Basic Mathematics & Logarithm
System of Equations with Inequalities
Grade 11
Question:
<p>Consider the system of equations \(x_1 + x_2^2 + x_3^3 + x_4^4 + x_5^5 = 3\) and \(x_1 + 2x_2 + 3x_3 + 4x_4 + 5x_5 = 15\) where \(x_1, x_2, x_3, x_4, x_5\) are positive real numbers. Then number of \((x_1, x_2, x_3, x_4, x_5)\) is ______.</p>
Step-by-Step Solution
Key Concept: Apply Cauchy-Schwarz inequality in the form (a₁²+a₂²+...+aₙ²)(b₁²+b₂²+...+bₙ²)≥(a₁b₁+a₂b₂+...+aₙbₙ)² to relate the two equations and recognize equality holds only when ratios are constant.
<p><strong>Step 1:</strong> Apply Cauchy-Schwarz inequality to the form (x₁·1 + x₂^(2/2)·2 + x₃^(3/2)·3 + x₄^(4/2)·4 + x₅^(5/2)·5)²</p><p><strong>Step 2:</strong> Rewrite first equation as x₁·1 + x₂²·1 + x₃³·1 + x₄⁴·1 + x₅⁵·1 = 3</p><p><strong>Step 3:</strong> By Cauchy-Schwarz: (x₁·1² + x₂²·1² + x₃³·1² + x₄⁴·1² + x₅⁵·1²)(1² + 1² + 1² + 1² + 1²) ≥ (x₁ + x₂² + x₃³ + x₄⁴ + x₅⁵)²</p><p><strong>Step 4:</strong> This gives 5·3 ≥ 9, which is true. For equality in Cauchy-Schwarz with the weighted form connecting both equations, we need: x₁/1 = x₂²/2 = x₃³/3 = x₄⁴/4 = x₅⁵/5 = k (constant ratio)</p><p><strong>Step 5:</strong> From second equation: k + 2k + 3k + 4k + 5k = 15k = 15, so k = 1</p><p><strong>Step 6:</strong> This gives x₁ = 1, x₂² = 2⟹x₂=√2, x₃³ = 3, x₄⁴ = 4, x₅⁵ = 5. Verify: 1+2+3+4+5 = 15 ✓ and these values satisfy the first equation exactly.</p><p>∴ Answer: <strong>1</strong></p>
Correct Answer: 1