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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?

danny
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  • Well... when would that ever not be real? There's only one way that expression wouldn't be real. When would that be. For what values of c will that never happen? – fleablood Aug 01 '16 at 05:05
  • What are you asking? Do you want $c$ such that the expression is defined for all real numbers? Or do you want $c$ such that the function is surjective onto $\mathbb{R}$? – MathematicsStudent1122 Aug 01 '16 at 05:10

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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:enter image description here 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 enter image description here and for positive values of $c$, the graph is enter image description here and for values within the range, the graph is enter image description here

  • Using c = -6 as example. Then, the denominator becomes $x^2 + x – 12 = (x – 3)(x + 4)$. This means x must not be 3 or -4. The 1st graph clearly shows a jump at x = -4. Shouldn’t there be a jump at x = 3 also? – Mick Aug 02 '16 at 08:35
  • $x-3$ is a factor of the numerator as well when $c=-6$ and hence cancels out. –  Aug 02 '16 at 13:38
  • Even if the numerator and the denominator have the same common factor (and can be cancelled out), we still cannot ignore its presence. – Mick Aug 02 '16 at 13:51
  • Yes, but as x tends to 3, the limiting value is finite and hence we do not see the jump. –  Aug 02 '16 at 14:07
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    @Muralidharan Why do we need to check at $c=0$ and $c=-6$? What did we miss in previous steps so that we need to check at upper bound and lower bound of $c$? – Mathematics Jul 21 '17 at 19:30