This exercise is a multiple choice questionnaire. For each of the six following questions, only one of the four proposed answers is correct.
A wrong answer, multiple answers, or absence of an answer to a question earns neither points nor deducts points.
- Consider the function $g$ defined and differentiable on $]0;+\infty[$ by: $$g(x) = \ln\left(x^2 + x + 1\right).$$ For every strictly positive real number $x$: a. $g^{\prime}(x) = \frac{1}{2x+1}$ b. $g^{\prime}(x) = \frac{1}{x^2+x+1}$ c. $g^{\prime}(x) = \ln(2x+1)$ d. $g^{\prime}(x) = \frac{2x+1}{x^2+x+1}$
- The function $x \longmapsto \ln(x)$ admits as an antiderivative on $]0;+\infty[$ the function: a. $x \longmapsto \ln(x)$ b. $x \longmapsto \frac{1}{x}$ c. $x \longmapsto x\ln(x) - x$ d. $x \longmapsto \frac{\ln(x)}{x}$
- Consider the sequence $(a_n)$ defined for all $n$ in $\mathbb{N}$ by: $$a_n = \frac{1 - 3^n}{1 + 2^n}.$$ The limit of the sequence $(a_n)$ is equal to: a. $-\infty$ b. $-1$ c. $1$ d. $+\infty$
- Consider a function $f$ defined and differentiable on $[-2;2]$. The variation table of the function $f^{\prime}$ derivative of the function $f$ on the interval $[-2;2]$ is given by:
| $x$ | $-2$ | $-1$ | $0$ | $2$ |
| variations of $f^{\prime}$ | $1$ | | | $>_{-2}^{-1}$ |
The function $f$ is: a. convex on $[-2;-1]$ b. concave on $[0;1]$ c. convex on $[-1;2]$ d. concave on $[-2;0]$
- The representative curve of the derivative $f^{\prime}$ of a function $f$ defined on the interval $[-2;4]$ is given above. By graphical reading of the curve of $f^{\prime}$, determine the correct statement for $f$: a. $f$ is decreasing on $[0;2]$ b. $f$ is decreasing on $[-1;0]$ c. $f$ admits a maximum at $1$ on $[0;2]$ d. $f$ admits a maximum at $3$ on $[2;4]$
- A stock is quoted at $57\,€$. Its value increases by $3\%$ every month. The Python function \texttt{seuil()} which returns the number of months to wait for its value to exceed $200\,€$ is: a. \begin{verbatim} def seuil() : m=0 v=57 while v < 200 : m=m+1 v = v*1.03 return m \end{verbatim} b. \begin{verbatim} def seuil() : m=0 v=57 while v > 200 : m=m+1 v = v*1.03 return m \end{verbatim} c. \begin{verbatim} def seuil() : v=57 for i in range (200) : v = v*1.03 return v \end{verbatim} d. \begin{verbatim} def seuil() : m=0 v=57 if v<200: m=m+1 else : v = v*1.03 return m \end{verbatim}