JEE MAIN - Mathematics (2010 - No. 20)
If $$\alpha $$ and $$\beta $$ are the roots of the equation $${x^2} - x + 1 = 0,$$ then $${\alpha ^{2009}} + {\beta ^{2009}} = $$
$$\, - 1$$
$$\, 1$$
$$\, 2$$
$$\, - 2$$
Explanation
$${x^2} - x + 1 = 0$$
$$ \Rightarrow x = {{1 \pm \sqrt {1 - 4} } \over 2}$$
$$x = {{1 \pm \sqrt 3 i} \over 2}$$
$$\alpha = {1 \over 2} + i{{\sqrt 3 } \over 2} = - {\omega ^2}$$
$$\beta = {1 \over 2} - {{i\sqrt 3 } \over 2} = - \omega $$
$${\alpha ^{2009}} + {\beta ^{2009}} = {\left( { - {\omega ^2}} \right)^{2009}} + {\left( { - \omega } \right)^{2009}}$$
$$ = - {\omega ^2} - \omega = 1$$
$$ \Rightarrow x = {{1 \pm \sqrt {1 - 4} } \over 2}$$
$$x = {{1 \pm \sqrt 3 i} \over 2}$$
$$\alpha = {1 \over 2} + i{{\sqrt 3 } \over 2} = - {\omega ^2}$$
$$\beta = {1 \over 2} - {{i\sqrt 3 } \over 2} = - \omega $$
$${\alpha ^{2009}} + {\beta ^{2009}} = {\left( { - {\omega ^2}} \right)^{2009}} + {\left( { - \omega } \right)^{2009}}$$
$$ = - {\omega ^2} - \omega = 1$$
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