MCQ
If the coefficient 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:

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

    We have,

    ${^\text{15}}\text{C}_{\text{r}}(\text{x}^{2})^{5-\text{r}}\ \Big(\frac{\lambda}{\text{x}}\Big)^{\text{r}}$

    $={^\text{15}}\text{C}_{\text{r}}\ \lambda^{\text{r}}\ \text{x}^{10-2\text{r}-\text{r}}$

    For it to contain x, we must have

    $10-3\text{r}=1$

    $\Rightarrow \text{r}=3$

    Coefficient of x in the given expansion,

    $={^\text{15}}\text{C}_{\text{3}}\ \lambda^{\text{3}}=10\lambda^{3}$

    Now, we have

    $10\lambda^{3}=270$

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

    $\Rightarrow \lambda=3$

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