QJEEM19Mad7e5: JEE Main 2019 Mathematics - Class 11 JEEM19Mad7e5 Solution: The positive value of for which the co-efficient of x 2 in the…

The general term in the expansion of the binomial expression $(a+b)^n$ is

$T_{r+1}={ }^n C_r a^{n-r} b^r$

Therefore, the general term in the expansion of the binomial expression

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

$\begin{aligned}T_{r+1} & =x^2\left({ }^{10} C_r(\sqrt{x})^{10-r}\left(\frac{\lambda}{x^2}\right)^r\right) \\\\& ={ }^{10} C_r x^2 \cdot x^{\frac{10-r}{2}} \lambda^r x^{-2 r} \\\\& ={ }^{10} C_r \lambda^r x^{2+\frac{10-r}{2}-2 r}\end{aligned}$

Now, for the coefficient of $x^2$ ,

$\begin{aligned}2+\frac{10-r}{2}-2 r =2 \\\\\Rightarrow \frac{10-r}{2}-2 r =0 \\\\\Rightarrow 10-r =4 r \Rightarrow r=2\end{aligned}$

So, the coefficient of $x^2$ is ${ }^{10} C_2 \lambda^2=720$

$\begin{aligned}\Rightarrow & \frac{10 !}{2 ! 8 !} \lambda^2 =720 \\\\\Rightarrow & \frac{10 \cdot 9 \cdot 8 !}{2 \cdot 8 !} \lambda^2 =720 \\\\\Rightarrow & 45 \lambda^2 =720 \\\\\Rightarrow & \lambda^2 =16 \\\\\Rightarrow & \lambda = \pm 4\end{aligned}$

Given the problem asks for the positive value of $\lambda$ , $\lambda = 4$ .

So, the correct option is (A) 4.