Complex Numbers And Quadratic Equations question 170
Question: If $ x $ is real and $ k=\frac{x^{2}-x+1}{x^{2}+x+1}, $ then [MNR 1992; RPET 1997]
Options:
A) $ \frac{1}{3}\le k\le 3 $
B) $ k\ge 5 $
C) $ k\le 0 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
From $ k=\frac{x^{2}-x+1}{x^{2}+x+1} $ We have $ x^{2}(k-1)+x(k+1)+k-1=0 $ As given, x is real
Þ $ {{(k+1)}^{2}}-4{{(k-1)}^{2}}\ge 0 $
Þ $ 3k^{2}-10k+3\ge 0 $ Which is possible only when the value of k lies between the roots of the equation $ 3k^{2}-10k+3=0 $ That is, when $ \frac{1}{3}\le k\le 3 $ {Since roots are $ \frac{1}{3} $ and 3}
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