Tr+1=Cr10(tx(1/5))10−r[t(1−x)101]r
=Cr10(t)10−2r(x)5(10−r)(1−x)10r
For the term independent of t,
⇒10−2r=0
⇒r=5
T6=f(x)=C510(x1−x); for maximum
{f}^{'}(x)=0\Rightarrow x=\frac{2}{3}&{f}^{''}(\frac{2}{3})<0
so f(x)max.=C510(32)⋅31
=33(5!)22.10!