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

    Answer

    1. 3

    Solution:

    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 $(\lambda)=3.$

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