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

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(A) Given $k = \frac{x^2 - x + 1}{x^2 + x + 1}$.Rearranging the terms,we get $k(x^2 + x + 1) = x^2 - x + 1$.$kx^2 + kx + k = x^2 - x + 1$.$(k - 1)x^2

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

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(A) Given $k = \frac{x^2 - x + 1}{x^2 + x + 1}$.Rearranging the terms,we get $k(x^2 + x + 1) = x^2 - x + 1$.$kx^2 + kx + k = x^2 - x + 1$.$(k - 1)x^2 + (k + 1)x + (k - 1) = 0$.Since $x$ is real,the discriminant $D \ge 0$.$D = (k + 1)^2 - 4(k - 1)(k - 1) \ge 0$.$(k + 1)^2 - 4(k - 1)^2 \ge 0$.Using $a^2 - b^2 = (a - b)(a + b)$,we get $((k + 1) - 2(k - 1))((k + 1) + 2(k - 1)) \ge 0$.$(-k + 3)(3k - 1) \ge 0$.Multiplying by $-1$,we get $(k - 3)(3k - 1) \le 0$.The roots are $k = \frac{1}{3}$ and $k = 3$.Thus,$\frac{1}{3} \le k \le 3$.

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

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