MCQ
If the co $-$ efficient of $x$ in $\Big(\text{x}^2+\frac{\lambda}{\text{x}}\Big)^5$is $270$, then $ \lambda=$
  • $3$
  • B
    $4$
  • C
    $5$
  • D
    None of these

Answer

Correct option: A.
$3$
Now, the co $-$ efficient of $x$ in $\Big(\text{x}^2+\frac{\lambda}{\text{x}}\Big)^5$is ${^6}\text{C}_{3}\cdot(\lambda)^3$or $10(\lambda)^3.$
According to the problem $10(\lambda)^{3}=270,$ or $(\lambda)=3.$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free