Complex Numbers And Quadratic Equations question 630
Question: Let $ x+\frac{1}{x}=1 $ and a, b and c are distinct positive integers such that $ ( x^{a}+\frac{1}{x^{a}} )+( x^{b}+\frac{1}{x^{b}} )+( x^{c}+\frac{1}{x^{c}} )=0. $ Then the minimum value of $ (a+b+c) $ is
Options:
A) 7
B) 8
C) 9
D) 10
Show Answer
Answer:
Correct Answer: C
Solution:
$ x+\frac{1}{x}=1 $ or $ x^{2}-x+1=0 $
$ \therefore x=\frac{1}{2}\pm i\frac{\sqrt{3}}{2} $ or $ x={e^{\frac{i\pi }{3}}} $
$ \therefore x^{a}+{x^{-a}}={e^{\frac{ia\pi }{3}}}+{e^{\frac{-ia}{3}}}=2\cos \frac{a\pi }{3} $ Hence, $ \cos \frac{a\pi }{3}+\cos \frac{b\pi }{3}+\cos \frac{c\pi }{3}=0 $ $ a,b,c\in I\therefore a+b+c{|_{\min }}=(1+3+5)=9 $
BETA
