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Worked solutions

IB Maths AA HL · Paper 2

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ3

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

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Section A · Short response

Questions 1–9 · 54 marks

Question 1

5 marks

A double angle identity and a trigonometric equation

Functions

(a)

2 marks

Use $cos 2 x = 1 - 2 sin^{2} x$ .

3 cos 2 x + 11 sin x = 3 (1 - 2 sin^{2} x) + 11 sin x = 3 + 11 sin x - 6 sin^{2} x

(b)

3 marks

Use part (a) in $3 cos 2 x + 11 sin x - 6 = 0$ .
Factorise (or use the quadratic formula).
Since $sin x \neq = \frac{3}{2}$ , we have $sin x = \frac{1}{3}$ .

3 + 11 sin x - 6 sin^{2} x - 6 = 0 \Rightarrow 6 sin^{2} x - 11 sin x + 3 = 0 (3 sin x - 1) (2 sin x - 3) = 0 x \approx 19. 5^{\circ} or x \approx 160. 5^{\circ}

Question 2

6 marks

A quiz median and finding two scores from the mean and range

Statistics and probability

(a)

1 mark

Aiden's score is below 10.5 and Brett's is above 10.5, so one new value is on each side of the median. The middle values stay the same, so the median is still 10.5.

(b)

5 marks

The total for the 28 pupils is $10.5 \times 28 = 294$ and the total for all 30 is $10.6 \times 30 = 318$ .
Aiden has the lowest score and Brett the highest, so the range is $B - A = 14$ .
Add the two equations.
Aiden scored 5 and Brett scored 19.

A + B = 318 - 294 = 24 2 B = 38 \Rightarrow B = 19, A = 24 - 19 = 5

Question 3

6 marks

A sector and a circle with related perimeter and area

Geometry and trigonometry

(a)

3 marks

The circumference of the circle is $2 π (8) = 16 π$ , so the perimeter of the sector is $1.5 \times 16 π = 24 π$ .

2 r + r θ = 24 π \Rightarrow r (2 + θ) = 24 π r = \frac{24 π}{2 + θ}

(b)

3 marks

Equate the areas and substitute for $r$ .
Simplify.
Solve this quadratic: $θ^{2} + (4 - 4.5 π) θ + 4 = 0$ . Its roots are about $0.411$ and $9.73$ . The angle is acute, so $θ < \frac{π}{2}$ .

\frac{1}{2} (\frac{24 π}{2 + θ})^{2} θ = 64 π 288 π^{2} θ = 64 π (2 + θ)^{2} \Rightarrow (2 + θ)^{2} = 4.5 π θ θ \approx 0.411

Question 4

7 marks

Normal heights of two sunflower varieties

Statistics and probability

(a)

3 marks

The height $180$ cm is exceeded by 25% of large sunflowers, so it is the upper quartile: $\frac{180 - m}{s} = z$ with $P (Z > z) = 0.25$ .

z = 0.6745 \Rightarrow m + 0.6745 s = 180 A \approx 0.674

(b)

4 marks

The giant sunflowers have mean $m + 35$ and standard deviation $2 s$ . Since 98% exceed 180 cm, 2% are below, so the standardised value of 180 is $- 2.0537$ .
Subtract this from the equation in part (a), $m + 0.6745 s = 180$ .

\frac{180 - ( m + 35 )}{2 s} = - 2.0537 \Rightarrow m - 4.1075 s = 145 (0.6745 + 4.1075) s = 180 - 145 = 35 \Rightarrow s \approx 7.32 m = 180 - 0.6745 (7.319) \approx 175

Question 5

5 marks

The volume of a solid of revolution about the y-axis

Number and algebra

5 marks

Rearrange $y = e^{3 x /4}$ to get $x$ in terms of $y$ .
The region runs from $y = 1$ (where $x = 0$ ) up to $y = 8$ . Rotate about the $y$ -axis.
Use $\int (ln y)^{2} d y = y (ln y)^{2} - 2 y ln y + 2 y$ .

ln y = \frac{3 x}{4} \Rightarrow x = \frac{4}{3} ln y V = π \int_{1}^{8} (\frac{4}{3} ln y)^{2} d y = \frac{16 π}{9} \int_{1}^{8} (ln y)^{2} d y V = \frac{16 π}{9} [y (ln y)^{2} - 2 y ln y + 2 y]_{1}^{8} = \frac{16 π}{9} (8 ln^{2} 8 - 16 ln 8 + 14) V \approx 85.6

Question 6

7 marks

A quadratic from the remainder theorem and the sum of squares of the roots

Number and algebra

7 marks

Use the remainder theorem.
For the roots, use $α^{2} + β^{2} = (α + β)^{2} - 2 α β$ .
Substitute $r = q + 10$ .
If $q = 11$ then $r = 21$ , and the discriminant is $121 - 168 < 0$ , so the roots would not be real. That is not allowed.
If $q = - 7$ then $r = 3$ , and the discriminant is $49 - 24 = 25 > 0$ , so the roots are real.

P (- 1) = 2 - q + r = 12 \Rightarrow r = q + 10 \frac{37}{4} = \frac{q ^{2}}{4} - r \Rightarrow q^{2} - 4 r = 37 q^{2} - 4 q - 40 = 37 \Rightarrow q^{2} - 4 q - 77 = 0 \Rightarrow (q - 11) (q + 7) = 0 q = - 7, r = 3

Question 7

6 marks

A system of equations: general solution, a unique solution and special cases

Number and algebra

(a)

2 marks

Let $z = t$ . Eliminate $y$ : the second equation gives $y = 3 x + 3 t - 12$ . Substitute in the first.
The general solution.

2 x + 6 (3 x + 3 t - 12) - 8 t = 13 \Rightarrow 20 x + 10 t = 85 \Rightarrow x = \frac{17}{4} - \frac{t}{2} y = \frac{3}{4} + \frac{3 t}{2} x = \frac{17}{4} - \frac{t}{2}, y = \frac{3}{4} + \frac{3 t}{2}, z = t

(b)

2 marks

Substitute into $3 x + 12 y - 16 z = - 3$ .
So $t = - \frac{99}{2}$ .

3 (\frac{17}{4} - \frac{t}{2}) + 12 (\frac{3}{4} + \frac{3 t}{2}) - 16 t = - 3 \frac{51}{4} + 9 - \frac{3 t}{2} + 18 t - 16 t = - 3 \Rightarrow \frac{t}{2} = - 3 - \frac{87}{4} = - \frac{99}{4} x = 29, y = - \frac{147}{2}, z = - \frac{99}{2}

(c)

2 marks

(i) The coefficients of $y$ and $z$ in the third equation are twice those in the first equation, so there is no unique solution when $a = 2 \times 2 = 4$ .

(ii) Twice the first equation is $4 x + 12 y - 16 z = 26$ . For infinitely many solutions the third equation must be this same equation.

a = 4 k = 26

Question 8

7 marks

Arranging and choosing recipe books

Number and algebra

(a)

1 mark

Arrange 15 different books in a line.

15! \approx 1.31 \times 1 0^{12}

(b)

3 marks

Arrange the six Asian books $6!$ ways, the five European books $5!$ ways and the four African books $4!$ ways. The three groups can be placed in any order, which is $3!$ ways.

6! \times 5! \times 4! \times 3! = 720 \times 120 \times 24 \times 6 = 12441600

(c)

3 marks

Choose 4 books from one continent. Add the three cases.

(4 6) + (4 5) + (4 4) = 15 + 5 + 1 = 21

Question 9

5 marks

Proof by contradiction of an inequality between a root and a mean

Number and algebra

5 marks

Assume the opposite: there are positive $x, y$ with $2 x y > \frac{1}{2} (2 x + y)$ .
Both sides are positive, so squaring keeps the direction of the inequality.
Rearrange.
A square cannot be negative, so this is a contradiction.
So the assumption is false, and $2 x y \leq \frac{1}{2} (2 x + y)$ for all $x, y > 0$ .

2 x y > \frac{1}{4} (2 x + y)^{2} \Rightarrow 8 x y > 4 x^{2} + 4 x y + y^{2} 0 > 4 x^{2} - 4 x y + y^{2} = (2 x - y)^{2}

Section B · Extended response

Questions 10–12 · 56 marks

Question 10

17 marks

A damped motion: maximum, minimum, total distance and a geometric series of peaks

Calculus

(a)

5 marks

(i) Find the first maximum of $s (t) = 2^{1 - t /5} sin \frac{2 π t}{3}$ on the calculator. It is at $t \approx 0.718$ .

(ii) The velocity is $\frac{d s}{d t}$ . The graph is steepest at $t = 0$ and the oscillations get smaller after that, so the largest speed is at $t = 0$ .

s_{ma x} \approx 1.81 cm v (0) = \frac{d s}{d t}_{t = 0} = 2 \cdot \frac{2 π}{3} cos 0 = \frac{4 π}{3} \approx 4.19 cm/s

(b)

3 marks

(i) The first minimum is at $t \approx 2.218$ .

(ii) Substitute $t = 3.5$ .

s_{min} \approx - 1.47 s (3.5) = 2^{1 - 0.7} sin \frac{7 π}{3} \approx 1.07

(c)

3 marks

In the first 3.5 seconds the particle goes out to $1.806$ and back to $O$ , then out to $1.467$ on the other side and back to $O$ , and then onwards to $1.066$ .

2 (1.806) + 2 (1.467) + 1.066 \approx 7.61 cm

(d)

1 mark

The sine factor is zero when $\frac{2 π t}{3} = π$ .

T = 1.5

(e)

5 marks

(i) The largest distances in each interval are $u_{1} \approx 1.806$ , $u_{2} \approx 1.467$ and $u_{3} \approx 1.192$ . The ratio.

Check: $\frac{u _{3}}{u _{2}} = \frac{1.192}{1.467} \approx 0.812$ . This is because the factor $2^{- t /5}$ shrinks by the same ratio $2^{- 1.5/5} = 2^{- 0.3}$ over every interval of length $T = 1.5$ .

(ii) In each interval the particle travels out and back, covering $2 u_{k}$ .

r = \frac{u _{2}}{u _{1}} = \frac{1.467}{1.806} \approx 0.812 Total = 2 S_{\infty} = \frac{2 ( 1.806 )}{1 - 0.8123} \approx 19.2 cm

Question 11

17 marks

A long jump density: constants, median against mode and a conditional probability

Calculus

(a)

4 marks

The total area under $f$ is 1.

\int_{3}^{9} [a (x - 3)^{3} + b (x - 3)^{2}] d x = [\frac{a ( x - 3 ) ^{4}}{4} + \frac{b ( x - 3 ) ^{3}}{3}]_{3}^{9} = \frac{6 ^{4} a}{4} + \frac{6 ^{3} b}{3} = 1 324 a + 72 b = 1

(b)

4 marks

(i) Use $f (9) = 0$ .

(ii) Substitute $b = - 6 a$ in part (a).

a (6)^{3} + b (6)^{2} = 216 a + 36 b = 0 \Rightarrow 6 a + b = 0 324 a - 432 a = 1 \Rightarrow a = - \frac{1}{108}, b = \frac{1}{18}

(c)

5 marks

The mode is the maximum of $f (x) = - \frac{( x - 3 ) ^{3}}{108} + \frac{( x - 3 ) ^{2}}{18}$ . Differentiate.
The mode is 7. Now find the area to the left of 7.
This is more than 0.5, so the median is less than 7. Solving $\int_{3}^{m} f (x) d x = 0.5$ gives $m \approx 6.69$ , which is below the mode of 7.

f^{'} (x) = - \frac{( x - 3 ) ^{2}}{36} + \frac{x - 3}{9} = 0 \Rightarrow x - 3 = 4 \Rightarrow x = 7 \int_{3}^{7} f (x) d x = [- \frac{( x - 3 ) ^{4}}{432} + \frac{( x - 3 ) ^{3}}{54}]_{3}^{7} = - \frac{256}{432} + \frac{64}{54} = \frac{16}{27} \approx 0.593

(d)

4 marks

Use conditional probability.

P (X > 8.52 ∣ X > 8) = \frac{P ( X > 8.52 )}{P ( X > 8 )} = \frac{\int _{8.52}^{9} f ( x ) d x}{\int _{8}^{9} f ( x ) d x} \frac{0.03443}{0.13194} \approx 0.261

Question 12

22 marks

A homogeneous differential equation: Euler's method, concavity and an exact solution

Calculus

(a)

6 marks

(i) Write $\frac{d y}{d x} = \frac{x ^{2} + x y - 3 y ^{2}}{x ^{2} + x y}$ and substitute $x = 1$ , $y = 1.5$ .

(ii) Use two steps of size $h = 0.2$ .

(iii) The gradient rises from $- 1.7$ to $- 0.4254$ , so the gradient is increasing and the solution curve is concave up.

\frac{d y}{d x} = \frac{1 + 1.5 - 6.75}{1 + 1.5} = - 1.7 y_{1} = 1.5 + 0.2 (- 1.7) = 1.16 \frac{d y}{d x}_{(1.2, 1.16)} = \frac{1.44 + 1.392 - 4.0368}{1.44 + 1.392} \approx - 0.4254 y_{2} = 1.16 + 0.2 (- 0.4254) \approx 1.07

(b)

6 marks

(i) Differentiate $(x^{2} + x y) \frac{d y}{d x} = x^{2} + x y - 3 y^{2}$ implicitly with respect to $x$ , using the product rule on the left.

Rearrange for the second derivative term.

(ii) Substitute $x = 1$ , $y = 1.5$ and $\frac{d y}{d x} = - 1.7$ .

(2 x + y + x \frac{d y}{d x}) \frac{d y}{d x} + (x^{2} + x y) \frac{d ^{2} y}{d x ^{2}} = 2 x + y + x \frac{d y}{d x} - 6 y \frac{d y}{d x} (x^{2} + x y) \frac{d ^{2} y}{d x ^{2}} = 2 x + y - x (\frac{d y}{d x})^{2} - (x + 7 y) \frac{d y}{d x} 2.5 \frac{d ^{2} y}{d x ^{2}} = 2 + 1.5 - 2.89 + 11.5 (1.7) = 20.16 \frac{d ^{2} y}{d x ^{2}} = 8.064 \approx 8.06

(c)

2 marks

Multiply through by the common denominator.
Put $v = \frac{1}{2}$ and $v = - \frac{1}{2}$ .

1 + v = A (1 + 2 v) + B (1 - 2 v) \frac{3}{2} = 2 A \Rightarrow A = \frac{3}{4}, \frac{1}{2} = 2 B \Rightarrow B = \frac{1}{4}

(d)

8 marks

Substitute $y = v x$ and $\frac{d y}{d x} = v + x \frac{d v}{d x}$ into $(x^{2} + x y) \frac{d y}{d x} = x^{2} + x y - 3 y^{2}$ , then divide by $x^{2}$ .
Expand and simplify.
Separate the variables.
Use part (c). Since $y > \frac{x}{2}$ we have $v > \frac{1}{2}$ , so $1 - 2 v < 0$ and the logarithm uses $2 v - 1$ .
Use $x = 1$ , $v = \frac{3}{2}$ .
Multiply by 8 and combine the logarithms.
Put $v = \frac{y}{x}$ , so $2 v + 1 = \frac{x + 2 y}{x}$ and $2 v - 1 = \frac{2 y - x}{x}$ .

(1 + v) (v + x \frac{d v}{d x}) = 1 + v - 3 v^{2} x (1 + v) \frac{d v}{d x} = 1 + v - 3 v^{2} - v - v^{2} = 1 - 4 v^{2} \int \frac{1 + v}{1 - 4 v ^{2}} d v = \int \frac{1}{x} d x - \frac{3}{8} ln (2 v - 1) + \frac{1}{8} ln (2 v + 1) = ln x + c - \frac{3}{8} ln 2 + \frac{1}{8} ln 4 = c \Rightarrow c = - \frac{1}{8} ln 2 ln \frac{2 v + 1}{( 2 v - 1 ) ^{3}} = ln x^{8} - ln 2 = ln \frac{x ^{8}}{2} 2 (2 v + 1) = x^{8} (2 v - 1)^{3} \frac{2 ( x + 2 y )}{x} = \frac{x ^{8} ( 2 y - x ) ^{3}}{x ^{3}} \Rightarrow x^{6} (2 y - x)^{3} = 2 (x + 2 y)

Our analysis

Where the marks are

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

Calculus56 marks · 51%
Q10, Q11, Q12
Number and algebra30 marks · 27%
Q5, Q6, Q7, Q8, Q9
Statistics and probability13 marks · 12%
Q2, Q4
Geometry and trigonometry6 marks · 5%
Q3
Functions5 marks · 5%
Q1

Mentor insights

Traps to watch for

01
Standard normal · Q4
$P (Z > z) = 0.25$ gives $z \approx 0.6745$ . $P (Z < z) = 0.02$ gives $z \approx - 2.0537$ .
02
Remainder theorem · Q6
The remainder on division by $(x + 1)$ is $P (- 1)$ .
03
Velocity · Q10
$v = \frac{d s}{d t}$ .
04
Density · Q11
The total area is 1 and the density is 0 at the upper limit.
05
Euler · Q12
$y_{n + 1} = y_{n} + h f (x_{n}, y_{n})$ .

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Worked solutions

IB Maths AA HL · Paper 2

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ3

Session: November 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 November 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 · 54 marks

Question 1

5 marks

A double angle identity and a trigonometric equation

Functions

(a)

2 marks

Use $cos 2 x = 1 - 2 sin^{2} x$ .

3 cos 2 x + 11 sin x = 3 (1 - 2 sin^{2} x) + 11 sin x = 3 + 11 sin x - 6 sin^{2} x

(b)

3 marks

Use part (a) in $3 cos 2 x + 11 sin x - 6 = 0$ .
Factorise (or use the quadratic formula).
Since $sin x \neq = \frac{3}{2}$ , we have $sin x = \frac{1}{3}$ .

3 + 11 sin x - 6 sin^{2} x - 6 = 0 \Rightarrow 6 sin^{2} x - 11 sin x + 3 = 0 (3 sin x - 1) (2 sin x - 3) = 0 x \approx 19. 5^{\circ} or x \approx 160. 5^{\circ}

Question 2

6 marks

A quiz median and finding two scores from the mean and range

Statistics and probability

(a)

1 mark

Aiden's score is below 10.5 and Brett's is above 10.5, so one new value is on each side of the median. The middle values stay the same, so the median is still 10.5.

(b)

5 marks

The total for the 28 pupils is $10.5 \times 28 = 294$ and the total for all 30 is $10.6 \times 30 = 318$ .
Aiden has the lowest score and Brett the highest, so the range is $B - A = 14$ .
Add the two equations.
Aiden scored 5 and Brett scored 19.

A + B = 318 - 294 = 24 2 B = 38 \Rightarrow B = 19, A = 24 - 19 = 5

Question 3

6 marks

A sector and a circle with related perimeter and area

Geometry and trigonometry

(a)

3 marks

The circumference of the circle is $2 π (8) = 16 π$ , so the perimeter of the sector is $1.5 \times 16 π = 24 π$ .

2 r + r θ = 24 π \Rightarrow r (2 + θ) = 24 π r = \frac{24 π}{2 + θ}

(b)

3 marks

Equate the areas and substitute for $r$ .
Simplify.
Solve this quadratic: $θ^{2} + (4 - 4.5 π) θ + 4 = 0$ . Its roots are about $0.411$ and $9.73$ . The angle is acute, so $θ < \frac{π}{2}$ .

\frac{1}{2} (\frac{24 π}{2 + θ})^{2} θ = 64 π 288 π^{2} θ = 64 π (2 + θ)^{2} \Rightarrow (2 + θ)^{2} = 4.5 π θ θ \approx 0.411

Question 4

7 marks

Normal heights of two sunflower varieties

Statistics and probability

(a)

3 marks

The height $180$ cm is exceeded by 25% of large sunflowers, so it is the upper quartile: $\frac{180 - m}{s} = z$ with $P (Z > z) = 0.25$ .

z = 0.6745 \Rightarrow m + 0.6745 s = 180 A \approx 0.674

(b)

4 marks

The giant sunflowers have mean $m + 35$ and standard deviation $2 s$ . Since 98% exceed 180 cm, 2% are below, so the standardised value of 180 is $- 2.0537$ .
Subtract this from the equation in part (a), $m + 0.6745 s = 180$ .

\frac{180 - ( m + 35 )}{2 s} = - 2.0537 \Rightarrow m - 4.1075 s = 145 (0.6745 + 4.1075) s = 180 - 145 = 35 \Rightarrow s \approx 7.32 m = 180 - 0.6745 (7.319) \approx 175

Question 5

5 marks

The volume of a solid of revolution about the y-axis

Number and algebra

5 marks

Rearrange $y = e^{3 x /4}$ to get $x$ in terms of $y$ .
The region runs from $y = 1$ (where $x = 0$ ) up to $y = 8$ . Rotate about the $y$ -axis.
Use $\int (ln y)^{2} d y = y (ln y)^{2} - 2 y ln y + 2 y$ .

ln y = \frac{3 x}{4} \Rightarrow x = \frac{4}{3} ln y V = π \int_{1}^{8} (\frac{4}{3} ln y)^{2} d y = \frac{16 π}{9} \int_{1}^{8} (ln y)^{2} d y V = \frac{16 π}{9} [y (ln y)^{2} - 2 y ln y + 2 y]_{1}^{8} = \frac{16 π}{9} (8 ln^{2} 8 - 16 ln 8 + 14) V \approx 85.6

Question 6

7 marks

A quadratic from the remainder theorem and the sum of squares of the roots

Number and algebra

7 marks

Use the remainder theorem.
For the roots, use $α^{2} + β^{2} = (α + β)^{2} - 2 α β$ .
Substitute $r = q + 10$ .
If $q = 11$ then $r = 21$ , and the discriminant is $121 - 168 < 0$ , so the roots would not be real. That is not allowed.
If $q = - 7$ then $r = 3$ , and the discriminant is $49 - 24 = 25 > 0$ , so the roots are real.

P (- 1) = 2 - q + r = 12 \Rightarrow r = q + 10 \frac{37}{4} = \frac{q ^{2}}{4} - r \Rightarrow q^{2} - 4 r = 37 q^{2} - 4 q - 40 = 37 \Rightarrow q^{2} - 4 q - 77 = 0 \Rightarrow (q - 11) (q + 7) = 0 q = - 7, r = 3

Question 7

6 marks

A system of equations: general solution, a unique solution and special cases

Number and algebra

(a)

2 marks

Let $z = t$ . Eliminate $y$ : the second equation gives $y = 3 x + 3 t - 12$ . Substitute in the first.
The general solution.

2 x + 6 (3 x + 3 t - 12) - 8 t = 13 \Rightarrow 20 x + 10 t = 85 \Rightarrow x = \frac{17}{4} - \frac{t}{2} y = \frac{3}{4} + \frac{3 t}{2} x = \frac{17}{4} - \frac{t}{2}, y = \frac{3}{4} + \frac{3 t}{2}, z = t

(b)

2 marks

Substitute into $3 x + 12 y - 16 z = - 3$ .
So $t = - \frac{99}{2}$ .

3 (\frac{17}{4} - \frac{t}{2}) + 12 (\frac{3}{4} + \frac{3 t}{2}) - 16 t = - 3 \frac{51}{4} + 9 - \frac{3 t}{2} + 18 t - 16 t = - 3 \Rightarrow \frac{t}{2} = - 3 - \frac{87}{4} = - \frac{99}{4} x = 29, y = - \frac{147}{2}, z = - \frac{99}{2}

(c)

2 marks

(i) The coefficients of $y$ and $z$ in the third equation are twice those in the first equation, so there is no unique solution when $a = 2 \times 2 = 4$ .

(ii) Twice the first equation is $4 x + 12 y - 16 z = 26$ . For infinitely many solutions the third equation must be this same equation.

a = 4 k = 26

Question 8

7 marks

Arranging and choosing recipe books

Number and algebra

(a)

1 mark

Arrange 15 different books in a line.

15! \approx 1.31 \times 1 0^{12}

(b)

3 marks

Arrange the six Asian books $6!$ ways, the five European books $5!$ ways and the four African books $4!$ ways. The three groups can be placed in any order, which is $3!$ ways.

6! \times 5! \times 4! \times 3! = 720 \times 120 \times 24 \times 6 = 12441600

(c)

3 marks

Choose 4 books from one continent. Add the three cases.

(4 6) + (4 5) + (4 4) = 15 + 5 + 1 = 21

Question 9

5 marks

Proof by contradiction of an inequality between a root and a mean

Number and algebra

5 marks

Assume the opposite: there are positive $x, y$ with $2 x y > \frac{1}{2} (2 x + y)$ .
Both sides are positive, so squaring keeps the direction of the inequality.
Rearrange.
A square cannot be negative, so this is a contradiction.
So the assumption is false, and $2 x y \leq \frac{1}{2} (2 x + y)$ for all $x, y > 0$ .

2 x y > \frac{1}{4} (2 x + y)^{2} \Rightarrow 8 x y > 4 x^{2} + 4 x y + y^{2} 0 > 4 x^{2} - 4 x y + y^{2} = (2 x - y)^{2}

Section B · Extended response

Questions 10–12 · 56 marks

Question 10

17 marks

A damped motion: maximum, minimum, total distance and a geometric series of peaks

Calculus

(a)

5 marks

(i) Find the first maximum of $s (t) = 2^{1 - t /5} sin \frac{2 π t}{3}$ on the calculator. It is at $t \approx 0.718$ .

(ii) The velocity is $\frac{d s}{d t}$ . The graph is steepest at $t = 0$ and the oscillations get smaller after that, so the largest speed is at $t = 0$ .

s_{ma x} \approx 1.81 cm v (0) = \frac{d s}{d t}_{t = 0} = 2 \cdot \frac{2 π}{3} cos 0 = \frac{4 π}{3} \approx 4.19 cm/s

(b)

3 marks

(i) The first minimum is at $t \approx 2.218$ .

(ii) Substitute $t = 3.5$ .

s_{min} \approx - 1.47 s (3.5) = 2^{1 - 0.7} sin \frac{7 π}{3} \approx 1.07

(c)

3 marks

In the first 3.5 seconds the particle goes out to $1.806$ and back to $O$ , then out to $1.467$ on the other side and back to $O$ , and then onwards to $1.066$ .

2 (1.806) + 2 (1.467) + 1.066 \approx 7.61 cm

(d)

1 mark

The sine factor is zero when $\frac{2 π t}{3} = π$ .

T = 1.5

(e)

5 marks

(i) The largest distances in each interval are $u_{1} \approx 1.806$ , $u_{2} \approx 1.467$ and $u_{3} \approx 1.192$ . The ratio.

(ii) In each interval the particle travels out and back, covering $2 u_{k}$ .

r = \frac{u _{2}}{u _{1}} = \frac{1.467}{1.806} \approx 0.812 Total = 2 S_{\infty} = \frac{2 ( 1.806 )}{1 - 0.8123} \approx 19.2 cm

Question 11

17 marks

A long jump density: constants, median against mode and a conditional probability

Calculus

(a)

4 marks

The total area under $f$ is 1.

\int_{3}^{9} [a (x - 3)^{3} + b (x - 3)^{2}] d x = [\frac{a ( x - 3 ) ^{4}}{4} + \frac{b ( x - 3 ) ^{3}}{3}]_{3}^{9} = \frac{6 ^{4} a}{4} + \frac{6 ^{3} b}{3} = 1 324 a + 72 b = 1

(b)

4 marks

(i) Use $f (9) = 0$ .

(ii) Substitute $b = - 6 a$ in part (a).

a (6)^{3} + b (6)^{2} = 216 a + 36 b = 0 \Rightarrow 6 a + b = 0 324 a - 432 a = 1 \Rightarrow a = - \frac{1}{108}, b = \frac{1}{18}

(c)

5 marks

The mode is the maximum of $f (x) = - \frac{( x - 3 ) ^{3}}{108} + \frac{( x - 3 ) ^{2}}{18}$ . Differentiate.
The mode is 7. Now find the area to the left of 7.
This is more than 0.5, so the median is less than 7. Solving $\int_{3}^{m} f (x) d x = 0.5$ gives $m \approx 6.69$ , which is below the mode of 7.

f^{'} (x) = - \frac{( x - 3 ) ^{2}}{36} + \frac{x - 3}{9} = 0 \Rightarrow x - 3 = 4 \Rightarrow x = 7 \int_{3}^{7} f (x) d x = [- \frac{( x - 3 ) ^{4}}{432} + \frac{( x - 3 ) ^{3}}{54}]_{3}^{7} = - \frac{256}{432} + \frac{64}{54} = \frac{16}{27} \approx 0.593

(d)

4 marks

Use conditional probability.

P (X > 8.52 ∣ X > 8) = \frac{P ( X > 8.52 )}{P ( X > 8 )} = \frac{\int _{8.52}^{9} f ( x ) d x}{\int _{8}^{9} f ( x ) d x} \frac{0.03443}{0.13194} \approx 0.261

Question 12

22 marks

A homogeneous differential equation: Euler's method, concavity and an exact solution

Calculus

(a)

6 marks

(i) Write $\frac{d y}{d x} = \frac{x ^{2} + x y - 3 y ^{2}}{x ^{2} + x y}$ and substitute $x = 1$ , $y = 1.5$ .

(ii) Use two steps of size $h = 0.2$ .

(iii) The gradient rises from $- 1.7$ to $- 0.4254$ , so the gradient is increasing and the solution curve is concave up.

\frac{d y}{d x} = \frac{1 + 1.5 - 6.75}{1 + 1.5} = - 1.7 y_{1} = 1.5 + 0.2 (- 1.7) = 1.16 \frac{d y}{d x}_{(1.2, 1.16)} = \frac{1.44 + 1.392 - 4.0368}{1.44 + 1.392} \approx - 0.4254 y_{2} = 1.16 + 0.2 (- 0.4254) \approx 1.07

(b)

6 marks

(i) Differentiate $(x^{2} + x y) \frac{d y}{d x} = x^{2} + x y - 3 y^{2}$ implicitly with respect to $x$ , using the product rule on the left.

Rearrange for the second derivative term.

(ii) Substitute $x = 1$ , $y = 1.5$ and $\frac{d y}{d x} = - 1.7$ .

(2 x + y + x \frac{d y}{d x}) \frac{d y}{d x} + (x^{2} + x y) \frac{d ^{2} y}{d x ^{2}} = 2 x + y + x \frac{d y}{d x} - 6 y \frac{d y}{d x} (x^{2} + x y) \frac{d ^{2} y}{d x ^{2}} = 2 x + y - x (\frac{d y}{d x})^{2} - (x + 7 y) \frac{d y}{d x} 2.5 \frac{d ^{2} y}{d x ^{2}} = 2 + 1.5 - 2.89 + 11.5 (1.7) = 20.16 \frac{d ^{2} y}{d x ^{2}} = 8.064 \approx 8.06

(c)

2 marks

Multiply through by the common denominator.
Put $v = \frac{1}{2}$ and $v = - \frac{1}{2}$ .

1 + v = A (1 + 2 v) + B (1 - 2 v) \frac{3}{2} = 2 A \Rightarrow A = \frac{3}{4}, \frac{1}{2} = 2 B \Rightarrow B = \frac{1}{4}

(d)

8 marks

Substitute $y = v x$ and $\frac{d y}{d x} = v + x \frac{d v}{d x}$ into $(x^{2} + x y) \frac{d y}{d x} = x^{2} + x y - 3 y^{2}$ , then divide by $x^{2}$ .
Expand and simplify.
Separate the variables.
Use part (c). Since $y > \frac{x}{2}$ we have $v > \frac{1}{2}$ , so $1 - 2 v < 0$ and the logarithm uses $2 v - 1$ .
Use $x = 1$ , $v = \frac{3}{2}$ .
Multiply by 8 and combine the logarithms.
Put $v = \frac{y}{x}$ , so $2 v + 1 = \frac{x + 2 y}{x}$ and $2 v - 1 = \frac{2 y - x}{x}$ .

(1 + v) (v + x \frac{d v}{d x}) = 1 + v - 3 v^{2} x (1 + v) \frac{d v}{d x} = 1 + v - 3 v^{2} - v - v^{2} = 1 - 4 v^{2} \int \frac{1 + v}{1 - 4 v ^{2}} d v = \int \frac{1}{x} d x - \frac{3}{8} ln (2 v - 1) + \frac{1}{8} ln (2 v + 1) = ln x + c - \frac{3}{8} ln 2 + \frac{1}{8} ln 4 = c \Rightarrow c = - \frac{1}{8} ln 2 ln \frac{2 v + 1}{( 2 v - 1 ) ^{3}} = ln x^{8} - ln 2 = ln \frac{x ^{8}}{2} 2 (2 v + 1) = x^{8} (2 v - 1)^{3} \frac{2 ( x + 2 y )}{x} = \frac{x ^{8} ( 2 y - x ) ^{3}}{x ^{3}} \Rightarrow x^{6} (2 y - x)^{3} = 2 (x + 2 y)

Our analysis

Where the marks are

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

Calculus56 marks · 51%
Q10, Q11, Q12
Number and algebra30 marks · 27%
Q5, Q6, Q7, Q8, Q9
Statistics and probability13 marks · 12%
Q2, Q4
Geometry and trigonometry6 marks · 5%
Q3
Functions5 marks · 5%
Q1

Mentor insights

Traps to watch for

01
Standard normal · Q4
$P (Z > z) = 0.25$ gives $z \approx 0.6745$ . $P (Z < z) = 0.02$ gives $z \approx - 2.0537$ .
02
Remainder theorem · Q6
The remainder on division by $(x + 1)$ is $P (- 1)$ .
03
Velocity · Q10
$v = \frac{d s}{d t}$ .
04
Density · Q11
The total area is 1 and the density is 0 at the upper limit.
05
Euler · Q12
$y_{n + 1} = y_{n} + h f (x_{n}, y_{n})$ .

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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