Q124 / 200 · TS EAMCET 2001 · Mathematics

If $\frac{x-4}{x^2-5 x-2 k}=\frac{2}{x-2}-\frac{1}{x+k^{\prime}}$, then $k$ is equal to

  • Correct answer
    A
    $-3$
  • B
    $-2$
  • C
    $2$
  • D
    $3$

Solution

We have, $\begin{aligned} \frac{x-4}{x^2-5 x-2 k} & =\frac{2}{x-2}-\frac{1}{x+k} \\ \Rightarrow \quad \frac{x-4}{x^2-5 x-2 k} & =\frac{2(x+k)-(x-2)}{(x-2)(x+k)} \\ & =\frac{x+2 k+2}{(x-2)(x+k)}\end{aligned}$ On equating, we get $\begin{aligned}-4 & =2 k+2 \\ 2 k & =-6 \Rightarrow k=-3\end{aligned}$

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

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