Question

The term independent of $x$ in the expansion ${(\frac{\sqrt{x}}{\sqrt{3}}+\frac{\sqrt{3}}{2x^2})}^{10}$ is equal to:

• A. $\frac{7}{12}$
• B. $\frac{1}{3}$
• C. $\frac{4}{7}$
• D. $\frac{5}{12}$

Answer 1

The correct option is D $\frac{5}{12}$

Let $T_{r+1}$ be the term independent of $x$.
Here, $T_{r+1}={}^{10}{C}_r\times {\left(\sqrt{\frac{x}{3}}\right)}^{10-r}\times {\left(\frac{\sqrt{3}}{2x^2}\right)}^r$
$\Rightarrow T_{r+1}={}^{10}{C}_r\times {\left(\frac{x}{3}\right)}^{\frac{10-r}{2}}\times 3^{\frac{r}{2}}\times \frac{1}{2^r{x}^{2r}}$
$={}^{10}{C}_r\times x^{\left(5-\frac{r}{2}-2r\right)}\times \frac{1}{3^{5-\frac{r}{2}}}\times 3^{\frac{r}{2}}\times 2^{-r}...\left(1\right)$
Now, $T_{r+1}$ to be independent of $x$
$5-\frac{5r}{2}=0\Rightarrow \frac{5r}{2}=5\Rightarrow r=2$
Thus, $T_3$ is independent of $x$.
Putting $r=2$ in $\left(1\right)$, we get
$T_3={}^{10}{C}_2\times 3^{2-5}\times 2^{-2}\times x^0=(\frac{10\times 9}{2\times 1}\times 3^{-3}\times 2^{-2})=\frac{5}{12}$
Hence, the term independent of $x$ in the given expansion is $\frac{5}{12}.$