Question

If the constants term in the expansions of ${\left(\sqrt{x}-\frac{k}{x^2}\right)}^{10}$ is $405$ , then what can be the value of $k$ ?

• A. $1$
• B. $9$
• C. $2$
• D. $3$

Answer 1

The correct option is D

$3$

Explanation for the correct option.

Step 1: Find the term

Given, ${\left(\sqrt{x}-\frac{k}{x^2}\right)}^{10}$the general term is given by:
$$\begin{gathered} {\mathrm{T}}_{\mathrm{r}+1}={}^{10}\mathrm{C}_{\mathrm{r}}(\sqrt{\mathrm{x}}{)}^{10-\mathrm{r}}{\left(-\frac{\mathrm{k}}{{\mathrm{x}}^2}\right)}^{\mathrm{r}} \\ ={}^{10}\mathrm{C}_{\mathrm{r}}(\mathrm{x}{)}^{\frac{1}{2}(10-\mathrm{r})}(-\mathrm{k}{)}^{\mathrm{r}}{\mathrm{x}}^{-2\mathrm{r}} \end{gathered}$$
For a constant term, the power of $x$must be $0$.

$$\begin{gathered} \Rightarrow \frac{10-5\mathrm{r}}{2}=0 \\ \Rightarrow r=2 \end{gathered}$$

Step 2: Find the value of $k$

$\begin{array}{rcl}{\mathrm{T}}_{2+1}& =& {}^{10}\mathrm{C}_2(-\mathrm{k}{)}^2\\ 405& =& {}^{10}\mathrm{C}_2(-\mathrm{k}{)}^2\\ \frac{10\times 9\times 8!}{2!\times 8!}(-\mathrm{k}{)}^2& =& 405\\ 45{\mathrm{k}}^2& =& 405\\ {\mathrm{k}}^2& =& 9\\ \mathrm{k}& =& \pm 3\end{array}$

Hence, option D is correct.