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

Find the set of values of x which satisfy the inequalities   (2/(x−1))≤(1/x)  and  x^2 −∣3x∣+2<0

$${Find}\:{the}\:{set}\:{of}\:{values}\:{of}\:{x}\:{which}\:{satisfy}\:{the}\:{inequalities}\: \\ $$ $$\frac{\mathrm{2}}{{x}−\mathrm{1}}\leqslant\frac{\mathrm{1}}{{x}}\:\:{and}\:\:{x}^{\mathrm{2}} −\mid\mathrm{3}{x}\mid+\mathrm{2}<\mathrm{0} \\ $$

Commented byPrithwish sen last updated on 30/Jun/19

(2/(x−1))≤(1/x) ⇒x≤−1  x^2 −∣3x∣ +2 <0  ⇒ x^2 +3x+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  ⇒ −2<x<−1              ⇒ −1<x<−2   ∵ x≤ −1  ∴  the only possible solution is                                       −2<x<−1  please check the answer.

$$\frac{\mathrm{2}}{\mathrm{x}−\mathrm{1}}\leqslant\frac{\mathrm{1}}{\mathrm{x}}\:\Rightarrow\mathrm{x}\leqslant−\mathrm{1} \\ $$ $$\mathrm{x}^{\mathrm{2}} −\mid\mathrm{3x}\mid\:+\mathrm{2}\:<\mathrm{0}\:\:\Rightarrow\:\mathrm{x}^{\mathrm{2}} +\mathrm{3x}+\mathrm{2}<\:\mathrm{0}\:\left(\:\because\:\mathrm{x}\leqslant−\mathrm{1}\right) \\ $$ $$\mathrm{i}.\mathrm{e}\:\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}+\mathrm{2}\right)<\:\mathrm{0} \\ $$ $$\mathrm{either}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{or} \\ $$ $$\:\:\left(\mathrm{x}+\mathrm{1}\right)<\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{x}+\mathrm{1}\right)>\mathrm{0} \\ $$ $$\mathrm{and}\left(\mathrm{x}+\mathrm{2}\right)>\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{and}\:\left(\mathrm{x}+\mathrm{2}\right)<\:\mathrm{0} \\ $$ $$\Rightarrow\:−\mathrm{2}<\mathrm{x}<−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:−\mathrm{1}<\mathrm{x}<−\mathrm{2}\: \\ $$ $$\because\:\mathrm{x}\leqslant\:−\mathrm{1}\:\:\therefore\:\:\mathrm{the}\:\mathrm{only}\:\mathrm{possible}\:\mathrm{solution}\:\mathrm{is} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2}<\mathrm{x}<−\mathrm{1} \\ $$ $$\mathrm{please}\:\mathrm{check}\:\mathrm{the}\:\mathrm{answer}. \\ $$

Commented byRio Michael last updated on 02/Jul/19

i think it makes sense..but explicit for O level students u know..  A level students can understand

$${i}\:{think}\:{it}\:{makes}\:{sense}..{but}\:{explicit}\:{for}\:{O}\:{level}\:{students}\:{u}\:{know}.. \\ $$ $${A}\:{level}\:{students}\:{can}\:{understand} \\ $$

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

(e_1 ) for x≠0 and x≠1    (2/(x−1)) ≤(1/x) ⇔(2/(x−1)) −(1/x) ≤0 ⇒((2x−x+1)/(x(x−1))) ≤0 ⇒  ((x+1)/(x(x−1))) ≤0  x          −∞                 −1                  0                1               +∞  x+1                     −        0        +                 +              +  x(x−1)              +                     +     0         −    0       +  ((x+1)/(x(x−1)    ))         −                     +               −                 +  ⇒ D_1 = ]−∞ ,−1] ∪  ]0,1[  (e_2 )           x^2 −3∣x∣ +2 <0 ⇒∣x∣−3∣x∣ +2 <0  Δ =9−8 =1  ⇒ ∣x∣_1 =((3+1)/2) =2 ⇒x =+^− 2  ∣x∣_2 =((3−1)/2) =1 ⇒x =+^− 1 ⇒x^2 −3∣x∣+2 =(∣x∣−1)(∣x∣−2)  (e_2 ) ⇒ (∣x∣−1)(∣x∣−2) <0  ⇒∣x∣ ∈]1,2[ ⇒ 1<∣x∣<2                 1<∣x∣ ⇒x>1 or x<−1 ⇒x ∈]−∞,−1[∪]1,+∞[  ∣x∣<2 ⇒−2<x<2 ⇒ ∩ =]−2,−1[∪]1,2[ =D_2  ⇒  D =D_1 ∩D_2 =....

$$\left({e}_{\mathrm{1}} \right)\:{for}\:{x}\neq\mathrm{0}\:{and}\:{x}\neq\mathrm{1}\:\:\:\:\frac{\mathrm{2}}{{x}−\mathrm{1}}\:\leqslant\frac{\mathrm{1}}{{x}}\:\Leftrightarrow\frac{\mathrm{2}}{{x}−\mathrm{1}}\:−\frac{\mathrm{1}}{{x}}\:\leqslant\mathrm{0}\:\Rightarrow\frac{\mathrm{2}{x}−{x}+\mathrm{1}}{{x}\left({x}−\mathrm{1}\right)}\:\leqslant\mathrm{0}\:\Rightarrow \\ $$ $$\frac{{x}+\mathrm{1}}{{x}\left({x}−\mathrm{1}\right)}\:\leqslant\mathrm{0} \\ $$ $${x}\:\:\:\:\:\:\:\:\:\:−\infty\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\infty \\ $$ $${x}+\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:+\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\:\:\:\:\:\:\:\:\:\:\:\:\:+ \\ $$ $${x}\left({x}−\mathrm{1}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:−\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:+ \\ $$ $$\frac{{x}+\mathrm{1}}{{x}\left({x}−\mathrm{1}\right)\:\:\:\:}\:\:\:\:\:\:\:\:\:−\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+ \\ $$ $$\left.\Rightarrow\left.\:\left.{D}_{\mathrm{1}} =\:\right]−\infty\:,−\mathrm{1}\right]\:\cup\:\:\right]\mathrm{0},\mathrm{1}\left[\right. \\ $$ $$\left({e}_{\mathrm{2}} \right)\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} −\mathrm{3}\mid{x}\mid\:+\mathrm{2}\:<\mathrm{0}\:\Rightarrow\mid{x}\mid−\mathrm{3}\mid{x}\mid\:+\mathrm{2}\:<\mathrm{0} \\ $$ $$\Delta\:=\mathrm{9}−\mathrm{8}\:=\mathrm{1}\:\:\Rightarrow\:\mid{x}\mid_{\mathrm{1}} =\frac{\mathrm{3}+\mathrm{1}}{\mathrm{2}}\:=\mathrm{2}\:\Rightarrow{x}\:=\overset{−} {+}\mathrm{2} \\ $$ $$\mid{x}\mid_{\mathrm{2}} =\frac{\mathrm{3}−\mathrm{1}}{\mathrm{2}}\:=\mathrm{1}\:\Rightarrow{x}\:=\overset{−} {+}\mathrm{1}\:\Rightarrow{x}^{\mathrm{2}} −\mathrm{3}\mid{x}\mid+\mathrm{2}\:=\left(\mid{x}\mid−\mathrm{1}\right)\left(\mid{x}\mid−\mathrm{2}\right) \\ $$ $$\left.\left({e}_{\mathrm{2}} \right)\:\Rightarrow\:\left(\mid{x}\mid−\mathrm{1}\right)\left(\mid{x}\mid−\mathrm{2}\right)\:<\mathrm{0}\:\:\Rightarrow\mid{x}\mid\:\in\right]\mathrm{1},\mathrm{2}\left[\:\Rightarrow\:\mathrm{1}<\mid{x}\mid<\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\right. \\ $$ $$\left.\mathrm{1}<\mid{x}\mid\:\Rightarrow{x}>\mathrm{1}\:{or}\:{x}<−\mathrm{1}\:\Rightarrow{x}\:\in\right]−\infty,−\mathrm{1}\left[\cup\right]\mathrm{1},+\infty\left[\right. \\ $$ $$\left.\mid{x}\mid<\mathrm{2}\:\Rightarrow−\mathrm{2}<{x}<\mathrm{2}\:\Rightarrow\:\cap\:=\right]−\mathrm{2},−\mathrm{1}\left[\cup\right]\mathrm{1},\mathrm{2}\left[\:={D}_{\mathrm{2}} \:\Rightarrow\right. \\ $$ $${D}\:={D}_{\mathrm{1}} \cap{D}_{\mathrm{2}} =.... \\ $$

Answered by ajfour last updated on 30/Jun/19

   1<∣x∣<2      { ((1<x<2)),((x ≤−1)) :}      or     { ((−2<x<−1)),((x ≤−1)) :}  ⇒  x∈φ            or   ⇒ −2< x <−1  hence  x∈(−1,−2).

$$\:\:\:\mathrm{1}<\mid{x}\mid<\mathrm{2} \\ $$ $$\:\:\:\begin{cases}{\mathrm{1}<{x}<\mathrm{2}}\\{{x}\:\leqslant−\mathrm{1}}\end{cases}\:\:\:\:\:\:{or}\:\:\:\:\begin{cases}{−\mathrm{2}<{x}<−\mathrm{1}}\\{{x}\:\leqslant−\mathrm{1}}\end{cases} \\ $$ $$\Rightarrow\:\:{x}\in\phi\:\:\:\:\:\:\:\:\:\:\:\:{or}\:\:\:\Rightarrow\:−\mathrm{2}<\:{x}\:<−\mathrm{1} \\ $$ $${hence}\:\:{x}\in\left(−\mathrm{1},−\mathrm{2}\right). \\ $$

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