What are all real values of p for which the inequality −3<((x^2 +px−2)/(x^2 −x+1))<2 is satisfied by all real values of x


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Question Number 115857 by bemath last updated on 29/Sep/20

$W h a t a r e a l l r e a l v a l u e s o f p f o r$ $w h i c h t h e i n e q u a l i t y$ $- 3 < \frac{x^{2} + p x - 2}{x^{2} - x + 1} < 2 i s s a t i s f i e d$ $b y a l l r e a l v a l u e s o f x$
	Answered by bobhans last updated on 29/Sep/20
	$consider x^2 −x+1 → { ((Δ=1−4.1<0)),((a=1)) :} so +ve definite inequality similar to ⇒ −3x^2 +3x−3<x^2 +px−2<2x^2 −2x+2 { ((−4x^2 +(3−p)x−1<0 ∧)),((−x^2 +(p+2)x−4<0)) :} { (((3−p)^2 −4.4.1<0 ∧)),(((p+2)^2 −4.1.4<0)) :} { (((3−p+4)(3−p−4)<0 ∧)),(((p+2+4)(p+2−4)<0)) :} { (((7−p)(−1−p)<0 ∧)),(((p+6)(p−2) <0)) :} ⇒ −1<p<7 ∧ −6<p<2 ∴ −1 < p < 2 .$
	$c o n s i d e r x^{2} - x + 1 \to {\begin{cases} Δ = 1 - 4.1 < 0 \\ a = 1 \end{cases}$ $s o + v e d e f i n i t e$ $i n e q u a l i t y s i m i l a r t o$ $\Rightarrow - 3 x^{2} + 3 x - 3 < x^{2} + p x - 2 < 2 x^{2} - 2 x + 2$ ${\begin{cases} - 4 x^{2} + (3 - p) x - 1 < 0 \land \\ - x^{2} + (p + 2) x - 4 < 0 \end{cases}$ ${\begin{cases} {(3 - p)}^{2} - 4.4.1 < 0 \land \\ {(p + 2)}^{2} - 4.1.4 < 0 \end{cases}$ ${\begin{cases} (3 - p + 4) (3 - p - 4) < 0 \land \\ (p + 2 + 4) (p + 2 - 4) < 0 \end{cases}$ ${\begin{cases} (7 - p) (- 1 - p) < 0 \land \\ (p + 6) (p - 2) < 0 \end{cases}$ $\Rightarrow - 1 < p < 7 \land - 6 < p < 2$ $∴ - 1 < p < 2 .$
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