Question

If the term free from $x$ in the expansion of ${(\sqrt{x}-\frac{k}{x^2})}^{10}$ is $405$, then the value of $k$ is

• A. $\pm 1$
• B. $\pm 3$
• C. $\pm 4$
• D. $\pm 2$

Answer 1

The correct option is B $\pm 3$

General term in the expansion of $(\sqrt{x}-\frac{k}{x^2})$

$T_{r+1}{=}^{10}{C}_r(\sqrt{x}{)}^{10-r}{\left(\frac{-k}{x^2}\right)}^r$

${=}^{10}{C}_r{x}^{\frac{10-r}{2}}\cdot (-k{)}^r\cdot x^{-2r}$

${=}^{10}{C}_r(-k{)}^r{x}^{\left(\frac{10-5r}{2}\right)}$

The term is free from $x$.

Put $\frac{10-5r}{2}=0$

$\Rightarrow r=2$

Now, ${}^{10}{C}_2(-k{)}^2=405$

$\Rightarrow \frac{10\times 9}{1\times 2}\cdot k^2=405$

$\Rightarrow k^2=\frac{405}{45}=9$

$\Rightarrow k=\pm 3$.