Using Binomial expansion, its (r+1) th term be, Tr+1=10Cr(tx1∕5)10−r{
(1−x)1∕10
t
}r =10Cr
(t)10−r
(t)r
(x1∕5)10−r(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 have 10−2r=0 gives, r=5 ∴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 (Maximum value) Thus, greatest term will be T6=10C5(