<p>In the expansion of \(\left(\dfrac{x^2}{y} + \dfrac{y^2}{x}\right)^{15}\), there is a term independent of \(x\) and a term independent of \(y\) but no term independent of \(x\) and \(y\) both. (State whether true or false.)</p>
Step-by-Step Solution
Key Concept: For the general term T_{r+1} = C(15,r)(x^2/y)^(15-r)(y^2/x)^r = C(15,r)x^(30-3r)y^(3r-15), we need to find when the power of x equals zero, when the power of y equals zero, and verify no term makes both zero simultaneously.
<p><strong>Step 1:</strong> Find the general term in the expansion of (x²/y + y²/x)^15:</p><p>T_{r+1} = C(15,r)(x²/y)^(15-r)(y²/x)^r = C(15,r)·x^(2(15-r))·y^(-15+r)·y^(2r)·x^(-r)</p><p>T_{r+1} = C(15,r)·x^(30-2r-r)·y^(-15+r+2r) = C(15,r)·x^(30-3r)·y^(3r-15)</p><p><strong>Step 2:</strong> Find term independent of x:</p><p>Power of x = 0 ⟹ 30-3r = 0 ⟹ r = 10 ✓ (valid, 0 ≤ r ≤ 15)</p><p>This gives T₁₁ = C(15,10)·y^(30-15) = C(15,10)·y^15 (independent of x)</p><p><strong>Step 3:</strong> Find term independent of y:</p><p>Power of y = 0 ⟹ 3r-15 = 0 ⟹ r = 5 ✓ (valid, 0 ≤ r ≤ 15)</p><p>This gives T₆ = C(15,5)·x^(30-15) = C(15,5)·x^15 (independent of y)</p><p><strong>Step 4:</strong> Check for term independent of both x and y:</p><p>We would need 30-3r = 0 AND 3r-15 = 0 simultaneously</p><p>But r = 10 from first equation and r = 5 from second equation ⟹ No common solution</p><p><strong>Conclusion:</strong> The statement is <strong>TRUE</strong>. There exists a term independent of x (when r=10) and a term independent of y (when r=5), but no single term is independent of both.</p><p>∴ Answer: A (True)</p>
Correct Answer: A