What-are-all-real-values-of-p-for-which-the-inequality-3-lt-x-2-px-2-x-2-x-1-lt-2-is-satisfied-by-all-real-values-of-x-

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

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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