If the maximum value of the term independent of t in the exp

Question

If the maximum value of the term independent of t in the expansion of ( t^2 x^(1 / 5)+frac(1-x)^(1 / 10) t)^15, x >= slant 0, is K, then 8 K is equal to ____________.

DivineJEE · Accepted Answer

General term of ( t^2x^1 / 5 + (1 - x)^1 / 10 / t )^15 is T_r + 1 = ^15C_r . ( t^2x^1 / 5 )^15 - r . ( (1 - x)^1 / 10 / t )^r = ^15C_r . t^30 - 2r . x^15 - r / 5 . ( 1 - x )^r / 10 . t^ - r = ^15C_r . t^30 - 3r . x^15 - r / 5 . ( 1 - x )^r / 10 Term will be independent of t when 30 - 3r = 0 => r = 10 Therefore T_10 + 1 = T_11 will be independent of t Therefore T_11 = ^15C_10 . x^15 - 10 / 5 . ( 1 - x )^10 / 10 = ^15C_10 . x^1 . ( 1 - x )^1 T_11 will be maximum when x(1 - x) is maximum. Let f(x) = x(1 - x) = x - x^2 f(x) is maximum or minimum when f'(x) = 0 Therefore f'(x) = 1 - 2x For maximum/minimum f'(x) = 0 Therefore 1 - 2x = 0 => x = 1 / 2 Now, f''(x) = - 2 Therefore At x = 1 / 2f(x) maximum Therefore Maximum value of T_11 is = ^15C_10 . 1 / 2( 1 - 1 / 2 ) = ^15C_10 . 1 / 4 Given K = ^15C_10 . 1 / 4 Now, 8K = 2( ^15C_10 ) = 6006

Explanation

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