Question

For what real values of $a$ does the range of the function $y=\frac{x-1}{a-x^2+1}$ not contain any value belonging to the interval $[-1,-\frac{1}{3}]?$

Answer 1

$y=\frac{x-1}{a-x^2+1}$

For range,

$\Rightarrow -y{x}^2+ay+y=x-1$

$\Rightarrow y{x}^2-ay-y+x-1=0$

$D\equiv 1^2-4\left(y\right)(-1)(ay+y+1)\ge 0$(for real values)

$\Rightarrow 1+4a{y}^2+4y^2+4y\ge 0$

$\Rightarrow 4(a+1)y^2+4y+1\ge 0$

$\Rightarrow (y-\frac{-4\pm \sqrt{16-16(a+1{)}^2}}{2\left(4\right)(a+1)}\ge 0$

$\Rightarrow (y-\frac{-1\pm \sqrt{1-a^2-1-2a}}{2(a+1)})\ge 0$

$\Rightarrow (y-\frac{-1\pm \sqrt{-a^2-2a}}{2(a+1)})\ge 0$

$\Rightarrow y\in (-\infty ,\frac{-1-\sqrt{-a^2-2a}}{2(a+1)}]\cup \left[\frac{-1-\sqrt{-a^2-2a}}{2(a+1)}\right)$

For, not belonging to $[-1,\frac{-1}{3}]$,

$\frac{-1-\sqrt{-a^2-2a}}{2(a+1)}<-1$

$\Rightarrow \frac{-1-\sqrt{-a^2-2a}}{<}-2(a+1)$

$\Rightarrow -a^2-2a>\left(2\right(a+1)-1{)}^2$ (On squaring both sides)

$\Rightarrow 4a^2+8a+4-4a-4+1+a^2+2a<0$

$\Rightarrow 5a^2+6a+1<0$

$\Rightarrow (a+\frac{1}{5})(a+1)<0$

$\Rightarrow a\in (-1,\frac{-1}{5})$

$\frac{-1-\sqrt{-a^2-2a}}{2(a+1)})>-\frac{1}{3}$

$\Rightarrow 3\sqrt{-a^2-2a}>-2(a+1)+3$

$\Rightarrow 9(-a^2-2a)>9+4a^2+8a=4-12a-12$

$\Rightarrow 13a^2+14a+1<0$

$\Rightarrow (a+\frac{1}{13})(a+1)<0$

$a\in (-1,-\frac{1}{13})$ Equation 2

From equation 1& 2, $a\in (-1,-\frac{1}{5})$