HL AI · Practice Paper 2 · Interactive

IB Maths Applications & Interpretation HL — Interactive Practice Paper 2

Paper 2 is the calculator-required paper — real-world modelling with your GDC in hand. Complex arithmetic, piecewise regression, graph algorithms, hypothesis testing, and calculus you can't shortcut with a formula sheet. Two modes: instant per-part feedback while you learn, or full timed exam simulation.

29 marks · 5 questions · GDC required · complex numbers, modelling, graphs, statistics, calculus
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Question 1

6 marksHLGDC required
Complex Numbers — modulus of a quotient

Let $z_1 = 3 + i$ and $z_2 = 1 - i$.

(a) Express $\dfrac{z_1}{z_2}$ in the form $a + bi$. Give as a+bi. [3]

Multiply top and bottom by the conjugate of the denominator, $1 + i$.

(b) Find the modulus $\left|\dfrac{z_1}{z_2}\right|$. Give an exact form. [2]

(c) Verify using $\left|\dfrac{z_1}{z_2}\right| = \dfrac{|z_1|}{|z_2|}$ — give $|z_1|^2 \cdot |z_2|^2$ as an integer. [1]

$|z_1|^2 = 10$, $|z_2|^2 = 2$.
(a) $\dfrac{3+i}{1-i} \cdot \dfrac{1+i}{1+i} = \dfrac{(3+i)(1+i)}{2} = \dfrac{3 + 3i + i + i^2}{2} = \dfrac{2 + 4i}{2} = \mathbf{1 + 2i}$.
(b) $|1 + 2i| = \sqrt{1 + 4} = \mathbf{\sqrt{5}}$.
(c) $|z_1|^2 = 9 + 1 = 10$, $|z_2|^2 = 1 + 1 = 2$, product $= \mathbf{20}$. (Consistency: $\sqrt{10/2} = \sqrt{5}$ ✓)

Question 2

6 marksHLGDC required
Advanced Modelling — piecewise continuity

A continuous piecewise function models a runner's speed (in m/s) as a function of time $t$ (in seconds):

$$S(t) = \begin{cases} 5t & 0 \le t \le 4 \\ at + b & 4 < t \le 10 \end{cases}$$

The runner's speed drops to $0$ at $t = 10$.

(a) Use continuity at $t = 4$ to write down an equation linking $a$ and $b$. State the value of $4a + b$. [2]

(b) Find the exact value of $a$. Give in the form $-p/q$ (lowest terms). [2]

(c) Find the exact value of $b$. Give in the form $p/q$. [2]

(a) Left limit at $t=4$: $S(4) = 5(4) = 20$. Right piece: $4a + b$. Continuity gives $\mathbf{4a + b = 20}$.
(b) & (c) Endpoint condition at $t=10$: $10a + b = 0$. Subtract: $6a = -20$ so $a = \mathbf{-\tfrac{10}{3}}$. Substitute: $b = -10a = \mathbf{\tfrac{100}{3}}$.

Question 3

5 marksHL
Graph Theory — trees & edge counts

A connected graph is called a tree if it contains no cycles.

(a) State the exact number of edges a tree with $15$ vertices must contain. [2]

A tree on $n$ vertices always has exactly $n - 1$ edges.

(b) A tree has $22$ edges. Find its number of vertices $n$. [2]

(c) A connected graph on $15$ vertices has $16$ edges. State the minimum number of cycles it must contain. [1]

(a) Trees satisfy $E = V - 1$. $E = 15 - 1 = \mathbf{14}$.
(b) $22 = n - 1$ so $n = \mathbf{23}$.
(c) Excess above a tree $= 16 - 14 = 2$. Each excess edge closes at least one cycle, so at least 2 cycles.

Question 4

6 marksHLGDC required
Hypothesis Testing — Type I error under a normal model

A quality-control machine rejects a metal bolt if its length is greater than $52.5$ mm. Under $H_0$ the machine is working properly and bolt lengths follow $X \sim N(\mu, \sigma^2)$ with $\mu = 50$ mm and $\sigma = 1.2$ mm.

(a) In this context, a Type I error is: A — Rejecting a good bolt. B — Accepting a bad bolt. C — Neither. Type A, B, or C. [1]

(b) Standardise: find the $z$-value at $X = 52.5$. Give to 3 s.f. [2]

(c) Using GDC Normal CDF, find $P(X > 52.5)$ under $H_0$. Give to 4 d.p. [3]

Casio: MENU 2 → DIST → NORM → Ncd, Lower=52.5, Upper=1E99, σ=1.2, μ=50.
(a) Rejecting $H_0$ when it's actually true — rejecting a bolt that's actually good. Answer A.
(b) $z = \dfrac{52.5 - 50}{1.2} = \dfrac{2.5}{1.2} \approx \mathbf{2.08}$ (to 3 s.f.).
(c) $P(X > 52.5) \approx \mathbf{0.0186}$. The machine wrongly rejects about $1.86\%$ of perfectly-good bolts.

Question 5

6 marksHLGDC required
Advanced Differentiation — stationary point of $x e^{-x}$

Consider the function $f(x) = x e^{-x}$.

(a) Find $f'(x)$. Given the form $e^{-x}(a - bx)$ with integers $a, b$, enter $a + b$. [3]

Product rule: $f'(x) = e^{-x} + x(-e^{-x}) = e^{-x}(1 - x)$.

(b) Find the exact $x$-coordinate of the stationary point. [1]

(c) Find the exact $y$-coordinate. Give as 1/e. [1]

(d) Classify the stationary point: max, min, or saddle. [1]

(a) Product rule: $f'(x) = e^{-x} + x(-e^{-x}) = e^{-x}(1 - x)$. So $a = 1, b = 1$ and $a + b = \mathbf{2}$.
(b) $e^{-x}$ never zero so $1 - x = 0$ gives $x = \mathbf{1}$.
(c) $f(1) = 1 \cdot e^{-1} = \mathbf{1/e}$.
(d) $f''(x) = e^{-x}(x - 2)$. At $x=1$: $f''(1) = -e^{-1} < 0$ so local maximum.

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