<s>[INST] The value of parameter $a$ for which $\dfrac{ax^2+3x-4}{a+3x-4x^2}$ takes all real values for $x\in R$ are: => The value of parameter $a$ for which $\dfrac{ax^2+3x-4}{a+3x-4x^2}$

takes all real values for $x\in R$ are:

My question is why we need to validate end points i.e. $1,7$ (Refer the last part of my attempt)
My attempt is as follows:-
$$y=\dfrac{ax^2+3x-4}{a+3x-4x^2}$$
$$ya+3yx-4yx^2=ax^2+3x-4$$
$$x^2(-4y-a)+x(3y-3)+ya+4=0$$
As $x$ can be any real, so $D\ge0$
$$9y^2+9-18y-4(ya+4)(-4y-a)\ge0$$
$$9y^2+9-18y+4(4y^2a+ya^2+16y+4a)\ge0$$
$$y^2(9+16a)+y(4a^2+64-18)+9+16a\ge0$$
$$y^2(9+16a)+y(4a^2+46)+9+16a\ge0$$
As range is $R$, so discriminant of quadratic in $y$ should be less than equal to zero
$$4(2a^2+23)^2-4(9+16a)^2\le0$$
$$(2a^2+23-9-16a)(2a^2+23+9+16a)\le0$$
$$(2a^2-16a+14)(2a^2+16a+32)\le0$$
$$(a^2-8a+7)(a^2+8a+16)\le0$$
$$a\in[1,7]$$
But in such type of questions, we always check at endpoints like here we need to check at $a=1$ and $a=7$. But I don't understand what is so special about endpoints.
From the above calculation I can only say at $a=1,7$ discriminant of quadratic in $y$ is zero, but what is so special about this. Please help me in this.
 [/INST] Let's denote $f(x)=ax^2+3x−4$, $g(x)=-4x^2+3x+a$, and $h(x)=f(x)/g(x)$.
Since $a>0>-9/16>-4$, $f(x)$ and $g(x)$ cannot have the same roots, and each of them has two distinct roots. However, they may have a common root. If $a=1,7$ then there exists $y_0\in \mathbb{\mathbb{R}}$ for which only one $x$ may give $h(x)=y_0$. This $x$ may be the root of the denominator, so we must check. Else, when $1<a<7$, there exists two distinct candidates $x_1,x_2$ which may give $h(x)=y$. In this case at least one of $x_1,x_2$ is not a root for $g$, since otherwise they will be the roots of $f$.
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