Question

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

• A. $2,-2$
• B. $3,-3$
• C. $4,-4$
• D. $1,-1$

Answer 1

The correct option is B $3,-3$

${(r+1)}^{th}$ term of ${(\sqrt{x}-\frac{k}{x^2})}^{10}$ is ${}^{10}{C}_r({\sqrt{x})}^{10-r}({-\frac{k}{x^2})}^r.$
$T_{r+1}{=}^{10}{C}_r({\sqrt{x})}^{10-r}({-\frac{k}{x^2})}^r$
If the (r+1)th term is independent of x, then
$\frac{10-r}{2}-2r=0$
$\Rightarrow r=2$
Hence, ${}^{10}{C}_2{(-k)}^2=405$
$\Rightarrow 45{(-k)}^2=405$
$\Rightarrow k=\pm 3$