Question:
The positive value of $\lambda$ for which the co-efficient of $x^{2}$ in the expression
$x^{2}\left(\sqrt{x}+\frac{\lambda}{x^{2}}\right)^{10}$ is 720, is :
Correct Option: , 2
Solution:
$x^{2}\left({ }^{10} C_{r}(\sqrt{x})^{10-r}\left(\frac{\lambda}{x^{2}}\right)^{r}\right)$
$x^{2}\left[{ }^{10} C_{r}(x)^{\frac{10-r}{2}}(\lambda)^{\mathrm{r}}(x)^{-2 r}\right]$
$\mathrm{X}^{2}\left[{ }^{10} \mathrm{C}_{\mathrm{r}} \lambda^{\mathrm{r}} \mathrm{X}^{\frac{10-5 \mathrm{r}}{2}}\right]$
$\therefore \quad \mathrm{r}=2$
Hence, ${ }^{10} \mathrm{C}_{2} \lambda^{2}=720$
$\lambda^{2}=16$
$\lambda=\pm 4$
Option (2)