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
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 $a$.
Given $$f(x)=\dfrac{x-1}{1-x^2-a}$$ for some $a$ notice that if for $x\ne0$ you multiply both the numerator and denominator by $\dfrac{1}{x}$ you obtain
$$ f(x)=\dfrac{1-\frac{1}{x}}{\frac{1}{x}-x-\frac{a}{x}}\text{ for }x\ne0 $$
Now notice what happens to the value of $f(x)$ for large values of $x$. Regardless of the size of $a$ one may make both $\frac{1}{x}$ and $\frac{a}{x}$ as small as one pleases, so small that they are negligible. Thus for sufficiently large positive or negative values of $x$ the graph of $f(x)$ approaches the graph of $y=-\frac{1}{x}$ which has the $x$ axis as a horizontal asymptote.
Thus, the $x$-axis is also the horizontal asymptote of $f(x)$. So regardless of the value of the variable $a$, the range of $f$ must always contain numbers in the interval $[-1,1]$.
Once you get to calculus there is an even easier way to show that the graph of $f$ approaches the $x$-axis as $x$ approaches either $+\infty$ or $-\infty$.
