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IB Maths AA HL · Paper 2

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1 TZ2TZ3

Session: May 2025
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 2025 TZ3 paper, which your school or the IB can provide.

Practice mode

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Section A · Short response

Questions 1–9 · 57 marks

Question 1

6 marks

A triangle with two sides and the included angle: the cosine and sine rules

Geometry and trigonometry

(a)

3 marks

Use the cosine rule.

B C^{2} = 1 2^{2} + 7^{2} - 2 (12) (7) cos 11 6^{\circ} \approx 266.6 \Rightarrow B C \approx 16.3

(b)

3 marks

Use the sine rule for angle $C$ , the angle opposite $A B = 7$ .
The angle is acute because the angle at $A$ is already obtuse.

\frac{sin A C B}{7} = \frac{sin 11 6 ^{\circ}}{16.33} \Rightarrow sin A C B = 0.3853 A C B \approx 22. 7^{\circ}

Question 2

5 marks

A quartic with a given factor and remainder

Number and algebra

5 marks

$f (x) = 2 x^{4} - 6 x^{3} + p x^{2} + q x - 2$ . Since $(x - 1)$ is a factor, $f (1) = 0$ .
The remainder on division by $(x - 3)$ is $f (3) = - 2$ .
Substitute into $p + q = 6$ .

2 - 6 + p + q - 2 = 0 \Rightarrow p + q = 6 162 - 162 + 9 p + 3 q - 2 = - 2 \Rightarrow 9 p + 3 q = 0 \Rightarrow q = - 3 p p - 3 p = 6 \Rightarrow p = - 3, q = 9

Question 3

6 marks

Supermarket visits: a probability distribution and a sampling method

Statistics and probability

(a)

3 marks

(i) The probabilities add to 1.

(ii) The most likely value is the one with the greatest probability. $P (X = 2) = 2 a = 0.308$ is the biggest.

1.5 a + 2 a + 0.281 + a + 0.026 + 0 = 1 \Rightarrow 4.5 a = 0.693 \Rightarrow a = 0.154 mode = 2

(b)

2 marks

Use $E (X) = \sum x P (X = x)$ with $P (X = 1) = 0.231$ , $P (X = 2) = 0.308$ , $P (X = 3) = 0.281$ , $P (X = 4) = 0.154$ , $P (X = 5) = 0.026$ .

E (X) = 0.231 + 0.616 + 0.843 + 0.616 + 0.130 = 2.436 \approx 2.44

(c)

1 mark

The manager chose the first 50 customers to arrive, which is whatever is easiest, so it is a convenience sample.

Question 4

5 marks

The arc length of a circle from the area of a triangle

Number and algebra

5 marks

The triangle $A O B$ has two sides equal to the radius $r$ , with included angle 2.51 radians.
The arc $A C B$ is the major arc with angle $2 π - 2.51 = 3.773$ radians.

\frac{1}{2} r^{2} sin 2.51 = 26 \Rightarrow r^{2} = 88.07 \Rightarrow r \approx 9.385 cm arc = r θ = 9.385 (3.773) \approx 35.4 cm

Question 5

7 marks

A rational function with a parameter: the derivative and a tangent angle

Calculus

(a)

3 marks

Differentiate with the quotient rule.
Factor out $2 (2 x + a)^{2} (x + 5)$ from the numerator and cancel one factor of $(x + 5)$ .

f^{'} (x) = \frac{6 ( 2 x + a ) ^{2} ( x + 5 ) ^{2} - 2 ( 2 x + a ) ^{3} ( x + 5 )}{( x + 5 ) ^{4}} f^{'} (x) = \frac{2 ( 2 x + a ) ^{2} [ 3 ( x + 5 ) - ( 2 x + a ) ]}{( x + 5 ) ^{3}} = \frac{2 ( 2 x + a ) ^{2} ( x + 15 - a )}{( x + 5 ) ^{3}}

(b)

4 marks

At $x = 1$ the gradient is $tan 7 0^{\circ} = 2.7475$ .
Solve for $a > 0$ with the calculator.

f^{'} (1) = \frac{2 ( 2 + a ) ^{2} ( 16 - a )}{216} = tan 7 0^{\circ} a \approx 2.73 or a \approx 15.0

Question 6

8 marks

Vectors with equal angles to a third vector

Number and algebra

(a)

3 marks

Use $a = (- 5, 7)$ , $b = (- 8, 9)$ and $c = (p, - 6)$ .

a \cdot c = - 5 p - 42 b \cdot c = - 8 p - 54

(b)

5 marks

The magnitudes are $∣ a ∣ = 74$ , $∣ b ∣ = 145$ and $∣ c ∣ = p^{2} + 36$ .
Cancel the common factor $p^{2} + 36$ and solve for $p$ .

cos θ = \frac{- 5 p - 42}{74 p ^{2} + 36} = \frac{- 8 p - 54}{145 p ^{2} + 36} \frac{5 p + 42}{74} = \frac{8 p + 54}{145} \Rightarrow p \approx 4.79

Question 7

6 marks

A binomial series with a given coefficient

Number and algebra

6 marks

Write $\frac{1}{q - x ^{2}} = q^{- 1/2} (1 - \frac{x ^{2}}{q})^{- 1/2}$ .
The term in $x^{6}$ comes from the third power of $\frac{x ^{2}}{q}$ , with $n = - \frac{1}{2}$ .
Include the factor $q^{- 1/2}$ .
Set the coefficient equal to 5120.

\frac{( - \frac{1}{2} ) ( - \frac{3}{2} ) ( - \frac{5}{2} )}{3 !} (- \frac{x ^{2}}{q})^{3} = \frac{5}{16} \cdot \frac{x ^{6}}{q ^{3}} \frac{5}{16} q^{- 7/2} x^{6} \frac{5}{16} q^{- 7/2} = 5120 \Rightarrow q^{- 7/2} = 16384 = 2^{14} \Rightarrow q = 2^{- 4} q = \frac{1}{16}

Question 8

5 marks

Counting games in a tic-tac-toe competition

Number and algebra

5 marks

Let $n$ be the number of students at the start. Stephen played 7 games, and all the games between the other $n - 1$ students were played.
The other $N = n - 1$ students play each pair twice.
The total number of games played is those games plus Stephen's 7.
Since $N > 0$ , $N = 23$ , so there were 24 students originally.

2 (2 N) = N (N - 1) N (N - 1) + 7 = 513 \Rightarrow N^{2} - N - 506 = 0 \Rightarrow (N - 23) (N + 22) = 0 n = 24

Question 9

9 marks

An aeroplane approaching a runway: the viewing angle and its rate of change

Calculus

(a)

2 marks

Let $α = A P G$ , so $tan α = \frac{x}{6}$ . The angle $B P G = α + θ$ and $tan (α + θ) = \frac{x + 2}{6}$ .

θ = arctan (\frac{x + 2}{6}) - α = arctan (\frac{x + 2}{6}) - arctan (\frac{x}{6})

(b)

7 marks

Solve $θ = 0.178$ for $x$ .
Differentiate $θ$ with respect to $x$ , using $\frac{d}{d x} arctan \frac{x}{6} = \frac{6}{36 + x ^{2}}$ .
At $x = 4.630$ this is
Use the chain rule with $\frac{d θ}{d t} = 12.5$ .
The plane is travelling towards the runway at constant speed.

arctan (\frac{x + 2}{6}) - arctan (\frac{x}{6}) = 0.178 \Rightarrow x \approx 4.630 \frac{d θ}{d x} = \frac{6}{36 + ( x + 2 ) ^{2}} - \frac{6}{36 + x ^{2}} \frac{d θ}{d x} \approx - 0.02942 \frac{d x}{d t} = \frac{12.5}{- 0.02942} \approx - 424.9 speed \approx 425 km h^{- 1}

Section B · Extended response

Questions 10–12 · 53 marks

Question 10

14 marks

A 200 metre race: velocity models for two athletes

Calculus

(a)

3 marks

Fiona's velocity is $v (t) = \frac{8.14 t}{t ^{2} + 0.2}$ .
(i) $v (1) = \frac{8.14}{1.2} \approx 7.43 m s^{- 1}$
(ii) Solve $v (t) = 5$ .

t \approx 0.348 seconds

(b)

2 marks

Solve $v^{'} (t) = 4$ with the calculator.

t \approx 0.591 seconds

(c)

3 marks

(i) For large $t$ , $t^{2} + 0.2 \approx t$ .

(ii) The race is over once Fiona has gone 200 m, so the value of $t$ is limited. Fiona never reaches the speed 8.14 (the graph only gets close to it), so this value is not valid for the race.

t \to \infty lim v (t) = 8.14 m s^{- 1}

(d)

6 marks

Find the time $t_{f}$ when Fiona has gone 200 m, by solving the equation for the distance.
In that time Lucy, with $w (t) = \frac{8 t}{t ^{2} + 0.3}$ , has gone
So Lucy is this far from the finishing line.

\int_{0}^{t_{f}} v (t) d t = 200 \Rightarrow t_{f} \approx 25.01 seconds \int_{0}^{25.01} w (t) d t \approx 195.77 m 200 - 195.77 \approx 4.23 m

Question 11

18 marks

Survey data entry errors: binomial, conditional probability and totals

Statistics and probability

(a)

5 marks

Let $E$ be the number of surveys entered inaccurately, $E \sim B (50, 0.08)$ .
(i) $P (E \leq 6) \approx 0.898$
(ii) Use conditional probability. If $E = 4$ then $E \leq 6$ .

P (E = 4 ∣ E \leq 6) = \frac{P ( E = 4 )}{P ( E \leq 6 )} = \frac{0.2037}{0.8981} \approx 0.227

(b)

3 marks

Find $n$ so that $P (E \leq 6) \approx 0.367$ with $E \sim B (n, 0.08)$ . Use a table of values of $n$ on the calculator.

n = 94 (P (E \leq 6) = 0.3667)

(c)

6 marks

(i) Add the inaccurate surveys from each person.

(ii) Use conditional probability.

P (I) = 0.55 (0.08) + 0.25 (0.06) + 0.20 (0.11) = 0.044 + 0.015 + 0.022 = 0.081 P (A ∣ I) = \frac{0.55 \times 0.08}{0.081} = \frac{0.044}{0.081} \approx 0.543

(d)

4 marks

Carmen's error probability is now $\frac{x}{100}$ , and it equals the overall probability of an inaccurate survey.

0.55 (0.08) + 0.25 (0.06) + 0.20 \cdot \frac{x}{100} = \frac{x}{100} 0.059 = 0.8 \cdot \frac{x}{100} \Rightarrow x = 7.375 \approx 7.38

Question 12

21 marks

Integration by parts, a first-order differential equation and a family of tangents

Calculus

(a)

6 marks

Integrate by parts with $u = x^{2} - 5$ and $d v = e^{x} d x$ .
Integrate by parts again.
Combine the results.

\int (x^{2} - 5) e^{x} d x = (x^{2} - 5) e^{x} - \int 2 x e^{x} d x \int 2 x e^{x} d x = 2 x e^{x} - \int 2 e^{x} d x = 2 x e^{x} - 2 e^{x} \int (x^{2} - 5) e^{x} d x = e^{x} (x^{2} - 5 - 2 x + 2) + c = e^{x} (x^{2} - 2 x - 3) + c

(b)

4 marks

Write the equation $\frac{d y}{d x} = x^{2} - 5 - y$ in the form $\frac{d y}{d x} + y = x^{2} - 5$ . The integrating factor is $e^{\int 1 d x} = e^{x}$ .
Integrate, using part (a).

\frac{d}{d x} (y e^{x}) = (x^{2} - 5) e^{x} y e^{x} = e^{x} (x^{2} - 2 x - 3) + C y = x^{2} - 2 x - 3 + C e^{- x}

(c)

5 marks

Use $y = 2$ at $x = - 3$ .
Sketch for $- 4 \leq x \leq 4$ : the curve has a local maximum at about $(- 2.70, 2.28)$ and a local minimum at about $(0.899, - 4.19)$ . It passes through $(- 3, 2)$ , with two negative $x$ -intercepts (near $- 3.47$ and $- 1.49$ ) and one positive $x$ -intercept, and reaches about $- 6.18$ at $x = - 4$ and about 4.99 at $x = 4$ .

2 = 9 + 6 - 3 + C e^{3} \Rightarrow C e^{3} = - 10 \Rightarrow C = - 10 e^{- 3} \approx - 0.498 y = x^{2} - 2 x - 3 - 0.498 e^{- x}

(d)

6 marks

The gradient at $x = - 3$ comes from the differential equation: $\frac{d y}{d x} = x^{2} - 5 - y = 4 - y$ . For the curves through $(- 3, p)$ and $(- 3, q)$ the gradients are $4 - p$ and $4 - q$ .
Subtract to find where they meet.
Since $p \neq = q$ , divide by $q - p$ .
Then $y - p = (4 - p) (1)$ , so $y = 4$ , whichever curve is used. All the tangents pass through

y - p = (4 - p) (x + 3), y - q = (4 - q) (x + 3) (q - p) = [(4 - p) - (4 - q)] (x + 3) = (q - p) (x + 3) x + 3 = 1 \Rightarrow x = - 2 (- 2, 4)

Our analysis

Where the marks are

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

Calculus51 marks · 46%
Q5, Q9, Q10, Q12
Number and algebra29 marks · 26%
Q2, Q4, Q6, Q7, Q8
Statistics and probability24 marks · 22%
Q3, Q11
Geometry and trigonometry6 marks · 5%
Q1

Mentor insights

Traps to watch for

01
Equal angles · Q6
If the angle between $a$ and $c$ equals the angle between $b$ and $c$ , the two cosines are equal, and $∣ c ∣$ cancels.
02
Difference of two angles · Q9
The viewing angle is the angle to $B$ minus the angle to $A$ , each measured from the vertical through the plane.
03
Distance from velocity · Q10
Distance is the integral of velocity.
04
Binomial · Q11
The number of inaccurate surveys is $B (n, 0.08)$ .
05
Integration by parts twice · Q12
Reduce the power of $x$ by one each time.

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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Worked solutions

IB Maths AA HL · Paper 2

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1 TZ2TZ3

Session: May 2025
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 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 · 57 marks

Question 1

6 marks

A triangle with two sides and the included angle: the cosine and sine rules

Geometry and trigonometry

(a)

3 marks

Use the cosine rule.

B C^{2} = 1 2^{2} + 7^{2} - 2 (12) (7) cos 11 6^{\circ} \approx 266.6 \Rightarrow B C \approx 16.3

(b)

3 marks

Use the sine rule for angle $C$ , the angle opposite $A B = 7$ .
The angle is acute because the angle at $A$ is already obtuse.

\frac{sin A C B}{7} = \frac{sin 11 6 ^{\circ}}{16.33} \Rightarrow sin A C B = 0.3853 A C B \approx 22. 7^{\circ}

Question 2

5 marks

A quartic with a given factor and remainder

Number and algebra

5 marks

$f (x) = 2 x^{4} - 6 x^{3} + p x^{2} + q x - 2$ . Since $(x - 1)$ is a factor, $f (1) = 0$ .
The remainder on division by $(x - 3)$ is $f (3) = - 2$ .
Substitute into $p + q = 6$ .

2 - 6 + p + q - 2 = 0 \Rightarrow p + q = 6 162 - 162 + 9 p + 3 q - 2 = - 2 \Rightarrow 9 p + 3 q = 0 \Rightarrow q = - 3 p p - 3 p = 6 \Rightarrow p = - 3, q = 9

Question 3

6 marks

Supermarket visits: a probability distribution and a sampling method

Statistics and probability

(a)

3 marks

(i) The probabilities add to 1.

(ii) The most likely value is the one with the greatest probability. $P (X = 2) = 2 a = 0.308$ is the biggest.

1.5 a + 2 a + 0.281 + a + 0.026 + 0 = 1 \Rightarrow 4.5 a = 0.693 \Rightarrow a = 0.154 mode = 2

(b)

2 marks

Use $E (X) = \sum x P (X = x)$ with $P (X = 1) = 0.231$ , $P (X = 2) = 0.308$ , $P (X = 3) = 0.281$ , $P (X = 4) = 0.154$ , $P (X = 5) = 0.026$ .

E (X) = 0.231 + 0.616 + 0.843 + 0.616 + 0.130 = 2.436 \approx 2.44

(c)

1 mark

The manager chose the first 50 customers to arrive, which is whatever is easiest, so it is a convenience sample.

Question 4

5 marks

The arc length of a circle from the area of a triangle

Number and algebra

5 marks

The triangle $A O B$ has two sides equal to the radius $r$ , with included angle 2.51 radians.
The arc $A C B$ is the major arc with angle $2 π - 2.51 = 3.773$ radians.

\frac{1}{2} r^{2} sin 2.51 = 26 \Rightarrow r^{2} = 88.07 \Rightarrow r \approx 9.385 cm arc = r θ = 9.385 (3.773) \approx 35.4 cm

Question 5

7 marks

A rational function with a parameter: the derivative and a tangent angle

Calculus

(a)

3 marks

Differentiate with the quotient rule.
Factor out $2 (2 x + a)^{2} (x + 5)$ from the numerator and cancel one factor of $(x + 5)$ .

f^{'} (x) = \frac{6 ( 2 x + a ) ^{2} ( x + 5 ) ^{2} - 2 ( 2 x + a ) ^{3} ( x + 5 )}{( x + 5 ) ^{4}} f^{'} (x) = \frac{2 ( 2 x + a ) ^{2} [ 3 ( x + 5 ) - ( 2 x + a ) ]}{( x + 5 ) ^{3}} = \frac{2 ( 2 x + a ) ^{2} ( x + 15 - a )}{( x + 5 ) ^{3}}

(b)

4 marks

At $x = 1$ the gradient is $tan 7 0^{\circ} = 2.7475$ .
Solve for $a > 0$ with the calculator.

f^{'} (1) = \frac{2 ( 2 + a ) ^{2} ( 16 - a )}{216} = tan 7 0^{\circ} a \approx 2.73 or a \approx 15.0

Question 6

8 marks

Vectors with equal angles to a third vector

Number and algebra

(a)

3 marks

Use $a = (- 5, 7)$ , $b = (- 8, 9)$ and $c = (p, - 6)$ .

a \cdot c = - 5 p - 42 b \cdot c = - 8 p - 54

(b)

5 marks

The magnitudes are $∣ a ∣ = 74$ , $∣ b ∣ = 145$ and $∣ c ∣ = p^{2} + 36$ .
Cancel the common factor $p^{2} + 36$ and solve for $p$ .

cos θ = \frac{- 5 p - 42}{74 p ^{2} + 36} = \frac{- 8 p - 54}{145 p ^{2} + 36} \frac{5 p + 42}{74} = \frac{8 p + 54}{145} \Rightarrow p \approx 4.79

Question 7

6 marks

A binomial series with a given coefficient

Number and algebra

6 marks

Write $\frac{1}{q - x ^{2}} = q^{- 1/2} (1 - \frac{x ^{2}}{q})^{- 1/2}$ .
The term in $x^{6}$ comes from the third power of $\frac{x ^{2}}{q}$ , with $n = - \frac{1}{2}$ .
Include the factor $q^{- 1/2}$ .
Set the coefficient equal to 5120.

\frac{( - \frac{1}{2} ) ( - \frac{3}{2} ) ( - \frac{5}{2} )}{3 !} (- \frac{x ^{2}}{q})^{3} = \frac{5}{16} \cdot \frac{x ^{6}}{q ^{3}} \frac{5}{16} q^{- 7/2} x^{6} \frac{5}{16} q^{- 7/2} = 5120 \Rightarrow q^{- 7/2} = 16384 = 2^{14} \Rightarrow q = 2^{- 4} q = \frac{1}{16}

Question 8

5 marks

Counting games in a tic-tac-toe competition

Number and algebra

5 marks

Let $n$ be the number of students at the start. Stephen played 7 games, and all the games between the other $n - 1$ students were played.
The other $N = n - 1$ students play each pair twice.
The total number of games played is those games plus Stephen's 7.
Since $N > 0$ , $N = 23$ , so there were 24 students originally.

2 (2 N) = N (N - 1) N (N - 1) + 7 = 513 \Rightarrow N^{2} - N - 506 = 0 \Rightarrow (N - 23) (N + 22) = 0 n = 24

Question 9

9 marks

An aeroplane approaching a runway: the viewing angle and its rate of change

Calculus

(a)

2 marks

Let $α = A P G$ , so $tan α = \frac{x}{6}$ . The angle $B P G = α + θ$ and $tan (α + θ) = \frac{x + 2}{6}$ .

θ = arctan (\frac{x + 2}{6}) - α = arctan (\frac{x + 2}{6}) - arctan (\frac{x}{6})

(b)

7 marks

Solve $θ = 0.178$ for $x$ .
Differentiate $θ$ with respect to $x$ , using $\frac{d}{d x} arctan \frac{x}{6} = \frac{6}{36 + x ^{2}}$ .
At $x = 4.630$ this is
Use the chain rule with $\frac{d θ}{d t} = 12.5$ .
The plane is travelling towards the runway at constant speed.

arctan (\frac{x + 2}{6}) - arctan (\frac{x}{6}) = 0.178 \Rightarrow x \approx 4.630 \frac{d θ}{d x} = \frac{6}{36 + ( x + 2 ) ^{2}} - \frac{6}{36 + x ^{2}} \frac{d θ}{d x} \approx - 0.02942 \frac{d x}{d t} = \frac{12.5}{- 0.02942} \approx - 424.9 speed \approx 425 km h^{- 1}

Section B · Extended response

Questions 10–12 · 53 marks

Question 10

14 marks

A 200 metre race: velocity models for two athletes

Calculus

(a)

3 marks

Fiona's velocity is $v (t) = \frac{8.14 t}{t ^{2} + 0.2}$ .
(i) $v (1) = \frac{8.14}{1.2} \approx 7.43 m s^{- 1}$
(ii) Solve $v (t) = 5$ .

t \approx 0.348 seconds

(b)

2 marks

Solve $v^{'} (t) = 4$ with the calculator.

t \approx 0.591 seconds

(c)

3 marks

(i) For large $t$ , $t^{2} + 0.2 \approx t$ .

(ii) The race is over once Fiona has gone 200 m, so the value of $t$ is limited. Fiona never reaches the speed 8.14 (the graph only gets close to it), so this value is not valid for the race.

t \to \infty lim v (t) = 8.14 m s^{- 1}

(d)

6 marks

Find the time $t_{f}$ when Fiona has gone 200 m, by solving the equation for the distance.
In that time Lucy, with $w (t) = \frac{8 t}{t ^{2} + 0.3}$ , has gone
So Lucy is this far from the finishing line.

\int_{0}^{t_{f}} v (t) d t = 200 \Rightarrow t_{f} \approx 25.01 seconds \int_{0}^{25.01} w (t) d t \approx 195.77 m 200 - 195.77 \approx 4.23 m

Question 11

18 marks

Survey data entry errors: binomial, conditional probability and totals

Statistics and probability

(a)

5 marks

Let $E$ be the number of surveys entered inaccurately, $E \sim B (50, 0.08)$ .
(i) $P (E \leq 6) \approx 0.898$
(ii) Use conditional probability. If $E = 4$ then $E \leq 6$ .

P (E = 4 ∣ E \leq 6) = \frac{P ( E = 4 )}{P ( E \leq 6 )} = \frac{0.2037}{0.8981} \approx 0.227

(b)

3 marks

Find $n$ so that $P (E \leq 6) \approx 0.367$ with $E \sim B (n, 0.08)$ . Use a table of values of $n$ on the calculator.

n = 94 (P (E \leq 6) = 0.3667)

(c)

6 marks

(i) Add the inaccurate surveys from each person.

(ii) Use conditional probability.

P (I) = 0.55 (0.08) + 0.25 (0.06) + 0.20 (0.11) = 0.044 + 0.015 + 0.022 = 0.081 P (A ∣ I) = \frac{0.55 \times 0.08}{0.081} = \frac{0.044}{0.081} \approx 0.543

(d)

4 marks

Carmen's error probability is now $\frac{x}{100}$ , and it equals the overall probability of an inaccurate survey.

0.55 (0.08) + 0.25 (0.06) + 0.20 \cdot \frac{x}{100} = \frac{x}{100} 0.059 = 0.8 \cdot \frac{x}{100} \Rightarrow x = 7.375 \approx 7.38

Question 12

21 marks

Integration by parts, a first-order differential equation and a family of tangents

Calculus

(a)

6 marks

Integrate by parts with $u = x^{2} - 5$ and $d v = e^{x} d x$ .
Integrate by parts again.
Combine the results.

\int (x^{2} - 5) e^{x} d x = (x^{2} - 5) e^{x} - \int 2 x e^{x} d x \int 2 x e^{x} d x = 2 x e^{x} - \int 2 e^{x} d x = 2 x e^{x} - 2 e^{x} \int (x^{2} - 5) e^{x} d x = e^{x} (x^{2} - 5 - 2 x + 2) + c = e^{x} (x^{2} - 2 x - 3) + c

(b)

4 marks

Write the equation $\frac{d y}{d x} = x^{2} - 5 - y$ in the form $\frac{d y}{d x} + y = x^{2} - 5$ . The integrating factor is $e^{\int 1 d x} = e^{x}$ .
Integrate, using part (a).

\frac{d}{d x} (y e^{x}) = (x^{2} - 5) e^{x} y e^{x} = e^{x} (x^{2} - 2 x - 3) + C y = x^{2} - 2 x - 3 + C e^{- x}

(c)

5 marks

Use $y = 2$ at $x = - 3$ .
Sketch for $- 4 \leq x \leq 4$ : the curve has a local maximum at about $(- 2.70, 2.28)$ and a local minimum at about $(0.899, - 4.19)$ . It passes through $(- 3, 2)$ , with two negative $x$ -intercepts (near $- 3.47$ and $- 1.49$ ) and one positive $x$ -intercept, and reaches about $- 6.18$ at $x = - 4$ and about 4.99 at $x = 4$ .

2 = 9 + 6 - 3 + C e^{3} \Rightarrow C e^{3} = - 10 \Rightarrow C = - 10 e^{- 3} \approx - 0.498 y = x^{2} - 2 x - 3 - 0.498 e^{- x}

(d)

6 marks

The gradient at $x = - 3$ comes from the differential equation: $\frac{d y}{d x} = x^{2} - 5 - y = 4 - y$ . For the curves through $(- 3, p)$ and $(- 3, q)$ the gradients are $4 - p$ and $4 - q$ .
Subtract to find where they meet.
Since $p \neq = q$ , divide by $q - p$ .
Then $y - p = (4 - p) (1)$ , so $y = 4$ , whichever curve is used. All the tangents pass through

y - p = (4 - p) (x + 3), y - q = (4 - q) (x + 3) (q - p) = [(4 - p) - (4 - q)] (x + 3) = (q - p) (x + 3) x + 3 = 1 \Rightarrow x = - 2 (- 2, 4)

Our analysis

Where the marks are

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

Calculus51 marks · 46%
Q5, Q9, Q10, Q12
Number and algebra29 marks · 26%
Q2, Q4, Q6, Q7, Q8
Statistics and probability24 marks · 22%
Q3, Q11
Geometry and trigonometry6 marks · 5%
Q1

Mentor insights

Traps to watch for

01
Equal angles · Q6
If the angle between $a$ and $c$ equals the angle between $b$ and $c$ , the two cosines are equal, and $∣ c ∣$ cancels.
02
Difference of two angles · Q9
The viewing angle is the angle to $B$ minus the angle to $A$ , each measured from the vertical through the plane.
03
Distance from velocity · Q10
Distance is the integral of velocity.
04
Binomial · Q11
The number of inaccurate surveys is $B (n, 0.08)$ .
05
Integration by parts twice · Q12
Reduce the power of $x$ by one each time.

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