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IB Maths AA HL · Paper 1

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1 TZ2TZ3

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

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2025 TZ3 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Section A · Short response

Questions 1–9 · 55 marks

Question 1

5 marks

Logarithms of 24 and base conversion

Functions

(a)

3 marks

Write $24 = 2^{3} \times 3$ .

lo g_{10} 24 = 3 lo g_{10} 2 + lo g_{10} 3 = 3 p + q

(b)

2 marks

Change the base to 10.

lo g_{3} 8 = \frac{lo g _{10} 8}{lo g _{10} 3} = \frac{3 lo g _{10} 2}{lo g _{10} 3} = \frac{3 p}{q}

Question 2

5 marks

Reading a function and its inverse from a graph

Functions

(a)

1 mark

Read the graph at $x = - 3$ .

f (- 3) = - 1

(b)

1 mark

The domain of $f^{- 1}$ is the range of $f$ , read from the graph of $f$ .

- 3 \leq x \leq 5

(c)

3 marks

$f^{- 1} (2 x - 7) = - 3$ means $f (- 3) = 2 x - 7$ .

- 1 = 2 x - 7 \Rightarrow x = 3

Question 3

5 marks

A trigonometric equation in cos 2 theta and cos theta

Number and algebra

5 marks

Use $cos 2 θ = 2 cos^{2} θ - 1$ .
Factorise.
Since $cos θ = \frac{5}{4}$ is impossible, $cos θ = 0$ . In the interval $π \leq θ \leq 2 π$ ,

2 (2 cos^{2} θ - 1) - 5 cos θ + 2 = 0 \Rightarrow 4 cos^{2} θ - 5 cos θ = 0 cos θ (4 cos θ - 5) = 0 θ = \frac{3 π}{2}

Question 4

7 marks

A line and a parabola: intersections and tangency

Calculus

(a)

2 marks

Equate the line and the curve.

x^{2} - x - 1 = m x - 3 \Rightarrow x^{2} - (m + 1) x + 2 = 0

(b)

5 marks

Tangency needs the discriminant to be zero.

Δ = (m + 1)^{2} - 4 (1) (2) = 0 \Rightarrow (m + 1)^{2} = 8 m + 1 = \pm 22 \Rightarrow m = - 1 + 22 or m = - 1 - 22

Question 5

6 marks

Two normal distributions with the same spread

Statistics and probability

(a)

2 marks

$Y$ has the same shape as $X$ with the mean 12 higher, so shift 22 down by 12.

b = 22 - (19 - 7) = 10

(b)

1 mark

About 68% of the values are within one standard deviation of the mean.

P (7 - a < X < 7 + a) \approx 0.68

(c)

3 marks

With $a = 3$ , the value 22 is $\frac{22 - 19}{3} = 1$ standard deviation above the mean of $Y$ . The area above is $\frac{1 - 0.68}{2} = 0.16$ .

P (Y < 22) = 1 - 0.16 = 0.84

Question 6

7 marks

Sketching |f(x)| and 15/f(x) from the key features of a graph

Functions

(a)

4 marks

The function $f$ is positive for $x > - \frac{1}{2}$ and negative for $x < - \frac{1}{2}$ , with no $x$ -intercepts.
Sketch: keep the right branch as it is, with its minimum at $B (0, \frac{15}{2})$ and the oblique asymptote $y = 5 x + 5$ . Reflect the left branch in the $x$ -axis. Its maximum at $A$ becomes a minimum at $(- 1, \frac{5}{2})$ , the oblique asymptote becomes $y = - 5 x - 5$ (for $x \to - \infty$ ), and both branches rise towards the vertical asymptote $x = - \frac{1}{2}$ .

(b)

3 marks

Sketch: the vertical asymptote $x = - \frac{1}{2}$ of $f$ turns into a point of the graph on the $x$ -axis, since $\frac{15}{f} \to 0$ there. As $x \to \pm \infty$ , $f \to \pm \infty$ so $\frac{15}{f} \to 0$ , giving the horizontal asymptote $y = 0$ .
The minimum $B$ of $f$ becomes a maximum of $\frac{15}{f}$ at $(0, \frac{15}{15/2}) = (0, 2)$ , which is also the $y$ -intercept.
The maximum $A$ of $f$ becomes a minimum of $\frac{15}{f}$ at $(- 1, \frac{15}{- 5/2}) = (- 1, - 6)$ .
So the graph is a hump with its top at $(0, 2)$ for $x > - \frac{1}{2}$ going down to $y = 0$ on the right, and a dip with its bottom at $(- 1, - 6)$ for $x < - \frac{1}{2}$ going back up to $y = 0$ on the left.

Question 7

6 marks

A solid of revolution about the y-axis for x squared + y to the fourth = r to the fourth

Number and algebra

6 marks

From $x^{2} + y^{4} = r^{4}$ get $x^{2} = r^{4} - y^{4}$ . The region extends from $y = - r$ to $y = r$ .
Integrate.
So $a = \frac{8}{5}$ and $b = 5$ .

V = π \int_{- r}^{r} (r^{4} - y^{4}) d y = 2 π \int_{0}^{r} (r^{4} - y^{4}) d y V = 2 π [r^{4} y - \frac{y ^{5}}{5}]_{0}^{r} = 2 π (r^{5} - \frac{r ^{5}}{5}) = \frac{8 π r ^{5}}{5}

Question 8

8 marks

A complex number in exponential form and cube roots of a quotient

Number and algebra

(a)

3 marks

The modulus of $z_{1} = 3 - 3 i$ .
The number is in the fourth quadrant with $tan θ = - \frac{3}{3} = - 3$ .

r = 3 + 9 = 12 = 23 θ = - \frac{π}{3} \Rightarrow z_{1} = 23 e^{- iπ /3}

(b)

5 marks

Divide the moduli and subtract the arguments.
All the arguments that give the same number are $- \frac{5 π}{6} + 2 π k$ . Take the cube root of the modulus (1) and divide the argument by 3.
For $k = 0, 1, 2$ and keeping the arguments in $(- π, π]$ :

\frac{z _{2}}{z _{1}} = \frac{2 3 e ^{i 5 π /6}}{2 3 e ^{- iπ /3}} = e^{i 7 π /6} = e^{- i 5 π /6} \frac{- \frac{5 π}{6} + 2 π k}{3} = - \frac{5 π}{18} + \frac{2 π k}{3} u = e^{- i 5 π /18}, v = e^{i 7 π /18}, w = e^{- i 17 π /18}

Question 9

6 marks

A limit with l'Hopital's rule: x sin x over 1 - cos x

Number and algebra

6 marks

At $x = 0$ , $x sin x \to 0$ and $1 - cos x \to 0$ , so use l'Hopital's rule.
This is still $\frac{0}{0}$ , so differentiate again.

x \to 0 lim \frac{x sin x}{1 - cos x} = x \to 0 lim \frac{sin x + x cos x}{sin x} x \to 0 lim \frac{cos x + cos x - x sin x}{cos x} = \frac{1 + 1 - 0}{1} = 2

Section B · Extended response

Questions 10–12 · 55 marks

Question 10

16 marks

A sequence that is arithmetic for one value of k and geometric for others

Number and algebra

(a)

5 marks

(i) Arithmetic: twice the middle term equals the sum of the other two.

(ii) Substitute $k = \frac{4}{5}$ into $u_{3}$ .

2 (3 - 2 k) = (k - 5) + (5 k + 3) \Rightarrow 6 - 4 k = 6 k - 2 \Rightarrow k = \frac{4}{5} u_{3} = 5 (\frac{4}{5}) + 3 = 7

(b)

4 marks

(i) With $k = 12$ the terms are $7$ , $- 21$ , $63$ . The ratios are equal.

So the terms form a geometric sequence with common ratio $- 3$ .

(ii) An infinite geometric series only has a sum when $∣ r ∣ < 1$ , but $∣ r ∣ = 3 \geq 1$ , so the series diverges and the sum does not exist.

\frac{- 21}{7} = - 3, \frac{63}{- 21} = - 3

(c)

7 marks

(i) For a geometric sequence, $u_{2}^{2} = u_{1} u_{3}$ .

(ii) Factorise.

The second value is $k = - 2$ , and the first three terms are

(iii) The terms alternate between $- 7$ and $7$ with ratio $- 1$ , so the sum of an even number of terms is zero.

(3 - 2 k)^{2} = (k - 5) (5 k + 3) \Rightarrow 9 - 12 k + 4 k^{2} = 5 k^{2} - 22 k - 15 k^{2} - 10 k - 24 = 0 (k - 12) (k + 2) = 0 u_{1} = - 7, u_{2} = 7, u_{3} = - 7 S_{2 m} = 0

Question 11

18 marks

A parallelogram in 3D, its plane and the angle with a second plane

Number and algebra

(a)

2 marks

$D = A + B C$ , with $B C = (2, 4, 2)$ .

D = (1 + 2, - 4 + 4, 0 + 2) = (3, 0, 2)

(b)

2 marks

$E$ is the midpoint of $[A C]$ .

E = (\frac{1 - 1}{2}, \frac{- 4 - 2}{2}, \frac{0 + 4}{2}) = (0, - 3, 2)

(c)

4 marks

(i) $A B = (- 4, - 2, 2)$ and $A D = (2, 4, 2)$ .

(ii) The area of the parallelogram is the length of the vector product.

A B \times A D = - 4 - 8 4 + 8 - 16 + 4 = - 12 12 - 12 = 12 - 1 1 - 1 m = 12 Area = 12 1 + 1 + 1 = 123

(d)

2 marks

The normal of $P_{1}$ is $(1, - 1, 1)$ , so $P_{1}$ is $x - y + z = d$ . Substitute $A (1, - 4, 0)$ .

d = 1 + 4 + 0 = 5 x - y + z = 5

(e)

3 marks

The normals are $n_{1} = (1, - 1, 1)$ and $n_{2} = (5, 1, - 7)$ , with $∣ n_{1} ∣ = 3$ and $∣ n_{2} ∣ = 75$ .
The scalar product is $- 3$ , which gives the obtuse angle between the normals, so the sign is dropped to get the acute angle between the planes.

cos θ = \frac{∣5 - 1 - 7∣}{3 75} = \frac{3}{225} = \frac{3}{15} = \frac{1}{5}

(f)

5 marks

The line $L$ passes through $E$ in the direction of the normal of $P_{1}$ .
Substitute $(λ, - 3 - λ, 2 + λ)$ into $5 x + y - 7 z = 1$ .

r = 0 - 3 2 + λ 1 - 1 1 5 λ + (- 3 - λ) - 7 (2 + λ) = 1 \Rightarrow - 3 λ - 17 = 1 \Rightarrow λ = - 6 F = (- 6, 3, - 4)

Question 12

21 marks

A circle in the complex plane, an argument condition and four roots

Number and algebra

(a)

3 marks

Write the modulus condition in terms of $x$ and $y$ .

∣ (x - 2) + (y - 1) i ∣ = 3 \Rightarrow (x - 2)^{2} + (y - 1)^{2} = 9 x^{2} - 4 x + 4 + y^{2} - 2 y + 1 = 9 \Rightarrow x^{2} + y^{2} - 4 x - 2 y - 4 = 0

(b)

7 marks

Write the quotient with real and imaginary parts, multiplying by the conjugate of the denominator.
Expand the numerator, $(x + p + y i) (x - 1 - y i)$ , into real and imaginary parts.
The denominator $(x - 1)^{2} + y^{2}$ is positive. An argument of $\frac{π}{4}$ means the real and imaginary parts are equal (and positive).

\frac{z + p}{z - 1} = \frac{( x + p ) + y i}{( x - 1 ) + y i} \times \frac{( x - 1 ) - y i}{( x - 1 ) - y i} Re = (x + p) (x - 1) + y^{2} = x^{2} + (p - 1) x - p + y^{2} Im = y (x - 1) - y (x + p) = - (p + 1) y x^{2} + (p - 1) x - p + y^{2} = - (p + 1) y x^{2} + y^{2} + (p - 1) x + (p + 1) y - p = 0

(c)

7 marks

With $p = 4$ the argument condition $Re = Im$ for $\frac{z + 4}{z - 1}$ gives the circle below. The roots also satisfy the circle from part (a).
Subtract the two equations.
Substitute $y = - x$ into the circle from part (a).

x^{2} + y^{2} + 3 x + 5 y - 4 = 0, x^{2} + y^{2} - 4 x - 2 y - 4 = 0 7 x + 7 y = 0 \Rightarrow y = - x x^{2} + x^{2} - 4 x + 2 x - 4 = 0 \Rightarrow x^{2} - x - 2 = 0 \Rightarrow (x - 2) (x + 1) = 0 x = 2, y = - 2 or x = - 1, y = 1 z_{1} = - 1 + i, z_{2} = 2 - 2 i

(d)

4 marks

The polynomial has real coefficients, so the conjugates of $z_{1}$ and $z_{2}$ are also roots, and these are the four roots.
The coefficient $a$ is minus the sum of the roots.

- 1 + i, - 1 - i, 2 - 2 i, 2 + 2 i a = - [(- 1 + i) + (- 1 - i) + (2 - 2 i) + (2 + 2 i)] = - 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 algebra80 marks · 73%
Q3, Q7, Q8, Q9, Q10, Q11, Q12
Functions17 marks · 15%
Q1, Q2, Q6
Calculus7 marks · 6%
Q4
Statistics and probability6 marks · 5%
Q5

Mentor insights

Traps to watch for

01
Tangency · Q4
A line is tangent to a curve when the intersection equation has a double root, so the discriminant is zero.
02
Cube roots · Q8
A cube root of $r e^{i ϕ}$ is $r^{1/3} e^{i (ϕ + 2 π k) /3}$ for $k = 0, 1, 2$ .
03
Arithmetic · Q10
$u_{2} - u_{1} = u_{3} - u_{2}$ , or $2 u_{2} = u_{1} + u_{3}$ .
04
Parallelogram · Q11
The diagonals bisect each other, so $D = A + C - B$ and $E$ is the midpoint of $[A C]$ .
05
Locus · Q12
$∣ z - (2 + i) ∣ = 3$ is a circle with centre $(2, 1)$ and radius 3.

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Worked solutions

IB Maths AA HL · Paper 1

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1 TZ2TZ3

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

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2025 TZ3 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Section A · Short response

Questions 1–9 · 55 marks

Question 1

5 marks

Logarithms of 24 and base conversion

Functions

(a)

3 marks

Write $24 = 2^{3} \times 3$ .

lo g_{10} 24 = 3 lo g_{10} 2 + lo g_{10} 3 = 3 p + q

(b)

2 marks

Change the base to 10.

lo g_{3} 8 = \frac{lo g _{10} 8}{lo g _{10} 3} = \frac{3 lo g _{10} 2}{lo g _{10} 3} = \frac{3 p}{q}

Question 2

5 marks

Reading a function and its inverse from a graph

Functions

(a)

1 mark

Read the graph at $x = - 3$ .

f (- 3) = - 1

(b)

1 mark

The domain of $f^{- 1}$ is the range of $f$ , read from the graph of $f$ .

- 3 \leq x \leq 5

(c)

3 marks

$f^{- 1} (2 x - 7) = - 3$ means $f (- 3) = 2 x - 7$ .

- 1 = 2 x - 7 \Rightarrow x = 3

Question 3

5 marks

A trigonometric equation in cos 2 theta and cos theta

Number and algebra

5 marks

Use $cos 2 θ = 2 cos^{2} θ - 1$ .
Factorise.
Since $cos θ = \frac{5}{4}$ is impossible, $cos θ = 0$ . In the interval $π \leq θ \leq 2 π$ ,

2 (2 cos^{2} θ - 1) - 5 cos θ + 2 = 0 \Rightarrow 4 cos^{2} θ - 5 cos θ = 0 cos θ (4 cos θ - 5) = 0 θ = \frac{3 π}{2}

Question 4

7 marks

A line and a parabola: intersections and tangency

Calculus

(a)

2 marks

Equate the line and the curve.

x^{2} - x - 1 = m x - 3 \Rightarrow x^{2} - (m + 1) x + 2 = 0

(b)

5 marks

Tangency needs the discriminant to be zero.

Δ = (m + 1)^{2} - 4 (1) (2) = 0 \Rightarrow (m + 1)^{2} = 8 m + 1 = \pm 22 \Rightarrow m = - 1 + 22 or m = - 1 - 22

Question 5

6 marks

Two normal distributions with the same spread

Statistics and probability

(a)

2 marks

$Y$ has the same shape as $X$ with the mean 12 higher, so shift 22 down by 12.

b = 22 - (19 - 7) = 10

(b)

1 mark

About 68% of the values are within one standard deviation of the mean.

P (7 - a < X < 7 + a) \approx 0.68

(c)

3 marks

With $a = 3$ , the value 22 is $\frac{22 - 19}{3} = 1$ standard deviation above the mean of $Y$ . The area above is $\frac{1 - 0.68}{2} = 0.16$ .

P (Y < 22) = 1 - 0.16 = 0.84

Question 6

7 marks

Sketching |f(x)| and 15/f(x) from the key features of a graph

Functions

(a)

4 marks

The function $f$ is positive for $x > - \frac{1}{2}$ and negative for $x < - \frac{1}{2}$ , with no $x$ -intercepts.
Sketch: keep the right branch as it is, with its minimum at $B (0, \frac{15}{2})$ and the oblique asymptote $y = 5 x + 5$ . Reflect the left branch in the $x$ -axis. Its maximum at $A$ becomes a minimum at $(- 1, \frac{5}{2})$ , the oblique asymptote becomes $y = - 5 x - 5$ (for $x \to - \infty$ ), and both branches rise towards the vertical asymptote $x = - \frac{1}{2}$ .

(b)

3 marks

Sketch: the vertical asymptote $x = - \frac{1}{2}$ of $f$ turns into a point of the graph on the $x$ -axis, since $\frac{15}{f} \to 0$ there. As $x \to \pm \infty$ , $f \to \pm \infty$ so $\frac{15}{f} \to 0$ , giving the horizontal asymptote $y = 0$ .
The minimum $B$ of $f$ becomes a maximum of $\frac{15}{f}$ at $(0, \frac{15}{15/2}) = (0, 2)$ , which is also the $y$ -intercept.
The maximum $A$ of $f$ becomes a minimum of $\frac{15}{f}$ at $(- 1, \frac{15}{- 5/2}) = (- 1, - 6)$ .
So the graph is a hump with its top at $(0, 2)$ for $x > - \frac{1}{2}$ going down to $y = 0$ on the right, and a dip with its bottom at $(- 1, - 6)$ for $x < - \frac{1}{2}$ going back up to $y = 0$ on the left.

Question 7

6 marks

A solid of revolution about the y-axis for x squared + y to the fourth = r to the fourth

Number and algebra

6 marks

From $x^{2} + y^{4} = r^{4}$ get $x^{2} = r^{4} - y^{4}$ . The region extends from $y = - r$ to $y = r$ .
Integrate.
So $a = \frac{8}{5}$ and $b = 5$ .

V = π \int_{- r}^{r} (r^{4} - y^{4}) d y = 2 π \int_{0}^{r} (r^{4} - y^{4}) d y V = 2 π [r^{4} y - \frac{y ^{5}}{5}]_{0}^{r} = 2 π (r^{5} - \frac{r ^{5}}{5}) = \frac{8 π r ^{5}}{5}

Question 8

8 marks

A complex number in exponential form and cube roots of a quotient

Number and algebra

(a)

3 marks

The modulus of $z_{1} = 3 - 3 i$ .
The number is in the fourth quadrant with $tan θ = - \frac{3}{3} = - 3$ .

r = 3 + 9 = 12 = 23 θ = - \frac{π}{3} \Rightarrow z_{1} = 23 e^{- iπ /3}

(b)

5 marks

Divide the moduli and subtract the arguments.
All the arguments that give the same number are $- \frac{5 π}{6} + 2 π k$ . Take the cube root of the modulus (1) and divide the argument by 3.
For $k = 0, 1, 2$ and keeping the arguments in $(- π, π]$ :

\frac{z _{2}}{z _{1}} = \frac{2 3 e ^{i 5 π /6}}{2 3 e ^{- iπ /3}} = e^{i 7 π /6} = e^{- i 5 π /6} \frac{- \frac{5 π}{6} + 2 π k}{3} = - \frac{5 π}{18} + \frac{2 π k}{3} u = e^{- i 5 π /18}, v = e^{i 7 π /18}, w = e^{- i 17 π /18}

Question 9

6 marks

A limit with l'Hopital's rule: x sin x over 1 - cos x

Number and algebra

6 marks

At $x = 0$ , $x sin x \to 0$ and $1 - cos x \to 0$ , so use l'Hopital's rule.
This is still $\frac{0}{0}$ , so differentiate again.

x \to 0 lim \frac{x sin x}{1 - cos x} = x \to 0 lim \frac{sin x + x cos x}{sin x} x \to 0 lim \frac{cos x + cos x - x sin x}{cos x} = \frac{1 + 1 - 0}{1} = 2

Section B · Extended response

Questions 10–12 · 55 marks

Question 10

16 marks

A sequence that is arithmetic for one value of k and geometric for others

Number and algebra

(a)

5 marks

(i) Arithmetic: twice the middle term equals the sum of the other two.

(ii) Substitute $k = \frac{4}{5}$ into $u_{3}$ .

2 (3 - 2 k) = (k - 5) + (5 k + 3) \Rightarrow 6 - 4 k = 6 k - 2 \Rightarrow k = \frac{4}{5} u_{3} = 5 (\frac{4}{5}) + 3 = 7

(b)

4 marks

(i) With $k = 12$ the terms are $7$ , $- 21$ , $63$ . The ratios are equal.

So the terms form a geometric sequence with common ratio $- 3$ .

(ii) An infinite geometric series only has a sum when $∣ r ∣ < 1$ , but $∣ r ∣ = 3 \geq 1$ , so the series diverges and the sum does not exist.

\frac{- 21}{7} = - 3, \frac{63}{- 21} = - 3

(c)

7 marks

(i) For a geometric sequence, $u_{2}^{2} = u_{1} u_{3}$ .

(ii) Factorise.

The second value is $k = - 2$ , and the first three terms are

(iii) The terms alternate between $- 7$ and $7$ with ratio $- 1$ , so the sum of an even number of terms is zero.

(3 - 2 k)^{2} = (k - 5) (5 k + 3) \Rightarrow 9 - 12 k + 4 k^{2} = 5 k^{2} - 22 k - 15 k^{2} - 10 k - 24 = 0 (k - 12) (k + 2) = 0 u_{1} = - 7, u_{2} = 7, u_{3} = - 7 S_{2 m} = 0

Question 11

18 marks

A parallelogram in 3D, its plane and the angle with a second plane

Number and algebra

(a)

2 marks

$D = A + B C$ , with $B C = (2, 4, 2)$ .

D = (1 + 2, - 4 + 4, 0 + 2) = (3, 0, 2)

(b)

2 marks

$E$ is the midpoint of $[A C]$ .

E = (\frac{1 - 1}{2}, \frac{- 4 - 2}{2}, \frac{0 + 4}{2}) = (0, - 3, 2)

(c)

4 marks

(i) $A B = (- 4, - 2, 2)$ and $A D = (2, 4, 2)$ .

(ii) The area of the parallelogram is the length of the vector product.

A B \times A D = - 4 - 8 4 + 8 - 16 + 4 = - 12 12 - 12 = 12 - 1 1 - 1 m = 12 Area = 12 1 + 1 + 1 = 123

(d)

2 marks

The normal of $P_{1}$ is $(1, - 1, 1)$ , so $P_{1}$ is $x - y + z = d$ . Substitute $A (1, - 4, 0)$ .

d = 1 + 4 + 0 = 5 x - y + z = 5

(e)

3 marks

The normals are $n_{1} = (1, - 1, 1)$ and $n_{2} = (5, 1, - 7)$ , with $∣ n_{1} ∣ = 3$ and $∣ n_{2} ∣ = 75$ .
The scalar product is $- 3$ , which gives the obtuse angle between the normals, so the sign is dropped to get the acute angle between the planes.

cos θ = \frac{∣5 - 1 - 7∣}{3 75} = \frac{3}{225} = \frac{3}{15} = \frac{1}{5}

(f)

5 marks

The line $L$ passes through $E$ in the direction of the normal of $P_{1}$ .
Substitute $(λ, - 3 - λ, 2 + λ)$ into $5 x + y - 7 z = 1$ .

r = 0 - 3 2 + λ 1 - 1 1 5 λ + (- 3 - λ) - 7 (2 + λ) = 1 \Rightarrow - 3 λ - 17 = 1 \Rightarrow λ = - 6 F = (- 6, 3, - 4)

Question 12

21 marks

A circle in the complex plane, an argument condition and four roots

Number and algebra

(a)

3 marks

Write the modulus condition in terms of $x$ and $y$ .

∣ (x - 2) + (y - 1) i ∣ = 3 \Rightarrow (x - 2)^{2} + (y - 1)^{2} = 9 x^{2} - 4 x + 4 + y^{2} - 2 y + 1 = 9 \Rightarrow x^{2} + y^{2} - 4 x - 2 y - 4 = 0

(b)

7 marks

Write the quotient with real and imaginary parts, multiplying by the conjugate of the denominator.
Expand the numerator, $(x + p + y i) (x - 1 - y i)$ , into real and imaginary parts.
The denominator $(x - 1)^{2} + y^{2}$ is positive. An argument of $\frac{π}{4}$ means the real and imaginary parts are equal (and positive).

\frac{z + p}{z - 1} = \frac{( x + p ) + y i}{( x - 1 ) + y i} \times \frac{( x - 1 ) - y i}{( x - 1 ) - y i} Re = (x + p) (x - 1) + y^{2} = x^{2} + (p - 1) x - p + y^{2} Im = y (x - 1) - y (x + p) = - (p + 1) y x^{2} + (p - 1) x - p + y^{2} = - (p + 1) y x^{2} + y^{2} + (p - 1) x + (p + 1) y - p = 0

(c)

7 marks

With $p = 4$ the argument condition $Re = Im$ for $\frac{z + 4}{z - 1}$ gives the circle below. The roots also satisfy the circle from part (a).
Subtract the two equations.
Substitute $y = - x$ into the circle from part (a).

x^{2} + y^{2} + 3 x + 5 y - 4 = 0, x^{2} + y^{2} - 4 x - 2 y - 4 = 0 7 x + 7 y = 0 \Rightarrow y = - x x^{2} + x^{2} - 4 x + 2 x - 4 = 0 \Rightarrow x^{2} - x - 2 = 0 \Rightarrow (x - 2) (x + 1) = 0 x = 2, y = - 2 or x = - 1, y = 1 z_{1} = - 1 + i, z_{2} = 2 - 2 i

(d)

4 marks

The polynomial has real coefficients, so the conjugates of $z_{1}$ and $z_{2}$ are also roots, and these are the four roots.
The coefficient $a$ is minus the sum of the roots.

- 1 + i, - 1 - i, 2 - 2 i, 2 + 2 i a = - [(- 1 + i) + (- 1 - i) + (2 - 2 i) + (2 + 2 i)] = - 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 algebra80 marks · 73%
Q3, Q7, Q8, Q9, Q10, Q11, Q12
Functions17 marks · 15%
Q1, Q2, Q6
Calculus7 marks · 6%
Q4
Statistics and probability6 marks · 5%
Q5

Mentor insights

Traps to watch for

01
Tangency · Q4
A line is tangent to a curve when the intersection equation has a double root, so the discriminant is zero.
02
Cube roots · Q8
A cube root of $r e^{i ϕ}$ is $r^{1/3} e^{i (ϕ + 2 π k) /3}$ for $k = 0, 1, 2$ .
03
Arithmetic · Q10
$u_{2} - u_{1} = u_{3} - u_{2}$ , or $2 u_{2} = u_{1} + u_{3}$ .
04
Parallelogram · Q11
The diagonals bisect each other, so $D = A + C - B$ and $E$ is the midpoint of $[A C]$ .
05
Locus · Q12
$∣ z - (2 + i) ∣ = 3$ is a circle with centre $(2, 1)$ and radius 3.

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