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mihaela_0205
Pe: 2013-05-25T18:27:45+03:00 In: In:

concurs

Sambata viitoare particip la un concurs la matematica, iar pentru luni am de rezolvat problemele de pe documentul :

Si ca de obicei, am facut tot, exceptand ultima parte.. as avea nevoie macar de o indicatie la exercitiile 25, 26, 28, 29 si 30. multumesc!

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3 raspunsuri

  1. dan
    expert (VI)

  2. dan
    expert (VI)

  4. gunty
    maestru (V)

    28.

        \[ \begin{array}{l} x \in \left( { - \infty ,\frac{3}{2}} \right] \Rightarrow \\ \left\{ \begin{array}{l} x - \sqrt 3 < 0 \Rightarrow \left| {x - \sqrt 3 } \right| = - \left( {x - \sqrt 3 } \right) \\ x - \frac{3}{2} \le 0 \Rightarrow \left| {x - \frac{3}{2}} \right| = - \left( {x - \frac{3}{2}} \right) \\ \end{array} \right. \Rightarrow - \left( {x - \sqrt 3 } \right) - \left( {x - \frac{3}{2}} \right) = x - 2 \Leftrightarrow x = \frac{{7 + 2\sqrt 3 }}{6} \notin \left( { - \infty ,\frac{3}{2}} \right]. \\ \\ x \in \left( {\frac{3}{2},\sqrt 3 } \right] \Rightarrow \\ \left\{ \begin{array}{l} x - \sqrt 3 \le 0 \Rightarrow \left| {x - \sqrt 3 } \right| = - \left( {x - \sqrt 3 } \right) \\ x - \frac{3}{2} > 0 \Rightarrow \left| {x - \frac{3}{2}} \right| = x - \frac{3}{2} \\ \end{array} \right. \Rightarrow - \left( {x - \sqrt 3 } \right) + x - \frac{3}{2} = x - 2 \Leftrightarrow x = \sqrt 3 + \frac{1}{2} \notin \left( {\frac{3}{2},\sqrt 3 } \right]. \\ \\ x \in \left( {\sqrt 3 , + \infty } \right) \Rightarrow \\ \left\{ \begin{array}{l} x - \sqrt 3 > 0 \Rightarrow \left| {x - \sqrt 3 } \right| = x - \sqrt 3 \\ x - \frac{3}{2} > 0 \Rightarrow \left| {x - \frac{3}{2}} \right| = x - \frac{3}{2} \\ \end{array} \right. \Rightarrow x - \sqrt 3 + x - \frac{3}{2} = x - 2 \Leftrightarrow x = \sqrt 3 - \frac{1}{2} \notin \left( {\sqrt 3 , + \infty } \right). \\ \end{array} \]

    In concluzie, multimea A are 0 elemente.

    29.

        \[ \begin{array}{l} x \le f\left( x \right) - 2 \Leftrightarrow f\left( x \right) \ge x + 2_{\left( {*1} \right)} \\ \left. \begin{array}{l} f\left( {x - 3} \right) + 1 \le x \Leftrightarrow f\left( {x - 3} \right) \le x - 1 \\ x - 3 \to x \\ \end{array} \right\} \Rightarrow f\left( x \right) \le x + 2_{\left( {*2} \right)} \\ \left( {*1} \right),\left( {*2} \right) \Rightarrow f\left( x \right) = x + 2. \\ \end{array} \]

    * Daca nu se cere rezolvarea completa, ci doar incercuirea rezultatului corect, o alta modalitate este sa incerci toate cele 4 cazuri si cea care verifica inegalitatea este raspunsul corect. (Consider ca asa e mai simplu).

    30.

        \[ \begin{array}{l} Fie...x = y. \Rightarrow 2f\left( {x - x} \right) = f\left( x \right) - f\left( x \right) + x - x + 2 \Leftrightarrow f\left( 0 \right) = 1. \\ \left. \begin{array}{l} Fie...x = 2013,y = 0. \Rightarrow 2f\left( {2013} \right) = f\left( {2013} \right) - f\left( 0 \right) + 2013 - 0 + 2 \\ f\left( 0 \right) = 1 \\ \end{array} \right\} \Rightarrow f\left( {2013} \right) = 2014. \\ \end{array} \]

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