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, $$f''(x) = - 2 < 0$$

$$\therefore$$ At $$ x = {1 \over 2}$$, $$f(x)$$ 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, $$8K = 2\left( {{}^{15}{C_{10}}} \right)$$

$$ = 6006$$