IIT JEE , AIEEE Forum - Mathematics , Algebra

deep01

given that the 4 th tem in expansion of(2+3x/8)¹⁰ has the max. numerical value find the range of value of x for which this will be true.

rik_mad

hi,
for 4th term to be maximum
it shud be greater than the fifth and third terms
so
third term< fourth term > fifth term
using formula for the rth term of binomial expansion
and using the two inequalities u get x to be between
[28/3 , 128/9]

pink_ele

T 4>T3

¹⁰C₃ X2⁷X3³X x³/8³>¹⁰C₂ X2⁸ X3² x²/8²

x>2

now also T4>T2 as done above

solving we get

x<64/21

2<x<64/21

shine

rewrite the exp as T4 is numerically largest then

|t4|>|t3|

=>|t4/t3| >1

=>{10C3(3x/16)³} / {10c2(3x/16⁾²} >1

=>on solving

10C2/10C3 <|3x/10|

|x|>2

also

|t4/t5|> 1

=>10C3/10C4>|3x/10|

=> |x|<64/21

thus finally

x

(-64/21 ,-2)

(2,64/21)

i hope its correct

plzzzz rate me if it is

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