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

  • $\begingroup$ Is the range for a $(-\infty,9/4)$ $\endgroup$ – Archis Welankar Oct 30 '16 at 14:20

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 $-\infty$.

  • $\begingroup$ Note that for $a\ge\dfrac{9}{4}$ the range of $f$ falls entirely within the interval $[-1,1]$. $\endgroup$ – John Wayland Bales Oct 31 '16 at 4:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.