Question

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

• A. $-3$
• B. $-2$
• C. $3$
• D. $2$

Answer 1

Answer: (A), (C)

$3$

$T_{r+1}={}^{10}{C}_r{\left(\sqrt{x}\right)}^{10-r}{\left(\frac{-k}{x^2}\right)}^r={}^{10}{C}_r{x}^{\left(\frac{10-r}{2}-2r\right)}(-k{)}^r$
For this to be independent of $x$, $\frac{10-r}{2}-2r=0\Rightarrow r=2$

$\therefore {}^{10}{C}_2(-k{)}^2=405\Rightarrow k=\pm 3$