The maximum value of the term independent of t in the expansion of t x 1 5 + 1 - x 1 10 t 10 where x ∈…

The maximum value of the term independent of t in the expansion of tx15+1-x110t10where x∈0,1 is:
  1. 10!35!2
  2. 10!35!2
  3. 2.10!335!2
  4. 2.10!35!2

Solution

Tr+1=Cr10tx1/510-r1-x110tr

=Cr10t10-2rx10-r51-xr10

For the term independent of t, 

⇒10-2r=0

⇒r=5

T6=fx=C510x1-x; for maximum

f'x=0⇒x=23&f''23<0

so fxmax.=C51023·13

=2.10!335!2

Asked in: JEE Main 2021 (26 Feb Shift 1)

Practice more Binomial Theorem questions on Aicharya