If the maximum value of the term independent of t in the expansion of (t² x^(1/5)+(1-x)^(1/10)/t)¹⁵, x ≥ 0, is K, then 8 K is equal to ____________.

Solution

Maths sir · 4 steps

Find the t-free term first, then it is just a quadratic in x to maximise.

1. General term. The (r+1)th term is C(15, r)(t² x^(1/5))¹⁵⁻ʳ((1-x)^(1/10)/t)ʳ. The power of t in it is 2(15-r) - r = 30 - 3r.
2. Remove t. Set 30 - 3r = 0, so r = 10. Then x gets the power (15-10)/5 = 1 and (1-x) gets the power 10/10 = 1.

   T₁₁ = C(15, 10) x(1-x) = 3003 x(1-x)

3. Maximise. For (1-x)^(1/10) to be real we need 0 ≤ x ≤ 1. The product x(1-x) is largest at x = 1/2, where it equals 1/4. So K = 3003/4.
4. So. 8K = 8 × 3003/4 = 2 × 3003 = 6006.

The common slip: 750.75. Stopping at K = 3003/4 gives 750.75. The question asks for 8K, so multiply by 8 at the end.

The idea to keep. For a term independent of a variable, write the general term, add up that variable's powers and set the total to zero. Any other variable left in the term is then a function you can maximise.

Answer from the NTA final answer key. Solution by PrepWiser.