Mathematics - Functions Question with Solution | TestHub

Let $y=f(x)=\frac{x-1}{c-x^2+1}$ Take $y=-t$, where $t\in \left[\frac{1}{3},1\right]$

$$\begin{array}{rlr}& \therefore -t=\frac{x-1}{c-x^2+1}& \Rightarrow x^2-c-1=\frac{x-1}{t}\Rightarrow x^2-\frac{1}{t}x+\frac{1}{t}-c-1=0\end{array}$$

As $-t\in \left[-1,-\frac{1}{3}\right]$, hence the above must not possess real solution

$$\therefore {\left(\frac{1}{t}\right)}^2-4\left(\frac{1}{t}-c-1\right)<0\Rightarrow \frac{1}{t^2}-\frac{4}{t}+4\le -4c$$

$$\Rightarrow c<-\frac{1}{4}{\left(\frac{1}{t}-2\right)}^2$$

Now, $\frac{1}{3}\le t\le l\Rightarrow 1\le \frac{1}{t}-2\le 1\Rightarrow -\frac{1}{4}\le \frac{1}{4}{\left(\frac{1}{t}-2\right)}^2\le 0$

Hence, $c\in \left(-\infty ,-\frac{1}{4}\right)$