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4-2.Quadratic Equations and Inequations
hard
If $x$ is real and $k = \frac{{{x^2} - x + 1}}{{{x^2} + x + 1}},$ then
A
$\frac{1}{3} \le k \le 3$
B
$k \ge 5$
C
$k \le 0$
D
None of these
Solution
(a) 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 ==> ${(k + 1)^2} – 4{(k – 1)^2} \ge 0$
==> $3{k^2} – 10k + 3 \ge 0$
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 $3$}
Standard 11
Mathematics
