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Question: Answered & Verified by Expert
The maximum value of the term independent of t in the expansion of tx15+1-x110t10where x0,1 is:
MathematicsBinomial TheoremJEE MainJEE Main 2021 (26 Feb Shift 1)
Options:
  • A 10!35!2
  • B 10!35!2
  • C 2.10!335!2
  • D 2.10!35!2
Solution:
1073 Upvotes Verified Answer
The correct answer is: 2.10!335!2

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=0x=23&f''23<0

so fxmax.=C51023·13

=2.10!335!2

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