Question

A function $R\to R$ is defined by $f\left(x\right)=\frac{\alpha x^2+6x-8}{\alpha +6x-8x^2}$. If $f$ is an onto function for $\alpha \in [P,Q]$ then $P+Q$ is

Answer 1

Since, $f:R\to R$ is an onto mapping.
$\therefore$ Range of f=R
$\Rightarrow$ $\frac{\alpha x^2+6x-8}{\alpha +6x-8x^2}$ assumes all real values of x.
Let $y=\frac{\alpha x^2+6x-8}{\alpha +6x-8x^2}$
Then, y assumes all real values for real values of x.
$\Rightarrow$ $\alpha y+6xy-8x^{2}y=\alpha x^2+6x-8$, $\forall yϵR$
$\Rightarrow$ $x^2(\alpha +8y)+6x(1-y)-(8+\alpha y)=0$, $\forall yϵR$
we know, above equation assumes all real values
$\Rightarrow$ $D\ge 0$
So, $36{(1-y)}^2+4(\alpha +8y)(8+\alpha y)\ge 0$
$\Rightarrow$ $4[9(1-2y+y^2)+(8\alpha +{\alpha }^{2}y+64y+8\alpha y^2)]\ge 0$
$\Rightarrow$ $[9-18y+9y^2+8\alpha +{\alpha }^{2}y+64y+8\alpha y^2]\ge 0$
$\Rightarrow$ $[y^2(8\alpha +9)+y({\alpha }^2+46)+(8\alpha +9)]\ge 0$
We know, if $a{x}^2+bx+c>0\forall x$ and a> 0$\Rightarrow$D< 0
So, ${({\alpha }^2+46)}^2-4(8\alpha +9)(8\alpha +9)\le 0$ and $(8\alpha +9)>0$
$\Rightarrow$ ${({\alpha }^2+46)}^2-{\left[2(8\alpha +9)\right]}^2\le 0$ and $\alpha >-9/8$
$\Rightarrow$ $({\alpha }^2+46-16\alpha -18)({\alpha }^2+46+16\alpha +18)\le 0$ and $\alpha >-9/8$
$({\alpha }^2-16\alpha +28)({\alpha }^2+16\alpha +64)\le 0$ and $\alpha >-9/8$
$\Rightarrow$ $(\alpha -14)(\alpha -2){(\alpha +8)}^2\le 0$ and $\alpha >-9/8$
$\alpha ϵ[2,14]\cup \{-8\}$ and $\alpha >-9/8$
Thus, $\alpha ϵ[2,14]$