Using Binomial expansion, its (r+1) th term be, Tr+1=10Ct(tx1∕5)10−r{
(1−x)1∕10
t
}r =10Cr
(t)10−r
(t)t
(x1∕5)10−t(1−x)r∕10 =10Cr(t)10−2r(x)
10−r
5
(1−x)r∕10 If this term is independent of ' t′, then we havel 0−2r=0 gives, r=5 ∴T6=10C5(x)1(1−x)1∕2 ∴T6=10C5(x)1(1−x)1∕2 Let f(x)=x(1−x)1∕2, to obtain its maximum value, we have to differentiate it and equate it to 0 . i.e. f′(x)=0⇒
x
2√1−x
(−1)+√1−x=0 ⇒−x+2(1−x)=0⇒−3x+2=0 ⇒x=
2
3
and f′′(
2
3
)<0 (Maximum value) Thus, greatest term will be T6=10C5(
