The problem consists of three exercises. Exercise 1 asks to determine the domain of definition of some functions, study the parity of some functions, and calculate the image and pre-images of a function. Exercise 2 asks to solve trigonometric equations and inequalities within specified intervals. Exercise 3 asks to find the domain of a function, verify if it is odd, prove an expression, study variations on intervals, deduce variations on intervals and write the variation table.

1. Problem Description

The problem consists of three exercises. Exercise 1 asks to determine the domain of definition of some functions, study the parity of some functions, and calculate the image and pre-images of a function. Exercise 2 asks to solve trigonometric equations and inequalities within specified intervals. Exercise 3 asks to find the domain of a function, verify if it is odd, prove an expression, study variations on intervals, deduce variations on intervals and write the variation table.

2. Solution Steps

Exercise 1:

1) Determine $D_f$ for the following functions:

a) $f(x) = \frac{3x}{x-9}$

The domain is all real numbers except where the denominator is zero.

$x - 9 = 0 \Rightarrow x = 9$.

Thus, $D_f = \mathbb{R} \setminus \{9\}$.

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

The domain is all real numbers except where the denominator is zero.

$x^2 + x - 2 = 0 \Rightarrow (x+2)(x-1) = 0 \Rightarrow x = -2$ or $x = 1$.

Thus, $D_f = \mathbb{R} \setminus \{-2, 1\}$.

c) $f(x) = \sqrt{3x-6}$

The domain requires that the expression under the square root be non-negative.

$3x - 6 \geq 0 \Rightarrow 3x \geq 6 \Rightarrow x \geq 2$.

Thus, $D_f = [2, +\infty)$.

d) $f(x) = \frac{2}{\sqrt{x-1}}$

The domain requires that the expression under the square root be positive.

$x - 1 > 0 \Rightarrow x > 1$.

Thus, $D_f = (1, +\infty)$.

2) Study the parity of the following functions:

a) $g(x) = |x| - \frac{1}{x^2}$

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

Thus, $g(x)$ is an even function.

b) $h(x) = x\sqrt{x^2 - 1}$

$h(-x) = (-x)\sqrt{(-x)^2 - 1} = -x\sqrt{x^2 - 1} = -h(x)$.

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

3) Given $h(x) = \frac{2x+5}{x+1}$:

a) Calculate the image of 3 and -2 by the function $h$.

$h(3) = \frac{2(3)+5}{3+1} = \frac{6+5}{4} = \frac{11}{4}$.

$h(-2) = \frac{2(-2)+5}{-2+1} = \frac{-4+5}{-1} = \frac{1}{-1} = -1$.

b) Determine the antecedents of number 5 by the function $h$.

$h(x) = 5 \Rightarrow \frac{2x+5}{x+1} = 5 \Rightarrow 2x+5 = 5(x+1) \Rightarrow 2x+5 = 5x+5 \Rightarrow 3x = 0 \Rightarrow x = 0$.

Thus, the antecedent of 5 is

0.

Exercise 3:

1) Determine $D_f$ for $f(x) = \frac{x}{x^2 - 1}$

$x^2 - 1 \neq 0 \Rightarrow x^2 \neq 1 \Rightarrow x \neq 1$ and $x \neq -1$.

$D_f = \mathbb{R} \setminus \{-1, 1\}$.

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

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

Thus, $f$ is an odd function.

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

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

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

Exercise 2:

1) Solve the equations in interval I:

a) $\cos(x) = \frac{1}{2}$, $I = \mathbb{R}$.

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

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

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

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

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

For $k = 0$, $x = \frac{7\pi}{36}$ and $x = \frac{13\pi}{12}$.

For $k = 1$, $x = \frac{7\pi}{36} + \frac{2\pi}{3} = \frac{7\pi + 24\pi}{36} = \frac{31\pi}{36}$ and $x = \frac{13\pi}{12} + 2\pi = \frac{13\pi + 24\pi}{12} = \frac{37\pi}{12}$.

For $k = 2$, $x = \frac{7\pi}{36} + \frac{4\pi}{3} = \frac{7\pi + 48\pi}{36} = \frac{55\pi}{36}$ and $x > 2\pi$.

So, $x = \{\frac{7\pi}{36}, \frac{31\pi}{36}, \frac{55\pi}{36}, \frac{13\pi}{12}, \frac{37\pi}{12}\}$.

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

$2x = \frac{\pi}{6} + 2k\pi$ or $2x = -\frac{\pi}{6} + 2k\pi$.

$x = \frac{\pi}{12} + k\pi$ or $x = -\frac{\pi}{12} + k\pi$.

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

For $k = 1$, $x = \frac{\pi}{12} + \pi = \frac{13\pi}{12} > \pi$ (not in I) and $x = -\frac{\pi}{12} + \pi = \frac{11\pi}{12}$.

For $k = -1$, $x = \frac{\pi}{12} - \pi = -\frac{11\pi}{12}$ and $x = -\frac{\pi}{12} - \pi = -\frac{13\pi}{12} < -\pi$ (not in I).

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

2) Solve the inequalities in interval I:

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

$\frac{\pi}{6} \leq x \leq \frac{5\pi}{6}$.

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

$\frac{\pi}{3} < x < \frac{5\pi}{3}$.

3. Final Answer

Solutions are provided in the step-by-step explanation above.

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