the question asks for the interval in which c lies so that $\frac{x^2-x+c}{x^2+x+2c}$ gives all real values for all x belongs to R.
how to proceed in this problem?
[Math] If ‘x’ is real then $\frac{x^2-x+c}{x^2+x+2c}$ can take all real values if
quadratics
Best Answer
Let $y = \frac{x^2-x+c}{x^2+x+2c} $ be a value taken by the given fraction. This gives \begin{equation*} x^2(y-1) +x(y+1) +c(2y-1) = 0 \end{equation*} should have a real root. Thus \begin{equation*} (y+1)^2 -4c(y-1)(2y-1) \geq 0 \end{equation*} and hence \begin{equation*} y^2(1-8c) + 2y(1+6c)+(1-4c) \geq 0 \end{equation*} If the given expression takes all values, then the above should be true for all $y$. Thus we have \begin{equation*} (1+6c)^2 - (1-8c)(1-4c) \leq 0 \end{equation*} and $8c \leq 1$. The above expression simplifies to $24c+4c^2 \leq 0$ and hence $-6 \leq c \leq 0$. when $c=-6$, the graph is:
Clearly, $y=1$ is an asymptote and hence the expression misses the value 1. Again, the same phenomenon is observed when $c=0$. Thus the required range is $-6 < c < 0$
For negative values outside the range, the graph is 
and for positive values of $c$, the graph is 
and for values within the range, the graph is 