Sets, Relations & Functions
Set Operations
Grade 11
Question:
<p>Let \(A = \{x \mid x^3 + x^2 - px + q = 0,\ p, q \in R\}\) and \(B = \{x \mid x^2 - qx + 2 = 0,\ q \in R\}\) be the sets. If \(n(A \cap B) = 2\) and \(x_0 \in (A - B)\), then find the value of \(|p - q + x_0|\).<br>[<strong>Note:</strong> \(n(P \cap Q)\) denotes number of common elements in set \(P\) and set \(Q\) and \(a \in (P - Q)\) denotes elements '\(a\)' lies in set \(P\) not in set \(Q\).]</p>
Step-by-Step Solution
Key Concept: Since n(A ∩ B) = 2, the cubic in A and quadratic in B share exactly 2 roots. If these 2 common roots are α and β, then both equations are satisfied by them, allowing us to use Vieta's formulas and the constraint that A has 3 roots (with one root x₀ not in B) to find p, q, and x₀.
<p><strong>Step 1: Set up the relationship between A and B</strong></p><p>Let the two common roots of the cubic and quadratic be α and β. Then B = {α, β} exactly.</p><p>For the quadratic x² - qx + 2 = 0: α + β = q and αβ = 2</p><p><strong>Step 2: Express the cubic</strong></p><p>Since the cubic has roots α, β, and x₀, we can write:</p><p>x³ + x² - px + q = (x - α)(x - β)(x - x₀)</p><p>Expanding: x³ + x² - px + q = x³ - (α + β + x₀)x² + (αβ + αx₀ + βx₀)x - αβx₀</p><p><strong>Step 3: Match coefficients</strong></p><p>Coefficient of x²: 1 = -(α + β + x₀) → α + β + x₀ = -1</p><p>Since α + β = q: q + x₀ = -1 ... (i)</p><p>Coefficient of x: -p = αβ + (α + β)x₀ = 2 + qx₀ ... (ii)</p><p>Constant term: q = -αβx₀ = -2x₀ ... (iii)</p><p><strong>Step 4: Solve the system</strong></p><p>From (iii): q = -2x₀</p><p>Substitute in (i): -2x₀ + x₀ = -1 → -x₀ = -1 → x₀ = 1</p><p>Therefore: q = -2(1) = -2</p><p>From (ii): -p = 2 + (-2)(1) = 2 - 2 = 0 → p = 0</p><p><strong>Step 5: Calculate the answer</strong></p><p>|p - q + x₀| = |0 - (-2) + 1| = |0 + 2 + 1| = |3| = 3</p><p><strong>∴ Answer: 3</strong></p>
Correct Answer: 3