<s>[INST] real values of $a$ for which the range of function $ f(x) = \frac{x-1}{1-x^2-a}$ does not contain value from $\left[-1,1\right]$ 
All real values of $a$ for which the range of function $\displaystyle f(x) = \frac{x-1}{1-x^2-a}$ does not
contain any value from $\left[-1,1\right]$

$\bf{My\; Try::}$ Let $\displaystyle y = \frac{x-1}{1-x^2-a}\Rightarrow y-x^2y-ay=x-1$
So $$x^2y+x+y(1-a)-1=0$$
Now for real values of $y<-1\cup y>1$ equation has real roots
So $$1-4y^2(1-a)\geq 0\Rightarrow 1\geq 4y^2(1-a)\Rightarrow y^2\leq \frac{1}{4(1-a)}\;, a\neq 1$$
Now how can i solve it after that, Help required, Thanks
 [/INST] We want to find $a$ such that $|f(x)|=\bigg|\frac{x-1}{1-x^2-a}\bigg|\gt 1$ for every $x$ in the domain.


*

*If $a\not=0$, then $f(1)=\frac{0}{-a}=0$.

*If $a=0$, then $f(-2)=\frac{-2-1}{1-4-0}=1$.
Therefore, there is no such $a$.
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