QJEEM22Maafb2: JEE Main 2022 Mathematics - Class 11 JEEM22Maafb2 Solution: Let be a root of the equation 1 + x 2 + x 4 = 0. Then, the…

Given, $\alpha$ is a root of the equation 1 + x² + x⁴ = 0

$\therefore$ $\alpha$ will satisfy the equation.

$\therefore$ 1 + $\alpha$ ² + $\alpha$ ⁴ = 0

${\alpha ^2} = {{ - 1 \pm \sqrt {1 - 4} } \over 2}$

$= {{ - 1 \pm \sqrt 3 i} \over 2}$

$\therefore$ ${\alpha ^2} = \omega \,ar\,{\omega ^2}$

Now,

${\alpha ^{1011}} + {\alpha ^{2022}} - {\alpha ^{3033}}$

$= \alpha \,.\,{({\alpha ^2})^{505}} + {({\alpha ^2})^{1011}} - \alpha \,.\,{({\alpha ^2})^{1516}}$

$= \alpha {(\omega )^{505}} + {(\omega )^{1011}} - \alpha \,.\,{(\omega )^{1516}}$

$= \alpha \,.\,{({\omega ^3})^{168}}\,.\,\omega + {({\omega ^3})^{337}} - \alpha \,.\,{({\omega ^3})^{505}}\,.\,\omega$

$= \alpha \,\omega + 1 - \alpha \,\omega$

$= 1$