Question

If the term independent of x in the expansion of $(2+5x+a{x}^3){\left(\frac{3}{2}{x}^2-\frac{1}{3x}\right)}^{9}is1,$

• A. 3
• B. 4
• C. 5
• D. -6

Answer 1

The correct option is D -6

$(2+5x+a{x}^3){(\frac{3}{2}{x}^2-\frac{1}{3x})}^9$

expanding this

$T_{r+1}={{}^{n}C}_r{a}^{r-l}{b}^r$

$=(2+5x+a{x}^3)[{{}^{9}C}_0{\left(\frac{3}{2}{x}^2\right)}^{9-0}{(-\frac{1}{3x})}^0+.........+{{}^{9}C}_6{\left(\frac{3}{2}{x}^2\right)}^3{(-\frac{1}{3x})}^6+{{}^{9}C}_7{\left(\frac{3}{2}{x}^2\right)}^2{(-\frac{1}{3x})}^7+......]$

To find term independent of $x$, (the term is as follows)

$=2\times {{}^{9}C}_6{\left(\frac{3}{2}\right)}^2{\left(x^2\right)}^2.{(-\frac{1}{3})}^6{\left(\frac{1}{x}\right)}^6+a{x}^3{{}^{9}C}_7{\left(\frac{3}{2}\right)}^2{\left(x^2\right)}^2.{(-\frac{1}{3})}^7{\left(\frac{1}{x}\right)}^7$

$=2\times \frac{9\times 8\times 7}{3\times 2\times 1}{\left(\frac{3}{2}\right)}^3\times x^6{(-\frac{1}{3})}^6\frac{1}{x^6}+a{x}^3\times \frac{9\times 8}{2\times 1}\times {\left(\frac{3}{2}\right)}^2.\left(x^4\right){(-\frac{1}{3})}^7\frac{1}{x^2}=+\frac{7}{9}+\frac{(-a)}{3^3}=1$

given in the question value is $1$)

$\Rightarrow \frac{3\times 7}{3\times 9}-1=\frac{9}{27}\Rightarrow \frac{21-27}{27}=\frac{9}{27}\Rightarrow \frac{-6}{27}=\frac{9}{27}$

$\therefore a=-6$