If x is real and k = frac{{{x^2} - x + 1}}{{{ | vedclass

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Question

(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} -

Vedclass · Accepted Answer

$\frac{1}{3} \le k \le 3$

Vedclass · Answer

(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 $3$}

If $x$ is real and $k = \frac{{{x^2} - x + 1}}{{{x^2} + x + 1}},$ then

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