A) \[\frac{1}{3}\le k\le 3\]
B) \[k\ge 5\]
C) \[k\le 0\]
D) None of these
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 \[3{{k}^{2}}-10k+3\ge 0\] Which is possible only when the value of k lies between the roots of the equation \[3{{k}^{2}}-10k+3=0\] That is, when \[\frac{1}{3}\le k\le 3\] {Since roots are \[\frac{1}{3}\] and 3}
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