# The positive value of

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. (1) 4

2. (2) $2 \sqrt{2}$

3. (3) $\sqrt{5}$

4. (4) 3

Correct Option: 1

Solution:

Since, coefficient of $x^{2}$ in the expression $x^{2}\left(\sqrt{x}+\frac{\lambda}{x^{2}}\right)$

is a constant term, then

Coefficient of $x^{2}$ in $x^{2}\left(\sqrt{x}+\frac{\lambda}{x^{2}}\right)^{10}$

$=$ co-efficient of constant term in $\left(\sqrt{x}+\frac{\lambda}{x^{2}}\right)^{10}$

General term in $\left(\sqrt{x}+\frac{\lambda}{x^{2}}\right)^{10}={ }^{10} C_{r}(\sqrt{x})^{10-r}\left(\frac{\lambda}{x^{2}}\right)^{r}$

$={ }^{10} C_{r}(x)^{\frac{10-r}{2}-2 r} \cdot \lambda^{2}$

Then, for constant term,

$={ }^{10} C_{r}(x)^{\frac{10-r}{2}-2 r} \cdot \lambda^{2}$

Then, for constant term,

$\frac{10-r}{2}-2 r=0 \Rightarrow r=2$

Co-efficient is $x^{2}$ in expression $={ }^{10} C_{2} \lambda^{2}=720$

$\Rightarrow \lambda^{2}=\frac{720}{5 \times 9}=16$

$\lambda=4$]

Hence, required value of $\lambda$ is 4 .