Question

If the coefficient of $x$ in ${(x^2+\frac{\lambda }{x})}^5$ is $270$, then $\lambda =$

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

Answer: (A)

$3$

The coefficient of $x$ in ${(x^2+\frac{\lambda }{x})}^5$ is ${}^5{C}_3.(\lambda {)}^3=10(\lambda {)}^3$.

According to the problem,

$10{\lambda }^3=270$

or, $\lambda =27$

or, $\lambda =3$.