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IB Maths AA HL · Paper 2

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2TZ3

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 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

4 marks

Play time and sleep time: a regression line and an estimate

Statistics and probability

(a)

2 marks

Enter the six pairs.

a \approx 1.98, b \approx 40.2

(b)

2 marks

Substitute $x = 20$ into $y = 1.9765 x + 40.229$ .

y = 1.9765 (20) + 40.229 \approx 79.8 hours

Question 2

6 marks

A square-based pyramid in 3D: height, diagonal and edge angle

Geometry and trigonometry

(a)

2 marks

With $V (1, 7, 0)$ and $X (- 3, 4, 2)$ .

V X = (1 + 3)^{2} + (7 - 4)^{2} + (0 - 2)^{2} = 29 \approx 5.39

(b)

2 marks

The diagonal of a square with side 5.

A C = 5^{2} + 5^{2} = 52 \approx 7.07 cm

(c)

2 marks

$X$ is the centre of the base, so $X C = \frac{5 2}{2}$ and the triangle $V X C$ has a right angle at $X$ .

tan θ = \frac{29}{5 2 /2} = \frac{5.385}{3.536} \Rightarrow θ \approx 56. 7^{\circ}

Question 3

6 marks

Where a function decreases and is concave up from its derivative

Calculus

(a)

3 marks

Solve $f^{'} (x) = 4 + 2 x - 3 e^{x} = 0$ with the calculator. The two roots are $x \approx - 1.736$ and $x \approx 0.518$ .
A graph of $f^{'}$ shows it is below the axis outside these roots.

x < - 1.74 and x > 0.518

(b)

3 marks

Differentiate $f^{'} (x)$ .
Concave up needs $f^{''} (x) > 0$ .

f^{''} (x) = 2 - 3 e^{x} 2 - 3 e^{x} > 0 \Rightarrow e^{x} < \frac{2}{3} \Rightarrow x < ln \frac{2}{3} \approx - 0.405

Question 4

6 marks

A depreciating car and an investment whose real value is adjusted for inflation

Number and algebra

(a)

2 marks

A value that falls 15% each year is multiplied by 0.85 each year.

30000 (0.85)^{10} \approx 5906.23 euros

(b)

4 marks

The real rate per month is $1.5 - 0.8 = 0.7%$ . Solve for $n$ .
After 13 months the value is 54 746.09 and after 14 months it is 55 129.31.

50000 (1.007)^{n} > 55000 \Rightarrow n > 13.66 n = 14 months

Question 5

7 marks

A particle with velocity e^(-sin t) cos 2t: speed, distance and acceleration

Calculus

(a)

2 marks

Sketch $v (t) = e^{- s i n t} cos 2 t$ on $0 \leq t \leq 5$ . The most extreme value of $v$ is the minimum at $t \approx 4.71 (\frac{3 π}{2})$ , where $v = - e$ .

maximum speed = e \approx 2.72 m s^{- 1}

(b)

2 marks

The distance travelled is the area under the graph of $∣ v ∣$ .

d = \int_{0}^{5} ∣ v (t) ∣ d t \approx 3.85 m

(c)

3 marks

The velocity is zero when $cos 2 t = 0$ : $t = \frac{π}{4}, \frac{3 π}{4}, \dots$ , and the sign of $v$ changes at each one. The second change of direction is at $t = \frac{3 π}{4} \approx 2.356$ .
Differentiate $v$ with the product rule.
At $t = \frac{3 π}{4}$ we have $cos 2 t = 0$ and $sin 2 t = - 1$ .

a = v^{'} (t) = e^{- s i n t} (- cos t cos 2 t - 2 sin 2 t) a = e^{- s i n (3 π /4)} (0 + 2) = 2 e^{- 1/ 2} \approx 0.986 m s^{- 2}

Question 6

6 marks

A circle sector split into a segment and a triangle in the ratio 3 : 5

Number and algebra

6 marks

The segment $P$ has area 12.8, so the triangle $Q$ has area $\frac{5}{3} \times 12.8 = 21.33$ and the sector has area $12.8 + 21.33 = 34.13$ .
Divide the second equation by the first.
Solve for $θ$ with the calculator (in radians).
Substitute into the sector area.

\frac{1}{2} r^{2} θ = 34.13, \frac{1}{2} r^{2} sin θ = 21.33 \frac{sin θ}{θ} = \frac{21.33}{34.13} = 0.625 θ \approx 1.599 \frac{1}{2} r^{2} (1.599) = 34.13 \Rightarrow r^{2} \approx 42.7 r \approx 6.53 cm

Question 7

7 marks

A geometric sequence linked to an arithmetic sequence: the greatest sum

Number and algebra

(a)

3 marks

Find the common ratio.

80 r^{3} = 0.74088 \Rightarrow r^{3} = 0.009261 \Rightarrow r = 0.21 u_{2} = 80 (0.21) = 16.8

(b)

4 marks

The arithmetic sequence has $u_{1} = 80$ and $u_{11} = 16.8$ .
The sum is largest when the last term added is positive.

80 + 10 d = 16.8 \Rightarrow d = - 6.32 80 + (n - 1) (- 6.32) > 0 \Rightarrow n < 13.66 \Rightarrow n = 13 S_{13} = \frac{13}{2} [2 (80) + 12 (- 6.32)] = 547.04

Question 8

7 marks

Quiz marks with a given mean and variance

Number and algebra

7 marks

The total frequency is $12 + q + 8 = 20 + q$ . Write the mean equation.
The variance is $\frac{\sum f x ^{2}}{\sum f} - \overset{x}{ˉ}^{2} = 124$ , so $\frac{\sum f x ^{2}}{\sum f} = 3 1^{2} + 124 = 1085$ . Here $\sum f x^{2} = 12 (20)^{2} + 1225 q + 8 p^{2}$ and $\sum f = 20 + q$ .
Substitute $q = 95 - 2 p$ .
Divide by 8 and factorise.
A mark must be positive, so $p = 45$ , and then $q = 95 - 90 = 5$ .

\frac{12 ( 20 ) + 35 q + 8 p}{20 + q} = 31 \Rightarrow 240 + 35 q + 8 p = 620 + 31 q \Rightarrow q = 95 - 2 p 4800 + 1225 (95 - 2 p) + 8 p^{2} = 1085 (115 - 2 p) \Rightarrow 8 p^{2} - 280 p - 3600 = 0 p^{2} - 35 p - 450 = (p - 45) (p + 10) = 0 p = 45, q = 5

Question 9

8 marks

A plane containing a line and a point, and three planes meeting in a line

Number and algebra

(a)

4 marks

The line $L_{1}$ has a point $(0, 0, 2)$ and direction $(1, 0, 1)$ . The vector from $(0, 0, 2)$ to $(2, 1, 5)$ is $(2, 1, 3)$ , which is also in the plane.
The normal is the vector product of the two vectors in the plane.
So the plane is $- x - y + z = d$ , and the point $(0, 0, 2)$ gives $d = 2$ . Multiplying by $- 1$ :

101 \times 213 = 0 - 1 2 - 3 1 - 0 = - 1 - 1 1 x + y - z = - 2

(b)

4 marks

The coefficient determinant must be zero.
Now eliminate $x$ . Subtract $2 P_{1}$ from $P_{2}$ , and subtract $P_{1}$ from $P_{3}$ .
From the first, $z = 7 + \frac{2}{3} y$ . Substitute into the second.

121 1 b - 1 - 1 - 1 2 = (2 b - 1) - (4 + 1) + (- 1) (- 2 - b) = 3 b - 4 = 0 b = \frac{4}{3} (b - 2) y + z = 7 \Rightarrow - \frac{2}{3} y + z = 7, - 2 y + 3 z = d + 2 - 2 y + 21 + 2 y = d + 2 \Rightarrow d = 19 b = \frac{4}{3}, d = 19

Section B · Extended response

Questions 10–12 · 53 marks

Question 10

17 marks

Apple weights: normal probabilities, premium apples in boxes and a second orchard

Statistics and probability

(a)

2 marks

$W \sim N (175, 8^{2})$ .

P (W < 170) \approx 0.266

(b)

2 marks

Use the inverse normal with 20% in the upper tail.

w \approx 181.7 grams

(c)

2 marks

Premium apples weigh between 170 and 185 grams.

P (170 < W < 185) \approx 0.628 = 62.8%

(d)

3 marks

Let $X$ be the number of premium apples in a box, $X \sim B (40, 0.6284)$ .

P (X \geq 30) \approx 0.0739

(e)

2 marks

Let $Y$ be the number of boxes (out of 10) with at least 30 premium apples, $Y \sim B (10, 0.0739)$ .

P (Y = 4) \approx 0.00394

(f)

6 marks

Of all the apples, 82% are premium, so 18% are not. The percentage below 170 is twice the percentage above 185, so $P (M < 170) = 0.12$ and $P (M > 185) = 0.06$ .
The z-scores are $z = - 1.175$ and $z = 1.555$ .
Subtract the first from the second: $\frac{15}{s} = 2.730$ , so $s = 5.495$ . Then $m = 170 + 1.175 s$ .

\frac{170 - m}{s} = - 1.175, \frac{185 - m}{s} = 1.555 m \approx 176.5 \approx 176 grams

Question 11

14 marks

Teams of students and a spinning top: integration by parts, a limit and a median

Number and algebra

(a)

3 marks

Choose the first team of 5, then the second team of 5 from the remaining 10, and the last 5 form the third team. The three teams are not labelled, so divide by $3!$ .

\frac{( 5 15 ) ( 5 10 )}{3 !} = \frac{3003 \times 252}{6} = 126126

(b)

4 marks

Integrate by parts with $u = t$ and $\frac{d v}{d t} = e^{- 3 t}$ , so $v = - \frac{1}{3} e^{- 3 t}$ .

\int_{0}^{a} k t e^{- 3 t} d t = k [- \frac{t}{3} e^{- 3 t}]_{0}^{a} + \frac{k}{3} \int_{0}^{a} e^{- 3 t} d t k [- \frac{t}{3} e^{- 3 t} - \frac{1}{9} e^{- 3 t}]_{0}^{a} = k (- \frac{a}{3} e^{- 3 a} - \frac{1}{9} e^{- 3 a} + \frac{1}{9}) \frac{k}{9} (1 - (3 a + 1) e^{- 3 a})

(c)

5 marks

(i) The form is $\frac{\infty}{\infty}$ , so use l'Hopital's rule.

(ii) The total probability is 1, which is the limit of the integral as $a \to \infty$ .

x \to \infty lim \frac{3 x + 1}{e ^{3 x}} = x \to \infty lim \frac{3}{3 e ^{3 x}} = 0 a \to \infty lim \frac{k}{9} (1 - (3 a + 1) e^{- 3 a}) = \frac{k}{9} (1 - 0) = 1 k = 9

(d)

2 marks

The median $m$ satisfies $\int_{0}^{m} f (t) d t = 0.5$ .
Solve with the calculator.

1 - (3 m + 1) e^{- 3 m} = 0.5 m \approx 0.559 minutes (about 33.6 seconds)

Question 12

22 marks

An ellipse given implicitly: gradient, domain, maximum, tangents and a volume

Calculus

(a)

4 marks

Differentiate $4 x^{2} + y^{2} - 24 x + 4 y + 20 = 0$ .

8 x + 2 y \frac{d y}{d x} - 24 + 4 \frac{d y}{d x} = 0 \frac{d y}{d x} (2 y + 4) = 24 - 8 x \Rightarrow \frac{d y}{d x} = \frac{12 - 4 x}{y + 2} = \frac{4 ( 3 - x )}{y + 2}

(b)

4 marks

The edges of the domain are where the curve has a vertical tangent, so $\frac{d y}{d x}$ is undefined, at $y = - 2$ .
Put $y = - 2$ into the equation of the curve.

4 x^{2} + 4 - 24 x - 8 + 20 = 0 \Rightarrow x^{2} - 6 x + 4 = 0 \Rightarrow x = 3 \pm 5 3 - 5 \leq x \leq 3 + 5, a = 5

(c)

3 marks

At the maximum $\frac{d y}{d x} = 0$ , so $x = 3$ .
Substitute into the equation of the curve.
$A$ is the maximum point, so take the larger value.

36 + y^{2} - 72 + 4 y + 20 = 0 \Rightarrow y^{2} + 4 y - 16 = 0 \Rightarrow y = - 2 \pm 25 (x_{A}, y_{A}) = (3, - 2 + 25) \approx (3, 2.47)

(d)

4 marks

Substitute $y = m x$ into the equation of the curve.
A tangent has a zero discriminant.

4 x^{2} + m^{2} x^{2} - 24 x + 4 m x + 20 = 0 \Rightarrow (4 + m^{2}) x^{2} + (4 m - 24) x + 20 = 0 (4 m - 24)^{2} - 80 (4 + m^{2}) = 0 \Rightarrow - 64 m^{2} - 192 m + 256 = 0 \Rightarrow m^{2} + 3 m - 4 = 0 (m + 4) (m - 1) = 0 \Rightarrow m = - 4 or m = 1

(e)

3 marks

Substitute $y = - 4 x$ , so $x = - \frac{y}{4}$ , into the equation of the curve.

4 (\frac{y ^{2}}{16}) + y^{2} + 6 y + 4 y + 20 = 0 \Rightarrow \frac{5}{4} y^{2} + 10 y + 20 = 0 \Rightarrow (y + 4)^{2} = 0 y_{B} = - 4

(f)

4 marks

Solve the equation of the curve for $x$ . It is quadratic in $x$ : $4 x^{2} - 24 x + (y^{2} + 4 y + 20) = 0$ , so
The region between the curve and the $y$ -axis (the left branch) from $y = y_{B} = - 4$ to $y = y_{A}$ is rotated about the $y$ -axis.

4 (x - 3)^{2} = 16 - 4 y - y^{2} \Rightarrow x = 3 - \frac{1}{2} 16 - 4 y - y^{2} V = π \int_{- 4}^{- 2 + 25} (3 - \frac{1}{2} 16 - 4 y - y^{2})^{2} d y \approx 29.7

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 algebra48 marks · 44%
Q4, Q6, Q7, Q8, Q9, Q11
Calculus35 marks · 32%
Q3, Q5, Q12
Statistics and probability21 marks · 19%
Q1, Q10
Geometry and trigonometry6 marks · 5%
Q2

Mentor insights

Traps to watch for

01
Maximum speed · Q5
The speed is $∣ v ∣$ , so look at the largest value of $∣ v ∣$ , including a negative minimum of $v$ .
02
Normal to a plane · Q9
The normal is the vector product of two direction vectors in the plane.
03
Normal and binomial · Q10
Premium apples in a box of 40 follow a binomial distribution with the probability from the normal model.
04
Teams · Q11
The order of the teams does not matter, so divide by $3!$ .
05
Implicit differentiation · Q12
Differentiate every term with respect to $x$ and treat $y$ as a function of $x$ .

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

TZ1TZ2TZ3

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 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

4 marks

Play time and sleep time: a regression line and an estimate

Statistics and probability

(a)

2 marks

Enter the six pairs.

a \approx 1.98, b \approx 40.2

(b)

2 marks

Substitute $x = 20$ into $y = 1.9765 x + 40.229$ .

y = 1.9765 (20) + 40.229 \approx 79.8 hours

Question 2

6 marks

A square-based pyramid in 3D: height, diagonal and edge angle

Geometry and trigonometry

(a)

2 marks

With $V (1, 7, 0)$ and $X (- 3, 4, 2)$ .

V X = (1 + 3)^{2} + (7 - 4)^{2} + (0 - 2)^{2} = 29 \approx 5.39

(b)

2 marks

The diagonal of a square with side 5.

A C = 5^{2} + 5^{2} = 52 \approx 7.07 cm

(c)

2 marks

$X$ is the centre of the base, so $X C = \frac{5 2}{2}$ and the triangle $V X C$ has a right angle at $X$ .

tan θ = \frac{29}{5 2 /2} = \frac{5.385}{3.536} \Rightarrow θ \approx 56. 7^{\circ}

Question 3

6 marks

Where a function decreases and is concave up from its derivative

Calculus

(a)

3 marks

Solve $f^{'} (x) = 4 + 2 x - 3 e^{x} = 0$ with the calculator. The two roots are $x \approx - 1.736$ and $x \approx 0.518$ .
A graph of $f^{'}$ shows it is below the axis outside these roots.

x < - 1.74 and x > 0.518

(b)

3 marks

Differentiate $f^{'} (x)$ .
Concave up needs $f^{''} (x) > 0$ .

f^{''} (x) = 2 - 3 e^{x} 2 - 3 e^{x} > 0 \Rightarrow e^{x} < \frac{2}{3} \Rightarrow x < ln \frac{2}{3} \approx - 0.405

Question 4

6 marks

A depreciating car and an investment whose real value is adjusted for inflation

Number and algebra

(a)

2 marks

A value that falls 15% each year is multiplied by 0.85 each year.

30000 (0.85)^{10} \approx 5906.23 euros

(b)

4 marks

The real rate per month is $1.5 - 0.8 = 0.7%$ . Solve for $n$ .
After 13 months the value is 54 746.09 and after 14 months it is 55 129.31.

50000 (1.007)^{n} > 55000 \Rightarrow n > 13.66 n = 14 months

Question 5

7 marks

A particle with velocity e^(-sin t) cos 2t: speed, distance and acceleration

Calculus

(a)

2 marks

Sketch $v (t) = e^{- s i n t} cos 2 t$ on $0 \leq t \leq 5$ . The most extreme value of $v$ is the minimum at $t \approx 4.71 (\frac{3 π}{2})$ , where $v = - e$ .

maximum speed = e \approx 2.72 m s^{- 1}

(b)

2 marks

The distance travelled is the area under the graph of $∣ v ∣$ .

d = \int_{0}^{5} ∣ v (t) ∣ d t \approx 3.85 m

(c)

3 marks

The velocity is zero when $cos 2 t = 0$ : $t = \frac{π}{4}, \frac{3 π}{4}, \dots$ , and the sign of $v$ changes at each one. The second change of direction is at $t = \frac{3 π}{4} \approx 2.356$ .
Differentiate $v$ with the product rule.
At $t = \frac{3 π}{4}$ we have $cos 2 t = 0$ and $sin 2 t = - 1$ .

a = v^{'} (t) = e^{- s i n t} (- cos t cos 2 t - 2 sin 2 t) a = e^{- s i n (3 π /4)} (0 + 2) = 2 e^{- 1/ 2} \approx 0.986 m s^{- 2}

Question 6

6 marks

A circle sector split into a segment and a triangle in the ratio 3 : 5

Number and algebra

6 marks

The segment $P$ has area 12.8, so the triangle $Q$ has area $\frac{5}{3} \times 12.8 = 21.33$ and the sector has area $12.8 + 21.33 = 34.13$ .
Divide the second equation by the first.
Solve for $θ$ with the calculator (in radians).
Substitute into the sector area.

\frac{1}{2} r^{2} θ = 34.13, \frac{1}{2} r^{2} sin θ = 21.33 \frac{sin θ}{θ} = \frac{21.33}{34.13} = 0.625 θ \approx 1.599 \frac{1}{2} r^{2} (1.599) = 34.13 \Rightarrow r^{2} \approx 42.7 r \approx 6.53 cm

Question 7

7 marks

A geometric sequence linked to an arithmetic sequence: the greatest sum

Number and algebra

(a)

3 marks

Find the common ratio.

80 r^{3} = 0.74088 \Rightarrow r^{3} = 0.009261 \Rightarrow r = 0.21 u_{2} = 80 (0.21) = 16.8

(b)

4 marks

The arithmetic sequence has $u_{1} = 80$ and $u_{11} = 16.8$ .
The sum is largest when the last term added is positive.

80 + 10 d = 16.8 \Rightarrow d = - 6.32 80 + (n - 1) (- 6.32) > 0 \Rightarrow n < 13.66 \Rightarrow n = 13 S_{13} = \frac{13}{2} [2 (80) + 12 (- 6.32)] = 547.04

Question 8

7 marks

Quiz marks with a given mean and variance

Number and algebra

7 marks

The total frequency is $12 + q + 8 = 20 + q$ . Write the mean equation.
The variance is $\frac{\sum f x ^{2}}{\sum f} - \overset{x}{ˉ}^{2} = 124$ , so $\frac{\sum f x ^{2}}{\sum f} = 3 1^{2} + 124 = 1085$ . Here $\sum f x^{2} = 12 (20)^{2} + 1225 q + 8 p^{2}$ and $\sum f = 20 + q$ .
Substitute $q = 95 - 2 p$ .
Divide by 8 and factorise.
A mark must be positive, so $p = 45$ , and then $q = 95 - 90 = 5$ .

\frac{12 ( 20 ) + 35 q + 8 p}{20 + q} = 31 \Rightarrow 240 + 35 q + 8 p = 620 + 31 q \Rightarrow q = 95 - 2 p 4800 + 1225 (95 - 2 p) + 8 p^{2} = 1085 (115 - 2 p) \Rightarrow 8 p^{2} - 280 p - 3600 = 0 p^{2} - 35 p - 450 = (p - 45) (p + 10) = 0 p = 45, q = 5

Question 9

8 marks

A plane containing a line and a point, and three planes meeting in a line

Number and algebra

(a)

4 marks

The line $L_{1}$ has a point $(0, 0, 2)$ and direction $(1, 0, 1)$ . The vector from $(0, 0, 2)$ to $(2, 1, 5)$ is $(2, 1, 3)$ , which is also in the plane.
The normal is the vector product of the two vectors in the plane.
So the plane is $- x - y + z = d$ , and the point $(0, 0, 2)$ gives $d = 2$ . Multiplying by $- 1$ :

101 \times 213 = 0 - 1 2 - 3 1 - 0 = - 1 - 1 1 x + y - z = - 2

(b)

4 marks

The coefficient determinant must be zero.
Now eliminate $x$ . Subtract $2 P_{1}$ from $P_{2}$ , and subtract $P_{1}$ from $P_{3}$ .
From the first, $z = 7 + \frac{2}{3} y$ . Substitute into the second.

121 1 b - 1 - 1 - 1 2 = (2 b - 1) - (4 + 1) + (- 1) (- 2 - b) = 3 b - 4 = 0 b = \frac{4}{3} (b - 2) y + z = 7 \Rightarrow - \frac{2}{3} y + z = 7, - 2 y + 3 z = d + 2 - 2 y + 21 + 2 y = d + 2 \Rightarrow d = 19 b = \frac{4}{3}, d = 19

Section B · Extended response

Questions 10–12 · 53 marks

Question 10

17 marks

Apple weights: normal probabilities, premium apples in boxes and a second orchard

Statistics and probability

(a)

2 marks

$W \sim N (175, 8^{2})$ .

P (W < 170) \approx 0.266

(b)

2 marks

Use the inverse normal with 20% in the upper tail.

w \approx 181.7 grams

(c)

2 marks

Premium apples weigh between 170 and 185 grams.

P (170 < W < 185) \approx 0.628 = 62.8%

(d)

3 marks

Let $X$ be the number of premium apples in a box, $X \sim B (40, 0.6284)$ .

P (X \geq 30) \approx 0.0739

(e)

2 marks

Let $Y$ be the number of boxes (out of 10) with at least 30 premium apples, $Y \sim B (10, 0.0739)$ .

P (Y = 4) \approx 0.00394

(f)

6 marks

Of all the apples, 82% are premium, so 18% are not. The percentage below 170 is twice the percentage above 185, so $P (M < 170) = 0.12$ and $P (M > 185) = 0.06$ .
The z-scores are $z = - 1.175$ and $z = 1.555$ .
Subtract the first from the second: $\frac{15}{s} = 2.730$ , so $s = 5.495$ . Then $m = 170 + 1.175 s$ .

\frac{170 - m}{s} = - 1.175, \frac{185 - m}{s} = 1.555 m \approx 176.5 \approx 176 grams

Question 11

14 marks

Teams of students and a spinning top: integration by parts, a limit and a median

Number and algebra

(a)

3 marks

Choose the first team of 5, then the second team of 5 from the remaining 10, and the last 5 form the third team. The three teams are not labelled, so divide by $3!$ .

\frac{( 5 15 ) ( 5 10 )}{3 !} = \frac{3003 \times 252}{6} = 126126

(b)

4 marks

Integrate by parts with $u = t$ and $\frac{d v}{d t} = e^{- 3 t}$ , so $v = - \frac{1}{3} e^{- 3 t}$ .

\int_{0}^{a} k t e^{- 3 t} d t = k [- \frac{t}{3} e^{- 3 t}]_{0}^{a} + \frac{k}{3} \int_{0}^{a} e^{- 3 t} d t k [- \frac{t}{3} e^{- 3 t} - \frac{1}{9} e^{- 3 t}]_{0}^{a} = k (- \frac{a}{3} e^{- 3 a} - \frac{1}{9} e^{- 3 a} + \frac{1}{9}) \frac{k}{9} (1 - (3 a + 1) e^{- 3 a})

(c)

5 marks

(i) The form is $\frac{\infty}{\infty}$ , so use l'Hopital's rule.

(ii) The total probability is 1, which is the limit of the integral as $a \to \infty$ .

x \to \infty lim \frac{3 x + 1}{e ^{3 x}} = x \to \infty lim \frac{3}{3 e ^{3 x}} = 0 a \to \infty lim \frac{k}{9} (1 - (3 a + 1) e^{- 3 a}) = \frac{k}{9} (1 - 0) = 1 k = 9

(d)

2 marks

The median $m$ satisfies $\int_{0}^{m} f (t) d t = 0.5$ .
Solve with the calculator.

1 - (3 m + 1) e^{- 3 m} = 0.5 m \approx 0.559 minutes (about 33.6 seconds)

Question 12

22 marks

An ellipse given implicitly: gradient, domain, maximum, tangents and a volume

Calculus

(a)

4 marks

Differentiate $4 x^{2} + y^{2} - 24 x + 4 y + 20 = 0$ .

8 x + 2 y \frac{d y}{d x} - 24 + 4 \frac{d y}{d x} = 0 \frac{d y}{d x} (2 y + 4) = 24 - 8 x \Rightarrow \frac{d y}{d x} = \frac{12 - 4 x}{y + 2} = \frac{4 ( 3 - x )}{y + 2}

(b)

4 marks

The edges of the domain are where the curve has a vertical tangent, so $\frac{d y}{d x}$ is undefined, at $y = - 2$ .
Put $y = - 2$ into the equation of the curve.

4 x^{2} + 4 - 24 x - 8 + 20 = 0 \Rightarrow x^{2} - 6 x + 4 = 0 \Rightarrow x = 3 \pm 5 3 - 5 \leq x \leq 3 + 5, a = 5

(c)

3 marks

At the maximum $\frac{d y}{d x} = 0$ , so $x = 3$ .
Substitute into the equation of the curve.
$A$ is the maximum point, so take the larger value.

36 + y^{2} - 72 + 4 y + 20 = 0 \Rightarrow y^{2} + 4 y - 16 = 0 \Rightarrow y = - 2 \pm 25 (x_{A}, y_{A}) = (3, - 2 + 25) \approx (3, 2.47)

(d)

4 marks

Substitute $y = m x$ into the equation of the curve.
A tangent has a zero discriminant.

4 x^{2} + m^{2} x^{2} - 24 x + 4 m x + 20 = 0 \Rightarrow (4 + m^{2}) x^{2} + (4 m - 24) x + 20 = 0 (4 m - 24)^{2} - 80 (4 + m^{2}) = 0 \Rightarrow - 64 m^{2} - 192 m + 256 = 0 \Rightarrow m^{2} + 3 m - 4 = 0 (m + 4) (m - 1) = 0 \Rightarrow m = - 4 or m = 1

(e)

3 marks

Substitute $y = - 4 x$ , so $x = - \frac{y}{4}$ , into the equation of the curve.

4 (\frac{y ^{2}}{16}) + y^{2} + 6 y + 4 y + 20 = 0 \Rightarrow \frac{5}{4} y^{2} + 10 y + 20 = 0 \Rightarrow (y + 4)^{2} = 0 y_{B} = - 4

(f)

4 marks

Solve the equation of the curve for $x$ . It is quadratic in $x$ : $4 x^{2} - 24 x + (y^{2} + 4 y + 20) = 0$ , so
The region between the curve and the $y$ -axis (the left branch) from $y = y_{B} = - 4$ to $y = y_{A}$ is rotated about the $y$ -axis.

4 (x - 3)^{2} = 16 - 4 y - y^{2} \Rightarrow x = 3 - \frac{1}{2} 16 - 4 y - y^{2} V = π \int_{- 4}^{- 2 + 25} (3 - \frac{1}{2} 16 - 4 y - y^{2})^{2} d y \approx 29.7

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 algebra48 marks · 44%
Q4, Q6, Q7, Q8, Q9, Q11
Calculus35 marks · 32%
Q3, Q5, Q12
Statistics and probability21 marks · 19%
Q1, Q10
Geometry and trigonometry6 marks · 5%
Q2

Mentor insights

Traps to watch for

01
Maximum speed · Q5
The speed is $∣ v ∣$ , so look at the largest value of $∣ v ∣$ , including a negative minimum of $v$ .
02
Normal to a plane · Q9
The normal is the vector product of two direction vectors in the plane.
03
Normal and binomial · Q10
Premium apples in a box of 40 follow a binomial distribution with the probability from the normal model.
04
Teams · Q11
The order of the teams does not matter, so divide by $3!$ .
05
Implicit differentiation · Q12
Differentiate every term with respect to $x$ and treat $y$ as a function of $x$ .

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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