International Baccalaureate Mathematics

Syllabus Content

The graphs of the functions, y=|f(x)|
and
y=f(|x|), y=1/f(x), y=f(ax+b), y=[f(x)]².
Solution of modulus equations and inequalities.

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Functions A relationship between two sets can be called a mapping. Elements of the first set (domain) are mapped to elements of the second set (range). A function is a special type of mapping for which one value in the domain maps to one, and only one value in the range.Pupils in Primary school will use the concept of function machines to perform calculations. They will then learn to ‘work backwards’ to find the inverse function. The study of functions becomes more formal as pupils become more proficient and able to cope with more complex mathematical ideas.

Furthermore

The modulus (or absolute value) of a number represents its distance from zero on the number line, regardless of its sign. When solving equations and inequalities involving modulus, it's essential to consider both the positive and negative cases. This often results in a system of equations or inequalities to solve.

Key Formulae:

$$ |a| = \begin{cases} a & \text{if } a \geq 0 \\ -a & \text{if } a < 0 \end{cases} $$

For any real numbers $ a $ and $ b $:

1) $ |a| = b $ is equivalent to $ a = b $ or $ a = -b $
2) $ |a| > b $ is equivalent to $ a > b $ or $ a < -b $
3) $ |a| < b $ is equivalent to $ -b < a < b $

Example 1:

Solve the equation $ |x - 3| = 5 $.

Using the key formulae, this equation can be broken down into two possible equations:

$$ x - 3 = 5 \quad \text{or} \quad x - 3 = -5 $$

Solving each equation, we get:

$ x = 8 $ or $ x = -2 $

Thus, the solutions to the equation $ |x - 3| = 5 $ are $ x = 8 $ and $ x = -2 $.

Example 2:

Solve the inequality $ |3x \arccos(x)| > 1 $ for $-1 \le x \le 1 $.

Using the key formulae for modulus inequalities, this inequality can be broken down into two possible inequalities:

$$ 3x \arccos(x) > 1 \quad \text{or} \quad 3x \arccos(x) < -1 $$

$$ x \arccos(x) > \frac{1}{3} $$

$$ x \arccos(x) < -\frac{1}{3} $$

Using the GDC graph plotting function the following solution is obtained:

$$ -1 \le x \lt -0.189 \text{ and } 0.254 \lt x \lt 0.937 $$

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