Q61 / 160 · TS EAMCET 2023 (14 May Shift 2) · Mathematics
The term independent of $x$ in the expansion of
$\left(1-3 x+2 x^3\right)\left(\frac{3 x^2}{2}-\frac{1}{3 x}\right)^9$ is
- A$\frac{7}{18}$
- B$\frac{5}{18}$
- C$\frac{19}{54}$
- Correct answerD$\frac{17}{54}$
Solution
$\left(1-3 x+2 x^3\right)\left(\frac{3 x^2}{2}-\frac{1}{3 x}\right)^9$
for $\left(\frac{3 x^2}{2}-\frac{1}{3 x}\right)^9$
$\mathrm{T}_{r+1}={ }^9 \mathrm{C}_{\mathrm{r}}\left(\frac{3 \mathrm{x}^2}{2}\right)^{9-\mathrm{r}}\left(\frac{-1}{3 \mathrm{x}}\right)^{\mathrm{r}}$
$={ }^9 \mathrm{C}_{\mathrm{r}}\left(\frac{3}{2}\right)^{9-\mathrm{r}}\left(\frac{-1}{3}\right)^{\mathrm{r}} \mathrm{x}^{18-3 \mathrm{r}}$
for coeff. of $x^0: 18-3 r=0 \Rightarrow r=6$
$\therefore$ Constant term $={ }^9 \mathrm{C}_6\left(\frac{3}{2}\right)^3\left(\frac{-1}{3}\right)^6=\frac{7}{18}$
Coeff. of $x^{-1}=18-3 r=-1$
$\begin{aligned} & 3 r=19 \\ & r=\frac{19}{3} \text { (not possible) }\end{aligned}$
Coeff. of $x^{-3}: 18-3 r=-3 \Rightarrow 3 r=21 \Rightarrow r=7$
$\therefore$ Term independent of $x=1 \times$ constant term $+2 \times$ Coeff. of $x^{-3}$
$=1 \times \frac{7}{18}+2 \times \frac{-1}{27}=\frac{21-4}{54}=\frac{17}{54}$
Hence, the correct answer is option: 'D'.