The maximum value of the term independent of 't' in the expansion
of $${\left( {t{x^{{1 \over 5}}} + {{{{(1 - x)}^{{1 \over {10}}}}} \over t}} \right)^{10}}$$ where x$\in$(0, 1) is :

$${T_{r + 1}} = {}^{10}{C_r}{(t{x^{1/5}})^{10 - r}}{\left[ {{{{{(1 - x)}^{1/10}}} \over t}} \right]^r}$$<br><br>$$ = {}^{10}{C_r}{t^{(10 - 2r)}} \times {x^{{{10 - r} \over 5}}} \times {(1 - x)^{{r \over {10}}}}$$<br><br>$\Rightarrow 10 - 2r = 0 \Rightarrow r = 5$ <br><br>$\therefore$ ${T_6} = {}^{10}{C_5} \times x\sqrt {1 - x}$<br><br>At maximum, $${{d{T_6}} \over {dx}} = {}^{10}{C_5}\left[ {\sqrt {1 - x} - {x \over {2\sqrt {1 - x} }}} \right] = 0$$<br><br>$\Rightarrow$ $1 - x = x/2 \Rightarrow 3x = 2 \Rightarrow x = 2/3$<br><br>${T_6}{|_{\max }} = {{10!} \over {5!5!}} \times {2 \over {3\sqrt 3 }}$ = ${{2.10!} \over {3\sqrt 3 {{(5!)}^2}}}$