Q167 / 270 · NDA 2019 (Phase 2) · Mathematics

What is the value of $\left[\frac{1+\sqrt{3}}{2}\right]^{2019}+\left[\frac{1-\sqrt{3}}{2}\right]^{2019} ?$

  • A
    1
  • B
    -1
  • Correct answer
    C
    $2 i$
  • D
    $-2 i$

Solution

$\begin{aligned} & \text {}\left[\frac{i+\sqrt{3}}{2}\right]^{2019}+\left[\frac{i-\sqrt{3}}{2}\right]^{2019} \\ & \because-i \times i=1 \\ & \therefore\left[\frac{-i(-1+i \sqrt{3})}{2}\right]^{2019}+\left[\frac{-i(-1-i \sqrt{3})}{2}\right]^{2019} \\ & =-i^{2019}\left[\omega^{2019}+\left(\omega^2\right)^{2019}\right] \\ & =-i^3[1+i] \\ & =-(-i) 2=2 i\end{aligned}$

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

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