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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 TZ2 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

Area between 1 - x^2 and e^(2x) on [-1, 0]

Calculus

(a)

3 marks

Solve $1 - x^{2} = e^{2 x}$ for $- 1 \leq x \leq 0$ on a graph.

a \approx - 0.917, b = 0

(b)

3 marks

On the interval $f (x) = 1 - x^{2}$ is above $g (x) = e^{2 x}$ .

Area = \int_{- 0.9166}^{0} (1 - x^{2} - e^{2 x}) d x \approx 0.240

Question 2

5 marks

A regression line through the mean point: finding two missing values

Statistics and probability

(a)

1 mark

Substitute $\overset{x}{ˉ} = 7$ into the regression line.

\overset{y}{ˉ} = 2.1875 (7) + 0.6875 = 15.3125 + 0.6875 = 16

(b)

4 marks

The mean of the five $y$ values is 16, so they add to 80.
With $q = p + 3$ , $2 p + 3 = 37$ .

9 + 13 + p + q + 21 = 80 \Rightarrow p + q = 37 p = 17, q = 20

Question 3

6 marks

Loudness in decibels: intensity, loudness and thunder

Number and algebra

(a)

1 mark

$S_{2}$ has twice the intensity of $S_{1}$ .

I = 2 \times 1 0^{- 6}

(b)

2 marks

Substitute into the formula.

L = 10 lo g_{10} (2 \times 1 0^{- 6} \times 1 0^{12}) = 10 lo g_{10} (2 \times 1 0^{6}) \approx 63.0 decibels

(c)

3 marks

Set $L = 115$ and solve for $I$ .

115 = 10 lo g_{10} (I \times 1 0^{12}) \Rightarrow 11.5 = lo g_{10} (I \times 1 0^{12}) I \times 1 0^{12} = 1 0^{11.5} \Rightarrow I = 1 0^{- 0.5} \approx 0.316

Question 4

6 marks

A particle with velocity 1 + e^(-t) - e^(-sin 2t): velocity, maximum and acceleration

Calculus

(a)

1 mark

Substitute $t = 2$ (radians).

v (2) = 1 + e^{- 2} - e^{- s i n 4} \approx - 0.996 m s^{- 1}

(b)

2 marks

Solve $v^{'} (t) = 0$ on a graph. The maximum occurs at $t \approx 0.406$ .

v_{m a x} = v (0.4058) \approx 1.18 m s^{- 1}

(c)

3 marks

The velocity is zero at $t \approx 1.658$ and changes sign there.
Evaluate the derivative at that time.

a = v^{'} (1.658) \approx - 2.53 m s^{- 2}

Question 5

5 marks

The least n so that a binomial variable is at least 1 with probability above 0.99

Number and algebra

5 marks

Use the complement.
Take logarithms, remembering that $ln 0.75 < 0$ reverses the inequality.
Check: for $n = 16$ , $P (X \geq 1) \approx 0.98998 < 0.99$ , and for $n = 17$ , $P (X \geq 1) \approx 0.99248 > 0.99$ .

1 - 0.7 5^{n} > 0.99 \Rightarrow 0.7 5^{n} < 0.01 n > \frac{ln 0.01}{ln 0.75} \approx 16.008 n = 17

Question 6

6 marks

A growing spherical bubble: related rates

Number and algebra

6 marks

Find the radius when the volume is 20.
Differentiate $V = \frac{4}{3} π r^{3}$ with respect to $r$ .
Use the chain rule with $\frac{d V}{d t} = 5$ .

\frac{4}{3} π r^{3} = 20 \Rightarrow r = (\frac{15}{π})^{1/3} \approx 1.684 cm \frac{d V}{d r} = 4 π r^{2} \frac{d r}{d t} = \frac{d V / d t}{d V / d r} = \frac{5}{4 π ( 1.684 ) ^{2}} \approx 0.140 cm s^{- 1}

Question 7

5 marks

A volume of revolution of 4 ln(x - 2) about the y-axis

Number and algebra

5 marks

Rearrange $y = 4 ln (x - 2)$ for $x$ .
Square it.
Integrate between $y = 0$ and $y = 4$ .

\frac{y}{4} = ln (x - 2) \Rightarrow x = 2 + e^{y /4} x^{2} = 4 + 4 e^{y /4} + e^{y /2} V = π \int_{0}^{4} (4 + 4 e^{y /4} + e^{y /2}) d y = π [4 y + 16 e^{y /4} + 2 e^{y /2}]_{0}^{4} V = π (16 + 16 e + 2 e^{2} - 16 - 2) = π (16 e + 2 e^{2} - 2) = 2 π (e^{2} + 8 e - 1) \approx 177

Question 8

10 marks

The complex number 1 + cos 2theta + i sin 2theta: argument, modulus and an equation

Number and algebra

(a)

7 marks

Rewrite with double angle identities.
Since $2 cos θ > 0$ for $- \frac{π}{2} < θ < \frac{π}{2}$ , this is modulus-argument form.
Check the modulus directly: $∣ z ∣^{2} = (1 + cos 2 θ)^{2} + sin^{2} 2 θ = 2 + 2 cos 2 θ = 4 cos^{2} θ$ .

z = 2 cos^{2} θ + 2 i sin θ cos θ = 2 cos θ (cos θ + i sin θ) ∣ z ∣ = 2 cos θ, ar g z = θ

(b)

3 marks

$ar g (z^{2}) = 2 θ$ and $z^{3} = (2 cos θ)^{3}$ .
Solve on a graph or with the solver on $- \frac{π}{2} < θ < \frac{π}{2}$ .

2 θ = 8 cos^{3} θ θ \approx 0.913

Question 9

8 marks

Finding a, b, c from a minimum point and an asymptote

Number and algebra

8 marks

The denominator is zero at $x = 1$ .
The point $(2, 1)$ is on the curve.
Differentiate with the quotient rule.
Put the numerator equal to 0 at $x = 2$ .
Solve the three linear equations. Subtract the first from the second: $3 a + b = - 2$ . Subtract the second from the third: $8 a + 2 b = 2$ , so $4 a + b = 1$ .
Check: $3 x^{2} - 11 x + 8 = (3 x - 8) (x - 1)$ , and $y (2) = \frac{- 2}{- 2} = 1$ .

a + b + c = 0 \frac{2 - 4}{4 a + 2 b + c} = 1 \Rightarrow 4 a + 2 b + c = - 2 \frac{d y}{d x} = \frac{( a x ^{2} + b x + c ) - ( x - 4 ) ( 2 a x + b )}{( a x ^{2} + b x + c ) ^{2}} (4 a + 2 b + c) + 2 (4 a + b) = 0 \Rightarrow 12 a + 4 b + c = 0 a = 3, b = - 11, c = 8

Section B · Extended response

Questions 10–12 · 53 marks

Question 10

15 marks

Chocolate weights: a density function, median and expected spend

Statistics and probability

(a)

2 marks

$f (x) = \frac{6}{85} (4 + 3 x - x^{2})$ is a downward parabola with its maximum at the axis of symmetry.

x = \frac{3}{2} \Rightarrow mode = 1.5 kg

(b)

2 marks

Integrate the density.

P (1 \leq X \leq 2) = \int_{1}^{2} \frac{6}{85} (4 + 3 x - x^{2}) d x = \frac{37}{85} \approx 0.435

(c)

3 marks

The median $m$ satisfies $\int_{0.5}^{m} f (x) d x = 0.5$ .
Solve with the calculator.

\frac{6}{85} [4 x + \frac{3 x ^{2}}{2} - \frac{x ^{3}}{3}]_{0.5}^{m} = 0.5 m \approx 1.69 kg

(d)

3 marks

A customer who buys under 0.75 kg pays at most $18.75. A customer who buys from 0.75 kg pays $24 x$ , which is at most $48 when $x \leq 2$ . So the condition is $0.5 \leq X \leq 2$ .

P = \int_{0.5}^{2} f (x) d x = \frac{54}{85} \approx 0.635

(e)

5 marks

The expected spend is the sum of the two pieces.
Evaluate each piece.
To the nearest cent the expected spend is $40.96.

E = \int_{0.5}^{0.75} 25 x f (x) d x + \int_{0.75}^{3} 24 x f (x) d x 1.5148 + 39.4428 \approx 40.9577

Question 11

17 marks

A rotating sprinkler beside a path: chord, timing and a turtle

Geometry and trigonometry

(a)

3 marks

Let $M$ be the midpoint of $[A B]$ . Then $S M = 14$ , $S A = 20$ and $S M A = 9 0^{\circ}$ .

A M = 2 0^{2} - 1 4^{2} = 204 \approx 14.283 A B = 2 A M = 2204 \approx 28.57 m

(b)

1 mark

One revolution is $2 π$ radians in 16 seconds.

\frac{2 π}{16} = \frac{π}{8} radians per second

(c)

4 marks

The angle $A S B$ is twice $A S M$ .
The time to turn through that angle at $\frac{π}{8}$ radians per second is

A S M = cos^{- 1} \frac{14}{20} \approx 0.7954, A S B \approx 1.5908 T = \frac{1.5908}{π /8} \approx 4.05 s

(d)

1 mark

The water has turned through $\frac{π}{8}$ radians every second since $t = 0$ .

α = \frac{π t}{8}

(e)

3 marks

In triangle $A S D$ , the angle at $D$ is $π - α - β = π - 0.7754 - α \approx 2.37 - α$ .
Use the sine rule: $\frac{d}{sin α} = \frac{20}{sin A D S}$ .

d = \frac{20 sin α}{sin ( 2.37 - α )} = \frac{20 sin \frac{π t}{8}}{sin ( 2.37 - \frac{π t}{8} )}

(f)

1 mark

At $t = 0$ , $g (0) = 18$ .

18 m

(g)

4 marks

The distance between the turtle and the point $D$ is the difference between the two distances.
The water reaches the turtle when $w = 0$ . The first solution is
Then the turtle is

w = 0.05 t^{2} + 1.1 t + 18 - \frac{20 sin \frac{π t}{8}}{sin ( 2.37 - \frac{π t}{8} )} t \approx 3.35 s g (3.349) = 0.05 (3.349)^{2} + 1.1 (3.349) + 18 \approx 22.2 m south of A

Question 12

21 marks

A first-order differential equation: Euler's method, integrating factor and a closed form

Calculus

(a)

3 marks

Rewrite as $\frac{d y}{d x} = y cosec 2 x + tan x$ with $h = \frac{π}{12}$ , $x_{0} = \frac{π}{4}$ , $y_{0} = \frac{π}{4}$ .

y_{1} = \frac{π}{4} + \frac{π}{12} (\frac{π}{4} \cdot 1 + 1) \approx 1.2528 (x_{1} = \frac{π}{3}) y_{2} = 1.2528 + \frac{π}{12} (1.2528 cosec \frac{2 π}{3} + tan \frac{π}{3}) \approx 1.98

(b)

4 marks

Differentiate $\frac{1}{2} ln (cot x)$ with the chain rule.

\frac{d}{d x} (\frac{1}{2} ln (cot x)) = \frac{1}{2} \cdot \frac{1}{cot x} \cdot (- cosec^{2} x) = - \frac{1}{2 sin x cos x} - \frac{1}{sin 2 x} = - cosec 2 x

(c)

4 marks

The equation is $\frac{d y}{d x} - cosec 2 x y = tan x$ , so $P (x) = - cosec 2 x$ .
The integrating factor is $e^{\int P d x}$ , and by part (b) $\int - cosec 2 x d x = \frac{1}{2} ln (cot x)$ .

I (x) = e^{\frac{1}{2} l n (c o t x)} = cot x

(d)

5 marks

Multiply the equation by $cot x$ .
Integrate.
Use $y = \frac{π}{4}$ at $x = \frac{π}{4}$ , where $cot \frac{π}{4} = 1$ : $\frac{π}{4} = \frac{π}{4} + C$ , so $C = 0$ .

\frac{d}{d x} (y cot x) = cot x tan x = 1 y cot x = x + C y = x tan x

(e)

3 marks

(i) At $x = \frac{5 π}{12}$ .

(ii) The gradient of the curve changes very quickly close to $x = \frac{5 π}{12}$ , as the curve heads towards $x = \frac{π}{2}$ , so the straight-line steps of Euler's method fall behind.

(iii) The curve is concave up on this interval ( $y^{''} > 0$ ), so each tangent line lies below the curve, and the approximation is less than the true value.

y = \frac{5 π}{12} tan \frac{5 π}{12} \approx 2.53

(f)

2 marks

Write the equation as $\frac{d y}{d x} = y cosec 2 x + tan x$ with $y = x tan x$ .
On $0 < x < \frac{π}{2}$ each of $x$ , $tan x$ and $cosec 2 x$ is positive, so both terms are positive. The gradient is positive.

\frac{d y}{d x} = x tan x cosec 2 x + tan x

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 algebra40 marks · 36%
Q3, Q5, Q6, Q7, Q8, Q9
Calculus33 marks · 30%
Q1, Q4, Q12
Statistics and probability20 marks · 18%
Q2, Q10
Geometry and trigonometry17 marks · 15%
Q11

Mentor insights

Traps to watch for

01
Double angles · Q8
$1 + cos 2 θ = 2 cos^{2} θ$ and $sin 2 θ = 2 sin θ cos θ$ .
02
Three conditions · Q9
A vertical asymptote at $x = 1$ makes the denominator zero there. The point $(2, 1)$ is on the curve. The gradient is zero at $x = 2$ .
03
Mode of a density · Q10
The mode is where $f$ is largest, at the vertex of the parabola.
04
Chord from Pythagoras · Q11
The perpendicular from the sprinkler to the path meets $[A B]$ at its midpoint $M$ .
05
Integrating factor · Q12
For $\frac{d y}{d x} + P (x) y = Q (x)$ the factor is $e^{\int P d x}$ . Here $P = - cosec 2 x$ .

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

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 TZ2 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

Area between 1 - x^2 and e^(2x) on [-1, 0]

Calculus

(a)

3 marks

Solve $1 - x^{2} = e^{2 x}$ for $- 1 \leq x \leq 0$ on a graph.

a \approx - 0.917, b = 0

(b)

3 marks

On the interval $f (x) = 1 - x^{2}$ is above $g (x) = e^{2 x}$ .

Area = \int_{- 0.9166}^{0} (1 - x^{2} - e^{2 x}) d x \approx 0.240

Question 2

5 marks

A regression line through the mean point: finding two missing values

Statistics and probability

(a)

1 mark

Substitute $\overset{x}{ˉ} = 7$ into the regression line.

\overset{y}{ˉ} = 2.1875 (7) + 0.6875 = 15.3125 + 0.6875 = 16

(b)

4 marks

The mean of the five $y$ values is 16, so they add to 80.
With $q = p + 3$ , $2 p + 3 = 37$ .

9 + 13 + p + q + 21 = 80 \Rightarrow p + q = 37 p = 17, q = 20

Question 3

6 marks

Loudness in decibels: intensity, loudness and thunder

Number and algebra

(a)

1 mark

$S_{2}$ has twice the intensity of $S_{1}$ .

I = 2 \times 1 0^{- 6}

(b)

2 marks

Substitute into the formula.

L = 10 lo g_{10} (2 \times 1 0^{- 6} \times 1 0^{12}) = 10 lo g_{10} (2 \times 1 0^{6}) \approx 63.0 decibels

(c)

3 marks

Set $L = 115$ and solve for $I$ .

115 = 10 lo g_{10} (I \times 1 0^{12}) \Rightarrow 11.5 = lo g_{10} (I \times 1 0^{12}) I \times 1 0^{12} = 1 0^{11.5} \Rightarrow I = 1 0^{- 0.5} \approx 0.316

Question 4

6 marks

A particle with velocity 1 + e^(-t) - e^(-sin 2t): velocity, maximum and acceleration

Calculus

(a)

1 mark

Substitute $t = 2$ (radians).

v (2) = 1 + e^{- 2} - e^{- s i n 4} \approx - 0.996 m s^{- 1}

(b)

2 marks

Solve $v^{'} (t) = 0$ on a graph. The maximum occurs at $t \approx 0.406$ .

v_{m a x} = v (0.4058) \approx 1.18 m s^{- 1}

(c)

3 marks

The velocity is zero at $t \approx 1.658$ and changes sign there.
Evaluate the derivative at that time.

a = v^{'} (1.658) \approx - 2.53 m s^{- 2}

Question 5

5 marks

The least n so that a binomial variable is at least 1 with probability above 0.99

Number and algebra

5 marks

Use the complement.
Take logarithms, remembering that $ln 0.75 < 0$ reverses the inequality.
Check: for $n = 16$ , $P (X \geq 1) \approx 0.98998 < 0.99$ , and for $n = 17$ , $P (X \geq 1) \approx 0.99248 > 0.99$ .

1 - 0.7 5^{n} > 0.99 \Rightarrow 0.7 5^{n} < 0.01 n > \frac{ln 0.01}{ln 0.75} \approx 16.008 n = 17

Question 6

6 marks

A growing spherical bubble: related rates

Number and algebra

6 marks

Find the radius when the volume is 20.
Differentiate $V = \frac{4}{3} π r^{3}$ with respect to $r$ .
Use the chain rule with $\frac{d V}{d t} = 5$ .

\frac{4}{3} π r^{3} = 20 \Rightarrow r = (\frac{15}{π})^{1/3} \approx 1.684 cm \frac{d V}{d r} = 4 π r^{2} \frac{d r}{d t} = \frac{d V / d t}{d V / d r} = \frac{5}{4 π ( 1.684 ) ^{2}} \approx 0.140 cm s^{- 1}

Question 7

5 marks

A volume of revolution of 4 ln(x - 2) about the y-axis

Number and algebra

5 marks

Rearrange $y = 4 ln (x - 2)$ for $x$ .
Square it.
Integrate between $y = 0$ and $y = 4$ .

\frac{y}{4} = ln (x - 2) \Rightarrow x = 2 + e^{y /4} x^{2} = 4 + 4 e^{y /4} + e^{y /2} V = π \int_{0}^{4} (4 + 4 e^{y /4} + e^{y /2}) d y = π [4 y + 16 e^{y /4} + 2 e^{y /2}]_{0}^{4} V = π (16 + 16 e + 2 e^{2} - 16 - 2) = π (16 e + 2 e^{2} - 2) = 2 π (e^{2} + 8 e - 1) \approx 177

Question 8

10 marks

The complex number 1 + cos 2theta + i sin 2theta: argument, modulus and an equation

Number and algebra

(a)

7 marks

Rewrite with double angle identities.
Since $2 cos θ > 0$ for $- \frac{π}{2} < θ < \frac{π}{2}$ , this is modulus-argument form.
Check the modulus directly: $∣ z ∣^{2} = (1 + cos 2 θ)^{2} + sin^{2} 2 θ = 2 + 2 cos 2 θ = 4 cos^{2} θ$ .

z = 2 cos^{2} θ + 2 i sin θ cos θ = 2 cos θ (cos θ + i sin θ) ∣ z ∣ = 2 cos θ, ar g z = θ

(b)

3 marks

$ar g (z^{2}) = 2 θ$ and $z^{3} = (2 cos θ)^{3}$ .
Solve on a graph or with the solver on $- \frac{π}{2} < θ < \frac{π}{2}$ .

2 θ = 8 cos^{3} θ θ \approx 0.913

Question 9

8 marks

Finding a, b, c from a minimum point and an asymptote

Number and algebra

8 marks

The denominator is zero at $x = 1$ .
The point $(2, 1)$ is on the curve.
Differentiate with the quotient rule.
Put the numerator equal to 0 at $x = 2$ .
Solve the three linear equations. Subtract the first from the second: $3 a + b = - 2$ . Subtract the second from the third: $8 a + 2 b = 2$ , so $4 a + b = 1$ .
Check: $3 x^{2} - 11 x + 8 = (3 x - 8) (x - 1)$ , and $y (2) = \frac{- 2}{- 2} = 1$ .

a + b + c = 0 \frac{2 - 4}{4 a + 2 b + c} = 1 \Rightarrow 4 a + 2 b + c = - 2 \frac{d y}{d x} = \frac{( a x ^{2} + b x + c ) - ( x - 4 ) ( 2 a x + b )}{( a x ^{2} + b x + c ) ^{2}} (4 a + 2 b + c) + 2 (4 a + b) = 0 \Rightarrow 12 a + 4 b + c = 0 a = 3, b = - 11, c = 8

Section B · Extended response

Questions 10–12 · 53 marks

Question 10

15 marks

Chocolate weights: a density function, median and expected spend

Statistics and probability

(a)

2 marks

$f (x) = \frac{6}{85} (4 + 3 x - x^{2})$ is a downward parabola with its maximum at the axis of symmetry.

x = \frac{3}{2} \Rightarrow mode = 1.5 kg

(b)

2 marks

Integrate the density.

P (1 \leq X \leq 2) = \int_{1}^{2} \frac{6}{85} (4 + 3 x - x^{2}) d x = \frac{37}{85} \approx 0.435

(c)

3 marks

The median $m$ satisfies $\int_{0.5}^{m} f (x) d x = 0.5$ .
Solve with the calculator.

\frac{6}{85} [4 x + \frac{3 x ^{2}}{2} - \frac{x ^{3}}{3}]_{0.5}^{m} = 0.5 m \approx 1.69 kg

(d)

3 marks

A customer who buys under 0.75 kg pays at most $18.75. A customer who buys from 0.75 kg pays $24 x$ , which is at most $48 when $x \leq 2$ . So the condition is $0.5 \leq X \leq 2$ .

P = \int_{0.5}^{2} f (x) d x = \frac{54}{85} \approx 0.635

(e)

5 marks

The expected spend is the sum of the two pieces.
Evaluate each piece.
To the nearest cent the expected spend is $40.96.

E = \int_{0.5}^{0.75} 25 x f (x) d x + \int_{0.75}^{3} 24 x f (x) d x 1.5148 + 39.4428 \approx 40.9577

Question 11

17 marks

A rotating sprinkler beside a path: chord, timing and a turtle

Geometry and trigonometry

(a)

3 marks

Let $M$ be the midpoint of $[A B]$ . Then $S M = 14$ , $S A = 20$ and $S M A = 9 0^{\circ}$ .

A M = 2 0^{2} - 1 4^{2} = 204 \approx 14.283 A B = 2 A M = 2204 \approx 28.57 m

(b)

1 mark

One revolution is $2 π$ radians in 16 seconds.

\frac{2 π}{16} = \frac{π}{8} radians per second

(c)

4 marks

The angle $A S B$ is twice $A S M$ .
The time to turn through that angle at $\frac{π}{8}$ radians per second is

A S M = cos^{- 1} \frac{14}{20} \approx 0.7954, A S B \approx 1.5908 T = \frac{1.5908}{π /8} \approx 4.05 s

(d)

1 mark

The water has turned through $\frac{π}{8}$ radians every second since $t = 0$ .

α = \frac{π t}{8}

(e)

3 marks

In triangle $A S D$ , the angle at $D$ is $π - α - β = π - 0.7754 - α \approx 2.37 - α$ .
Use the sine rule: $\frac{d}{sin α} = \frac{20}{sin A D S}$ .

d = \frac{20 sin α}{sin ( 2.37 - α )} = \frac{20 sin \frac{π t}{8}}{sin ( 2.37 - \frac{π t}{8} )}

(f)

1 mark

At $t = 0$ , $g (0) = 18$ .

18 m

(g)

4 marks

The distance between the turtle and the point $D$ is the difference between the two distances.
The water reaches the turtle when $w = 0$ . The first solution is
Then the turtle is

w = 0.05 t^{2} + 1.1 t + 18 - \frac{20 sin \frac{π t}{8}}{sin ( 2.37 - \frac{π t}{8} )} t \approx 3.35 s g (3.349) = 0.05 (3.349)^{2} + 1.1 (3.349) + 18 \approx 22.2 m south of A

Question 12

21 marks

A first-order differential equation: Euler's method, integrating factor and a closed form

Calculus

(a)

3 marks

Rewrite as $\frac{d y}{d x} = y cosec 2 x + tan x$ with $h = \frac{π}{12}$ , $x_{0} = \frac{π}{4}$ , $y_{0} = \frac{π}{4}$ .

y_{1} = \frac{π}{4} + \frac{π}{12} (\frac{π}{4} \cdot 1 + 1) \approx 1.2528 (x_{1} = \frac{π}{3}) y_{2} = 1.2528 + \frac{π}{12} (1.2528 cosec \frac{2 π}{3} + tan \frac{π}{3}) \approx 1.98

(b)

4 marks

Differentiate $\frac{1}{2} ln (cot x)$ with the chain rule.

\frac{d}{d x} (\frac{1}{2} ln (cot x)) = \frac{1}{2} \cdot \frac{1}{cot x} \cdot (- cosec^{2} x) = - \frac{1}{2 sin x cos x} - \frac{1}{sin 2 x} = - cosec 2 x

(c)

4 marks

The equation is $\frac{d y}{d x} - cosec 2 x y = tan x$ , so $P (x) = - cosec 2 x$ .
The integrating factor is $e^{\int P d x}$ , and by part (b) $\int - cosec 2 x d x = \frac{1}{2} ln (cot x)$ .

I (x) = e^{\frac{1}{2} l n (c o t x)} = cot x

(d)

5 marks

Multiply the equation by $cot x$ .
Integrate.
Use $y = \frac{π}{4}$ at $x = \frac{π}{4}$ , where $cot \frac{π}{4} = 1$ : $\frac{π}{4} = \frac{π}{4} + C$ , so $C = 0$ .

\frac{d}{d x} (y cot x) = cot x tan x = 1 y cot x = x + C y = x tan x

(e)

3 marks

(i) At $x = \frac{5 π}{12}$ .

(ii) The gradient of the curve changes very quickly close to $x = \frac{5 π}{12}$ , as the curve heads towards $x = \frac{π}{2}$ , so the straight-line steps of Euler's method fall behind.

(iii) The curve is concave up on this interval ( $y^{''} > 0$ ), so each tangent line lies below the curve, and the approximation is less than the true value.

y = \frac{5 π}{12} tan \frac{5 π}{12} \approx 2.53

(f)

2 marks

Write the equation as $\frac{d y}{d x} = y cosec 2 x + tan x$ with $y = x tan x$ .
On $0 < x < \frac{π}{2}$ each of $x$ , $tan x$ and $cosec 2 x$ is positive, so both terms are positive. The gradient is positive.

\frac{d y}{d x} = x tan x cosec 2 x + tan x

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 algebra40 marks · 36%
Q3, Q5, Q6, Q7, Q8, Q9
Calculus33 marks · 30%
Q1, Q4, Q12
Statistics and probability20 marks · 18%
Q2, Q10
Geometry and trigonometry17 marks · 15%
Q11

Mentor insights

Traps to watch for

01
Double angles · Q8
$1 + cos 2 θ = 2 cos^{2} θ$ and $sin 2 θ = 2 sin θ cos θ$ .
02
Three conditions · Q9
A vertical asymptote at $x = 1$ makes the denominator zero there. The point $(2, 1)$ is on the curve. The gradient is zero at $x = 2$ .
03
Mode of a density · Q10
The mode is where $f$ is largest, at the vertex of the parabola.
04
Chord from Pythagoras · Q11
The perpendicular from the sprinkler to the path meets $[A B]$ at its midpoint $M$ .
05
Integrating factor · Q12
For $\frac{d y}{d x} + P (x) y = Q (x)$ the factor is $e^{\int P d x}$ . Here $P = - cosec 2 x$ .

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