7.Binomial Theorem
hard

$(2 -x^2)$ અને  $((1 + 2x + 3x^2)^6 +(1 -4x^2)^6)$ ના ગુણાકારમાં $x^2$ નો સહગુણક મેળવો. 

 

A

$106$

B

$107$

C

$155$

D

$108$

(JEE MAIN-2018)

Solution

$\text { Let } a=\left(\left(1+2 x+3 x^{2}\right)^{6}+\left(1-4 x^{2}\right)^{6}\right)$

$\therefore $ Coefficient of $x^{2}$ in the expansion of the product

$\left(2-x^{2}\right)\left(\left(1+2 x+3 x^{2}\right)^{6}+\left(1-4 x^{2}\right)^{6}\right)$

$=2\left(\text { Coefficient of } x^{2} \text { in a }\right)-1$ (Constant of expansion)

In the expansion of

$\left(\left(1+2 x+3 x^{2}\right)^{6}+\left(1-4 x^{2}\right)^{6}\right)$

Constant $=1+1=2$

Coefficient of $x^{2}=$

[Coefficient of $x^2$ in ${(^6}{C_0}{(1 + 2x)^6}{(3{x^2})^0}$ ] $+$ [Coefficient of $x^2$ in ${(^6}{C_1}{(1 + 2x)^5}{(3{x^2})^1}$ ] $-$ ${[^6}{C_1}(4{x^2})]$ 

$=60+6 \times 3-24=54$

$\therefore \quad$ The coefficient of $x^{2}$ in $\left(2-x^{2}\right)$

${\left(\left(1+2 x+3 x^{2}\right)^{6}+\left(1-4 x^{2}\right)^{6}\right)}$

${=2 \times 54-1(2)=108-2=106}$

Standard 11
Mathematics

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