Question Number 115857 by bemath last updated on 29/Sep/20
$${What}\:{are}\:{all}\:{real}\:{values}\:{of}\:{p}\:{for} \\ $$ $${which}\:{the}\:{inequality}\: \\ $$ $$−\mathrm{3}<\frac{{x}^{\mathrm{2}} +{px}−\mathrm{2}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}<\mathrm{2}\:{is}\:{satisfied}\: \\ $$ $${by}\:{all}\:{real}\:{values}\:{of}\:{x} \\ $$
Answered by bobhans last updated on 29/Sep/20
$$\:{consider}\:{x}^{\mathrm{2}} −{x}+\mathrm{1}\:\rightarrow\begin{cases}{\Delta=\mathrm{1}−\mathrm{4}.\mathrm{1}<\mathrm{0}}\\{{a}=\mathrm{1}}\end{cases} \\ $$ $${so}\:+{ve}\:{definite} \\ $$ $${inequality}\:{similar}\:{to}\: \\ $$ $$\Rightarrow\:−\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{3}<{x}^{\mathrm{2}} +{px}−\mathrm{2}<\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2} \\ $$ $$\begin{cases}{−\mathrm{4}{x}^{\mathrm{2}} +\left(\mathrm{3}−{p}\right){x}−\mathrm{1}<\mathrm{0}\:\:\wedge}\\{−{x}^{\mathrm{2}} +\left({p}+\mathrm{2}\right){x}−\mathrm{4}<\mathrm{0}}\end{cases} \\ $$ $$\begin{cases}{\left(\mathrm{3}−{p}\right)^{\mathrm{2}} −\mathrm{4}.\mathrm{4}.\mathrm{1}<\mathrm{0}\:\wedge}\\{\left({p}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{4}.\mathrm{1}.\mathrm{4}<\mathrm{0}}\end{cases} \\ $$ $$\begin{cases}{\left(\mathrm{3}−{p}+\mathrm{4}\right)\left(\mathrm{3}−{p}−\mathrm{4}\right)<\mathrm{0}\:\wedge}\\{\left({p}+\mathrm{2}+\mathrm{4}\right)\left({p}+\mathrm{2}−\mathrm{4}\right)<\mathrm{0}}\end{cases} \\ $$ $$\begin{cases}{\left(\mathrm{7}−{p}\right)\left(−\mathrm{1}−{p}\right)<\mathrm{0}\:\wedge}\\{\left({p}+\mathrm{6}\right)\left({p}−\mathrm{2}\right)\:<\mathrm{0}}\end{cases} \\ $$ $$\Rightarrow\:−\mathrm{1}<{p}<\mathrm{7}\:\wedge\:−\mathrm{6}<{p}<\mathrm{2} \\ $$ $$\therefore\:−\mathrm{1}\:<\:{p}\:<\:\mathrm{2}\:. \\ $$ $$ \\ $$