Xavier Aptitude Test 2019 Paper

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Question : 65 of 98

Marks: +1, -0

If

x^2 + x + 1 = 0

, then

x^2018 + x^2019

equals which of the following:

x
- x
None of the others
x - 1
x + 1

Solution:

x^2+ x +1 = 0

⇒x^2 = - x - 1

⇒x^3 = - x ^2- x =1

Also,

x^2+ x +1 = 0

⇒x^2+ 1 =-x

Now,

x^2018 + x^2019 = x^2016

(x^2 + x^3)

=(x^3)^{672}(x^2+1)

=(1)^672(-x)=-x

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