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← JEE Main 2015 · 11 Apr · Shift 1
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JEE Main 2015 · 11 Apr · Shift 1, Mathematics question 67

The term independent of xx in the binomial expansion of
(1−1x+3x5)(2x2−1x)8\left(1-\frac1x+3x^5\right)\left(2x^2-\frac1x\right)^8
is :

Choose one option
Answer and worked solution

Official answer: A

Tr+1=8Cr28−r(−1)rx16−3rT_{r+1}={}^8C_r2^{8-r}(-1)^rx^{16-3r}.

  • Times 11: 16−3r=016-3r=0 has no integer solution.
  • Times −1x-\frac1x: need 16−3r=116-3r=1, so r=5r=5: −1×(−8C5⋅23)=448-1\times(-{}^8C_5\cdot2^3)=448.
  • Times 3x53x^5: need 16−3r=−516-3r=-5, so r=7r=7: 3×(−8C7⋅2)=−483\times(-{}^8C_7\cdot2)=-48.

Total 400400.

Topics:General and middle terms
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