The greatest value of the term independent of t in the expansion of ${\left\{t{x}^{1/5}+\frac{(1-x{)}^{1/10}}{t}\right\}}^{10},x\in (0,1)$ is 

The general term $T_{r+1}$is given by

$T_{r+1}{=}^{10}{C}_r{\left(t{x}^{1/5}\right)}^{10-r}{\left\{\frac{(1-x{)}^{1/10}}{t}\right\}}^r$

$\Rightarrow T_{r+1}{=}^{10}{C}_r{t}^{10-2r}{x}^{r/5}(1-x{)}^{r/10}$           (i)

For this term to be independent of t, we must have             

$10-2r=0\text{ i.e. }r=5$                        

Putting $r =5$ in (i), we obtain          

$T_6{=}^{10}{C}_{5}x(1-x{)}^{1/2}{=}^{10}{C}_{5}f(x),$where $f\left(x\right)=x(1-x{)}^{1/2}$ 

$\therefore f^{\mathrm{\prime }}\left(x\right)=(1-x{)}^{1/2}-\frac{x}{2(1-x{)}^{1/2}}=\frac{2-3x}{(1-x{)}^{1/2}}$

For maximum or minimum values of $f\left(x\right)$, we must have                 

$\therefore f^{\mathrm{\prime }}\left(x\right)=0\Rightarrow x=\frac{2}{3}$                 

Clearly,$f\text{'} \left(x\right)$ changes its sign from positive to negative as $x$ increases through $2/3$ So, $f\left(x\right)$attains its maximum value at $x=\frac{2}{3}$ The maximum value of$f\left(x\right)$  is $f\left(\frac{2}{3}\right)=\frac{2}{3\sqrt{3}}$

$\therefore$ Greatest value of   $T_6$is ${ }^{10}{C}_5\times \frac{2}{3\sqrt{3}}=\frac{2}{3\sqrt{3}}\cdot \frac{10!}{(5!{)}^2}$