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Worked solutions

IB Maths AA HL · Paper 2

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2024
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 May 2024 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 · 55 marks

Question 1

7 marks

A car losing value: 15% then 11% a year

Number and algebra

(a)

2 marks

A 15% loss leaves 85%.

0.85 \times 35000 = $29 750

(b)

2 marks

Nine further years of 11% loss after the first.

29750 \times 0.8 9^{9} \approx $10 423

(c)

3 marks

10% of the original price is $3500. Solve for the number of years.
The value after 19 years is $3250 and after 18 years is $3652, so the first time below $3500 is

29750 \times 0.8 9^{n - 1} < 3500 \Rightarrow n - 1 > 18.36 n = 20

Question 2

5 marks

A cone-shaped monument seen at an angle: radius and volume

Geometry and trigonometry

(a)

3 marks

Use the right triangle from Oliver's eye to the vertex.

tan 5 8^{\circ} = \frac{18.2}{5 + r} \Rightarrow 5 + r = 11.373 r \approx 6.37 m

(b)

2 marks

Volume of a cone.

V = \frac{1}{3} π r^{2} h = \frac{1}{3} π (6.3726)^{2} (20) \approx 851 m^{3}

Question 3

4 marks

A normal variable: standard deviations above the mean

Statistics and probability

(a)

2 marks

$1.5$ standard deviations above the mean is $X = 10 + 3 = 13$ .

P (X > 13) = P (Z > 1.5) \approx 0.0668

(b)

2 marks

Use the inverse normal with $P (Z < k) = 0.9$ .

k \approx 1.28

Question 4

6 marks

A particle with sinusoidal velocity: direction change, displacement and distance

Calculus

(a)

2 marks

Solve $v = 2 sin (0.5 t) + 0.3 t - 2 = 0$ on a graph or with the solver.

t \approx 1.69 s

(b)

2 marks

Displacement increases while $v > 0$ . The zeros of $v$ in $[0, 10]$ are $t \approx 1.687$ and $t \approx 6.119$ , and $v$ is positive between them.

1.69 < t < 6.12

(c)

2 marks

The displacement at $t = 10$ is the definite integral of $v$ .
The particle ends up 2.13 m behind $O$ .

\int_{0}^{10} (2 sin (0.5 t) + 0.3 t - 2) d t \approx - 2.13 m

Question 5

7 marks

Two test scores: correlation, extrapolation and predicting x from y

Statistics and probability

(a)

2 marks

Enter the ten pairs in the calculator.

r \approx 0.901

(b)

1 mark

The Test A scores in the data run from 52 to 100, and 10 is far below this range. Using the line there is extrapolation, so the estimate of 27 is not reliable.

(c)

4 marks

(i) The line of $y$ on $x$ is for predicting $y$ from a given $x$ . It is the wrong line for predicting $x$ from $y$ .

(ii) Use the regression line of $x$ on $y$ from the same data.

Substitute $y = 90$ .

x = 0.987 y - 3.22 x = 0.987124 (90) - 3.21971 \approx 85.6 \approx 86

Question 6

6 marks

Rainy days in May: binomial and geometric probabilities

Statistics and probability

(a)

2 marks

Let $X$ be the number of rainy days, $X \sim B (31, 0.2)$ .

P (X = 10) = (10 31) (0.2)^{10} (0.8)^{21} \approx 0.0419

(b)

2 marks

Use the complement.

P (X \geq 10) = 1 - P (X \leq 9) = 1 - 0.9254 \approx 0.0746

(c)

2 marks

Nine dry days followed by a rainy day, all independent.

(0.8)^{9} (0.2) \approx 0.0268

Question 7

7 marks

Solving dy/dx = x + y with y(0) = 2

Number and algebra

7 marks

Rearrange to the standard linear form.
The integrating factor is $e^{\int - 1 d x} = e^{- x}$ . Multiply through.
Integrate the right side by parts: $\int x e^{- x} d x = - x e^{- x} + \int e^{- x} d x$ .
Use $y = 2$ at $x = 0$ : $2 = - 1 + c$ , so $c = 3$ .

\frac{d y}{d x} - y = x \frac{d}{d x} (y e^{- x}) = x e^{- x} y e^{- x} = - x e^{- x} - e^{- x} + c y = - x - 1 + c e^{x} y = 3 e^{x} - x - 1

Question 8

6 marks

A piecewise density function: finding k and the median

Calculus

(a)

2 marks

The two pieces integrate to
Total area is 1.

\int_{0}^{k} k x d x = \frac{k ^{3}}{2} \int_{k}^{2 k} (2 k x - x^{2}) d x = [k x^{2} - \frac{x ^{3}}{3}]_{k}^{2 k} = (4 k^{3} - \frac{8 k ^{3}}{3}) - (k^{3} - \frac{k ^{3}}{3}) = \frac{2 k ^{3}}{3} \frac{k ^{3}}{2} + \frac{2 k ^{3}}{3} = \frac{7 k ^{3}}{6} = 1 \Rightarrow 7 k^{3} = 6

(b)

4 marks

From $k^{3} = \frac{6}{7}$ we get $k \approx 0.9499$ and $P (X \leq k) = \frac{k ^{3}}{2} = \frac{3}{7} \approx 0.4286 < 0.5$ . So the median $m$ is greater than $k$ .
The remaining probability needed is $\frac{1}{2} - \frac{3}{7} = \frac{1}{14}$ .
Solve for $m$ with the calculator.

\int_{k}^{m} (2 k x - x^{2}) d x = k m^{2} - \frac{m ^{3}}{3} - \frac{2 k ^{3}}{3} = \frac{1}{14} m \approx 1.03

Question 9

7 marks

Seating ten children at two rows of desks, with pairs together and apart

Number and algebra

(a)

3 marks

There are 4 adjacent pairs of desks in each row, so 8 pairs in all, and the brothers can swap places.
The other 8 children fill the remaining 8 desks in $8!$ ways.

8 \times 2 = 16 16 \times 8! = 645120

(b)

4 marks

First count the seatings in which the sisters are also next to each other.
Of the 8 desk pairs for the brothers, 4 are at the end of a row. They leave 3 joined desks in that row (2 adjacent pairs) and 4 adjacent pairs in the other row, so the sisters have 6 adjacent pairs to choose from.
The other 4 desk pairs for the brothers are in the middle of a row. They leave 1 desk and 2 joined desks in that row (1 adjacent pair) and 4 adjacent pairs in the other row, so the sisters have 5 adjacent pairs.
Each placement can be done in $2 \times 2$ orders (brothers, sisters), and the remaining 6 children sit in $6!$ ways.
Subtract from the total in (a).

4 \times 6 + 4 \times 5 = 44 placements 44 \times 2 \times 2 \times 6! = 126720 645120 - 126720 = 518400

Section B · Extended response

Questions 10–12 · 55 marks

Question 10

16 marks

Tide heights at two islands: a sine model and fitting a second model

Calculus

(a)

3 marks

The first low tide and first high tide of $H (t)$ are at $t \approx 5.138$ and $t \approx 11.262$ .

11.262 - 5.138 = 6.124 hours = 6 h 7.4 min m = 7

(b)

2 marks

Solve $H (t) = 1$ either side of the low tide at $t = 5.138$ .

t \approx 3.569 and t \approx 6.707 6.707 - 3.569 \approx 3.14 hours

(c)

2 marks

Differentiate and evaluate at $t = 13$ .

H^{'} (t) = 1.63 (0.513) cos (0.513 (t - 8.20)) \Rightarrow H^{'} (13) \approx - 0.651 m h^{- 1}

(d)

7 marks

Half the range gives $a$ and the midpoint gives $d$ .
Low tide at 02:41 and high tide at 09:02 are half a period apart.
The curve is at its midline, rising, halfway between the low tide and the high tide.

a = \frac{2.74 - 0.40}{2} = 1.17, d = \frac{2.74 + 0.40}{2} = 1.57 9 \frac{2}{60} - 2 \frac{41}{60} = 6.35 h \Rightarrow period = 12.7 h b = \frac{2 π}{12.7} \approx 0.495 c = \frac{2.6833 + 9.0333}{2} \approx 5.86

(e)

2 marks

Plot $H (t)$ and $h (t)$ and find the first intersection.

T \approx 4.16

Question 11

19 marks

An implicit curve e^(x+y) = x^2 + y^2: gradients and tangent points

Calculus

(a)

5 marks

Differentiate $e^{x + y} = x^{2} + y^{2}$ .
Collect the $\frac{d y}{d x}$ terms.

e^{x + y} (1 + \frac{d y}{d x}) = 2 x + 2 y \frac{d y}{d x} \frac{d y}{d x} (e^{x + y} - 2 y) = 2 x - e^{x + y} \frac{d y}{d x} = \frac{2 x - e ^{x + y}}{e ^{x + y} - 2 y}

(b)

9 marks

(i) A horizontal tangent needs $2 x - e^{x + y} = 0$ , so $e^{x + y} = 2 x$ and

The point is on the curve, so $x^{2} + y^{2} = e^{x + y} = 2 x$ . Substitute $y$ .

(ii) Solve this equation with the calculator and use $y = ln (2 x) - x$ .

x + y = ln (2 x) \Rightarrow y = ln (2 x) - x x^{2} + (ln (2 x) - x)^{2} = 2 x \Rightarrow 2 x^{2} + (ln (2 x))^{2} - 2 x ln (2 x) - 2 x = 0 x \approx 0.331, y \approx - 0.743 and x \approx 1.84, y \approx - 0.538 P, Q = (0.331, - 0.743) and (1.84, - 0.538)

(c)

1 mark

The curve is symmetric about $y = x$ , so swap the coordinates of the two points. Horizontal tangents become vertical tangents.

(- 0.743, 0.331) and (- 0.538, 1.84)

(d)

4 marks

Set $\frac{d y}{d x} = - 1$ .
With $y = x$ the curve equation becomes $e^{2 x} = 2 x^{2}$ . Solve with the calculator.
The point is $(- 0.451, - 0.451)$ .

2 x - e^{x + y} = - e^{x + y} + 2 y \Rightarrow x = y x = y \approx - 0.451

Question 12

20 marks

Vectors: Lagrange identity, a triangle with given area and a perpendicular vector

Number and algebra

(a)

2 marks

Square the two definitions and add.

(u \cdot v)^{2} + ∣ u \times v ∣^{2} = ∣ u ∣^{2} ∣ v ∣^{2} cos^{2} θ + ∣ u ∣^{2} ∣ v ∣^{2} sin^{2} θ = ∣ u ∣^{2} ∣ v ∣^{2}

(b)

13 marks

(i) The area is $\frac{1}{2} ∣ u \times v ∣ = 6$ .

(ii) Here $v = A C = (3, 1, - 1)$ , so $∣ v ∣^{2} = 11$ . Use part (a).

(iii) With $u = A B = (p, q - 1, 1)$ , the dot product condition is

The length condition is $p^{2} + (q - 1)^{2} + 1 = 3$ . Substitute $q - 1 = 4 - 3 p$ .

∣ u \times v ∣ = 26 3^{2} + (26)^{2} = 33 = 11∣ u ∣^{2} \Rightarrow ∣ u ∣^{2} = 3 \Rightarrow ∣ u ∣ = 3 3 p + (q - 1) - 1 = 3 \Rightarrow q = 5 - 3 p p^{2} + (4 - 3 p)^{2} = 2 \Rightarrow 10 p^{2} - 24 p + 14 = 0 \Rightarrow (p - 1) (5 p - 7) = 0 p = 1, q = 2 or p = \frac{7}{5}, q = \frac{4}{5}

(c)

5 marks

With $p = 1$ , $q = 2$ we have $u = (1, 1, 1)$ and $v = (3, 1, - 1)$ .
Since $w$ is perpendicular to both, $w = λ (1, - 2, 1)$ .
$w$ is perpendicular to $v$ , so the area of triangle ACD is $\frac{1}{2} ∣ v ∣∣ w ∣ = 5$ .

u \times v = (- 2, 4, - 2) 11 ∣ λ ∣ 6 = 10 \Rightarrow ∣ λ ∣ = \frac{10}{66} = \frac{5 66}{33} w = \pm \frac{5 66}{33} 1 - 2 1 \approx \pm 1.23 - 2.46 1.23

Our analysis

Where the marks are

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

Calculus47 marks · 43%
Q4, Q8, Q10, Q11
Number and algebra41 marks · 37%
Q1, Q7, Q9, Q12
Statistics and probability17 marks · 15%
Q3, Q5, Q6
Geometry and trigonometry5 marks · 5%
Q2

Mentor insights

Traps to watch for

01
First year is different · Q1
The first-year factor is 0.85, the later years use 0.89.
02
Which line to use · Q5
The line of $y$ on $x$ is for predicting $y$ from $x$ . To predict $x$ from $y$ use the line of $x$ on $y$ .
03
Tide times · Q10
Low and high tide are half a period apart, so the period is twice the gap.
04
Implicit differentiation · Q11
Differentiate both sides with respect to $x$ , remembering $\frac{d y}{d x}$ on every $y$ term.
05
Dot and cross magnitudes · Q12
$u \cdot v = ∣ u ∣∣ v ∣ cos θ$ and $∣ u \times v ∣ = ∣ u ∣∣ v ∣ sin θ$ .

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Worked solutions

IB Maths AA HL · Paper 2

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2024
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 May 2024 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 · 55 marks

Question 1

7 marks

A car losing value: 15% then 11% a year

Number and algebra

(a)

2 marks

A 15% loss leaves 85%.

0.85 \times 35000 = $29 750

(b)

2 marks

Nine further years of 11% loss after the first.

29750 \times 0.8 9^{9} \approx $10 423

(c)

3 marks

10% of the original price is $3500. Solve for the number of years.
The value after 19 years is $3250 and after 18 years is $3652, so the first time below $3500 is

29750 \times 0.8 9^{n - 1} < 3500 \Rightarrow n - 1 > 18.36 n = 20

Question 2

5 marks

A cone-shaped monument seen at an angle: radius and volume

Geometry and trigonometry

(a)

3 marks

Use the right triangle from Oliver's eye to the vertex.

tan 5 8^{\circ} = \frac{18.2}{5 + r} \Rightarrow 5 + r = 11.373 r \approx 6.37 m

(b)

2 marks

Volume of a cone.

V = \frac{1}{3} π r^{2} h = \frac{1}{3} π (6.3726)^{2} (20) \approx 851 m^{3}

Question 3

4 marks

A normal variable: standard deviations above the mean

Statistics and probability

(a)

2 marks

$1.5$ standard deviations above the mean is $X = 10 + 3 = 13$ .

P (X > 13) = P (Z > 1.5) \approx 0.0668

(b)

2 marks

Use the inverse normal with $P (Z < k) = 0.9$ .

k \approx 1.28

Question 4

6 marks

A particle with sinusoidal velocity: direction change, displacement and distance

Calculus

(a)

2 marks

Solve $v = 2 sin (0.5 t) + 0.3 t - 2 = 0$ on a graph or with the solver.

t \approx 1.69 s

(b)

2 marks

Displacement increases while $v > 0$ . The zeros of $v$ in $[0, 10]$ are $t \approx 1.687$ and $t \approx 6.119$ , and $v$ is positive between them.

1.69 < t < 6.12

(c)

2 marks

The displacement at $t = 10$ is the definite integral of $v$ .
The particle ends up 2.13 m behind $O$ .

\int_{0}^{10} (2 sin (0.5 t) + 0.3 t - 2) d t \approx - 2.13 m

Question 5

7 marks

Two test scores: correlation, extrapolation and predicting x from y

Statistics and probability

(a)

2 marks

Enter the ten pairs in the calculator.

r \approx 0.901

(b)

1 mark

The Test A scores in the data run from 52 to 100, and 10 is far below this range. Using the line there is extrapolation, so the estimate of 27 is not reliable.

(c)

4 marks

(i) The line of $y$ on $x$ is for predicting $y$ from a given $x$ . It is the wrong line for predicting $x$ from $y$ .

(ii) Use the regression line of $x$ on $y$ from the same data.

Substitute $y = 90$ .

x = 0.987 y - 3.22 x = 0.987124 (90) - 3.21971 \approx 85.6 \approx 86

Question 6

6 marks

Rainy days in May: binomial and geometric probabilities

Statistics and probability

(a)

2 marks

Let $X$ be the number of rainy days, $X \sim B (31, 0.2)$ .

P (X = 10) = (10 31) (0.2)^{10} (0.8)^{21} \approx 0.0419

(b)

2 marks

Use the complement.

P (X \geq 10) = 1 - P (X \leq 9) = 1 - 0.9254 \approx 0.0746

(c)

2 marks

Nine dry days followed by a rainy day, all independent.

(0.8)^{9} (0.2) \approx 0.0268

Question 7

7 marks

Solving dy/dx = x + y with y(0) = 2

Number and algebra

7 marks

Rearrange to the standard linear form.
The integrating factor is $e^{\int - 1 d x} = e^{- x}$ . Multiply through.
Integrate the right side by parts: $\int x e^{- x} d x = - x e^{- x} + \int e^{- x} d x$ .
Use $y = 2$ at $x = 0$ : $2 = - 1 + c$ , so $c = 3$ .

\frac{d y}{d x} - y = x \frac{d}{d x} (y e^{- x}) = x e^{- x} y e^{- x} = - x e^{- x} - e^{- x} + c y = - x - 1 + c e^{x} y = 3 e^{x} - x - 1

Question 8

6 marks

A piecewise density function: finding k and the median

Calculus

(a)

2 marks

The two pieces integrate to
Total area is 1.

\int_{0}^{k} k x d x = \frac{k ^{3}}{2} \int_{k}^{2 k} (2 k x - x^{2}) d x = [k x^{2} - \frac{x ^{3}}{3}]_{k}^{2 k} = (4 k^{3} - \frac{8 k ^{3}}{3}) - (k^{3} - \frac{k ^{3}}{3}) = \frac{2 k ^{3}}{3} \frac{k ^{3}}{2} + \frac{2 k ^{3}}{3} = \frac{7 k ^{3}}{6} = 1 \Rightarrow 7 k^{3} = 6

(b)

4 marks

From $k^{3} = \frac{6}{7}$ we get $k \approx 0.9499$ and $P (X \leq k) = \frac{k ^{3}}{2} = \frac{3}{7} \approx 0.4286 < 0.5$ . So the median $m$ is greater than $k$ .
The remaining probability needed is $\frac{1}{2} - \frac{3}{7} = \frac{1}{14}$ .
Solve for $m$ with the calculator.

\int_{k}^{m} (2 k x - x^{2}) d x = k m^{2} - \frac{m ^{3}}{3} - \frac{2 k ^{3}}{3} = \frac{1}{14} m \approx 1.03

Question 9

7 marks

Seating ten children at two rows of desks, with pairs together and apart

Number and algebra

(a)

3 marks

There are 4 adjacent pairs of desks in each row, so 8 pairs in all, and the brothers can swap places.
The other 8 children fill the remaining 8 desks in $8!$ ways.

8 \times 2 = 16 16 \times 8! = 645120

(b)

4 marks

First count the seatings in which the sisters are also next to each other.
Of the 8 desk pairs for the brothers, 4 are at the end of a row. They leave 3 joined desks in that row (2 adjacent pairs) and 4 adjacent pairs in the other row, so the sisters have 6 adjacent pairs to choose from.
The other 4 desk pairs for the brothers are in the middle of a row. They leave 1 desk and 2 joined desks in that row (1 adjacent pair) and 4 adjacent pairs in the other row, so the sisters have 5 adjacent pairs.
Each placement can be done in $2 \times 2$ orders (brothers, sisters), and the remaining 6 children sit in $6!$ ways.
Subtract from the total in (a).

4 \times 6 + 4 \times 5 = 44 placements 44 \times 2 \times 2 \times 6! = 126720 645120 - 126720 = 518400

Section B · Extended response

Questions 10–12 · 55 marks

Question 10

16 marks

Tide heights at two islands: a sine model and fitting a second model

Calculus

(a)

3 marks

The first low tide and first high tide of $H (t)$ are at $t \approx 5.138$ and $t \approx 11.262$ .

11.262 - 5.138 = 6.124 hours = 6 h 7.4 min m = 7

(b)

2 marks

Solve $H (t) = 1$ either side of the low tide at $t = 5.138$ .

t \approx 3.569 and t \approx 6.707 6.707 - 3.569 \approx 3.14 hours

(c)

2 marks

Differentiate and evaluate at $t = 13$ .

H^{'} (t) = 1.63 (0.513) cos (0.513 (t - 8.20)) \Rightarrow H^{'} (13) \approx - 0.651 m h^{- 1}

(d)

7 marks

Half the range gives $a$ and the midpoint gives $d$ .
Low tide at 02:41 and high tide at 09:02 are half a period apart.
The curve is at its midline, rising, halfway between the low tide and the high tide.

a = \frac{2.74 - 0.40}{2} = 1.17, d = \frac{2.74 + 0.40}{2} = 1.57 9 \frac{2}{60} - 2 \frac{41}{60} = 6.35 h \Rightarrow period = 12.7 h b = \frac{2 π}{12.7} \approx 0.495 c = \frac{2.6833 + 9.0333}{2} \approx 5.86

(e)

2 marks

Plot $H (t)$ and $h (t)$ and find the first intersection.

T \approx 4.16

Question 11

19 marks

An implicit curve e^(x+y) = x^2 + y^2: gradients and tangent points

Calculus

(a)

5 marks

Differentiate $e^{x + y} = x^{2} + y^{2}$ .
Collect the $\frac{d y}{d x}$ terms.

e^{x + y} (1 + \frac{d y}{d x}) = 2 x + 2 y \frac{d y}{d x} \frac{d y}{d x} (e^{x + y} - 2 y) = 2 x - e^{x + y} \frac{d y}{d x} = \frac{2 x - e ^{x + y}}{e ^{x + y} - 2 y}

(b)

9 marks

(i) A horizontal tangent needs $2 x - e^{x + y} = 0$ , so $e^{x + y} = 2 x$ and

The point is on the curve, so $x^{2} + y^{2} = e^{x + y} = 2 x$ . Substitute $y$ .

(ii) Solve this equation with the calculator and use $y = ln (2 x) - x$ .

x + y = ln (2 x) \Rightarrow y = ln (2 x) - x x^{2} + (ln (2 x) - x)^{2} = 2 x \Rightarrow 2 x^{2} + (ln (2 x))^{2} - 2 x ln (2 x) - 2 x = 0 x \approx 0.331, y \approx - 0.743 and x \approx 1.84, y \approx - 0.538 P, Q = (0.331, - 0.743) and (1.84, - 0.538)

(c)

1 mark

The curve is symmetric about $y = x$ , so swap the coordinates of the two points. Horizontal tangents become vertical tangents.

(- 0.743, 0.331) and (- 0.538, 1.84)

(d)

4 marks

Set $\frac{d y}{d x} = - 1$ .
With $y = x$ the curve equation becomes $e^{2 x} = 2 x^{2}$ . Solve with the calculator.
The point is $(- 0.451, - 0.451)$ .

2 x - e^{x + y} = - e^{x + y} + 2 y \Rightarrow x = y x = y \approx - 0.451

Question 12

20 marks

Vectors: Lagrange identity, a triangle with given area and a perpendicular vector

Number and algebra

(a)

2 marks

Square the two definitions and add.

(u \cdot v)^{2} + ∣ u \times v ∣^{2} = ∣ u ∣^{2} ∣ v ∣^{2} cos^{2} θ + ∣ u ∣^{2} ∣ v ∣^{2} sin^{2} θ = ∣ u ∣^{2} ∣ v ∣^{2}

(b)

13 marks

(i) The area is $\frac{1}{2} ∣ u \times v ∣ = 6$ .

(ii) Here $v = A C = (3, 1, - 1)$ , so $∣ v ∣^{2} = 11$ . Use part (a).

(iii) With $u = A B = (p, q - 1, 1)$ , the dot product condition is

The length condition is $p^{2} + (q - 1)^{2} + 1 = 3$ . Substitute $q - 1 = 4 - 3 p$ .

∣ u \times v ∣ = 26 3^{2} + (26)^{2} = 33 = 11∣ u ∣^{2} \Rightarrow ∣ u ∣^{2} = 3 \Rightarrow ∣ u ∣ = 3 3 p + (q - 1) - 1 = 3 \Rightarrow q = 5 - 3 p p^{2} + (4 - 3 p)^{2} = 2 \Rightarrow 10 p^{2} - 24 p + 14 = 0 \Rightarrow (p - 1) (5 p - 7) = 0 p = 1, q = 2 or p = \frac{7}{5}, q = \frac{4}{5}

(c)

5 marks

With $p = 1$ , $q = 2$ we have $u = (1, 1, 1)$ and $v = (3, 1, - 1)$ .
Since $w$ is perpendicular to both, $w = λ (1, - 2, 1)$ .
$w$ is perpendicular to $v$ , so the area of triangle ACD is $\frac{1}{2} ∣ v ∣∣ w ∣ = 5$ .

u \times v = (- 2, 4, - 2) 11 ∣ λ ∣ 6 = 10 \Rightarrow ∣ λ ∣ = \frac{10}{66} = \frac{5 66}{33} w = \pm \frac{5 66}{33} 1 - 2 1 \approx \pm 1.23 - 2.46 1.23

Our analysis

Where the marks are

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

Calculus47 marks · 43%
Q4, Q8, Q10, Q11
Number and algebra41 marks · 37%
Q1, Q7, Q9, Q12
Statistics and probability17 marks · 15%
Q3, Q5, Q6
Geometry and trigonometry5 marks · 5%
Q2

Mentor insights

Traps to watch for

01
First year is different · Q1
The first-year factor is 0.85, the later years use 0.89.
02
Which line to use · Q5
The line of $y$ on $x$ is for predicting $y$ from $x$ . To predict $x$ from $y$ use the line of $x$ on $y$ .
03
Tide times · Q10
Low and high tide are half a period apart, so the period is twice the gap.
04
Implicit differentiation · Q11
Differentiate both sides with respect to $x$ , remembering $\frac{d y}{d x}$ on every $y$ term.
05
Dot and cross magnitudes · Q12
$u \cdot v = ∣ u ∣∣ v ∣ cos θ$ and $∣ u \times v ∣ = ∣ u ∣∣ v ∣ sin θ$ .

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