if the constant term in the binomial expansion
of $\left(\sqrt{x}-\frac{k}{x^{2}}\right)^{10}$ is 405, then $|k|$ equals :
Correct Option: , 3
$\left(\sqrt{\mathrm{x}}-\frac{\mathrm{k}}{\mathrm{x}^{2}}\right)^{10}$
$\mathrm{T}_{\mathrm{r}+1}={ }^{10} \mathrm{C}_{\mathrm{r}}(\sqrt{\mathrm{x}})^{10-\mathrm{r}}\left(\frac{-\mathrm{k}}{\mathrm{x}^{2}}\right)^{\mathrm{r}}$
$\mathrm{T}_{\mathrm{r}+1}={ }^{10} \mathrm{C}_{\mathrm{r}} \cdot \mathrm{x}^{\frac{10-\mathrm{r}}{2}} \cdot(-\mathrm{k})^{\mathrm{r}} \cdot \mathrm{x}^{-2 \mathrm{r}}$
$\mathrm{T}_{\mathrm{r}+1}={ }^{10} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{\frac{10-5 \mathrm{r}}{2}}(-\mathrm{k})^{\mathrm{r}}$
Constant term : $\frac{10-5 \mathrm{r}}{2}=0 \Rightarrow \mathrm{r}=2$
$\mathrm{T}_{3}={ }^{10} \mathrm{C}_{2} \cdot(-\mathrm{k})^{2}=405$
$\mathrm{k}^{2}=\frac{405}{45}=9$
$\mathrm{k}=\pm 3 \Rightarrow|\mathrm{k}|=3$