Question

Find the term independent of $x$ in the expansion of ${\left(\frac{\sqrt{x}}{\sqrt{3}}+\frac{\sqrt{3}}{2x^2}\right)}^{10}$.

• A. $\frac{1}{12}$
• B. $\frac{7}{12}$
• C. $\frac{5}{12}$
• D. $\frac{11}{12}$

Answer 1

Answer: (C)

Let $(r+1{)}^{th}$ term be independent of $x$.
$T_{r+1}={}^{10}{C}_r{\left(\sqrt{\frac{x}{3}}\right)}^{10-r}{\left(\frac{\sqrt{3}}{2x^2}\right)}^r$
$={}^{10}{C}_r{\left(\frac{x}{3}\right)}^{\frac{10-r}{2}}{3}^{\frac{r}{2}}\left(\frac{1}{2^r{x}^{2r}}\right)$
$={}^{10}{C}_r{3}^{\frac{r}{2}-\frac{10-r}{2}}{2}^{-r}{x}^{\frac{10-r}{2}-2r}$
Since the term is independent of $x$, we have
$\frac{10-r}{2}-2r=0$
$\Rightarrow r=2$
$\therefore T_3={}^{10}{C}_2\frac{3^{-3}}{4}$
$=\frac{10\times 9}{2\times 1}\times \frac{1}{9\times 12}=\frac{5}{12}$