Question

The coefficient of the term independent of $x$ in ${\left(\left[\sqrt{\left(\frac{x}{3}\right)}+\frac{\sqrt{3}}{x^2}\right]\right)}^{10}is$

• A. $\frac{5}{3}$
• B. $\frac{4}{5}$
• C. $6$
• D. $\frac{1}{2}$

Answer 1

Answer: (A)

Given, ${\left[\sqrt{\frac{x}{3}}+\frac{\sqrt{3}}{x^2}\right]}^{10}$
General term, $T_{r+1}={\left(\right)}^{10}{C}_r{\left(\frac{x}{3}\right)}^{\frac{1}{2}\left(10-r\right)}{\left(\frac{\sqrt{3}}{x^2}\right)}^r$
$⟹T_{r+1}={\left(\right)}^{10}{C}_r{\left(\frac{1}{3}\right)}^{\frac{10-r}{2}}{\left(\sqrt{3}\right)}^r{x}^{\frac{1}{2}\left(10-r\right)-2r}$
For the term independent of $x$ , put
$\frac{1}{2}\left(10-r\right)-2r=0$
$⟹r=2$
$\therefore T_{2+1}=T_3={\left(\right)}^{10}{C}_2{\left(\frac{1}{3}\right)}^{\frac{8}{2}}{\left(\sqrt{3}\right)}^2$
$=45\times \frac{1\times 3}{81}=\frac{5}{3}$