Q55 / 150 · MHT CET 2026 (13 April Shift 2) · Mathematics
If $x = \sqrt{-1-\sqrt{-1-\sqrt{-1-\ldots\infty}}}$, where $\omega$ is a non-real complex cube root of unity, then the value of $x$ is...
- A$1$
- B$-1$
- C$-\omega$
- Correct answerD$\omega^2$
Solution
Given $x = \sqrt{-1-\sqrt{-1-\sqrt{-1-\ldots\infty}}}$
Squaring both sides, we get:
$x^2 = -1 - \sqrt{-1-\sqrt{-1-\ldots\infty}}$
$x^2 = -1 - x$
$x^2 + x + 1 = 0$
Solving this quadratic equation, we get:
$x = \dfrac{-1 \pm \sqrt{1 - 4}}{2} = \dfrac{-1 \pm i\sqrt{3}}{2}$
Thus, $x = \omega$ or $x = \omega^2$, where $\omega$ is a non-real complex cube root of unity.
Among the given options, $\omega^2$ is the correct choice.
Answer: $\omega^2$
Squaring both sides, we get:
$x^2 = -1 - \sqrt{-1-\sqrt{-1-\ldots\infty}}$
$x^2 = -1 - x$
$x^2 + x + 1 = 0$
Solving this quadratic equation, we get:
$x = \dfrac{-1 \pm \sqrt{1 - 4}}{2} = \dfrac{-1 \pm i\sqrt{3}}{2}$
Thus, $x = \omega$ or $x = \omega^2$, where $\omega$ is a non-real complex cube root of unity.
Among the given options, $\omega^2$ is the correct choice.
Answer: $\omega^2$
Hence, the correct answer is option: 'D'.