<p>If \(\begin{vmatrix} 4a^2 & 4a & 1 \\ 4b^2 & 4b & 1 \\ 4c^2 & 4c & 1 \end{vmatrix} \begin{bmatrix} f(-1) \\ f(1) \\ f(2) \end{bmatrix} = \begin{bmatrix} 3a^2 + 3a \\ 3b^2 + 3b \\ 3c^2 + 3c \end{bmatrix}\), \(f(x)\) is a quadratic function and its maximum value occurs at a point \(V\). \(A\) is a point of intersection of \(y = f(x)\) with \(x\)-axis and point \(B\) is such that chord \(AB\) subtends a right angle at \(V\). Find the area enclosed by \(f(x)\) and chord \(AB\) (up to two decimal places).</p>
Step-by-Step Solution
Key Concept: The Vandermonde-like determinant on the left creates a system that determines f(x) uniquely; the determinant equals 16(a-b)(b-c)(c-a), allowing us to extract f(-1), f(1), f(2) values and reconstruct the quadratic function through Lagrange interpolation or system solving.
<p><strong>Step 1: Recognize the determinant structure.</strong></p><p>The determinant is a Vandermonde-type: $\begin{vmatrix} 4a^2 & 4a & 1 \\ 4b^2 & 4b & 1 \\ 4c^2 & 4c & 1 \end{vmatrix} = 16(a-b)(b-c)(c-a)$</p><p>For the equation to hold for all distinct a, b, c, we need: $16(a-b)(b-c)(c-a) \cdot \begin{bmatrix} f(-1) \\ f(1) \\ f(2) \end{bmatrix} = \begin{bmatrix} 3a^2 + 3a \\ 3b^2 + 3b \\ 3c^2 + 3c \end{bmatrix}$</p><p><strong>Step 2: Determine f(x).</strong></p><p>Since this must hold for arbitrary a, b, c, comparing coefficients: The RHS suggests $3x(x+1)$ evaluated at x = a, b, c. Thus: $f(-1) = 0$, $f(1) = 6$, $f(2) = 15$</p><p>Alternatively, the Vandermonde structure implies $f(x) = 3x^2 + 3x = 3x(x+1)$</p><p><strong>Step 3: Apply the right angle constraint.</strong></p><p>For $f(x) = 3x^2 + 3x$: vertex $V = (-\frac{1}{2}, -\frac{3}{4})$</p><p>Roots: $x = 0$ or $x = -1$. Let $A = (0, 0)$.</p><p>For chord AB where $\angle AVB = 90°$ with B on the parabola: Using the perpendicularity condition $\vec{VA} \cdot \vec{VB} = 0$, we get $B = (-1, 0)$</p><p><strong>Step 4: Calculate enclosed area.</strong></p><p>Area $= \int_{-1}^{0} |3x^2 + 3x| \, dx = \int_{-1}^{0} (3x^2 + 3x) \, dx = [x^3 + \frac{3x^2}{2}]_{-1}^{0} = 0 - (-1 + \frac{3}{2}) = -\frac{1}{2}$ (absolute value)</p><p>∴ Answer: <strong>0.50</strong></p>
Correct Answer: 0