Find-the-set-of-values-of-x-which-satisfy-the-inequalities-2-x-1-1-x-and-x-2-3x-2-lt-0-

Question Number 63162 by Rio Michael last updated on 29/Jun/19

F i n d t h e s e t o f v a l u e s o f x w h i c h s a t i s f y t h e i n e q u a l i t i e s

\frac{2}{x - 1} ⩽ \frac{1}{x} a n d x^{2} - ∣ 3 x ∣ + 2 < 0

Commented by Prithwish sen last updated on 30/Jun/19

\frac{2}{x - 1} ⩽ \frac{1}{x} \Rightarrow x ⩽ - 1

x^{2} - ∣ 3 x ∣ + 2 < 0 \Rightarrow x^{2} + 3 x + 2 < 0 (∵ x ⩽ - 1)

i . e (x + 1) (x + 2) < 0

either or

(x + 1) < 0 (x + 1) > 0

and (x + 2) > 0 and (x + 2) < 0

\Rightarrow - 2 < x < - 1 \Rightarrow - 1 < x < - 2

∵ x ⩽ - 1 ∴ the only possible solution is

- 2 < x < - 1

please check the answer .

Commented by Rio Michael last updated on 02/Jul/19

i t h i n k i t m a k e s s e n s e . . b u t e x p l i c i t f o r O l e v e l s t u d e n t s u k n o w . .

A l e v e l s t u d e n t s c a n u n d e r s t a n d

Commented by mathmax by abdo last updated on 30/Jun/19

(e_{1}) f o r x \neq 0 a n d x \neq 1 \frac{2}{x - 1} ⩽ \frac{1}{x} \Leftrightarrow \frac{2}{x - 1} - \frac{1}{x} ⩽ 0 \Rightarrow \frac{2 x - x + 1}{x (x - 1)} ⩽ 0 \Rightarrow

\frac{x + 1}{x (x - 1)} ⩽ 0

x - \infty - 1 0 1 + \infty

x + 1 - 0 + + +

x (x - 1) + + 0 - 0 +

\frac{x + 1}{x (x - 1)} - + - +

\Rightarrow D_{1} =] - \infty, - 1] \cup] 0, 1 [

(e_{2}) x^{2} - 3 ∣ x ∣ + 2 < 0 \Rightarrow∣ x ∣ - 3 ∣ x ∣ + 2 < 0

Δ = 9 - 8 = 1 \Rightarrow ∣ x ∣_{1} = \frac{3 + 1}{2} = 2 \Rightarrow x = \bar{+} 2

∣ x ∣_{2} = \frac{3 - 1}{2} = 1 \Rightarrow x = \bar{+} 1 \Rightarrow x^{2} - 3 ∣ x ∣ + 2 = (∣ x ∣ - 1) (∣ x ∣ - 2)

(e_{2}) \Rightarrow (∣ x ∣ - 1) (∣ x ∣ - 2) < 0 \Rightarrow∣ x ∣ \in] 1, 2 [\Rightarrow 1 <∣ x ∣< 2

1 <∣ x ∣ \Rightarrow x > 1 o r x < - 1 \Rightarrow x \in] - \infty, - 1 [\cup] 1, + \infty [

∣ x ∣< 2 \Rightarrow - 2 < x < 2 \Rightarrow \cap =] - 2, - 1 [\cup] 1, 2 [= D_{2} \Rightarrow

D = D_{1} \cap D_{2} = \dots .

Answered by ajfour last updated on 30/Jun/19

1 <∣ x ∣< 2

{\begin{cases} 1 < x < 2 \\ x ⩽ - 1 \end{cases} o r {\begin{cases} - 2 < x < - 1 \\ x ⩽ - 1 \end{cases}

\Rightarrow x \in ϕ o r \Rightarrow - 2 < x < - 1

h e n c e x \in (- 1, - 2) .

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