If the maximum value of the term independent of t in the expansion of (t^2 x^15+(1-x)^110t)^15, x 0, is K, then 8 ~K is equal to ____________.

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 T_10 + 1 = T

If the maximum value of the term independent of t in the exp: null (Mathematics)

General term of ${\left( {{t^2}{x^{{1 \over 5}}} + {{{{(1 - x)}^{{1 \over {10}}}}} \over t}} \right)^{15}}$ is

{T_{r + 1}} = {}^{15}{C_r}\,.\,{\left( {{t^2}{x^{{1 \over 5}}}} \right)^{15 - r}}\,.\,{\left( {{{{{(1 - x)}^{{1 \over {10}}}}} \over t}} \right)^r}

= {}^{15}{C_r}\,.\,{t^{30 - 2r}}\,.\,{x^{{{15 - r} \over 5}}}\,.\,{\left( {1 - x} \right)^{{r \over {10}}}}\,.\,{t^{ - r}}

= {}^{15}{C_r}\,.\,{t^{30 - 3r}}\,.\,{x^{{{15 - r} \over 5}}}\,.\,{\left( {1 - x} \right)^{{r \over {10}}}}

Term will be independent of $\mathrm{t}$ when $30 - 3r = 0 \Rightarrow r = 10$

\therefore {T_{10 + 1}} = {T_{11}}

will be independent of

\mathrm{t}

\therefore {T_{11}} = {}^{15}{C_{10}}\,.\,{x^{{{15 - 10} \over 5}}}\,.\,{\left( {1 - x} \right)^{{{10} \over {10}}}}

= {}^{15}{C_{10}}\,.\,{x^1}\,.\,{\left( {1 - x} \right)^1}

$\mathrm{T_{11}}$ 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

\Rightarrow x = {1 \over 2}

Now,

f''(x) = - 2 < 0

\therefore \text{At } x = {1 \over 2}f(x) \text{ maximum}

$\therefore$ Maximum value of $\mathrm{T_{11}}$ is

= {}^{15}{C_{10}}\,.\,{1 \over 2}\left( {1 - {1 \over 2}} \right)

= {}^{15}{C_{10}}\,.\,{1 \over 4}

Given

K = {}^{15}{C_{10}}\,.\,{1 \over 4}

Now,

8K = 2\left( {{}^{15}{C_{10}}} \right)

= 6006

Explanation

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