### 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 $