Question

The term independent of $x$ in the expansion of $\left(1+x+2x^2\right){\left(3x^2-\frac{1}{3x^2}\right)}^4$ is

• A. 10
• B. 2
• C. 0
• D. 6

Answer 1

Answer: (D)

general term $-{\left(1+x+2x^2\right)}^4{C}_r{\left(3x^2\right)}^{4-r}{\left(\frac{-1}{3x^2}\right)}^r$
$={\left(1+x+2x^2\right)}^4{C}_r{3}^{4-r}{\left(\frac{-1}{3}\right)}^r{x}^{8-4r}$
$={}^4{C}_r{3}^{4-r}{\left(\frac{-1}{3}\right)}^r{x}^{0-4r}+{}^4{C}_r{3}^{4-r}{\left(\frac{-1}{3}\right)}^r{x}^{9-4r}+{}^4{C}_r{3}^{4-r}2{\left(\frac{-1}{3}\right)}^r{x}^{10-4r}$
For independent term of $x$
$8-4r=0$
$\Rightarrow r=2$
and $9-4r=0$
$r=\frac{9}{4}$ Not possible
and $10-4r=0$
$r=\frac{5}{2}$ Not possible
term$={}^4{C}_2\cdot 3^2\times \frac{1}{3^2}=6$