Functions - Modulus equations and inequalities (HL)

Solve the equation $|2x - 5| = x + 1$.

Case 1: $2x - 5 = x + 1 \implies x = 6$. Check: $|2(6) - 5| = |7| = 7$; RHS: $6 + 1 = 7$. (Valid) Case 2: $2x - 5 = -(x + 1) \implies 2x - 5 = -x - 1 \implies 3x = 4 \implies x = \frac{4}{3}$. Check: $|2(\frac{4}{3}) - 5| = |\frac{8}{3} - \frac{15}{3}| = |-\frac{7}{3}| = \frac{7}{3}$; RHS: $\frac{4}{3} + 1 = \frac{7}{3}$. (Valid) Final solutions: $x = 6, x = \frac{4}{3}$.

We split the modulus into its positive and negative cases. Each result must be substituted back into the original equation to ensure the right-hand side is not negative, which would make the equality impossible.

Solve the inequality $|x - 3| < |2x + 1|$.

Since both sides are non-negative, square both sides: $(x - 3)^2 < (2x + 1)^2$ $x^2 - 6x + 9 < 4x^2 + 4x + 1$ $0 < 3x^2 + 10x - 8$ Solve the quadratic equation $3x^2 + 10x - 8 = 0$ using the quadratic formula or factoring: $(3x - 2)(x + 4) = 0 \implies x = \frac{2}{3}, x = -4$. Test intervals for $3x^2 + 10x - 8 > 0$: If $x < -4$, $3(-5)^2 + 10(-5) - 8 = 75 - 50 - 8 = 17 > 0$. (True) If $-4 < x < \frac{2}{3}$, $3(0)^2 + 10(0) - 8 = -8 < 0$. (False) If $x > \frac{2}{3}$, $3(1)^2 + 10(1) - 8 = 5 > 0$. (True) Solution: $x < -4$ or $x > \frac{2}{3}$.

Squaring is the most efficient algebraic method for inequalities involving two modulus expressions. After squaring, we solve the resulting quadratic inequality by finding the roots and testing intervals.

Sketch the graph of $f(x) = |x^2 - 4|$. Find the values of $k$ for which $|x^2 - 4| = k$ has exactly 3 solutions.

1. Start with $y = x^2 - 4$ (a parabola with vertex $(0, -4)$ and roots $x = \pm 2$).
2. Apply the modulus: Reflect the portion between $x = -2$ and $x = 2$ (where $y$ is negative) across the $x$-axis. The new vertex is $(0, 4)$.
3. To have exactly 3 solutions, the horizontal line $y = k$ must intersect the graph at exactly 3 points. Looking at the graph, the line $y = 4$ passes through the local maximum $(0, 4)$ and two other points on the outer arms of the parabola. Therefore, $k = 4$.

Graphical analysis is often required in IB HL. By reflecting the negative parts of the parabola, we see a 'W' shaped curve. The line $y=k$ intersects the 'peaks' or 'troughs' to change the number of solutions.

1. Set up the two cases: Case 1: $3x - 2 = 2x + 3$ $x = 5$ Case 2: $3x - 2 = -(2x + 3)$ $3x - 2 = -2x - 3$ $5x = -1$ $x = -0.2$

2. Check validity: For $x = 5$: $|3(5)-2| = |13| = 13$; $2(5)+3 = 13$. Valid. For $x = -0.2$: $|3(-0.2)-2| = |-2.6| = 2.6$; $2(-0.2)+3 = 2.6$. Valid.

The solutions are $x = 5$ and $x = -0.2$.

To solve a modulus equation, we branch into the positive and negative possibilities of the expression inside the modulus. Each solution must be checked against the right-hand side function to ensure it doesn't result in a negative value, which is impossible for a modulus output.

1. To solve $|2x + 4| \geq |x - 1|$, we square both sides (since both sides are non-negative): $(2x + 4)^2 \geq (x - 1)^2$
2. Expand the expressions: $4x^2 + 16x + 16 \geq x^2 - 2x + 1$
3. Rearrange into a quadratic inequality: $3x^2 + 18x + 15 \geq 0$
4. Divide by 3: $x^2 + 6x + 5 \geq 0$
5. Factor the quadratic: $(x + 5)(x + 1) \geq 0$
6. Find the critical values: $x = -5$ and $x = -1$. Testing intervals or observing the upward-opening parabola, the expression is non-negative when: $x \leq -5 \text{ or } x \geq -1$

Thus, the solution set is $x \in (-\infty, -5] \cup [-1, \infty)$.

Squaring both sides is an effective method for inequalities involving two absolute values because $|a|^2 = a^2$. The resulting quadratic inequality defines the regions where the distance of $2x$ from $-4$ is greater than or equal to the distance of $x$ from $1$. The graph shows the intersections of the two V-shaped functions at $x = -5$ and $x = -1$.