Solve this following

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 :

  1. $\sqrt{5}$

  2. 4

  3. $2 \sqrt{2}$

  4. 3


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)

 

 

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