Let 𝐴 = ( 3 1 2 − 1 ) .

(a) det(A)=3(−1)−(1)(2)=−5. (b) A−1=1−5(−1−1−23)=(0.20.20.4−0.6). (c) x=A−1b=(0.20.20.4−0.6)(50)=(12). Check: 3(1)+2=5 ✓, 2(1)−2=0 ✓.

The differential equation 𝑑 𝑦 𝑑 𝑥 = 𝑥 + 𝑦 has initial condition 𝑦 ( 0 ) = 1 . Use Euler's method with step ℎ = 0.2 to estimate the following:

Euler recursion: yn+1=yn+h⋅f(xn,yn) with f(x,y)=x+y. (a) y(0.2)=1+0.2(0+1)=1.200. (b) y(0.4)=1.2+0.2(0.2+1.2)=1.2+0.28=1.480.

HL AI · Practice Paper 1 · Interactive

IB Maths Applications & Interpretation HL — Interactive Practice Paper 1

HL AI Paper 1 is GDC-allowed short-response — matrices, chi-squared testing, graph theory, coupled & differential equations, financial maths. Two modes: per-part feedback, or a full timed exam.

30 marks · 5 questions · GDC required · 3 s.f. answers

Choose your mode

Duration:

Progress: 0 / 19 correct

Question 1

6 marks

Matrices — determinant, inverse, linear systems

Let $A = (\begin{matrix} 3 & 1 \\ 2 & - 1 \end{matrix})$ .

(a) Find $det (A)$ . [1]

det (A) =

(b) Write down the entries of $A^{- 1}$ as decimals. Reading left-to-right, top-to-bottom, type each entry. [2]

top-left

=

top-right

=

bottom-left

=

bottom-right

=

(c) Use $A^{- 1}$ to solve the system $3 x + y = 5, 2 x - y = 0$ . [3]

x =

y =

(a)

det (A) = 3 (- 1) - (1) (2) = - 5

.
(b)

A^{- 1} = \frac{1}{- 5} (\begin{matrix} - 1 & - 1 \\ - 2 & 3 \end{matrix}) = (\begin{matrix} 0.2 & 0.2 \\ 0.4 & - 0.6 \end{matrix})

.
(c)

x = A^{- 1} b = (\begin{matrix} 0.2 & 0.2 \\ 0.4 & - 0.6 \end{matrix}) (\begin{matrix} 5 \\ 0 \end{matrix}) = (\begin{matrix} 1 \\ 2 \end{matrix})

. Check:

3 (1) + 2 = 5

✓,

2 (1) - 2 = 0

✓.

Question 2

7 marks

Chi-squared goodness-of-fit test

A researcher surveys 200 IB students about their favourite Maths paper. She hypothesises that preferences are equally split between Paper 1, Paper 2 and Paper 3. The observed counts are:

	P1	P2	P3
Observed	55	78	67

(a) State the expected count for each paper under the null hypothesis. Give your answer to 3 s.f. [1]

expected

\approx

(b) Calculate the chi-squared test statistic $χ_{calc}^{2}$ . Give to 3 s.f. [3]

χ_{calc}^{2} \approx

(c) State the number of degrees of freedom. [1]

d.o.f.

=

(d) Given the critical value at the 5% level with 2 d.o.f. is $χ_{crit}^{2} = 5.991$ , state whether the null hypothesis should be rejected. Type reject or accept. [2]

decision =

(a) Under

H_{0}

(uniform), expected

= 200 / 3 \approx 66.7

.
(b)

χ_{calc}^{2} = \frac{(55 - 66.67)^{2}}{66.67} + \frac{(78 - 66.67)^{2}}{66.67} + \frac{(67 - 66.67)^{2}}{66.67} \approx 2.04 + 1.93 + 0.00 = 3.97

.
(c)

d . o . f . = 3 - 1 = 2

.
(d)

3.97 < 5.991

, so accept (fail to reject)

H_{0}

— no significant evidence against uniform preference.

Question 3

6 marks

Graph theory — minimum spanning tree & shortest path

Four cities $A, B, C, D$ have the following direct road lengths (km):

edge	AB	AC	BC	BD	CD
km	12	8	5	10	15

(a) Using Kruskal's algorithm, find the total length of the minimum spanning tree. [3]

MST total

=

(b) Find the length of the shortest path from $A$ to $D$ . [2]

shortest

=

(c) Type the vertex sequence of that shortest path (e.g. ABD). [1]

path =

(a) Sorted edges:

B C (5), A C (8), B D (10), A B (12), C D (15)

. Add

B C

(5),

A C

(8),

B D

(10) — all four vertices connected. Total

= 5 + 8 + 10 = 23

km.
(b) Possible

A \to D

routes:

A \to B \to D = 22

;

A \to C \to D = 23

;

A \to B \to C \to D = 32

;

A \to C \to B \to D = 23

. Shortest

= 22

km.
(c) Path:

A \to B \to D

Question 4

5 marks

Differential equations — Euler's method

The differential equation $\frac{d y}{d x} = x + y$ has initial condition $y (0) = 1$ . Use Euler's method with step $h = 0.2$ to estimate the following:

(a) $y (0.2)$ . Give your answer to 3 d.p. [2]

y (0.2) \approx

(b) $y (0.4)$ . Give your answer to 3 d.p. [3]

y (0.4) \approx

Euler recursion:

y_{n + 1} = y_{n} + h \cdot f (x_{n}, y_{n})

with

f (x, y) = x + y

.
(a)

y (0.2) = 1 + 0.2 (0 + 1) = 1.200

.
(b)

y (0.4) = 1.2 + 0.2 (0.2 + 1.2) = 1.2 + 0.28 = 1.480

Question 5

6 marks

Financial mathematics — compound & continuous

Alex invests €5000 in a savings account paying 4% per year, compounded quarterly.

(a) Find the value of the investment after 3 years, to the nearest cent (2 d.p.). [3]

value

=

€

(b) Find the interest earned. [1]

interest

=

€

(c) Suppose instead the €5000 earns 4% p.a. compounded continuously. Find the value after 3 years, to the nearest cent. [2]

value

=

€

(a)

Math input error

(GDC TVM:

N = 12, I = 4, P / Y = C / Y = 4

).
(b) Interest

Math input error

.
(c)

Math input error

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