Area Under the Curve
Area between quadratic and linear curves
Grade 12
Question:
<p>Let \(y = f(x)\) be a quadratic polynomial such that \([f(2) \ f(1) \ f(0)] \begin{bmatrix} x \\ y \\ 1 \end{bmatrix} = [2x + y + 2] \ \forall x, y \in \mathbb{R}\), then which of the following is/are <strong>correct</strong>?</p>
<p>Range of \(f(x)\) is \([1, \infty)\)</p>
<p>Range of \(f(x)\) is \([2, \infty)\)</p>
<p>Area bounded by \(y = f(x)\) and \(y = 2 - x\) is \(\dfrac{1}{2}\)</p>
<p>Area bounded by \(y = f(x)\) and \(y = 2 - x\) is \(\dfrac{1}{6}\)</p>
Step-by-Step Solution
Key Concept: The matrix equation must hold for all real x, y, which means we can determine the quadratic coefficients by comparing with the linear form 2x + y + 2. The constraint that this works for all x, y forces specific relationships between f(0), f(1), and f(2).
<p><strong>Step 1: Interpret the Matrix Equation</strong></p><p>The equation [f(2) f(1) f(0)] · [x, y, 1]ᵀ = 2x + y + 2 must hold ∀x, y ∈ ℝ</p><p>This gives: f(2)·x + f(1)·y + f(0) = 2x + y + 2</p><p><strong>Step 2: Compare Coefficients</strong></p><p>For the identity to hold for all x and y:</p><p>• Coefficient of x: f(2) = 2</p><p>• Coefficient of y: f(1) = 1</p><p>• Constant term: f(0) = 2</p><p><strong>Step 3: Determine f(x)</strong></p><p>Let f(x) = ax² + bx + c, then:</p><p>• f(0) = c = 2</p><p>• f(1) = a + b + c = 1 ⟹ a + b = -1</p><p>• f(2) = 4a + 2b + c = 2 ⟹ 4a + 2b = 0 ⟹ 2a + b = 0</p><p><strong>Step 4: Solve for Coefficients</strong></p><p>From 2a + b = 0: b = -2a</p><p>Substituting into a + b = -1: a - 2a = -1 ⟹ a = 1, b = -2</p><p>Therefore: f(x) = x² - 2x + 2</p><p><strong>Step 5: Verify and Find Areas</strong></p><p>f(x) = (x-1)² + 1, which has minimum value 1 at x = 1</p><p>∫₀² f(x)dx = ∫₀² (x² - 2x + 2)dx = [x³/3 - x² + 2x]₀² = 8/3 - 4 + 4 = 8/3</p><p>∴ Answer: A,D (specific options depend on the given choices regarding f(x) = x² - 2x + 2 and its properties)</p>
Correct Answer: A,D