JEE Main On 16 April 2018 Question 10
Question: The coefficient of $ x^{2} $ in the expansion of the product $ (2-x^{2}).({{(1+2x+3x^{2})}^{6}}+{{(1-4x^{2})}^{6}}) $ is? [JEE Main 16-4-2018]
Options:
A) 106
B) 107
C) 155
D) 108
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ a=({{(1+2x+3x^{2})}^{6}}+{{(1-4x^{2})}^{6}}) $
$ \therefore $ Coefficient of $ x^{2} $ in the expansion of the product $ {{(2-x)}^{2}}({{(1+2x+3x^{2})}^{6}}+{{(1-4x^{2})}^{6}}) $
$ =2 (Coefficient,of,x^{2}in,a)-1(Constant,of,expansion) $ IN the expansion of $ ({{(1+2x+3x^{2})}^{6}}+{{(1-4x^{2})}^{6}}), $
Constant = 1 + 1 = 2 Coefficient of $ x^{2}= $ Coefficient of $ x^{2} $ in $ ({{,}^{6}}C _0{{(1+2x)}^{6}}{{(3x^{2})}^{0}})+ $
Coefficient of $ x^{2} $ in $ ({{,}^{6}}C _1{{(1+2x)}^{5}}3)-{{,}^{6}}C _1(4) $ $ ={{,}^{6}}C _24+6\times 3-24 $ $ =6=+18-24=54 $
Then, coefficient of $ x^{2} $ in $ {{(2-x)}^{2}}({{(1+2x+3x^{2})}^{6}}+{{(1-4x^{2})}^{6}}) $ $ =2\times 54-1(2)=108-2=106 $
