Question

Am mare nevoie de ajutor la acestea.. intervale analiza clasa 11.. Rog​

Answer 1

 

Explicatii:

La fiecare din subpunctele problemei A1) avem cate un interval inchis la stanga si la dreapta, care contine si la stanga si la dreapta cate o expresie E1 si E2, in functie de x.

Se cere sa gasim valorile lui x pentru care intervalele sunt multimi nevide.

Conditia pentru a fi multime nevida este:

E1 < E2

Nu putem accepta intervalul  [5; 3]

Nici nu poate fi desenat pe axa Ox.

Putem accepta conditia

E1 ≤ E2

La egalitate vom avea de exemplu intervalul

[5; 5] care este egal cu  {5} si care are un singur element.

Rezulta ca e o multime nevida.

Rezolvare:

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$\displaystyle\bf\\a)~~I=[-2x+1;~1+3x]\\\\-2x+1\leq1+3x\\\\-2x+1-1-3x\leq0\\\\ -2x-3x\leq0\\\\-5x\leq0~~\Big|:(-5)~~(Se~schimba~sensul~inegalitatii)\\\\x\geq0\\\\x\in[0,~\infty)$

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$\displaystyle\bf\\\\b)~~I=\left[\frac{x}{x+1},~\frac{1}{x} \right]\\\\\\\frac{x}{x+1}\leq\frac{1}{x}\\\\x^2\leq1(x+1)\\\\x^2\leq x+1\\\\x^2-x-1<0\\\\a=1>0 \implies~x^2-x-1~~este~negativa~intre~radacini.\\\\x_{12}=\frac{1\pm\sqrt{1+4}}{2}=\frac{1\pm\sqrt{5}}{2}\\\\x_1=\frac{1-\sqrt{5}}{2}\\\\x_2=\frac{1+\sqrt{5}}{2}\\\\x\in\left[\frac{1-\sqrt{5}}{2},~~\frac{1+\sqrt{5}}{2}\right]-\{0;~-1\}$

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$\displaystyle\bf\\c)~~I=\left[\frac{x-1}{x+1},~\frac{2}{x+2}\right]\\\\\\\frac{x-1}{x+1}\leq\frac{2}{x+2}\\\\(x-1)(x+2)\leq2(x+1)\\\\x^2-x+2x-2\leq2x+2\\\\x^2-x+2x-2-2x-2\leq0\\\\x^2-x-4\leq0\\\\a=1>0 \implies~x^2-x-1~~este~negativa~intre~radacini.\\\\x_{12}=\frac{1\pm\sqrt{1+16}}{2}\\\\x_1=\frac{1-\sqrt{17}}{2}\\\\x_2=\frac{1+\sqrt{17}}{2}\\\\\\x\in\left[\frac{1-\sqrt{17}}{2},~~\frac{1+\sqrt{17}}{2}\right]-\{-1;~-2\}$