If the maximum value of the term independent of $t$ in the expansion of $$\left(\mathrm{t}^{2} x^{\frac{1}{5}}+\frac{(1-x)^{\frac{1}{10}}}{\mathrm{t}}\right)^{15}, x \geqslant 0$$, is $\mathrm{K}$, then $8 \mathrm{~K}$ is equal to ____________.

<p>General term of $${\left( {{t^2}{x^{{1 \over 5}}} + {{{{(1 - x)}^{{1 \over {10}}}}} \over t}} \right)^{15}}$$ is</p> <p>$${T_{r + 1}} = {}^{15}{C_r}\,.\,{\left( {{t^2}{x^{{1 \over 5}}}} \right)^{15 - r}}\,.\,{\left( {{{{{(1 - x)}^{{1 \over {10}}}}} \over t}} \right)^r}$$</p> <p>$$ = {}^{15}{C_r}\,.\,{t^{30 - 2r}}\,.\,{x^{{{15 - r} \over 5}}}\,.\,{\left( {1 - x} \right)^{{r \over {10}}}}\,.\,{t^{ - r}}$$</p> <p>$$ = {}^{15}{C_r}\,.\,{t^{30 - 3r}}\,.\,{x^{{{15 - r} \over 5}}}\,.\,{\left( {1 - x} \right)^{{r \over {10}}}}$$</p> <p>Term will be independent of $\mathrm{t}$ when $30 - 3r = 0 \Rightarrow r = 10$</p> <p>$\therefore$ ${T_{10 + 1}} = {T_{11}}$ will be independent of $\mathrm{t}$</p> <p>$\therefore$ $${T_{11}} = {}^{15}{C_{10}}\,.\,{x^{{{15 - 10} \over 5}}}\,.\,{\left( {1 - x} \right)^{{{10} \over {10}}}}$$</p> <p>$= {}^{15}{C_{10}}\,.\,{x^1}\,.\,{\left( {1 - x} \right)^1}$</p> <p>$\mathrm{T_{11}}$ will be maximum when $x(1 - x)$ is maximum.</p> <p>Let $f(x) = x(1 - x) = x - {x^2}$</p> <p>$f(x)$ is maximum or minimum when $f'(x) = 0$</p> <p>$\therefore$ $f'(x) = 1 - 2x$</p> <p>For maximum/minimum $f'(x) = 0$</p> <p>$\therefore$ $1 - 2x = 0$</p> <p>$\Rightarrow x = {1 \over 2}$</p> <p>Now, $f''(x) = - 2 < 0$</p> <p>$\therefore$ At $x = {1 \over 2}$, $f(x)$ maximum</p> <p>$\therefore$ Maximum value of $\mathrm{T_{11}}$ is</p> <p>$= {}^{15}{C_{10}}\,.\,{1 \over 2}\left( {1 - {1 \over 2}} \right)$</p> <p>$= {}^{15}{C_{10}}\,.\,{1 \over 4}$</p> <p>Given $K = {}^{15}{C_{10}}\,.\,{1 \over 4}$</p> <p>Now, $8K = 2\left( {{}^{15}{C_{10}}} \right)$</p> <p>$= 6006$</p>