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>(a) Range of \(f(x)\) is \([1, \infty)\)</p>
<p>(b) Range of \(f(x)\) is \([2, \infty)\)</p>
<p>(c) Area bounded by \(y = f(x)\) and \(y = 2 - x\) is \(\dfrac{1}{2}\)</p>
<p>(d) Area bounded by \(y = f(x)\) and \(y = 2 - x\) is \(\dfrac{1}{6}\)</p>
Step-by-Step Solution
Key Concept: Treat the matrix equation as an identity in variables x and y to extract the coefficients of the quadratic f(x), then use the constraint that coefficients must satisfy the matrix equation for all x,y values simultaneously.
<p><strong>Step 1:</strong> Let f(x) = ax² + bx + c (quadratic polynomial).</p><p><strong>Step 2:</strong> Write out the matrix equation: [f(2) f(1) f(0)] · [x y 1]ᵀ = 2x + y + 2</p><p>This gives: f(2)·x + f(1)·y + f(0)·1 = 2x + y + 2</p><p><strong>Step 3:</strong> Since this must hold ∀x,y ∈ ℝ, equate coefficients:</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 4:</strong> Set up system with f(x) = ax² + bx + c:</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 5:</strong> Solve: From 2a + b = 0 and a + b = -1:</p><p>Subtracting: a = 1, then b = -2</p><p>∴ f(x) = x² - 2x + 2</p><p><strong>Step 6:</strong> Verify key properties (minimum value, derivative, etc.) to identify correct statements A and D.</p><p>∴ Answer: A, D</p>
Correct Answer: A,D