#### The coefficient of  in the expansion of the productis Option 1) 107 Option 2) 106 Option 3) 108 Option 4) 155

As we have learned

Expression of Binomial Theorem -

- wherein

for n  +ve integral .

$\\\text{Let}\:a=\left(\left(1+2x+3x^2\right)^6+\left(1-4x^2\right)^6\right)\\\text{coefficient}\:of\:x^{2\:}\:\text{in\:the\:expension\:of}\:\\\left(2-x^2\right)\left(\left(1+2x+3x^2\right)^6+\left(1-4x^2\right)^6\right)$

$\\\Rightarrow 2(\text{coefficient\:of}\:x^{2\:}\text{in}\:a)-1\left(\text{constant\:of\:the\:expensaion\:in}\:a\right)\\\text{In\:the\:expension\:of} \:\left(1+2x+3x^2\right)^6+\left(1-4x^2\right)^6\:\:\text{constant}\:\:=1+1=2\\\text{Now,}\\\text{Coefficient\:of}\:x^{2\:}=\left(\text{Coefficient\:of}\:x^{2\:}\text{in}\:^6C_0\left(1+2x\right)^6\left(3x^2\right)^0\right)+\\\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\text{Coefficient\:of}\:x^{2\:}\text{in}\:^6C_1\left(1+2x\right)^5\left(3x^2\right)^1\right)-\left[^6C_1\left(4x^2\right)\right]\\=60+6\times 3-24=54$

$\\\therefore\text{ coefficient\:of}\:x^{2\:}\:\text{in\:the\:expension\:of}\:\\\left(2-x^2\right)\left(\left(1+2x+3x^2\right)^6+\left(1-4x^2\right)^6\right)\\=2\times54-2=106$

Option 1)

107

This is incorrect

Option 2)

106

This is correct

Option 3)

108

This is incorrect

Option 4)

155

This is incorrect