Binomial Theorem
General term
Grade 11
Question:
<p>Given \((1+x^{\log_2 x})^5\), its third term is \(T_3 = 2560\). Find the value(s) of \(x\).</p>
<p>\(x = 4, \dfrac{1}{4}\)</p>
<p>\(x = 2, \dfrac{1}{2}\)</p>
<p>\(x = 8, \dfrac{1}{8}\)</p>
<p>\(x = 4, \dfrac{1}{2}\)</p>
Step-by-Step Solution
Key Concept: The third term in binomial expansion is T₃ = C(5,2)·(1)³·(x^(log₂x))² = 10·x^(2log₂x). Recognize that x^(2log₂x) = (2^(log₂x))^(2log₂x) = 2^((log₂x)²), which can be solved by setting (log₂x)² = k.
<p><strong>Step 1:</strong> Find T₃ using binomial theorem.</p><p>T₃ = C(5,2)·(1)^(5-2)·(x^(log₂x))² = 10·x^(2log₂x)</p><p><strong>Step 2:</strong> Convert x^(2log₂x) to exponential form with base 2.</p><p>Let log₂x = t, so x = 2^t</p><p>x^(2log₂x) = (2^t)^(2t) = 2^(2t²)</p><p><strong>Step 3:</strong> Set up equation from T₃ = 2560.</p><p>10·2^(2t²) = 2560</p><p>2^(2t²) = 256 = 2⁸</p><p>2t² = 8</p><p>t² = 4</p><p>t = ±2</p><p><strong>Step 4:</strong> Solve for x.</p><p>When t = 2: log₂x = 2 ⟹ x = 2² = 4</p><p>When t = -2: log₂x = -2 ⟹ x = 2^(-2) = 1/4</p><p><strong>Step 5:</strong> Verify validity (x > 0, x ≠ 1).</p><p>Both x = 4 and x = 1/4 satisfy the domain requirements.</p><p>∴ Answer: x = 4 or x = 1/4</p>
Correct Answer: A