Confusing $c$

Rearranging (forming a quadratic in $x$ ), we get:

$(y-1)x^2+2(2y-1)x+3cy-c=0$

Since $x\in\mathbb R$ , discriminant of this quadratic must be non negative i.e:

$\not4(2y-1)^2-\not4(3cy-c)(y-1)\ge 0$

$\iff (4-3c)y^2+4(c-1)y-(c-1)\ge 0$

Now for this to hold for all $y\in \mathbb R$ , we need to insure leading coefficient is positive (i.e $c<\dfrac 43$ ) and discriminant is non positive i.e:

$4(c-1)^2+(c-1)(4-3c)\le 0$

$\iff (c-1)(4c-4+4-3c)\le 0$

$\iff (c-1)(c)\le 0$

$\iff c\in[0,1]$

$\color{#3D99F6}{(I)c=0}$

$y=\dfrac{\cancel x(x+2)}{\cancel x(x+4)}=1-\dfrac{2}{x+4},x\ne 0$

Its apparent that $y$ can take all real values except $1~(\small{\because \dfrac{2}{x+4}\ne 0})$ and also $\frac 12$ since $x\ne 0$ , so that range is not $\mathbb R$ . Hence $c\ne 0$

$\color{#3D99F6}{(II)c=1}$

$y=\dfrac{(x+1)^{\cancel 2}}{\cancel{(x+1)}(x+3)}=1-\dfrac{2}{x+3},x\ne -1$

Its apparent that $y$ can take all real values except $1~(\small{\because\dfrac{2}{x+3}\ne 0})$ and also $0$ since $x\ne -1$ , so that range is not $\mathbb R$ . Hence $c\ne 1$

Finally we conclude $\boxed{c\in(0,1)}$

$\large 0+1=\boxed 1$

Yo! akshat similar problem of kj sir