JEE Main 25 June 2022 Shift 2 Solved Paper

Given, Binomial Expansion
(2x3+‌

3

x

)10
General term
Tr+1=‌10Cr⋅(2x3)10−r⋅(‌

3

x

)r
=‌10Cr⋅210−r⋅3r⋅x30−3r⋅x−r
=‌10Cr⋅210−r⋅3r⋅x30−4r
For positive even power of x,30−4r should be even and positive.
For r=0,30−4×0=30 (even and positive)
For r=1,30−4×1=26 (even and positive)
For r=2,30−4×2=22 (even and positive)
For r=3,30−4×3=18 (even and positive)
For r=4,30−4×4=14 (even and positive)
For r=5,30−4×5=10 (even and positive)
For r=6,30−4×6=6 (even and positive)
For r=7,30−4×7=2 (even and positive)
For r=8,30−4×8=−2 (even but not positive)
So, for r=1,2,3,4,5,6 and 7 we can get positive even power of x .
∴ Sum of coefficient for positive even power of x=‌10C0⋅210⋅30+‌10C1.29⋅31+‌10C2⋅28⋅32+‌10C3⋅27⋅33+‌10C4⋅26⋅34+‌10C5⋅25⋅35+‌10C6⋅24⋅36+‌10C7⋅23⋅37
=‌10C10⋅210⋅30+‌10C1.29⋅31+......+‌10C10⋅20⋅310−[‌10C8⋅22⋅38+‌10C9⋅2⋅39+‌10C10⋅20⋅310] =(2+3)10−[45.4.38+10.2.39+1.1.310]
=510−[60×39+20⋅39+3⋅39]
=510−(60+20+3)39
=510−83.39
∴β=83