Loading…

Worked solutions

IB Maths AA HL · Paper 2

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: November 2023
Marks: 110
Duration: 2 hours
Calculator: 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 November 2023 TZ1 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 · 57 marks

Question 1

6 marks

A pyramid on a rectangular base: a length and an angle

Geometry and trigonometry

(a)

2 marks

$B V = (3 - 6, 4 - 8, 9 - 0) = (- 3, - 4, 9)$ .

B V = 9 + 16 + 81 = 106 \approx 10.3

(b)

4 marks

$V B = (3, 4, - 9)$ and $V C = (3, - 4, - 9)$ .

V B \cdot V C = 9 - 16 + 81 = 74, ∣ V B ∣ = ∣ V C ∣ = 106 cos B V C = \frac{74}{106} \Rightarrow B V C = 45. 7^{\circ}

Question 2

5 marks

f(x) = e^x - 3x - 4: a sketch and transformations

Functions

(a)

3 marks

Features on $- 4 \leq x \leq 3$ : $y$ -intercept $f (0) = - 3$ ; roots near $x = - 1.4$ and $x = 2.4$ ; a minimum at $x = ln 3 = 1.10$ with $f = - 4.30$ ; end points $f (- 4) = 8.02$ and $f (3) = 7.09$ .
Sketch: the curve starts at $(- 4, 8.0)$ , falls through a root in $(- 2, - 1)$ and the $y$ -intercept $(0, - 3)$ to the minimum near $(1.1, - 4.3)$ , then rises through a root in $(2, 3)$ to $(3, 7.1)$ .

(b)

2 marks

Compare $g (x) = e^{2 x} - 6 x - 7$ with $f (2 x) = e^{2 x} - 6 x - 4$ .

g (x) = f (2 x) - 3 k = \frac{1}{2}, c = - 3

Question 3

7 marks

A letter C between two circles: area and perimeter

Geometry and trigonometry

(a)

2 marks

Area of a sector is $\frac{1}{2} R^{2} θ$ with $θ = 5.2$ .

A = \frac{1}{2} (10)^{2} (5.2) - \frac{1}{2} r^{2} (5.2) = 260 - 2.6 r^{2}

(b)

5 marks

Set $260 - 2.6 r^{2} = 64$ .
The perimeter: outer arc $10 \times 5.2 = 52$ , inner arc $5.2 r = 45.1$ , and two straight ends each of length $10 - r = 1.32$ .

r^{2} = \frac{196}{2.6} = 75.38 \Rightarrow r = 8.68 P = 52 + 45.1 + 2 (1.32) \approx 99.8 cm

Question 4

5 marks

A particle with a square-root sine displacement: rest and distance

Calculus

(a)

2 marks

The displacement is $s (t) = 4.3 sin (3 t + 5)$ . The velocity is zero where $cos (3 t + 5) = 0$ .
(The value $3 (0) + 5 = 2.24$ is already past $\frac{π}{2}$ , so $\frac{3 π}{2}$ is the first one reached.)

3 t + 5 = \frac{3 π}{2} \Rightarrow 3 t + 5 = \frac{9 π ^{2}}{4} \Rightarrow q = 5.74

(b)

3 marks

Distance is the integral of speed from 0 to $q$ .
Check: $s (0) = 3.38$ and $s (q) = - 4.3$ , and the particle moves in one direction in between, so the distance is $3.38 + 4.30 = 7.68$ .

\int_{0}^{5.74} ∣ v (t) ∣ d t = 7.68 m

Question 5

5 marks

Finding a from an expected value and total probability

Number and algebra

5 marks

The probabilities are $k, k^{2}, a, k^{3}$ for $x = 1, 2, 3, 4$ .
Eliminate $a$ using $a = 1 - k - k^{2} - k^{3}$ .
Solve for $k$ with $0 < k < 1$ .

k + k^{2} + a + k^{3} = 1 E (X) = k + 2 k^{2} + 3 a + 4 k^{3} = 2.3 k + 2 k^{2} + 3 (1 - k - k^{2} - k^{3}) + 4 k^{3} = 2.3 \Rightarrow k^{3} - k^{2} - 2 k + 0.7 = 0 k = 0.316 a = 1 - 0.3159 - 0.0998 - 0.0315 \approx 0.553

Question 6

5 marks

Variance of a binomial and of a linear transformation

Statistics and probability

(a)

3 marks

For a binomial, $Var (X) = n p (1 - p)$ .

25 p (1 - p) = 5.75 \Rightarrow p^{2} - p + 0.23 = 0 p = \frac{1 \pm 1 - 0.92}{2} = 0.641 or 0.359

(b)

2 marks

Use $Var (1 - 2 X) = (- 2)^{2} Var (X)$ .

Var (Y) = 4 (5.75) = 23

Question 7

6 marks

Arranging a baseball team and choosing five players

Number and algebra

(a)

3 marks

With no restrictions there are 9 people in a line.
With the three girls together, treat them as one block: $7!$ ways to arrange the 7 items and $3!$ ways to order the girls.

9! = 362880 7! \times 3! = 5040 \times 6 = 30240

(b)

3 marks

All selections of 5 from 9:
No girls: $(5 6) = 6$ . Exactly one girl: $3 \times (4 6) = 45$ .

(5 9) = 126 126 - 6 - 45 = 75

Question 8

9 marks

Vector product with a parameter and the smallest triangle area

Geometry and trigonometry
Calculus

(a)

4 marks

Find the two vectors from A.
Take the vector product.

A B = 1 1 - p - 1, A C = p - p 2 A B \times A C = 2 (1 - p) - p - p - 2 - p - p (1 - p) = 2 - 3 p - p - 2 p^{2} - 2 p

(b)

3 marks

The squared length is
Find the minimum for $p > 0$ with the calculator.

∣ A B \times A C ∣^{2} = (2 - 3 p)^{2} + (p + 2)^{2} + (p^{2} - 2 p)^{2} = p^{4} - 4 p^{3} + 14 p^{2} - 8 p + 8 p = 0.326, min = 6.75

(c)

2 marks

The area is half the length.

Area_{m i n} = \frac{1}{2} 6.7526 \approx 1.30

Question 9

9 marks

Euler's method and the exact solution of dy/dx = (4 - y)/10

Calculus

(a)

3 marks

The slope is $\frac{d y}{d x} = \frac{4 - y}{10}$ . Start at $(0, 2)$ and take five steps.

y_{1} = 2.02, y_{2} = 2.0398, y_{3} = 2.0594, y_{4} = 2.0788, y_{5} = 2.0980 y (0.5) \approx 2.098

(b)

5 marks

Separate the variables.
Use $y = 2$ at $x = 0$ .

\int \frac{d y}{4 - y} = \int \frac{d x}{10} \Rightarrow - ln ∣4 - y ∣ = \frac{x}{10} + c - ln 2 = c \Rightarrow ln (4 - y) = ln 2 - \frac{x}{10} 4 - y = 2 e^{- x /10} \Rightarrow y = 4 - 2 e^{- x /10}

(c)

1 mark

The exact value at $x = 0.5$ is $4 - 2 e^{- 0.05} = 2.09754$ .

∣2.09802 - 2.09754∣ \approx 0.000479

Section B · Extended response

Questions 10–12 · 53 marks

Question 10

16 marks

Wheat heights: finding a normal distribution and conditional probability

Statistics and probability

(a)

2 marks

The three regions add to 1.

P (94.6 < H < 98.1) = 1 - 0.288 - 0.434 = 0.278

(b)

5 marks

The z-scores are $z = - 0.5592$ for $P (H < 94.6) = 0.288$ and $z = 0.1662$ for $P (H > 98.1) = 0.434$ .
Solve the pair of equations.

\frac{94.6 - μ}{σ} = - 0.5592, \frac{98.1 - μ}{σ} = 0.1662 μ = 97.3, σ = 4.82

(c)

6 marks

A plant is ready with probability 0.434. Let $X \sim B (100, 0.434)$ .
Given fewer than 49:

P (X = 34) \approx 0.0133 P (X = 34 ∣ X < 49) = \frac{P ( X = 34 )}{P ( X \leq 48 )} = \frac{0.01332}{0.8482} \approx 0.0157

(d)

3 marks

The interquartile range of a normal distribution is $2 (0.6745) σ$ .

2 (0.6745) d = 4.82 \Rightarrow d = 3.57

Question 11

19 marks

A rational function with an oblique asymptote: sketch, range and inequality

Functions

(a)

1 mark

The denominator $2 x + 6$ is zero at $x = - 3$ .

x = - 3

(b)

2 marks

Factorise the numerator: $x^{2} - 14 x + 24 = (x - 2) (x - 12)$ .

(2, 0) and (12, 0)

(c)

4 marks

Divide $x^{2} - 14 x + 24$ by $2 x + 6$ .

f (x) = \frac{1}{2} x - \frac{17}{2} + \frac{75}{2 x + 6} a = \frac{1}{2}, b = - \frac{17}{2}

(d)

4 marks

Sketch: the vertical asymptote $x = - 3$ , and the oblique asymptote $y = \frac{1}{2} x - \frac{17}{2}$ . The graph crosses the $x$ -axis at $(2, 0)$ and $(12, 0)$ and the $y$ -axis at $(0, 4)$ . Right of $x = - 3$ it comes down from $+ \infty$ , dips below the axis between the roots and then follows the oblique line upward. Left of $x = - 3$ it falls to $- \infty$ near the asymptote, rising from far left below the oblique line.

(e)

4 marks

Write $y (2 x + 6) = x^{2} - 14 x + 24$ as $x^{2} - (14 + 2 y) x + 24 - 6 y = 0$ . It has real roots when the discriminant is non-negative.

(14 + 2 y)^{2} - 4 (24 - 6 y) = 4 y^{2} + 80 y + 100 \geq 0 \Rightarrow y^{2} + 20 y + 25 \geq 0 y \leq - 10 - 53 or y \geq - 10 + 53

(f)

4 marks

Solve $f (x) = x$ : $x^{2} - 14 x + 24 = 2 x^{2} + 6 x$ , so $x^{2} + 20 x - 24 = 0$ and $x = - 10 \pm 231 = - 21.1, 1.14$ .
For $x > - 3$ the denominator is positive and $f (x) > x$ when $x^{2} + 20 x - 24 < 0$ , that is $- 3 < x < 1.14$ . For $x < - 3$ the denominator is negative, so $f (x) > x$ when $x^{2} + 20 x - 24 > 0$ , that is $x < - 21.1$ .

x < - 10 - 231 or - 3 < x < - 10 + 231

Question 12

18 marks

Two lines meeting at a point, a plane, and a reflection

Geometry and trigonometry

(a)

5 marks

Equate the coordinates of $L$ and $M$ .
The second equation gives $t = 3 s - 7$ . In the third, $- 3 + 6 s = 11 + 6 s - 14$ , which is true for every $s$ , so this equation is equivalent to the second. Substitute into the first.
The two lines meet at

1 + 2 s = 9 + 4 t, 2 + 3 s = 9 + t, - 3 + 6 s = 11 + 2 t 1 + 2 s = 9 + 12 s - 28 \Rightarrow 10 s = 20 \Rightarrow s = 2, t = - 1 A = 589

(b)

3 marks

The normal of the plane is $n = (0, 2, - 1)$ . Check the directions are perpendicular to $n$ and one point of each line is on the plane.
Both lines are parallel to the plane and each contains a point of the plane, so both lie in the plane.

236 \cdot n = 6 - 6 = 0, 412 \cdot n = 2 - 2 = 0 12 - 3 \cdot n = 4 + 3 = 7, 9911 \cdot n = 18 - 11 = 7

(c)

7 marks

The line through B perpendicular to the plane: $r = - 3 122 + λ 02 - 1$ . Put it into the plane.

2 (12 + 2 λ) - (2 - λ) = 7 \Rightarrow 22 + 5 λ = 7 \Rightarrow λ = - 3 C = - 3 65, ∣ B C ∣ = 3∣ n ∣ = 35 \approx 6.71

(d)

3 marks

The reflection of B is the same distance beyond the plane, so it is $2 O C - O B$ .

B^{'} = 2 - 3 65 - - 3 122 = - 3 08

Our analysis

Where the marks are

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

Geometry and trigonometry37 marks · 34%
Q1, Q3, Q8, Q12
Functions24 marks · 22%
Q2, Q11
Statistics and probability21 marks · 19%
Q6, Q10
Calculus17 marks · 15%
Q4, Q8, Q9
Number and algebra11 marks · 10%
Q5, Q7

Mentor insights

Traps to watch for

01
Area from the vector product · Q8
$Area = \frac{1}{2} ∣ A B \times A C ∣$ . Minimise the square to avoid the square root.
02
Euler with a small step · Q9
Use $y_{n + 1} = y_{n} + h f (x_{n}, y_{n})$ with $h = 0.1$ .
03
Two inverse-normal equations · Q10
Each given probability gives a z-score and an equation in $μ$ and $σ$ .
04
Oblique asymptote by division · Q11
Divide the numerator by the denominator and keep the quotient.
05
Intersection · Q12
Equate the three components and solve for $s$ and $t$ . One equation is redundant here.

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

Want someone to go through your paper with you?

Our maths mentors took these exams recently. One-to-one, they'll find where your marks are going and fix it before May.

Find your maths mentor More past papers

Worked solutions

IB Maths AA HL · Paper 2

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: November 2023
Marks: 110
Duration: 2 hours
Calculator: 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 November 2023 TZ1 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 · 57 marks

Question 1

6 marks

A pyramid on a rectangular base: a length and an angle

Geometry and trigonometry

(a)

2 marks

$B V = (3 - 6, 4 - 8, 9 - 0) = (- 3, - 4, 9)$ .

B V = 9 + 16 + 81 = 106 \approx 10.3

(b)

4 marks

$V B = (3, 4, - 9)$ and $V C = (3, - 4, - 9)$ .

V B \cdot V C = 9 - 16 + 81 = 74, ∣ V B ∣ = ∣ V C ∣ = 106 cos B V C = \frac{74}{106} \Rightarrow B V C = 45. 7^{\circ}

Question 2

5 marks

f(x) = e^x - 3x - 4: a sketch and transformations

Functions

(a)

3 marks

Features on $- 4 \leq x \leq 3$ : $y$ -intercept $f (0) = - 3$ ; roots near $x = - 1.4$ and $x = 2.4$ ; a minimum at $x = ln 3 = 1.10$ with $f = - 4.30$ ; end points $f (- 4) = 8.02$ and $f (3) = 7.09$ .
Sketch: the curve starts at $(- 4, 8.0)$ , falls through a root in $(- 2, - 1)$ and the $y$ -intercept $(0, - 3)$ to the minimum near $(1.1, - 4.3)$ , then rises through a root in $(2, 3)$ to $(3, 7.1)$ .

(b)

2 marks

Compare $g (x) = e^{2 x} - 6 x - 7$ with $f (2 x) = e^{2 x} - 6 x - 4$ .

g (x) = f (2 x) - 3 k = \frac{1}{2}, c = - 3

Question 3

7 marks

A letter C between two circles: area and perimeter

Geometry and trigonometry

(a)

2 marks

Area of a sector is $\frac{1}{2} R^{2} θ$ with $θ = 5.2$ .

A = \frac{1}{2} (10)^{2} (5.2) - \frac{1}{2} r^{2} (5.2) = 260 - 2.6 r^{2}

(b)

5 marks

Set $260 - 2.6 r^{2} = 64$ .
The perimeter: outer arc $10 \times 5.2 = 52$ , inner arc $5.2 r = 45.1$ , and two straight ends each of length $10 - r = 1.32$ .

r^{2} = \frac{196}{2.6} = 75.38 \Rightarrow r = 8.68 P = 52 + 45.1 + 2 (1.32) \approx 99.8 cm

Question 4

5 marks

A particle with a square-root sine displacement: rest and distance

Calculus

(a)

2 marks

The displacement is $s (t) = 4.3 sin (3 t + 5)$ . The velocity is zero where $cos (3 t + 5) = 0$ .
(The value $3 (0) + 5 = 2.24$ is already past $\frac{π}{2}$ , so $\frac{3 π}{2}$ is the first one reached.)

3 t + 5 = \frac{3 π}{2} \Rightarrow 3 t + 5 = \frac{9 π ^{2}}{4} \Rightarrow q = 5.74

(b)

3 marks

Distance is the integral of speed from 0 to $q$ .
Check: $s (0) = 3.38$ and $s (q) = - 4.3$ , and the particle moves in one direction in between, so the distance is $3.38 + 4.30 = 7.68$ .

\int_{0}^{5.74} ∣ v (t) ∣ d t = 7.68 m

Question 5

5 marks

Finding a from an expected value and total probability

Number and algebra

5 marks

The probabilities are $k, k^{2}, a, k^{3}$ for $x = 1, 2, 3, 4$ .
Eliminate $a$ using $a = 1 - k - k^{2} - k^{3}$ .
Solve for $k$ with $0 < k < 1$ .

k + k^{2} + a + k^{3} = 1 E (X) = k + 2 k^{2} + 3 a + 4 k^{3} = 2.3 k + 2 k^{2} + 3 (1 - k - k^{2} - k^{3}) + 4 k^{3} = 2.3 \Rightarrow k^{3} - k^{2} - 2 k + 0.7 = 0 k = 0.316 a = 1 - 0.3159 - 0.0998 - 0.0315 \approx 0.553

Question 6

5 marks

Variance of a binomial and of a linear transformation

Statistics and probability

(a)

3 marks

For a binomial, $Var (X) = n p (1 - p)$ .

25 p (1 - p) = 5.75 \Rightarrow p^{2} - p + 0.23 = 0 p = \frac{1 \pm 1 - 0.92}{2} = 0.641 or 0.359

(b)

2 marks

Use $Var (1 - 2 X) = (- 2)^{2} Var (X)$ .

Var (Y) = 4 (5.75) = 23

Question 7

6 marks

Arranging a baseball team and choosing five players

Number and algebra

(a)

3 marks

With no restrictions there are 9 people in a line.
With the three girls together, treat them as one block: $7!$ ways to arrange the 7 items and $3!$ ways to order the girls.

9! = 362880 7! \times 3! = 5040 \times 6 = 30240

(b)

3 marks

All selections of 5 from 9:
No girls: $(5 6) = 6$ . Exactly one girl: $3 \times (4 6) = 45$ .

(5 9) = 126 126 - 6 - 45 = 75

Question 8

9 marks

Vector product with a parameter and the smallest triangle area

Geometry and trigonometry
Calculus

(a)

4 marks

Find the two vectors from A.
Take the vector product.

A B = 1 1 - p - 1, A C = p - p 2 A B \times A C = 2 (1 - p) - p - p - 2 - p - p (1 - p) = 2 - 3 p - p - 2 p^{2} - 2 p

(b)

3 marks

The squared length is
Find the minimum for $p > 0$ with the calculator.

∣ A B \times A C ∣^{2} = (2 - 3 p)^{2} + (p + 2)^{2} + (p^{2} - 2 p)^{2} = p^{4} - 4 p^{3} + 14 p^{2} - 8 p + 8 p = 0.326, min = 6.75

(c)

2 marks

The area is half the length.

Area_{m i n} = \frac{1}{2} 6.7526 \approx 1.30

Question 9

9 marks

Euler's method and the exact solution of dy/dx = (4 - y)/10

Calculus

(a)

3 marks

The slope is $\frac{d y}{d x} = \frac{4 - y}{10}$ . Start at $(0, 2)$ and take five steps.

y_{1} = 2.02, y_{2} = 2.0398, y_{3} = 2.0594, y_{4} = 2.0788, y_{5} = 2.0980 y (0.5) \approx 2.098

(b)

5 marks

Separate the variables.
Use $y = 2$ at $x = 0$ .

\int \frac{d y}{4 - y} = \int \frac{d x}{10} \Rightarrow - ln ∣4 - y ∣ = \frac{x}{10} + c - ln 2 = c \Rightarrow ln (4 - y) = ln 2 - \frac{x}{10} 4 - y = 2 e^{- x /10} \Rightarrow y = 4 - 2 e^{- x /10}

(c)

1 mark

The exact value at $x = 0.5$ is $4 - 2 e^{- 0.05} = 2.09754$ .

∣2.09802 - 2.09754∣ \approx 0.000479

Section B · Extended response

Questions 10–12 · 53 marks

Question 10

16 marks

Wheat heights: finding a normal distribution and conditional probability

Statistics and probability

(a)

2 marks

The three regions add to 1.

P (94.6 < H < 98.1) = 1 - 0.288 - 0.434 = 0.278

(b)

5 marks

The z-scores are $z = - 0.5592$ for $P (H < 94.6) = 0.288$ and $z = 0.1662$ for $P (H > 98.1) = 0.434$ .
Solve the pair of equations.

\frac{94.6 - μ}{σ} = - 0.5592, \frac{98.1 - μ}{σ} = 0.1662 μ = 97.3, σ = 4.82

(c)

6 marks

A plant is ready with probability 0.434. Let $X \sim B (100, 0.434)$ .
Given fewer than 49:

P (X = 34) \approx 0.0133 P (X = 34 ∣ X < 49) = \frac{P ( X = 34 )}{P ( X \leq 48 )} = \frac{0.01332}{0.8482} \approx 0.0157

(d)

3 marks

The interquartile range of a normal distribution is $2 (0.6745) σ$ .

2 (0.6745) d = 4.82 \Rightarrow d = 3.57

Question 11

19 marks

A rational function with an oblique asymptote: sketch, range and inequality

Functions

(a)

1 mark

The denominator $2 x + 6$ is zero at $x = - 3$ .

x = - 3

(b)

2 marks

Factorise the numerator: $x^{2} - 14 x + 24 = (x - 2) (x - 12)$ .

(2, 0) and (12, 0)

(c)

4 marks

Divide $x^{2} - 14 x + 24$ by $2 x + 6$ .

f (x) = \frac{1}{2} x - \frac{17}{2} + \frac{75}{2 x + 6} a = \frac{1}{2}, b = - \frac{17}{2}

(d)

4 marks

(e)

4 marks

Write $y (2 x + 6) = x^{2} - 14 x + 24$ as $x^{2} - (14 + 2 y) x + 24 - 6 y = 0$ . It has real roots when the discriminant is non-negative.

(14 + 2 y)^{2} - 4 (24 - 6 y) = 4 y^{2} + 80 y + 100 \geq 0 \Rightarrow y^{2} + 20 y + 25 \geq 0 y \leq - 10 - 53 or y \geq - 10 + 53

(f)

4 marks

Solve $f (x) = x$ : $x^{2} - 14 x + 24 = 2 x^{2} + 6 x$ , so $x^{2} + 20 x - 24 = 0$ and $x = - 10 \pm 231 = - 21.1, 1.14$ .
For $x > - 3$ the denominator is positive and $f (x) > x$ when $x^{2} + 20 x - 24 < 0$ , that is $- 3 < x < 1.14$ . For $x < - 3$ the denominator is negative, so $f (x) > x$ when $x^{2} + 20 x - 24 > 0$ , that is $x < - 21.1$ .

x < - 10 - 231 or - 3 < x < - 10 + 231

Question 12

18 marks

Two lines meeting at a point, a plane, and a reflection

Geometry and trigonometry

(a)

5 marks

Equate the coordinates of $L$ and $M$ .
The second equation gives $t = 3 s - 7$ . In the third, $- 3 + 6 s = 11 + 6 s - 14$ , which is true for every $s$ , so this equation is equivalent to the second. Substitute into the first.
The two lines meet at

1 + 2 s = 9 + 4 t, 2 + 3 s = 9 + t, - 3 + 6 s = 11 + 2 t 1 + 2 s = 9 + 12 s - 28 \Rightarrow 10 s = 20 \Rightarrow s = 2, t = - 1 A = 589

(b)

3 marks

The normal of the plane is $n = (0, 2, - 1)$ . Check the directions are perpendicular to $n$ and one point of each line is on the plane.
Both lines are parallel to the plane and each contains a point of the plane, so both lie in the plane.

236 \cdot n = 6 - 6 = 0, 412 \cdot n = 2 - 2 = 0 12 - 3 \cdot n = 4 + 3 = 7, 9911 \cdot n = 18 - 11 = 7

(c)

7 marks

The line through B perpendicular to the plane: $r = - 3 122 + λ 02 - 1$ . Put it into the plane.

2 (12 + 2 λ) - (2 - λ) = 7 \Rightarrow 22 + 5 λ = 7 \Rightarrow λ = - 3 C = - 3 65, ∣ B C ∣ = 3∣ n ∣ = 35 \approx 6.71

(d)

3 marks

The reflection of B is the same distance beyond the plane, so it is $2 O C - O B$ .

B^{'} = 2 - 3 65 - - 3 122 = - 3 08

Our analysis

Where the marks are

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

Geometry and trigonometry37 marks · 34%
Q1, Q3, Q8, Q12
Functions24 marks · 22%
Q2, Q11
Statistics and probability21 marks · 19%
Q6, Q10
Calculus17 marks · 15%
Q4, Q8, Q9
Number and algebra11 marks · 10%
Q5, Q7

Mentor insights

Traps to watch for

01
Area from the vector product · Q8
$Area = \frac{1}{2} ∣ A B \times A C ∣$ . Minimise the square to avoid the square root.
02
Euler with a small step · Q9
Use $y_{n + 1} = y_{n} + h f (x_{n}, y_{n})$ with $h = 0.1$ .
03
Two inverse-normal equations · Q10
Each given probability gives a z-score and an equation in $μ$ and $σ$ .
04
Oblique asymptote by division · Q11
Divide the numerator by the denominator and keep the quotient.
05
Intersection · Q12
Equate the three components and solve for $s$ and $t$ . One equation is redundant here.

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

Want someone to go through your paper with you?

Our maths mentors took these exams recently. One-to-one, they'll find where your marks are going and fix it before May.

Find your maths mentor More past papers