Question

If the coefficient of x in ${\left(x^2+\frac{\lambda }{x}\right)}^5$ is 270, then $\lambda =$
(a) 3
(b) 4
(c) 5
(d) none of these

Answer 1

(a) 3

$$\begin{gathered} \mathrm{The}\mathrm{coefficient}\mathrm{of}x\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{expansion}\mathrm{where}x\mathrm{occurs}\mathrm{at}\mathrm{the}(r+1)\mathrm{th}\mathrm{term}. \\ \mathrm{We}\mathrm{have} \\ {}^{5}C_r(x^2{)}^{5-r}{\left(\frac{\lambda }{x}\right)}^r \\ ={}^{5}C_r{\lambda }^r{x}^{10-2r-r} \\ \mathrm{For}\mathrm{it}\mathrm{to}\mathrm{contain}x,\mathrm{we}\mathrm{must}\mathrm{have}: \\ 10-3r=1 \\ \Rightarrow r=3 \\ \therefore \mathrm{Coefficient}\mathrm{of}x\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{expansion}: \\ {}^{5}C_3{\lambda }^3=10{\lambda }^3 \\ \mathrm{Now},\mathrm{we}\mathrm{have} \\ 10{\lambda }^3=270 \\ \Rightarrow {\lambda }^3=27 \\ \Rightarrow \lambda =3 \end{gathered}$$