Q48 / 160 · TS EAMCET 2023 (12 May Shift 1) · Mathematics

If the term independent of $x$ in the expansion of $\left(\sqrt{\mathrm{x}}-\frac{\mathrm{k}}{\mathrm{x}^2}\right)^{10}$ is 405 , then $k=$

  • A
    $\pm 1$
  • B
    $0$
  • Correct answer
    C
    $\pm 3$
  • D
    $\pm 5$

Solution

$\left(\sqrt{x}-\frac{K}{x^2}\right)^{10}$ $\begin{aligned} & T_{r+1}={ }^{10} C_r(\sqrt{x})^{10-r}\left(\frac{-K}{x^2}\right)^r \\ & ={ }^{10} C_r(-K)^r x^{\frac{10-r}{2}-2 r} \Rightarrow \frac{10-r}{2}-2 r=0 \\ & r=2 \\ & \therefore T_3={ }^{10} C_2(\sqrt{x})^8\left(\frac{-K}{x^2}\right)^2 \\ & \quad T_3={ }^{10} C_2 \cdot K^2=405 \\ & \Rightarrow 45 K^2=405 \\ & \Rightarrow K^2=9 \Rightarrow K= \pm 3 .\end{aligned}$

Hence, the correct answer is option: 'C'.

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