Absolute value and piecewise-defined functions: the piecewise definition of |x|, graphing y = |ax + b|, solving |ax + b| = k with cases (with Extension 1 inequalities and variable right-hand sides flagged), evaluating and graphing piecewise functions and finding where they are continuous (the piecewise definition, domain, range, symmetry and piecewise functions are new in the 2024 syllabus)

The graph is a V-shape with vertex $(2, 0)$ and $y$-intercept $4$: for $x \geq 2$, $y = 2x - 4$, and for $x < 2$, $y = 4 - 2x$. Setting $2x - 4 = x$ gives $x = 4$ (valid since $4 \geq 2$). Setting $4 - 2x = x$ gives $x = \frac{4}{3}$ (valid since $\frac{4}{3} < 2$). So $x = \frac{4}{3}$ or $x = 4$; the line $y = x$ crosses the V twice.

Markers award one mark for the sketch with vertex and intercept, one for each solution with its case checked.

$f(-x) = |-x| + (-x)^2 = |x| + x^2 = f(x)$, so $f$ is even and the graph is symmetric about the $y$-axis. For $x \geq 0$, $f(x) = x + x^2$ with $f(0) = 0$, $f(1) = 2$ and $f(2) = 6$. The left half is the mirror image: $f(-1) = 2$ and $f(-2) = 6$. There is a corner at the origin where the two halves meet.

Markers award one mark for the algebraic test, one for the right half and one for the reflected left half.

An absolute value equals $7$ when the inside equals $7$ or $-7$.

Check: $|2(5) - 3| = 7$ and $|2(-2) - 3| = |-7| = 7$. Both work, so $x = 5$ or $x = -2$.

Marker's note: one mark for each solution. Forgetting the negative case loses a mark.

"Absolute value at most $4$" means the inside lies between $-4$ and $4$.

Marker's note: one mark for the double inequality, one for the solution. The graph of $|3x - 2|$ lies on or below $y = 4$ between its two crossings.

(a) $-2 < 1$, so $f(-2) = -2 + 3 = 1$. $f(1) = 1^2 = 1$ (the rule with $x \geq 1$ applies). $f(3) = 9$.

(b) As $x$ approaches $1$ from the left, $f(x) = x + 3$ approaches $4$. But $f(1) = 1$. The graph jumps from height $4$ to height $1$ at $x = 1$, so $f$ is not continuous there (a jump discontinuity of size $3$).

Marker's note: one mark for the three values, one for the left-hand approach value and one for the conclusion.

Match the values at the join. From the left, $f(3) = 2(3) + k = 6 + k$. From the right, $x^2 - 1 \to 9 - 1 = 8$. For continuity, $6 + k = 8$, so $k = 2$.

Shape. The graph is the line $y = 2x + 2$ up to and including the point $(3, 8)$, then the parabola $y = x^2 - 1$ continuing from that same point with gradient $6$. The line has gradient $2$, so there is a corner at $(3, 8)$, though no break.

Marker's note: one mark for equating the two values, one for $k = 2$ and one for the description. Joining at a point makes a function continuous, not smooth.

(a) Split at $x = -1$ and $x = 2$, where each modulus changes.

• $x < -1$: $|x - 2| = 2 - x$ and $|x + 1| = -x - 1$, so $f(x) = 1 - 2x$.
• $-1 \leq x \leq 2$: $|x - 2| = 2 - x$ and $|x + 1| = x + 1$, so $f(x) = 3$.
• $x > 2$: $|x - 2| = x - 2$ and $|x + 1| = x + 1$, so $f(x) = 2x - 1$.

(b) Solve in each part. For $x < -1$: $1 - 2x = 5$ gives $x = -2$, which is less than $-1$, so it is valid. For $-1 \leq x \leq 2$: $f(x) = 3 \neq 5$, no solution. For $x > 2$: $2x - 1 = 5$ gives $x = 3 > 2$, valid.

The solutions are $x = -2$ and $x = 3$.

Marker's note: two marks for the three pieces correctly simplified and two for the solutions with their interval checks. Each answer must fit its own interval.

Case 1: $x \geq 1$

Then $|x - 1| = x - 1$, so $x - 1 = 3x - 5$, giving $4 = 2x$ and $x = 2$. This satisfies $x \geq 1$.

Case 2: $x < 1$

Then $|x - 1| = 1 - x$, so $1 - x = 3x - 5$, giving $6 = 4x$ and $x = \frac{3}{2}$. This does not satisfy $x < 1$, so it is rejected.

Check the survivor

At $x = 2$: $|2 - 1| = 1$ and $3(2) - 5 = 1$. The solution is $x = 2$ only.

Another way to see why $x = \frac{3}{2}$ fails: the right side $3x - 5 = -\frac{1}{2}$ is negative, but an absolute value cannot equal a negative number.

Marker's note: one mark for each case set up, one for solving both and one for the justified rejection.