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. $4$
• C. $2\sqrt{2}$
• D. $3$

Answer 1

Answer: (B)

$4$

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

the general term$=x^2\left({}^{10}{C}_r\right(\sqrt{x}{)}^{10-r}{\left(\frac{\lambda }{x^2}\right)}^2)$

$=x^2\left[{}^{10}{C}_r\right(x{)}^{\frac{10-r}{2}}\left(\lambda {)}^r\right(x{)}^{-2r}]$

$=\left[{}^{10}{C}_r{\lambda }^r{x}^{\frac{10-3r}{2}}\right]$

for the coefficient of $x^2$ is $\frac{10-3r}{2}=2$

$\therefore r=2$

Hence, ${}^{10}{C}_2{\lambda }^2=720$

${\lambda }^2=16$

$\lambda =\pm 4$