Q59 / 160 · AP EAMCET 2016 · Mathematics

To find the coefficient of $x^4$ in the expansion of $\frac{3 x}{(x-2)(x-1)}$, the interval in which the expansion isvalid, is

  • A
    $-2 < x < \infty$
  • B
    $-\frac{1}{2} < x < \frac{1}{2}$
  • Correct answer
    C
    $-1 < x < 1$
  • D
    $-\infty < x < \infty$

Solution

Given, $\frac{3 x}{(x-2)(x-1)}$ can be written as $\frac{6}{x-2}-\frac{3}{x-1}$ $\begin{aligned} & \therefore \frac{3 x}{(x-2)(x-1)}=\frac{6}{x-2}-\frac{3}{x-1} \\ & =6(x-2)^{-1}-3(x-1)^{-1} \\ & =-3\left(1-\frac{x}{2}\right)+3(x-1)^{-1} \end{aligned}$ It is valid iff $\left|\frac{x}{2}\right| < 1$ and $|x| < 1$ $\begin{aligned} & \Rightarrow \quad|x| < 2 \text { and }|x| < 1 \\ & \Rightarrow \quad x \in(-2,2) \text { and } x \in(-1,1) \quad \therefore \quad x \in(-1,1) \end{aligned}$ Hence, $-1 < x < 1$

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

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