Question

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

• A. $m=2$
• B. $m=-2$
• C. $m=3$
• D. $m=-3$

Answer 1

Answer: (C), (D)

$m=-3$

The general term in the given expansion is given by
$T_{r+1}=(-1{)}^r\times {}^{10}{C}_r\times {\left(\sqrt{x}\right)}^{10-r}\times {\left(\frac{m}{x^2}\right)}^r$
$=(-1{)}^r\times {}^{10}{C}_r\times x^{(5-\frac{r}{2})}\times \frac{m^r}{x^{2r}}$
$=(-1{)}^r\times {}^{10}{C}_r\times x^{(5-\frac{r}{2}-2r)}\times m^r$
$=(-1{)}^r\times {}^{10}{C}_r\times x^{(5-\frac{5r}{2})}\times m^r.$

Let $T_{r+1}$ be free from $x$.

Then, the power of $x$ in $T_{r+1}$ must be $0$.

$\therefore 5-\frac{5r}{2}=0\Rightarrow \frac{5r}{2}=5\Rightarrow r=2\Rightarrow r+1=3.$
So, $T_3$ will be free from $x$.

Now, $T_3=T_{(2+1)}$
$=(-1{)}^2\times {}^{10}{C}_2\times x^0\times m^2$
$=(\frac{10\times 9}{2}\times m^2)=45m^2.$
But, it is given that the term independent of $x$ is $405$.

$\therefore 45m^2=405\Rightarrow m^2=9\Rightarrow m=\pm 3.$

Hence, $m=\pm 3.$