<p>If <em>T</em><sub><em>r</em>+1</sub> is the general term of <span>\(\left[ax^2 + \frac{1}{bx}\right]^{11}\)</span>, and the coefficient of <span>\(x^7\)</span> in <span>\(\left[ax^2 + \frac{1}{bx}\right]^{11}\)</span> equals the coefficient of <span>\(x^{-7}\)</span> in <span>\(\left[ax - \frac{1}{bx^2}\right]^{11}\)</span>, then <span>\(ab\)</span> equals:</p>
Step-by-Step Solution
Key Concept: The power of x in the general term depends on the binomial index and the individual term exponents. You must equate powers of x to find which term gives x^7 in the first expansion and x^(-7) in the second, then use the condition that their coefficients are equal.
<p><strong>Step 1: Find the general term and power of x in first expansion.</strong></p><p>For [ax² + 1/(bx)]^11, the general term is: T_{r+1} = C(11,r)(ax²)^(11-r)(1/(bx))^r = C(11,r)a^(11-r)b^(-r)x^(22-3r)</p><p>For coefficient of x^7: 22 - 3r = 7 ⟹ r = 5</p><p>Coefficient of x^7 = C(11,5)a^6b^(-5)</p><p><strong>Step 2: Find the general term and power of x in second expansion.</strong></p><p>For [ax - 1/(bx²)]^11, the general term is: T_{s+1} = C(11,s)(ax)^(11-s)(-1/(bx²))^s = C(11,s)(-1)^s a^(11-s)b^(-s)x^(11-3s)</p><p>For coefficient of x^(-7): 11 - 3s = -7 ⟹ s = 6</p><p>Coefficient of x^(-7) = C(11,6)(-1)^6 a^5 b^(-6) = C(11,6)a^5 b^(-6)</p><p><strong>Step 3: Equate the coefficients.</strong></p><p>C(11,5)a^6 b^(-5) = C(11,6)a^5 b^(-6)</p><p>Since C(11,5) = C(11,6) = 462:</p><p>a^6 b^(-5) = a^5 b^(-6)</p><p>a^6 · b^6 = a^5 · b^5</p><p>ab = 1</p><p>∴ Answer: A</p>
Correct Answer: A