<p>Let \([f(2)\quad f(1)\quad f(0)]\begin{bmatrix} x \\ y \\ 1 \end{bmatrix} = [2x + y + 2]\) for all \(x, y \in \mathbb{R}\). Find the area enclosed between \(y = f(x)\) and \(y = 2 - x\).</p>
<p>(a) \(f(x) = x^2 - 2x + 2\)</p>
<p>(b) \(f(2) = 2,\; f(1) = 1,\; f(0) = 2\)</p>
<p>(c) Area = \(\dfrac{1}{6}\)</p>
<p>(d) Area = \(\dfrac{1}{3}\)</p>
Step-by-Step Solution
Key Concept: First, extract f(x) by substituting specific values into the matrix equation to find coefficients f(2), f(1), f(0), then find the intersection points of f(x) with y = 2 - x to set integration bounds.
<p><strong>Step 1: Extract f(x) from the matrix equation</strong></p><p>The equation [f(2) f(1) f(0)]·[x y 1]ᵀ = 2x + y + 2 must hold for all x, y ∈ ℝ.</p><p>Expanding: f(2)·x + f(1)·y + f(0)·1 = 2x + y + 2</p><p>Comparing coefficients:</p><ul><li>Coefficient of x: f(2) = 2</li><li>Coefficient of y: f(1) = 1</li><li>Constant term: f(0) = 2</li></ul><p><strong>Step 2: Determine f(x)</strong></p><p>With three points: (0, 2), (1, 1), (2, 2), assume f(x) = ax² + bx + c</p><p>From f(0) = 2: c = 2</p><p>From f(1) = 1: a + b + 2 = 1 ⟹ a + b = -1</p><p>From f(2) = 2: 4a + 2b + 2 = 2 ⟹ 4a + 2b = 0 ⟹ 2a + b = 0</p><p>Solving: a = 1, b = -2</p><p>∴ f(x) = x² - 2x + 2</p><p><strong>Step 3: Find intersection points with y = 2 - x</strong></p><p>x² - 2x + 2 = 2 - x</p><p>x² - x = 0</p><p>x(x - 1) = 0</p><p>∴ x = 0 or x = 1</p><p><strong>Step 4: Determine which function is above</strong></p><p>At x = 0.5: f(0.5) = 0.25 - 1 + 2 = 1.25 and 2 - 0.5 = 1.5</p><p>So y = 2 - x is above f(x) on [0, 1]</p><p><strong>Step 5: Calculate area</strong></p><p>Area = ∫₀¹ [(2 - x) - (x² - 2x + 2)] dx</p><p>= ∫₀¹ [2 - x - x² + 2x - 2] dx</p><p>= ∫₀¹ [x - x²] dx</p><p>= [x²/2 - x³/3]₀¹</p><p>= 1/2 - 1/3 = 1/6</p><p>∴ Answer: <strong>1/6</strong></p>
Correct Answer: a,d