The problem is divided into two exercises. Exercise 3 is about the function $f(x) = \frac{x}{x^2 - 1}$. We need to: 1) Determine the domain of definition $D_f$ of $f$. 2) Verify that $f$ is an odd function. 3) Show that for all distinct real numbers $a$ and $b$ in $D_f$, $T = \frac{f(a) - f(b)}{a - b} = -\frac{ab+1}{(a^2 - 1)(b^2 - 1)}$. 4) Study the variations of $f$ on $[0, 1[$ and $]1, +\infty[$. 5) Deduce the variations of $f$ on $]-1, 0]$ and $]-\infty, -1[$. 6) Draw the variation table of $f$ on $D_f$. Exercise 2 asks us to solve some trigonometric equations and inequalities. 1) Solve the following equations in the given interval: - $cos(x) = \frac{1}{2}, I = R$ - $sin(2x - \frac{\pi}{3}) = sin(\frac{\pi}{4} - x), I = [0, 2\pi]$ - $cos(2x) = \frac{\sqrt{3}}{2}, I = ]-\pi, \pi]$ 2) Solve the following inequalities in the given interval: - $sin(x) \geq \frac{1}{2}, I = ]-\pi, \pi]$ - $cos(x) < \frac{1}{2}, I = [0, 2\pi]$

1. Problem Description

The problem is divided into two exercises. Exercise 3 is about the function $f(x) = \frac{x}{x^2 - 1}$. We need to:

1) Determine the domain of definition $D_f$ of $f$.

2) Verify that $f$ is an odd function.

3) Show that for all distinct real numbers $a$ and $b$ in $D_f$, $T = \frac{f(a) - f(b)}{a - b} = -\frac{ab+1}{(a^2 - 1)(b^2 - 1)}$.

4) Study the variations of $f$ on $[0, 1[$ and $]1, +\infty[$.

5) Deduce the variations of $f$ on $]-1, 0]$ and $]-\infty, -1[$.

6) Draw the variation table of $f$ on $D_f$.

Exercise 2 asks us to solve some trigonometric equations and inequalities.

1) Solve the following equations in the given interval:

- $cos(x) = \frac{1}{2}, I = R$

- $sin(2x - \frac{\pi}{3}) = sin(\frac{\pi}{4} - x), I = [0, 2\pi]$

- $cos(2x) = \frac{\sqrt{3}}{2}, I = ]-\pi, \pi]$

2) Solve the following inequalities in the given interval:

- $sin(x) \geq \frac{1}{2}, I = ]-\pi, \pi]$

- $cos(x) < \frac{1}{2}, I = [0, 2\pi]$

2. Solution Steps

Exercise 3:

1) Domain of definition:

$f(x) = \frac{x}{x^2 - 1}$. $f(x)$ is defined when the denominator is not zero.

$x^2 - 1 \neq 0 \implies x^2 \neq 1 \implies x \neq \pm 1$.

So $D_f = R \setminus \{-1, 1\} = ]-\infty, -1[ \cup ]-1, 1[ \cup ]1, +\infty[$.

2) Odd function:

A function $f$ is odd if $f(-x) = -f(x)$.

$f(-x) = \frac{-x}{(-x)^2 - 1} = \frac{-x}{x^2 - 1} = - \frac{x}{x^2 - 1} = -f(x)$.

Thus, $f(x)$ is an odd function.

3) Showing the expression for T:

$f(a) = \frac{a}{a^2 - 1}$ and $f(b) = \frac{b}{b^2 - 1}$.

$f(a) - f(b) = \frac{a}{a^2 - 1} - \frac{b}{b^2 - 1} = \frac{a(b^2 - 1) - b(a^2 - 1)}{(a^2 - 1)(b^2 - 1)} = \frac{ab^2 - a - ba^2 + b}{(a^2 - 1)(b^2 - 1)} = \frac{ab^2 - ba^2 - a + b}{(a^2 - 1)(b^2 - 1)} = \frac{ab(b - a) + (b - a)}{(a^2 - 1)(b^2 - 1)} = \frac{(b - a)(ab + 1)}{(a^2 - 1)(b^2 - 1)}$.

$T = \frac{f(a) - f(b)}{a - b} = \frac{(b - a)(ab + 1)}{(a - b)(a^2 - 1)(b^2 - 1)} = \frac{-(a - b)(ab + 1)}{(a - b)(a^2 - 1)(b^2 - 1)} = - \frac{ab + 1}{(a^2 - 1)(b^2 - 1)}$.

4) Variations of f on $[0, 1[$ and $]1, +\infty[$:

Let $x \in [0, 1[$ and $x \in ]1, +\infty[$.

$f'(x) = \frac{(x^2 - 1) - x(2x)}{(x^2 - 1)^2} = \frac{x^2 - 1 - 2x^2}{(x^2 - 1)^2} = \frac{-x^2 - 1}{(x^2 - 1)^2} = - \frac{x^2 + 1}{(x^2 - 1)^2}$.

Since $x^2 + 1 > 0$ and $(x^2 - 1)^2 > 0$, we have $f'(x) < 0$ for all $x \in D_f$.

Therefore, $f$ is strictly decreasing on $[0, 1[$ and $]1, +\infty[$.

5) Variations of f on $]-1, 0]$ and $]-\infty, -1[$:

Since f is an odd function, if f is strictly decreasing on $]1, +\infty[$, then it is strictly decreasing on $]-\infty, -1[$. If f is strictly decreasing on $[0, 1[$, then it is strictly decreasing on $]-1, 0]$.

Therefore, $f$ is strictly decreasing on $]-1, 0]$ and $]-\infty, -1[$.

6) Variation Table of f on $D_f$:

Since f is strictly decreasing on $]-\infty, -1[$, $]-1, 0]$, $[0, 1[$ and $]1, +\infty[$, the variation table is as follows.

x | -inf | -1 | -1 | 0 | 1 | 1 | +inf

---|-------|--------|--------|-------|-------|-------|------

f'(x)| - | - | - | - | - | - | -

f(x) | 0 | -inf | +inf | 0 | -inf | +inf | 0

Exercise 2:

1)

- $cos(x) = \frac{1}{2}, I = R$

$x = \pm \frac{\pi}{3} + 2k\pi$, where $k \in Z$.

- $sin(2x - \frac{\pi}{3}) = sin(\frac{\pi}{4} - x), I = [0, 2\pi]$

$2x - \frac{\pi}{3} = \frac{\pi}{4} - x + 2k\pi$ or $2x - \frac{\pi}{3} = \pi - (\frac{\pi}{4} - x) + 2k\pi$.

$3x = \frac{\pi}{4} + \frac{\pi}{3} + 2k\pi = \frac{7\pi}{12} + 2k\pi \implies x = \frac{7\pi}{36} + \frac{2k\pi}{3}$.

$x = \frac{7\pi}{36}, \frac{7\pi}{36} + \frac{2\pi}{3} = \frac{31\pi}{36}, \frac{7\pi}{36} + \frac{4\pi}{3} = \frac{55\pi}{36}$.

$2x - \frac{\pi}{3} = \pi - \frac{\pi}{4} + x + 2k\pi$

$x = \frac{3\pi}{4} + \frac{\pi}{3} + 2k\pi = \frac{13\pi}{12} + 2k\pi$.

$x = \frac{13\pi}{12}, \frac{13\pi}{12} > 2\pi = \frac{24\pi}{12}$ so we don't need more values for k >

0. The solutions are: $x = \frac{7\pi}{36}, \frac{31\pi}{36}, \frac{55\pi}{36}, \frac{13\pi}{12}$.

- $cos(2x) = \frac{\sqrt{3}}{2}, I = ]-\pi, \pi]$

$2x = \pm \frac{\pi}{6} + 2k\pi \implies x = \pm \frac{\pi}{12} + k\pi$.

For $k = 0$, $x = \pm \frac{\pi}{12}$.

For $k = 1$, $x = \frac{\pi}{12} + \pi = \frac{13\pi}{12}$ and $x = - \frac{\pi}{12} + \pi = \frac{11\pi}{12}$.

For $k = -1$, $x = \frac{\pi}{12} - \pi = - \frac{11\pi}{12}$ and $x = - \frac{\pi}{12} - \pi = - \frac{13\pi}{12}$.

Since $x \in ]-\pi, \pi]$, the solutions are: $x = - \frac{11\pi}{12}, - \frac{\pi}{12}, \frac{\pi}{12}, \frac{11\pi}{12}$.

2)

- $sin(x) \geq \frac{1}{2}, I = ]-\pi, \pi]$

$x \in [\frac{\pi}{6}, \frac{5\pi}{6}]$.

- $cos(x) < \frac{1}{2}, I = [0, 2\pi]$

$x \in (\frac{\pi}{3}, \frac{5\pi}{3})$.

3. Final Answer

Exercise 3:

1) $D_f = R \setminus \{-1, 1\} = ]-\infty, -1[ \cup ]-1, 1[ \cup ]1, +\infty[$.

2) $f(x)$ is an odd function.

3) $T = - \frac{ab + 1}{(a^2 - 1)(b^2 - 1)}$.

4) $f$ is strictly decreasing on $[0, 1[$ and $]1, +\infty[$.

5) $f$ is strictly decreasing on $]-1, 0]$ and $]-\infty, -1[$.

Exercise 2:

1)

- $x = \pm \frac{\pi}{3} + 2k\pi$, where $k \in Z$.

- $x = \frac{7\pi}{36}, \frac{31\pi}{36}, \frac{55\pi}{36}, \frac{13\pi}{12}$.

- $x = - \frac{11\pi}{12}, - \frac{\pi}{12}, \frac{\pi}{12}, \frac{11\pi}{12}$.

2)

- $x \in [\frac{\pi}{6}, \frac{5\pi}{6}]$.

- $x \in (\frac{\pi}{3}, \frac{5\pi}{3})$.