Question

A function $f:R-\left\{\beta \right\}\to R,$ where $\beta =\frac{3\pm \sqrt{9+8\alpha }}{8},$ is defined by $f\left(x\right)=\frac{\alpha x^2+6x-8}{\alpha +6x-8x^2}$. Then the possible value(s) of $\alpha$ for which $f$ is onto.

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $3$
B $4$
C $5$
D $6$

$y=\frac{\alpha x^2+6x-8}{\alpha +6x-8x^2}$
$\Rightarrow (\alpha +8y)x^2+6(1-y)x-8-y\alpha =0$
$x\in R\Rightarrow \Delta \ge 0$
$36(y-1{)}^2+4(\alpha +8y\left)\right(\alpha y+8)\ge 0$
$\Rightarrow (8\alpha +9)y^2+({\alpha }^2+46)y+(8\alpha +9)\ge 0$
$y\in R\Rightarrow 8\alpha +9\ge 0,\Delta \le 0$
$(\alpha +8{)}^2(\alpha -2\left)\right(\alpha -14)\le 0$
$\Rightarrow \alpha \in [2,14]\cup \{-8\}$