Question

Let $f(x)$ be a polynomial of degree 3. If the curve $y=f(x)$ has relative extrema at $x=\frac{\pm 2}{\sqrt{3}}$ and passes through (0, 0) and (1, -2) dividing the circle $x^2 + y^2 = 4$ in two parts, then the area bounded by $x^2 + y^2 = 4$ and $y \geq f(x)$ is $\frac{k\pi}{2}$. Find the value of $k$.

Answer 1

4

Answer 2

The polynomial is found to be $f(x) = \frac{2}{3}(x^3 - 4x)$. This is an odd function, meaning $f(-x) = -f(x)$. The circle $x^2+y^2=4$ is symmetric with respect to the origin. Let $A$ be the area inside the circle where $y \geq f(x)$, and $A'$ be the area inside the circle where $y \leq f(x)$.

Due to the odd symmetry of $f(x)$ and the origin symmetry of the circle, the area inside the circle above $f(x)$ is equal to the area inside the circle below $-f(x)$. Since $f(x)$ is odd, $-f(x) = f(-x)$. Thus, the area inside the circle where $y \geq f(x)$ is equal to the area inside the circle where $y \geq f(-x)$. By substituting $X=-x$, this implies the area inside the circle where $y \geq f(x)$ is equal to the area inside the circle where $y \geq f(X)$, which is the same area $A$.

Alternatively, consider the area inside the circle where $y \leq f(x)$. Let this area be $A'$. For any point $(x,y)$ inside the circle such that $y \geq f(x)$, consider the point $(-x, -y)$. Since $x^2+y^2 \leq 4$, we have $(-x)^2+(-y)^2 \leq 4$, so $(-x,-y)$ is also inside the circle. The condition $y \geq f(x)$ implies $-y \leq -f(x)$. Since $f$ is odd, $-f(x) = f(-x)$. So, $-y \leq f(-x)$. Let $X=-x$ and $Y=-y$. The condition becomes $Y \leq f(X)$. This shows that for every point $(x,y)$ in the region $y \geq f(x)$ within the circle, there is a corresponding point $(-x,-y)$ in the region $y \leq f(x)$ within the circle. This symmetry implies that the area of the region where $y \geq f(x)$ inside the circle is equal to the area of the region where $y \leq f(x)$ inside the circle. Therefore, $A = A'$.

The union of the region where $y \geq f(x)$ and the region where $y \leq f(x)$ covers the entire area of the circle. Thus, $A + A' = \text{Area of the Circle}$. The area of the circle $x^2+y^2=4$ is $\pi R^2 = \pi (2^2) = 4\pi$. So, $A + A' = 4\pi$. Since $A = A'$, we have $2A = 4\pi$, which means $A = 2\pi$.

The problem states that the area bounded by $x^2 + y^2 = 4$ and $y \geq f(x)$ is $\frac{k\pi}{2}$. Equating our calculated area with the given form: $2\pi = \frac{k\pi}{2}$ Multiplying both sides by 2: $4\pi = k\pi$ Dividing by $\pi$: $k = 4$.