We consider the function $f$ defined on $]0; 8]$ by $$f ( x ) = \frac { 10 \ln \left( - x ^ { 2 } + 7 x + 9 \right) } { x }$$ Let $C _ { f }$ be the graphical representation of the function $f$ in an orthonormal coordinate system $(\mathrm{O}; \vec{\imath}, \vec{\jmath})$.
Part A
- Solve in $\mathbb { R }$ the inequality $- x ^ { 2 } + 7 x + 8 \geqslant 0$.
- Deduce that for all $x \in ] 0 ; 8 ]$, we have $f ( x ) \geqslant 0$.
- Interpret this result graphically.
Part B
The curve $C _ { f }$ is represented below. Let $M$ be the point of $C _ { f }$ with abscissa $x$ where $x \in ] 0; 8]$. We call $N$ and $P$ the orthogonal projections of the point $M$ respectively on the abscissa axis and on the ordinate axis. In this part, we are interested in the area $\mathscr { A } ( x )$ of the rectangle $\mathrm{O}NMP$.
- Give the coordinates of points $N$ and $P$ as a function of $x$.
- Show that for all $x$ belonging to the interval $] 0 ; 8 ]$, $$\mathscr { A } ( x ) = 10 \ln \left( - x ^ { 2 } + 7 x + 9 \right)$$
- Does there exist a position of the point $M$ for which the area of the rectangle $\mathrm{O}NMP$ is maximum? If it exists, determine this position.
Part C
We consider a strictly positive real number $k$. We wish to determine the smallest value of $x$, approximated to the nearest tenth, belonging to $[ 3.5; 8 ]$ for which the area $\mathscr { A } ( x )$ becomes less than or equal to $k$. To do this, we consider the algorithm below. As a reminder, in Python language, $\ln ( x )$ is written log$(x)$.
\begin{verbatim} from math import * def A(x) : return 10*log (- 1* x**2 + 7*x + 9) def pluspetitevaleur(k) : x = 3.5 while A(x).........: x = x + 0.1 return ........... \end{verbatim}
- Copy and complete lines 8 and 10 of the algorithm.
- What number does the instruction \texttt{pluspetitevaleur(30)} then return?
- What happens when $k = 35$? Justify.