Question

If the range of the function $f(x)=\frac{x-1}{p-x^2+1}$ does not contain any values belonging to the interval $\left[-1, \frac{-1}{3}\right]$ then the true set of values of $p$, is

• A. $(-\infty,-1)$
• B. $\left(-\infty, \frac{-1}{4}\right)$
• C. $(0, \infty)$
• D. $(-\infty, 0)$

Answer 1

Answer: (B)

We have $y=\frac{x-1}{p-x^2+1} \Rightarrow p y-x^2 y+y=x-1 \Rightarrow x^2 y+x-y(p+1)-1=0$
As, $x \in R$, so $D \geq 0 \Rightarrow 1+4 y(y(p+1)+1) \geq 0 \Rightarrow 4 y^2(p+1)+4 y+1 \geq 0$
Since $y \neq\left[-1, \frac{-1}{3}\right]$
So, $ 4 y ^2( p +1)+4 y +1< 0 \forall y \in\left[-1, \frac{-1}{3}\right] $
$\Rightarrow(2 y +1)^2+4 y ^2 p < 0$
$\Rightarrow p <-1\left(\frac{2 y +1}{2 y }\right)^2 \forall y \in\left[-1, \frac{-1}{3}\right]$
Hence $p< -\left.\left(\frac{2 y+1}{2 y}\right)^2\right|_{\min }$
Now max. value of $\left(\frac{2 y+1}{2 y}\right)^2$ occurs at $y=-1$ and is equal to $\frac{-1}{4}$.
$\therefore p <\frac{-1}{4} \Rightarrow \text { B }$