تمارين - 2BACSEF

التانية باكالوريا العلوم التجريبية – خيار فرنسي

درس : Fonctions Exponentielles

Exercice 1

  1. Simplifier les expressions :

    A=(ex)5e−2xA = (e^x)^5 e^{-2x}
    B=e2x+3e2x−1B = \frac{e^{2x+3}}{e^{2x-1}}
    C=ex+e−xe−xC = \frac{e^x + e^{-x}}{e^{-x}}

  1. Démontrer que pour tout réel xx :

    a) (ex+e−x)2−(ex−e−x)2=4(e^x + e^{-x})^2 - (e^x - e^{-x})^2 = 4

    b) ex−1ex+1=1−e−x1+e−x\dfrac{e^x-1}{e^x+1} = \dfrac{1 - e^{-x}}{1 + e^{-x}}

    c) e−x−e−2x=ex−1e2xe^{-x} - e^{-2x} = \dfrac{e^x - 1}{e^{2x}}

1/

A=(ex)5e−2x=e5x.e−2x=e5x−2x=e3x\begin{aligned} A &= (e^x)^5 e^{-2x}=e^{5x}.e^{-2x}\\ &=e^{5x-2x}=e^{3x} \end{aligned}

B=e2x+3e2x−1=e2x+3−2x+1=e4\begin{aligned} B &=\frac{e^{2x+3}}{e^{2x-1}}=e^{2x+3-2x+1}=e^4 \end{aligned}

C=ex+e−xe−x=exe−x+1=e2x+1\begin{aligned} C=\frac{e^x + e^{-x}}{e^{-x}}=\frac{e^x}{e^{-x}}+1=e^{2x}+1 \end{aligned}

2/a/

∙(ex+e−x)2=e2x+2ex.e−x+e−2x=e2x+2+e−2x ∙(ex−e−x)2=e2x−2ex.e−x+e−2x=e2x−2+e−2x\begin{aligned} \bullet\quad (e^x + e^{-x})^2&=e^{2x}+2e^x.e^{-x}+e^{-2x}\\ &=e^{2x}+2+e^{-2x}\\~\\ \bullet\quad (e^x - e^{-x})^2&=e^{2x}-2e^x.e^{-x}+e^{-2x}\\ &=e^{2x}-2+e^{-2x} \end{aligned}

et donc

(ex+e−x)2−(ex−e−x)2=2+2=4\begin{aligned} (e^x + e^{-x})^2 - (e^x - e^{-x})^2 =2+2=4 \end{aligned}

b/

ex−1ex+1=ex(1−1ex)ex(1+1ex)=1−e−x1+e−x\begin{aligned} \dfrac{e^x-1}{e^x+1} = \dfrac{e^{x}\left(1 - \dfrac{1}{e^x}\right)}{e^{x}\left(1 + \dfrac{1}{e^x}\right)} =\dfrac{1-e^{-x}}{1+e^{-x}} \end{aligned}

c/

e−x−e−2x=e2x(e−x−e−2x)e2x=ex−1e2x\begin{aligned} e^{-x} - e^{-2x} &= \dfrac{e^{2x}(e^{-x} - e^{-2x})}{e^{2x}} =\dfrac{e^x-1}{e^{2x}} \end{aligned}