Worked solutions

IB Maths AA HL · Paper 1

Mathematics: Analysis & Approaches, Higher Level

Timezone

Session: May 2025
Marks: 110
Duration: 2 hours
Calculator: Not allowed
Questions: 12

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2025 TZ2 paper, which your school or the IB can provide.

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Section A · Short response

Questions 1–9 · 57 marks

Question 1

5 marks

A rational function: value, asymptote and the range of its reflection

Functions

(a)

1 mark

Put $x = 0$ in $f (x) = \frac{3 x - 2}{2 x + 1}$ .

f (0) = - 2

(b)

1 mark

As $x \to \infty$ , $f (x) \to \frac{3}{2}$ .

y = \frac{3}{2}

(c)

3 marks

For $x \geq 0$ the function $f$ increases from $f (0) = - 2$ towards, but never reaching, the asymptote $\frac{3}{2}$ . So $- 2 \leq f (x) < \frac{3}{2}$ .
Negate to get $g = - f$ .

- \frac{3}{2} < g (x) \leq 2

Question 2

5 marks

A line from its Cartesian form and its intersection with another line

Number and algebra

(a)

2 marks

Put $\frac{x - 1}{2} = \frac{y + 2}{3} = z = λ$ .

x = 1 + 2 λ, y = - 2 + 3 λ, z = λ r = 1 - 2 0 + λ 231

(b)

3 marks

Equate the components of $L_{1}$ and $L_{2} = 04 - 8 + t 102$ .
Check the third component: $λ = 2$ and $- 8 + 2 (5) = 2$ .
The point of intersection is

- 2 + 3 λ = 4 \Rightarrow λ = 2 1 + 2 λ = t \Rightarrow t = 5 (5, 4, 2)

Question 3

7 marks

A triangle with angles theta and 2 theta: the sine rule and an area

Geometry and trigonometry

(a)

3 marks

The side $B C = 62$ is opposite the angle $2 θ$ at $A$ , and $A B = 5$ is opposite the angle $θ$ at $C$ .
Use $sin 2 θ = 2 sin θ cos θ$ and divide by $sin θ$ (which is not zero).

\frac{sin 2 θ}{6 2} = \frac{sin θ}{5} \frac{2 cos θ}{6 2} = \frac{1}{5} \Rightarrow cos θ = \frac{6 2}{10} = \frac{3 2}{5}

(b)

2 marks

Use $sin^{2} θ + cos^{2} θ = 1$ with $θ$ acute.

sin^{2} θ = 1 - \frac{18}{25} = \frac{7}{25} \Rightarrow sin θ = \frac{7}{5}

(c)

2 marks

In triangle $B C D$ the angle at $C$ is $θ$ and $D$ is on $[A C]$ .

\frac{1}{2} (62) (D C) sin θ = 214 \Rightarrow \frac{3 14}{5} D C = 214 D C = \frac{10}{3}

Question 4

7 marks

Two events: finding P(B) and a conditional probability under independence

Statistics and probability

(a)

3 marks

The union is the part of $A$ outside $B$ together with all of $B$ .

\frac{5}{8} = \frac{7}{24} + P (B) \Rightarrow P (B) = \frac{15 - 7}{24} = \frac{1}{3}

(b)

4 marks

If $A$ and $B$ are independent then $P (A \cup B) = P (A) + P (B) - P (A) P (B)$ .
Independence of $A^{'}$ and $B$ gives $P (A^{'} ∣ B) = P (A^{'})$ .

\frac{5}{8} = P (A) + \frac{1}{3} - \frac{1}{3} P (A) \Rightarrow \frac{2}{3} P (A) = \frac{7}{24} \Rightarrow P (A) = \frac{7}{16} P (A^{'} ∣ B) = 1 - \frac{7}{16} = \frac{9}{16}

Question 5

7 marks

A quadratic with two distinct roots of the same sign

Number and algebra

(a)

5 marks

Compute the discriminant of $x^{2} + k x + 15 - k$ .
Two distinct real roots need $Δ > 0$ .

Δ = k^{2} - 4 (15 - k) = k^{2} + 4 k - 60 = (k - 6) (k + 10) k < - 10 or k > 6

(b)

2 marks

The product of the roots is $15 - k$ , which must be positive.
Combine with part (a).

15 - k > 0 \Rightarrow k < 15 k < - 10 or 6 < k < 15

Question 6

7 marks

The point of a graph closest to the origin

Calculus

(a)

1 mark

The distance from the origin to the point $(x, f (x))$ is $x^{2} + f (x)^{2}$ .

l = x^{2} + (x^{2} ln x + 4 - x^{2}) = x^{2} ln x + 4

(b)

6 marks

The distance $l$ is smallest when $x^{2} ln x + 4$ is smallest. Differentiate with the product rule.
Set the derivative to zero (with $x > 0$ ).
The derivative is negative for $x < e^{- 1/2}$ and positive after it, so this is a minimum.

\frac{d}{d x} (x^{2} ln x + 4) = 2 x ln x + x^{2} \cdot \frac{1}{x} = 2 x ln x + x = x (2 ln x + 1) 2 ln x + 1 = 0 \Rightarrow ln x = - \frac{1}{2} x = e^{- 1/2}

Question 7

5 marks

A quartic exactly divisible by (x + 1) squared

Number and algebra

5 marks

Let $P (x) = x^{4} + p x^{3} - 2 x^{2} + q x - 3$ . Then $P (- 1) = 0$ .
Differentiate: $P^{'} (x) = 4 x^{3} + 3 p x^{2} - 4 x + q$ , and $P^{'} (- 1) = 0$ .
Subtract the first equation from the second.
Check: $(x + 1)^{2} (x^{2} - 3) = (x^{2} + 2 x + 1) (x^{2} - 3) = x^{4} + 2 x^{3} - 2 x^{2} - 6 x - 3$ .

1 - p - 2 - q - 3 = 0 \Rightarrow p + q = - 4 - 4 + 3 p + 4 + q = 0 \Rightarrow 3 p + q = 0 2 p = 4 \Rightarrow p = 2, q = - 6

Question 8

8 marks

The arccos graph and an equation with two arccos terms

Functions

(a)

2 marks

The graph of $y = arccos x$ on $- 1 \leq x \leq 1$ is decreasing. It starts at $(- 1, π)$ , crosses the $y$ -axis at $(0, \frac{π}{2})$ and ends at $(1, 0)$ .
Sketch: a smooth decreasing curve through those three points.

(b)

6 marks

Let $A = arccos x$ and $B = arccos (x 3)$ , so $cos A = x$ and $cos B = 3 x$ , and $sin A = 1 - x^{2}$ , $sin B = 1 - 3 x^{2}$ (both positive since $A, B \in [0, π]$ ).
Take the cosine of both sides, using $cos \frac{3 π}{2} = 0$ .
Rearrange and square.
Check $x = \frac{1}{2}$ : $arccos \frac{1}{2} + arccos \frac{3}{2} = \frac{π}{3} + \frac{π}{6} = \frac{π}{2}$ , which is not $\frac{3 π}{2}$ .
Check $x = - \frac{1}{2}$ : $arccos (- \frac{1}{2}) + arccos (- \frac{3}{2}) = \frac{2 π}{3} + \frac{5 π}{6} = \frac{3 π}{2}$ .

cos (A + B) = 3 x^{2} - 1 - x^{2} 1 - 3 x^{2} = 0 3 x^{4} = (1 - x^{2}) (1 - 3 x^{2}) = 1 - 4 x^{2} + 3 x^{4} \Rightarrow 4 x^{2} = 1 \Rightarrow x = \pm \frac{1}{2} x = - \frac{1}{2}

Question 9

6 marks

Proof by contradiction that 1/(x(1 - x)) is at least 4

Number and algebra

6 marks

Assume that for some $x$ with $0 < x < 1$ , $\frac{1}{x ( 1 - x )} < 4$ .
For $0 < x < 1$ both $x$ and $1 - x$ are positive, so $x (1 - x) > 0$ and multiplying by it keeps the direction of the inequality.
Rearrange.
This is a contradiction, since a square is never negative. So the assumption is false, and $\frac{1}{x ( 1 - x )} \geq 4$ for $0 < x < 1$ .

\frac{1}{x ( 1 - x )} < 4 \Rightarrow 1 < 4 x (1 - x) 4 x^{2} - 4 x + 1 < 0 \Rightarrow (2 x - 1)^{2} < 0

Section B · Extended response

Questions 10–12 · 53 marks

Question 10

17 marks

Inverse functions, transformations, a composition and the area under a rational function

Functions

(a)

3 marks

Solve $f^{- 1} (8) = x$ , where $4^{x} = 8$ . Use the same base 2.

2^{2 x} = 2^{3} \Rightarrow 2 x = 3 \Rightarrow f^{- 1} (8) = \frac{3}{2}

(b)

4 marks

(i) Interchange $x$ and $y$ in $y = 1 + lo g_{2} x$ .

(ii) We want to go from $y = 2^{x - 1}$ to $y = 4^{x} = 2^{2 x}$ . First translate the graph 1 unit to the left, to get $y = 2^{x}$ . Then apply a horizontal stretch with scale factor $\frac{1}{2}$ (a horizontal compression), to get $y = 2^{2 x} = 4^{x}$ .

x = 1 + lo g_{2} y \Rightarrow y = 2^{x - 1} \Rightarrow g^{- 1} (x) = 2^{x - 1}

(c)

3 marks

Substitute $g (x)$ into $f$ .

(f \circ g) (x) = 4^{1 + l o g_{2} x} = 2^{2 + 2 l o g_{2} x} = 2^{2} \times 2^{l o g_{2} x^{2}} = 4 x^{2}

(d)

7 marks

(i) Put the left side over the common denominator $2 x + 1$ .

(ii) The region $R$ is under the graph of $h$ between $x = 1$ and $x = 3$ , so integrate using part (i).

2 x - 1 + \frac{1}{2 x + 1} = \frac{( 2 x - 1 ) ( 2 x + 1 ) + 1}{2 x + 1} = \frac{4 x ^{2} - 1 + 1}{2 x + 1} = \frac{4 x ^{2}}{2 x + 1} Area = \int_{1}^{3} (2 x - 1 + \frac{1}{2 x + 1}) d x = [x^{2} - x + \frac{1}{2} ln (2 x + 1)]_{1}^{3} (9 - 3 + \frac{1}{2} ln 7) - (1 - 1 + \frac{1}{2} ln 3) = 6 + \frac{1}{2} ln 7 - \frac{1}{2} ln 3 6 + \frac{1}{2} ln \frac{7}{3}

Question 11

17 marks

A binomial expansion and a rational approximation to a square root

Calculus

(a)

4 marks

Use $n = \frac{1}{2}$ with $y = 5 x$ .

1 + 5 x = 1 + \frac{1}{2} (5 x) + \frac{\frac{1}{2} ( - \frac{1}{2} )}{2} (5 x)^{2} + \frac{\frac{1}{2} ( - \frac{1}{2} ) ( - \frac{3}{2} )}{6} (5 x)^{3} + \dots 1 + \frac{5}{2} x - \frac{25}{8} x^{2} + \frac{125}{16} x^{3} + \dots

(b)

3 marks

Expand $(1 + q x)^{- 1} = 1 - q x + q^{2} x^{2} - \dots$ and multiply by $(1 + p x)$ .

(1 + p x) (1 - q x + q^{2} x^{2} - \dots) = 1 + (p - q) x + (q^{2} - pq) x^{2} + \dots

(c)

4 marks

Equate the coefficients of $x$ and $x^{2}$ with those from part (a).
The second equation is $q (q - p) = - \frac{25}{8}$ , and from the first $q - p = - \frac{5}{2}$ .
Then $p = q + \frac{5}{2} = \frac{15}{4}$ .

p - q = \frac{5}{2}, q^{2} - pq = - \frac{25}{8} q (- \frac{5}{2}) = - \frac{25}{8} \Rightarrow q = \frac{5}{4}

(d)

6 marks

(i) For $1.2$ we need $1 + 5 x = 1.2$ , so $x = 0.04 = \frac{1}{25}$ .

(ii) Since $\frac{5}{2} = \frac{5}{4} = 1.25$ , we need $1 + 5 x = 1.25$ , so $x = \frac{1}{20}$ . The approximation is better the closer $x$ is to 0, and $\frac{1}{20}$ is further from 0 than $\frac{1}{25}$ , so the approximation for $\frac{5}{2}$ is less accurate than the one for $1.2$ .

1.2 \approx \frac{1 + \frac{15}{4} \cdot \frac{1}{25}}{1 + \frac{5}{4} \cdot \frac{1}{25}} = \frac{1.15}{1.05} = \frac{23}{21}

Question 12

19 marks

Square roots of complex numbers, a rectangle and reciprocals

Number and algebra

(a)

4 marks

$- 1 - 3 i$ has modulus 2 and argument $- \frac{2 π}{3}$ , so $z^{2} = 2 (cos (- \frac{2 π}{3}) + i sin (- \frac{2 π}{3}))$ .
Take the square root and add $π$ for the other root.

z = 2 (cos (- \frac{π}{3}) + i sin (- \frac{π}{3})) or z = 2 (cos \frac{2 π}{3} + i sin \frac{2 π}{3})

(b)

4 marks

(i) Convert to Cartesian form.

(ii) $z_{3}$ and $z_{4}$ are the square roots of $- 1 + 3 i$ , which is the conjugate of $- 1 - 3 i$ . Conjugating $z_{1}^{2} = - 1 - 3 i$ gives $\overset{z}{ˉ}_{1}^{2} = - 1 + 3 i$ , so the roots are the conjugates of $z_{1}$ and $z_{2}$ , with $Re (z_{3}) > 0$ .

z_{1} = \frac{2}{2} - \frac{6}{2} i, z_{2} = - \frac{2}{2} + \frac{6}{2} i z_{3} = \frac{2}{2} + \frac{6}{2} i, z_{4} = - \frac{2}{2} - \frac{6}{2} i

(c)

4 marks

(i) Sketch: the four points $A (\frac{2}{2}, - \frac{6}{2})$ , $B (- \frac{2}{2}, \frac{6}{2})$ , $C (\frac{2}{2}, \frac{6}{2})$ , $D (- \frac{2}{2}, - \frac{6}{2})$ . They are the corners of a rectangle centred on the origin.

(ii) The width is $2 \times \frac{2}{2} = 2$ and the height is $2 \times \frac{6}{2} = 6$ .

Area = 2 \times 6 = 12 = 23

(d)

3 marks

Put $w = \frac{1}{z}$ , so $z = \frac{1}{w}$ in $z^{4} + 2 z^{2} + 4 = 0$ .
Multiply by $w^{4}$ .

\frac{1}{w ^{4}} + \frac{2}{w ^{2}} + 4 = 0 1 + 2 w^{2} + 4 w^{4} = 0 \Rightarrow 4 w^{4} + 2 w^{2} + 1 = 0 p = 4, q = 2, r = 1

(e)

4 marks

(i) Use $\frac{1}{z} = \frac{z ˉ}{∣ z ∣ ^{2}}$ with $∣ z_{1} ∣^{2} = 2$ .

(ii) The modulus of every root has changed from $2$ to $\frac{1}{2}$ and the points are reflected in the real axis, so the new rectangle is half the size in each direction.

\frac{1}{z _{1}} = \frac{\frac{2}{2} + \frac{6}{2} i}{2} = \frac{2}{4} + \frac{6}{4} i Area = \frac{1}{4} \times 23 = \frac{3}{2}

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Number and algebra carries the most marks on this paper.

Number and algebra42 marks · 38%
Q2, Q5, Q7, Q9, Q12
Functions30 marks · 27%
Q1, Q8, Q10
Calculus24 marks · 22%
Q6, Q11
Geometry and trigonometry7 marks · 6%
Q3
Statistics and probability7 marks · 6%
Q4

Mentor insights

Traps to watch for

01
Double angle · Q3
$sin 2 θ = 2 sin θ cos θ$ , then cancel $sin θ$ .
02
Compound angle · Q8
Take the cosine of both sides and use $cos (A + B) = cos A cos B - sin A sin B$ .
03
Inverse of an exponential · Q10
$f^{- 1} (8) = lo g_{4} 8$ .
04
Binomial series · Q11
$(1 + y)^{n} = 1 + n y + \frac{n ( n - 1 )}{2} y^{2} + \frac{n ( n - 1 ) ( n - 2 )}{6} y^{3} + \dots$ for $∣ y ∣ < 1$ .
05
Square roots · Q12
Use de Moivre's theorem: halve the argument and take the square root of the modulus, then add $π$ for the second root.

Solutions and tips are GeorgeTheTutor's own; the questions are not reproduced here. Not affiliated with or endorsed by the International Baccalaureate Organization.

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