Q49 / 160 · TS EAMCET 2020 (14 Sep Shift 1) · Mathematics

The roots of the equation $(x-1)^5=32(x+1)^5$ are

  • Correct answer
    A
    $\frac{1+2 e^{\frac{2 k \pi i}{5}}}{1-2 e^{\frac{2 k \pi i}{5}}}, k=1,2,3,4,5$
  • B
    $\frac{1-2 e^{\frac{2 k \pi i}{5}}}{1+2 e^{\frac{2 k \pi i}{5}}}, k=0,1,2,3,4$
  • C
    $1,2 \omega, 3 \omega^2, 2 \omega+3 \omega^2, 5 \omega^2+7$
  • D
    $\frac{3+2 e^{\frac{2(k+1) \pi i}{5}}}{3-2 e^{\frac{2(k+1) \pi i}{5}}}, k=0,1,2,3,4$

Solution

We have, $(x-1)^5=32(x+1)^5$ $\Rightarrow\left(\frac{x-1}{x+1}\right)^5=32 \Rightarrow \frac{x-1}{x+1}=(32)^{1 / 5} \Rightarrow \frac{x-1}{x+1}=2 e^{\frac{2 k \pi i}{5}}$ Apply componendo and dividendo $\frac{2}{2 x}=\frac{1-2 e^{\frac{2 k \pi i}{5}}}{1+2 e^{\frac{2 k \pi i}{5}}} \Rightarrow x=\frac{1+2 e^{\frac{2 k \pi i}{5}}}{1-2 e^{\frac{2 k \pi i}{5}}}$

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

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