JEE MAIN - Mathematics (2018 (Offline) - No. 3)

If $$\alpha ,\beta \in C$$ are the distinct roots of the equation
x2 - x + 1 = 0, then $${\alpha ^{101}} + {\beta ^{107}}$$ is equal to :
2
-1
0
1

Explanation

Given equation,

x2 $$-$$ x + 1 = 0

Roots of this equation

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

$$\therefore\,\,\,$$ $$ \propto \, = \,{{1 + \sqrt 3 \,i} \over 2}$$

and $$\beta = \,{{1 - \sqrt 3 \,i} \over 2}$$

We know;

$$\omega = {{ - 1 + \sqrt 3 \,i} \over 2} = - \left( {{{1 - \sqrt 3 \,i} \over 2}} \right) = - \beta $$

and $${\omega ^2} = {{ - 1 - \sqrt 3 \,i} \over 2} = - \left( {{{1 + \sqrt 3 \,i} \over 2}} \right) = - \propto $$

$$\therefore\,\,\,$$ $$ \propto \, = - {\omega ^2}$$ and $$\beta \, = \, - \omega $$

$$\therefore\,\,\,$$ $${ \propto ^{101}} + {\beta ^{107}}$$

$$ = {\left( { - {\omega ^2}} \right)^{101}} + {\left( { - \omega } \right)^{107}}$$

$$ = {\left( { - 1} \right)^{101}}.{\left( {{\omega ^2}} \right)^{101}} + {\left( { - 1} \right)^{107}}.{\left( \omega \right)^{107}}$$

$$ = - 1.{\left( {{\omega ^2}} \right)^{101}} - {\omega ^{107}}$$

$$ = - \left( {{\omega ^{202}} + {\omega ^{107}}} \right)$$

$$ = - \left( {{\omega ^{3.67}}.\omega + {\omega ^{3.35}}.{\omega ^2}} \right)$$

$$ = - \left( {\omega + {\omega ^2}} \right)\,\,\,$$ [ as $$\,\,\,$$ $${\omega ^{3n}} = 1$$]

$$ = - \left( { - 1} \right)$$ $$\,\,\,\,\,\,$$ [as $$\,\,\,$$ $$1 + \omega + {\omega ^2} = 0$$ ]

$$ = 1$$

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