\geq2, x-y\geq2, -x-y\geq2\). Each of these equations represent lines that form a square. 2. \(x^{2}+y^{2} \leq 4 \implies (x^2+y^2)=r^2\). This equation represents a circle with radius \(r\) and centered at the origin. With \(r^2=4\), the radius is \(r=2\).

Now, we need to find the intersection points between the four lines of the square and the circle. We'll find intersection points between the first line \(x+y=2\) and the circle \(x^2+y^2=4\). Substituting \(x+y-2\) into the circle equation: \((x+y-2)^2 = (x^2+y^2-4x-4y+4) = 4\) Simplifying and solving for \(y\) in terms of \(x\): \(2x+2y = 4y+4\) \(y = \frac{x}{2}\) Plugging this back into \(x+y=2\): \(x+\frac{x}{2}=2\) \(\frac{3}{2}x=2\) \(x=\frac{4}{3}\) Now we can plug this \(x\) value back into \(y = \frac{x}{2}\) to find the corresponding \(y\) value: \(y = \frac{4}{3}\cdot\frac{1}{2}\) \(y = \frac{2}{3}\) Since the other three lines are symmetric to the first one, we can find the intersection points for the other lines by rotating the point (\(\frac{4}{3}, \frac{2}{3}\)). The intersection points between the circle and the square are: 1. \((\frac{4}{3}, \frac{2}{3})\) 2. \((-\frac{2}{3}, \frac{4}{3})\) 3. \((-\frac{4}{3}, -\frac{2}{3})\) 4. \((\frac{2}{3}, -\frac{4}{3})\)

To find the area of intersection, we can divide the region into 4 sections based on the intersection points. We can then find the area of one of the sections and multiply it by 4 to get the total area. Let's find the area of the sector in the first quadrant (bounded by the x-axis, \(x+y=2\), and the circle \(x^2+y^2=4\)). Area = area of sector - area of triangle The angle at the intersection point is given by: \(\theta = \arctan\left(\frac{y}{x}\right) = \arctan\left(\frac{\frac{2}{3}}{\frac{4}{3}}\right) = \arctan(\frac{1}{2})\) Central angle of the sector = \(2\theta\) Area of sector: \(A_\text{sector} = \frac{1}{2}r^2\cdot(2\theta) = 2\cdot\left(\frac{1}{2}\right)\cdot(2)\theta = 4\theta\) Area of triangle: \(A_\text{triangle} = \frac{1}{2}(\text{base})\cdot(\text{height}) = \frac{1}{2}(2x)(y) = \frac{1}{2}(2\cdot\frac{4}{3})(\frac{2}{3})\) \(A_\text{triangle} = \frac{4}{3}\) Area of the first quadrant region: \(A_1 = A_\text{sector} - A_\text{triangle} = 4\theta - \frac{4}{3}\) Since there are 4 symmetric parts, the total area bounded by the curves: \(A_\text{total} = 4A_1 = 4(4\theta-\frac{4}{3}) = 16\theta - 16/3\) Now, we are given four options: (a) \(4\pi - 4\) (b) \(4\pi - 2\) (c) \(4(\pi - 2)\) (d) \(4(\pi - 1)\) To check which one matches our result, plug in the value for \(\theta\): \(16\arctan(\frac{1}{2}) - \frac{16}{3}\) Among the given options, only option (a) seems to fit the result: \(4\pi - 4 = 4\left(\pi - 1\right)\) So the answer is (a) \(4\pi - 4\).