Question

If the coefficient 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

The correct option is A

3

The coefficient of x in the given expansion where x occurs at the (r+1)th term.

We have,

${}^5{C}_r(x^2{)}^{5-r}{\left(\frac{\lambda }{x}\right)}^r$

$={}^5{C}_r{\lambda }^r{x}^{10-2r-r}$

For it to contain x, we must have:

$10-3r=1\Rightarrow r=3$

$\therefore$ Coefficient of x in the given expansion:

${}^5{C}_3{\lambda }^3=10{\lambda }^3$

Now, we have

$10{\lambda }^3=270$

$\Rightarrow {\lambda }^3=27$

$\Rightarrow \lambda =3$