Question

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

• A. $3$
• B. $4$
• C. $5$
• D. None of these

Answer 1

Answer: (A)

$3$

Now, the co-efficient of $x$ in ${(x^2+\frac{\lambda }{x})}^5$ is ${}^5{C}_3.(\lambda {)}^3$ or $10(\lambda {)}^3$.

According to the problem

$10(\lambda {)}^3=270$

or, $\lambda =3$.