Question

The positive value of $\lambda$ for which the co-efficient of $x^2$ in the expression $x^2{(\sqrt{x}+\frac{\lambda }{x^2})}^{10}$ is $720$, is :

• A. $\sqrt{5}$
• B. $2\sqrt{2}$
• C. $3$
• D. $4$

Answer 1

The correct option is D $4$

$x^2{(\sqrt{x}+\frac{\lambda }{x^2})}^{10}$
$\Rightarrow x^2\left\{{}^{10}{C}_r\right(x{)}^{\frac{(10-r)}{2}}\cdot {\lambda }^r{x}^{-2r}\}$
$\Rightarrow x^2[{}^{10}{C}_r\cdot {\lambda }^r\cdot x^{\frac{(10-5r)}{2}}]$
coefficient of $x^2$ means that $x^{\frac{(10-5r)}{2}}$ has to be $x^0$
$\therefore r=2$
$\Rightarrow {}^{10}{C}_2{\lambda }^2=720$
$\Rightarrow {\lambda }^2=16$
$\Rightarrow \lambda =\pm 4$
$\therefore$ positive value of $\lambda$ is 4.