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TS EAMCET 4 May 2018 Shift 2 Solved Paper

Section: Mathematics

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Question : 10 of 160

Marks: +1, -0

If

\alpha

z^{2}-z+1=0

\left(\alpha^{2014}+\frac{1}{\alpha^{2014}}\right)+\left(\alpha^{2015}+\frac{1}{\alpha^{2015}}\right)^{2}

+\left(\alpha^{2016}+\frac{1}{\alpha^{2016}}\right)^{3}

+\left(\alpha^{2017}+\frac{1}{\alpha^{2017}}\right)^{4}+\left(\alpha^{2018}+\frac{1}{\alpha^{2018}}\right)^{5}=

8
5
3
−5

Solution:

Consider the equation,

z^{2}-z+1=0

This gives,

z=-\omega

Now,

\{\left((-\omega)^{2014}+\frac{1}{(-\omega)^{2014}}\right]+\left[(-\omega)^{2015}+\frac{1}{(-\omega)^{2015}}\right]^{2}

+\left[(-\omega)^{2016}+\frac{1}{(-\omega)^{2016}}\right]^{3}

+\left[(-\omega)^{2017}+\frac{1}{(-\omega)^{2017}}\right]^{4}

+\left[(-\omega)^{2018}+\frac{1}{(-\omega)^{2018}}\right]^{5})

=\left[-\omega-\frac{1}{\omega}\right]+\left[\omega^{2}+\frac{1}{\omega^{2}}\right]^{2}

+8+\left[-\omega+\frac{1}{-\omega}\right]^{4}+\left[\omega^{2}+\frac{1}{\omega^{2}}\right]^{5}

=\left[\omega+\omega^{2}\right]+\left[\omega^{2}+\omega\right]^{2}

+8+\left[-\omega-\omega^{2}\right]^{4}+\left[\omega^{2}+\omega\right]^{5}

=-1+1+8+1-1

=8

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