Q54 / 160 · AP EAMCET 2023 (17 May Shift 2) · Mathematics

The coefficient of $x^2$ in the expansion of $(1-3 x)^{\frac{1}{3}}(1+2 x)^{-\frac{1}{2}}$ is

  • A
    $-\frac{3}{2}$
  • Correct answer
    B
    $\frac{3}{2}$
  • C
    $\frac{1}{2}$
  • D
    $-\frac{1}{2}$

Solution

$(1-3 x)^{\frac{1}{3}}(1+2 x)^{-\frac{1}{2}}$ $\begin{aligned} & =\left(1+\frac{1}{3}(-3 x)+\frac{\frac{1}{3}\left(\frac{1}{3}-1\right)}{2 !}(-3 x)^2+\ldots\right) \\ & \left.\qquad 1+\left(-\frac{1}{2}\right)(2 x)+\frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)}{2 !}(2 x)^2+\ldots\right) \\ & =\left(1-x-x^2+\ldots\right)\left(1-x+\frac{3}{2} x^2+\ldots\right)\end{aligned}$ The coefficient of $x^2=1 \times \frac{3}{2}+(-1) \times(-1)+1 \times(-1)$ $=\frac{3}{2}+1-1=\frac{3}{2}$

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

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