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Q.
If ' $x$ ' is real, then $\frac{x^2-x+c}{x^2+x+2 c}$ can take all real values if :
Complex Numbers and Quadratic Equations
Solution:
Let $y=\frac{x^2-x+c}{x^2+x+2 c} ; x \in R$ and $y \in R_{\text {. }}$
$\Rightarrow (y-1) x^2+(y+1) x+2 y c-c=0$
$\because x \in R$
$\Rightarrow D \geq 0 $
$\Rightarrow (y+1)^2-4 c(y-1)(2 y-1) \geq 0$
$\Rightarrow y^2+1+2 y-4 c\left[2 y^2-3 y+1\right] \geq 0$
$\Rightarrow (1-8 c) y^2+(2+12 c) y+1-4 c \geq 0$........(1)
Now for all $y \in R(1)$ will be true if
$1-8 c>0 \Rightarrow c<\frac{1}{8} \text { and } D \leq 0$
$\Rightarrow 4(1+6 c)^2-4(1-8 c)(1-4 c) \leq 0 $
$ \Rightarrow 1+36 c^2+12 c-1-32 c^2+12 c \leq 0 $
$ \Rightarrow 4 c^2+24 c \leq 0 $
$ \Rightarrow -6 \leq c \leq 0$
But $c=-6$ and $c=0$ will not satisfy given condition
$\therefore c \in(-6,0)$