Q179 / 270 · NDA 2019 (Phase 2) · Mathematics
If the constant term in the expansion of $\left(\sqrt{x}-\frac{k}{x^2}\right)^{10}$ is 405 , then what can be the values of $k$ ?
- A$\pm 2$
- Correct answerB$\pm 3$
- C$\pm 5$
- D$\pm 9$
Solution
$\left(\sqrt{x}-\frac{k}{x^2}\right)^{10}$
according to question. Term independent of $x$ can be calculated as
$\begin{aligned}
& { }^{10} C_r(\sqrt{x})^{10-r} \cdot\left(-\frac{k}{x^2}\right)^r \\
& \Rightarrow x^{5-r / 2-2 r}=x^0 \\
& \Rightarrow 10-4 r-r=0 \\
& \Rightarrow r=2 \\
& \therefore \quad{ }^{10} C_2 \cdot(\sqrt{x})^8 \cdot\left(\frac{-k}{x^2}\right)^2=405 \\
& \Rightarrow \quad{ }^{10} C_2 \cdot k^2=405 \\
& \Rightarrow \frac{10!}{2!\cdot 8!} \cdot k^2=405 \\
& \Rightarrow k^2=9 \\
& \Rightarrow k= \pm 3
\end{aligned}$
Hence, the correct answer is option: 'B'.