JEE MAIN - Mathematics (2022 - 29th June Evening Shift - No. 1)
Let $$\alpha$$ be a root of the equation 1 + x2 + x4 = 0. Then, the value of $$\alpha$$1011 + $$\alpha$$2022 $$-$$ $$\alpha$$3033 is equal to :
1
$$\alpha$$
1 + $$\alpha$$
1 + 2$$\alpha$$
Explanation
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,
$${\alpha ^{1011}} + {\alpha ^{2022}} - {\alpha ^{3033}}$$
$$ = \alpha \,.\,{({\alpha ^2})^{505}} + {({\alpha ^2})^{1011}} - \alpha \,.\,{({\alpha ^2})^{1516}}$$
$$ = \alpha {(\omega )^{505}} + {(\omega )^{1011}} - \alpha \,.\,{(\omega )^{1516}}$$
$$ = \alpha \,.\,{({\omega ^3})^{168}}\,.\,\omega + {({\omega ^3})^{337}} - \alpha \,.\,{({\omega ^3})^{505}}\,.\,\omega $$
$$ = \alpha \,\omega + 1 - \alpha \,\omega $$
$$ = 1$$
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