Predicted paper walkthrough — May 2027 SL AA style questions

Predicted paper walkthrough — May 2027 SL AA style questions

I have spent over a decade teaching IB Maths. Every year, I guide students through their final exams. What I have learned is that practice is key, especially with a focus on potential exam content. Today, I want to walk through some questions that reflect what I predict could appear in the May 2027 SL AA exams. This isn't about guessing the exact questions, but about understanding the types of problems, the common pitfalls, and the strategies that work.

My goal here is to give you a framework for tackling these kinds of questions. We will look at how to approach them, what mathematical concepts are being tested, and how to maximize your marks. This is a direct approach, focusing on what you need to do to succeed. Let's get straight into it.

Section A: Paper 1 - Non-calculator style questions

Paper 1 for SL AA is about demonstrating your fundamental understanding without relying on a calculator. This means algebra, trigonometry, and calculus basics need to be solid. Let's consider a question covering multiple topics, which is common.

Question 1: Functions and Transformations

The function $f(x) = 2x^2 - 8x + 5$ is defined for $x \in \mathbb{R}$.

1. Write $f(x)$ in the form $a(x-h)^2 + k$.
2. Hence, find the coordinates of the vertex of the graph of $y = f(x)$.
3. The graph of $y = f(x)$ is transformed by a horizontal stretch with a scale factor of $\frac{1}{2}$ and a vertical translation of 3 units downwards. Write down the equation of the transformed graph, $g(x)$.

My approach: Part (a) is a classic completing the square problem. I always tell my students to factor out the coefficient of $x^2$ first. So, $f(x) = 2(x^2 - 4x) + 5$. Then complete the square for the term inside the parenthesis: $x^2 - 4x = (x-2)^2 - 4$. Substituting back, we get $f(x) = 2((x-2)^2 - 4) + 5 = 2(x-2)^2 - 8 + 5 = 2(x-2)^2 - 3$. This is in the required form $a(x-h)^2 + k$, where $a=2$, $h=2$, $k=-3$.

For part (b), once the function is in vertex form, the vertex is straightforward $(h, k)$. So, the vertex is $(2, -3)$. This is a common point of confusion; remember the sign convention for $h$.

Part (c) tests transformations. A horizontal stretch with a scale factor of $\frac{1}{2}$ means we replace $x$ with $2x$. A vertical translation of 3 units downwards means we subtract 3 from the entire function. So, we apply these transformations to the original function: $g(x) = f(2x) - 3$. Using the vertex form we found: $g(x) = 2((2x)-2)^2 - 3 - 3 = 2(2(x-1))^2 - 6 = 2 \cdot 4(x-1)^2 - 6 = 8(x-1)^2 - 6$. It is important to apply transformations correctly and in the right order. For more on functions, check out my study notes.

Section B: Paper 2 - Calculator active style questions

Paper 2 allows a calculator, which means questions can be more complex computationally. However, it doesn't mean you avoid showing working. Your calculator is a tool to evaluate, not a substitute for understanding. Many questions still require setup, interpretation, and conceptual understanding.

Question 2: Calculus - Optimisation

A farmer has 120 m of fencing to enclose a rectangular area. One side of the rectangular area is against an existing straight wall, so no fencing is needed along that side. Let the width of the rectangle be $x$ metres and the length be $y$ metres.

1. Show that the area $A$ of the rectangle can be expressed as $A(x) = 120x - 2x^2$.
2. Find the maximum area the farmer can enclose.

My approach: For part (a), I always encourage drawing a diagram. If one side is against a wall, then the fencing covers $x$ (width), $y$ (length), and another $x$ (width). So, the total fencing is $2x + y = 120$. We need to express area $A = xy$ in terms of $x$ only. From the perimeter equation, $y = 120 - 2x$. Substitute this into the area formula: $A(x) = x(120 - 2x) = 120x - 2x^2$. This matches the given expression.

Part (b) is an optimisation problem, which screams calculus. To find the maximum area, we need to find the derivative of $A(x)$ with respect to $x$ and set it to zero. $A'(x) = \frac{dA}{dx} = 120 - 4x$. Setting $A'(x) = 0$: $120 - 4x = 0 \implies 4x = 120 \implies x = 30$. My students sometimes stop here, but this is only the $x$ value. We need the maximum area. Substitute $x=30$ back into $A(x)$: $A(30) = 120(30) - 2(30)^2 = 3600 - 2(900) = 3600 - 1800 = 1800$ m$^2$. Remember to include units in your final answer when applicable.

Also, it's good practice to confirm it's a maximum. The second derivative $A''(x) = -4$, which is less than 0, confirming it's a maximum. While this isn't strictly necessary for full marks if you're confident, it's a good check. For more on calculus applications, refer to my CG50 calculator guide for how to use your calculator to find derivatives and solve equations.

Section C: Mixed Topics - Bridging the Gap

IB exams often combine concepts. These questions test your ability to connect different areas of the syllabus. I often see questions that blend trigonometry with coordinate geometry, or probability with statistics. It's about seeing the bigger picture.

Question 3: Trigonometry and Modelling

The depth of water, $D$ metres, in a harbour can be modelled by the function $D(t) = P \cos(Qt) + R$, where $t$ is the time in hours after midnight. The maximum depth is 15 m at 02:00 and the minimum depth is 7 m at 08:00.

1. Find the value of $P$, $Q$, and $R$.
2. Find the first time in the day when the depth of water is 12 m.

My approach: Part (a) involves finding the parameters of a trigonometric model. The amplitude $P$ is half the difference between maximum and minimum depths: $P = \frac{15 - 7}{2} = \frac{8}{2} = 4$. The vertical shift $R$ (or mean depth) is the average of the maximum and minimum depths: $R = \frac{15 + 7}{2} = \frac{22}{2} = 11$.

To find $Q$, we need the period. The time from maximum (02:00) to minimum (08:00) is 6 hours. This represents half a period. So, the full period is $2 \times 6 = 12$ hours. The period of $\cos(Qt)$ is $\frac{2\pi}{Q}$. So, $\frac{2\pi}{Q} = 12 \implies Q = \frac{2\pi}{12} = \frac{\pi}{6}$. Therefore, $D(t) = 4\cos\left(\frac{\pi}{6}t\right) + 11$. Always check if the model fits the given max/min points. At $t=2$, $D(2) = 4\cos\left(\frac{\pi}{6} \cdot 2\right) + 11 = 4\cos\left(\frac{\pi}{3}\right) + 11 = 4\left(\frac{1}{2}\right) + 11 = 2 + 11 = 13$. This is not 15. I need to adjust the phase shift. Since the maximum is at $t=2$, I need the argument of the cosine function to be $0$ or a multiple of $2\pi$ when $t=2$. This means I should use $D(t) = P \cos(Q(t-C)) + R$. $Q(t-C)=0 \implies \frac{\pi}{6}(2-C)=0$. This would mean $C=2$. So $D(t) = 4\cos\left(\frac{\pi}{6}(t-2)\right) + 11$. Let's re-check $D(2) = 4\cos(0) + 11 = 4(1) + 11 = 15$. This is correct. $D(8) = 4\cos\left(\frac{\pi}{6}(8-2)\right) + 11 = 4\cos(\pi) + 11 = 4(-1) + 11 = -4 + 11 = 7$. This is also correct. The phase shift is critical for cosine/sine modelling.

For part (b), we need to solve $D(t) = 12$. So, $4\cos\left(\frac{\pi}{6}(t-2)\right) + 11 = 12$. Subtract 11: $4\cos\left(\frac{\pi}{6}(t-2)\right) = 1$. Divide by 4: $\cos\left(\frac{\pi}{6}(t-2)\right) = \frac{1}{4}$. Now use the inverse cosine function: $\frac{\pi}{6}(t-2) = \arccos\left(\frac{1}{4}\right)$. Using a calculator (in radians mode!), $\arccos(0.25) \approx 1.318$ radians. So $\frac{\pi}{6}(t-2) \approx 1.318$. Multiply by $\frac{6}{\pi}$: $t-2 \approx 1.318 \times \frac{6}{\pi} \approx 2.517$. So $t \approx 2.517 + 2 = 4.517$ hours. The first time the depth is 12 m is approximately at 04:31. Remember to consider the domain of the function and periodicity for other solutions, but for the "first time", this solution usually suffices if it's within $0 \le t < 24$. This is a crucial skill for Paper 2 SL AI as well, although the context might differ.

Next Steps: Consolidate and Practice

The examples above highlight common themes: algebraic manipulation, calculus applications, and trigonometric modelling. My advice is always the same: understand the underlying concepts, not just the steps to a solution. These questions are designed to test your understanding, not just your memorisation.

Review these types of problems, go through past papers, and identify your weak areas. Don't shy away from revisiting earlier topics. For further practice, consider using my flashcards or going through more Paper 1 SL AA practice questions. Consistent effort, focused practice, and a clear understanding of problem-solving strategies are what will ultimately lead to success in your IB Maths exams.