If the range of the real valued function f(x) | vedclass

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(C) Let $y = \frac{x^2+x+k}{x^2-x+k}$.Then $y(x^2-x+k) = x^2+x+k$.$x^2(y-1) - x(y+1) + k(y-1) = 0$.Since $x$ is real,the discriminant $D \ge 0$.$D = (

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$1$

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(C) Let $y = \frac{x^2+x+k}{x^2-x+k}$.Then $y(x^2-x+k) = x^2+x+k$.$x^2(y-1) - x(y+1) + k(y-1) = 0$.Since $x$ is real,the discriminant $D \ge 0$.$D = (y+1)^2 - 4(y-1)(k(y-1)) \ge 0$.$(y+1)^2 - 4k(y-1)^2 \ge 0$.$(y+1)^2 \ge 4k(y-1)^2$.For the range $\left[\frac{1}{3}, 3\right]$,the boundary values $y = \frac{1}{3}$ and $y = 3$ must satisfy the equality $(y+1)^2 = 4k(y-1)^2$.Substituting $y = 3$: $(3+1)^2 = 4k(3-1)^2 \implies 16 = 4k(4) \implies 16 = 16k \implies k = 1$.Checking for $y = \frac{1}{3}$: $(\frac{1}{3}+1)^2 = 4k(\frac{1}{3}-1)^2 \implies (\frac{4}{3})^2 = 4k(-\frac{2}{3})^2 \implies \frac{16}{9} = 4k(\frac{4}{9}) \implies \frac{16}{9} = \frac{16k}{9} \implies k = 1$.

If the range of the real valued function $f(x) = \frac{x^2+x+k}{x^2-x+k}$ is $\left[\frac{1}{3}, 3\right]$,then $k=$

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