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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 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Section A · Short response

Questions 1–9 · 54 marks

Question 1

6 marks

A regular pentagon inside a circle: arc, radius and the area outside the pentagon

Geometry and trigonometry

(a)

3 marks

(i) The five equal angles at the centre make a full turn.

(ii) Use arc length.

θ = \frac{2 π}{5} \approx 1.26 12 = r θ \Rightarrow r = \frac{12}{2 π /5} = \frac{30}{π} \approx 9.55 cm

(b)

3 marks

The shaded region is the five equal segments between the pentagon and the circle.
Multiply by 5.

Segment = \frac{1}{2} (9.549)^{2} (1.2566 - sin 1.2566) = 57.30 - 43.36 \approx 13.93 5 \times 13.93 \approx 69.7 cm^{2}

Question 2

4 marks

Choosing a committee with three named roles

Number and algebra

4 marks

Choose the chairperson, vice-chairperson and treasurer in order from 10 people.
Choose the four additional members from the remaining 7.
Multiply.

^{10} P_{3} = 10 \times 9 \times 8 = 720 (4 7) = 35 720 \times 35 = 25200

Question 3

7 marks

A regression line through the mean point and a missing data value

Statistics and probability

(a)

2 marks

Add the $x$ -values and divide by 8.

\overset{x}{ˉ} = \frac{3.8 + 6.3 + 2.5 + 2.4 + 8.9 + 10.7 + 6.7 + 10.3}{8} = \frac{51.6}{8} = 6.45

(b)

4 marks

(i) Substitute $\overset{x}{ˉ}$ into the regression line.

(ii) The eight $y$ -values add up to $8 \overset{y}{ˉ}$ .

\overset{y}{ˉ} = 0.014375 + 2.1625 \times 6.45 = 13.9625 \approx 14.0 11.8 + 9.7 + 6.1 + 4.1 + p + 23.3 + 13.7 + 24.5 = 93.2 + p = 8 \times 13.9625 = 111.7 p = 18.5

(c)

1 mark

Enter the eight pairs on the calculator, with $p = 18.5$ .

r \approx 0.954

Question 4

7 marks

A cosine graph from its extreme points, its period and a sine form

Functions

(a)

3 marks

(i) The maximum is 2 and the minimum is $- 8$ . The graph starts at a minimum, so $a$ is negative.

(ii) The minimum is at $x = 0$ and the maximum at $x = 16$ , which is half a period, so the period is 32.

∣ a ∣ = \frac{2 - ( - 8 )}{2} = 5 \Rightarrow a = - 5 b = \frac{2 π}{32} = \frac{π}{16} \approx 0.196

(b)

1 mark

The graph repeats after one period.

k = 32

(c)

3 marks

(i) The sine form must start its rising midline crossing where the cosine graph does. The graph of $g$ rises through the midline $y = - 3$ at $x = 8$ , so the sine curve is shifted 8 to the right.

(ii) The sine graph has amplitude 5, and a positive $p$ gives a rising start from the midline.

q = 8 p = 5, g (x) = 5 sin (\frac{π}{16} (x - 8)) - 3

Question 5

5 marks

Grouped data: mean, standard deviation and an interquartile range from a normal model

Statistics and probability

(a)

2 marks

Use the midpoints 17.5, 22.5, 27.5, 32.5 and 37.5 with the frequencies 31, 42, 61, 46 and 29 (209 values in all).
(i) The mean.
(ii) The standard deviation, from the calculator.

\overset{y}{ˉ} = 27.5 σ \approx 6.26

(b)

3 marks

The quartiles of $N (27.5, 6.263 7^{2})$ .
Check using $IQR = 2 \times 0.6745 σ$ .

Q_{3} \approx 31.72, Q_{1} \approx 23.28 IQR \approx 31.72 - 23.28 \approx 8.4 2 \times 0.6745 \times 6.2637 \approx 8.4

Question 6

5 marks

A repeating decimal as an infinite geometric series

Number and algebra

(a)

3 marks

The first term is $\frac{53}{1000}$ (when $k = 0$ ) and the common ratio is $\frac{1}{100}$ .

S_{\infty} = \frac{0.053}{1 - 0.01} = \frac{0.053}{0.99} = \frac{53}{990}

(b)

2 marks

Split $0.2 \overline{53}$ as $0.2 + 0.0535353 \dots$ and notice the second part is the series from part (a).
The numbers 251 (prime) and 990 have no common factor, so this is in lowest terms.

0.2 + \frac{53}{990} = \frac{198}{990} + \frac{53}{990} = \frac{251}{990}

Question 7

7 marks

A defective semiconductor from one of two machines: a binomial and Bayes

Statistics and probability

(a)

4 marks

Given the machine, the number of defective items is binomial with $n = 10$ . For machine A, $p = 0.1$ ; for machine B, $p = 0.05$ .
Each machine is chosen with probability $\frac{1}{2}$ .

P (2 ∣ A) = (2 10) (0.1)^{2} (0.9)^{8} \approx 0.19371, P (2 ∣ B) = (2 10) (0.05)^{2} (0.95)^{8} \approx 0.07463 P (2) = 0.5 (0.19371) + 0.5 (0.07463) \approx 0.134

(b)

3 marks

Use the conditional probability formula.

P (A ∣ 2) = \frac{0.5 \times 0.19371}{0.13417} \approx 0.722

Question 8

6 marks

Volume of revolution of a region between two curves about the y-axis

Number and algebra

6 marks

Write both curves with $x^{2}$ as a function of $y$ .
Rotating about the $y$ -axis, the region for $0 \leq y \leq 3$ reaches out to the curve $f$ . For $2 \leq y \leq 3$ there is also a hollow part inside the curve $g$ , so subtract it.
Evaluate the first integral.
Evaluate the second.

y = x^{2} - 16 \Rightarrow x^{2} = y^{2} + 16, y = 2 + \frac{x ^{2}}{25} \Rightarrow x^{2} = 25 y - 50 V = π \int_{0}^{3} (y^{2} + 16) d y - π \int_{2}^{3} (25 y - 50) d y π [\frac{y ^{3}}{3} + 16 y]_{0}^{3} = π (9 + 48) = 57 π π [\frac{25 y ^{2}}{2} - 50 y]_{2}^{3} = π [(112.5 - 150) - (50 - 100)] = \frac{25 π}{2} V = 57 π - \frac{25 π}{2} = \frac{89 π}{2} \approx 140 cubic units

Question 9

7 marks

Related rates in a triangle with a shrinking angle

Number and algebra

7 marks

The area is $\frac{1}{2} (15) (20) sin θ = 140$ .
Since $θ$ is acute, $cos θ = 1 - \frac{196}{225} = \frac{29}{15}$ and $θ \approx 1.2036$ .
Differentiate $x^{2} = 625 - 600 cos θ$ with respect to $t$ .
Find $x$ .
Substitute $\frac{d θ}{d t} = - \frac{π}{60}$ .
So $A C$ is decreasing at about $0.724$ cm per minute.

150 sin θ = 140 \Rightarrow sin θ = \frac{14}{15} 2 x \frac{d x}{d t} = 600 sin θ \frac{d θ}{d t} \Rightarrow \frac{d x}{d t} = \frac{300 sin θ}{x} \frac{d θ}{d t} x = 625 - 4029 \approx 20.24 \frac{d x}{d t} = \frac{300 \times \frac{14}{15}}{20.24} \times (- \frac{π}{60}) \approx - 0.724

Section B · Extended response

Questions 10–12 · 56 marks

Question 10

17 marks

An airplane and a car: velocities, displacement and the gap between them

Calculus

(a)

2 marks

Put $t = 0$ .
(i) The airplane: $v = 60 e^{0} = 60$ m/s.
(ii) The car: $v = 5 (0) = 0$ m/s.

(b)

3 marks

(i) Equal speeds.

(ii) Substitute.

60 e^{- 0.1 t} = 5 t \Rightarrow t \approx 6.36 s v = 5 (6.3556) \approx 31.8 m/s

(c)

7 marks

Integrate each velocity to get a displacement.
The gap is the plane's position minus the car's.
Use $d (0) = 100$ .

s_{ai r} (t) = \int 60 e^{- 0.1 t} d t = - 600 e^{- 0.1 t} + C, s_{c a r} (t) = \int 5 t d t = \frac{5}{2} t^{2} d (t) = - 600 e^{- 0.1 t} - \frac{5}{2} t^{2} + C 100 = - 600 + C \Rightarrow C = 700 d (t) = 700 - 600 e^{- 0.1 t} - \frac{5}{2} t^{2}

(d)

2 marks

The car reaches the back of the airplane when $d (t) = 0$ .

700 - 600 e^{- 0.1 t} - 2.5 t^{2} = 0 \Rightarrow t \approx 15.1 s

(e)

3 marks

The distance the car has travelled is its displacement.

\frac{5}{2} (15.0586)^{2} = 2.5 \times 226.76 \approx 567 m

Question 11

18 marks

A density with arccos: integration by parts, finding k, mean, variance and a probability

Calculus

(a)

4 marks

Let $u = arccos x$ and $d v = d x$ , so $d u = - \frac{d x}{1 - x ^{2}}$ and $v = x$ .
The last integral is $- 1 - x^{2}$ , which can be found by inspection or by the substitution $w = 1 - x^{2}$ .

\int arccos x d x = x arccos x + \int \frac{x}{1 - x ^{2}} d x \int arccos x d x = x arccos x - 1 - x^{2} + C

(b)

6 marks

(i) The total probability is 1. Use $u = x^{2}$ with $d u = 2 x d x$ and then part (a).

Rearrange.

(ii) Solve on the calculator for $0 < k < 1$ .

\int_{0}^{k} 3 x arccos (x^{2}) d x = \frac{3}{2} \int_{0}^{k^{2}} arccos u d u = \frac{3}{2} [u arccos u - 1 - u^{2}]_{0}^{k^{2}} \frac{3}{2} [k^{2} arccos (k^{2}) - 1 - k^{4} - (0 - 1)] = 1 k^{2} arccos (k^{2}) - 1 - k^{4} + 1 = \frac{2}{3} \Rightarrow k^{2} arccos (k^{2}) - 1 - k^{4} + \frac{1}{3} = 0 k \approx 0.713250

(c)

5 marks

(i) The mean.

(ii) The mean of the squares.

The variance.

E (X) = \int_{0}^{0.71325} x \cdot 3 x arccos (x^{2}) d x \approx 0.456 E (X^{2}) = \int_{0}^{0.71325} x^{2} \cdot 3 x arccos (x^{2}) d x \approx 0.2372 Var (X) = 0.2372 - 0.456 3^{2} \approx 0.0290

(d)

3 marks

The standard deviation.
Integrate the density over $(μ - σ, μ + σ) = (0.28607, 0.62655)$ .

σ = 0.028980 \approx 0.17024 P (μ - σ < X < μ + σ) = \int_{0.28607}^{0.62655} 3 x arccos (x^{2}) d x \approx 0.620

Question 12

21 marks

Two lines, the plane containing them, a triangle and a pyramid

Number and algebra

(a)

3 marks

Write $L_{2}$ in parametric form using the parameter $t$ : $(x, y, z) = (- 1 + 2 t, 7 + t, - 5 + 3 t)$ . Equate with $L_{1}$ , which is $(5 + s, 4 - s, 2 + s)$ .
From the second equation $s = - 3 - t$ . Substitute in the first.
Check the third: $2 - 4 = - 2$ and $- 5 + 3 = - 2$ . They agree. Substitute $s = - 4$ in $L_{1}$ .

5 + s = - 1 + 2 t, 4 - s = 7 + t, 2 + s = - 5 + 3 t 5 - 3 - t = - 1 + 2 t \Rightarrow t = 1, s = - 4 5 - 4 4 + 4 2 - 4 = 18 - 2

(b)

3 marks

(i) The plane contains the intersection point and the two direction vectors.

(ii) The normal is $b \times c = - 4 - 1 3$ , so the plane is $- 4 x - y + 3 z = d$ . Substitute the point $(1, 8, - 2)$ .

r = 18 - 2 + λ 1 - 1 1 + μ 213 - 4 - 8 - 6 = - 18 \Rightarrow - 4 x - y + 3 z = - 18 \Rightarrow 4 x + y - 3 z = 18

(c)

2 marks

Put the other two variables to zero in $4 x + y - 3 z = 18$ .
(i) On the $y$ -axis: $y = 18$ , so $q = 18$ .
(ii) On the $z$ -axis: $- 3 z = 18$ , so $r = - 6$ .

(d)

5 marks

The vertices are $P (4.5, 0, 0)$ , $Q (0, 18, 0)$ and $R (0, 0, - 6)$ .
Take the cross product.
The area is half its magnitude.

P Q = - 4.5 180, P R = - 4.5 0 - 6 P Q \times P R = - 108 - 27 81 Area = \frac{1}{2} 10 8^{2} + 2 7^{2} + 8 1^{2} = \frac{1}{2} 18954 \approx 68.8

(e)

2 marks

$L_{3}$ passes through $(1, 8, - 2)$ and has the normal $(4, 1, - 3)$ as its direction.

r_{3} = 18 - 2 + γ 41 - 3

(f)

2 marks

Equate with $S (- 11, 5, 7)$ .
Check: $8 + (- 3) = 5$ and $- 2 - 3 (- 3) = 7$ .

1 + 4 γ = - 11 \Rightarrow γ = - 3

(g)

4 marks

The point $(1, 8, - 2)$ is in the plane, and $L_{3}$ is perpendicular to the plane there. So the height of the pyramid is the distance from it to $S$ .
Use the volume formula with the base area from part (d).

N S = - 12 - 3 9, h = 144 + 9 + 81 = 234 = 326 V = \frac{1}{3} \times \frac{1}{2} 18954 \times 234 = \frac{1}{3} \times \frac{1}{2} \times 2726 \times 326 = 351

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 algebra43 marks · 39%
Q2, Q6, Q8, Q9, Q12
Calculus35 marks · 32%
Q10, Q11
Statistics and probability19 marks · 17%
Q3, Q5, Q7
Functions7 marks · 6%
Q4
Geometry and trigonometry6 marks · 5%
Q1

Mentor insights

Traps to watch for

01
Mean point · Q3
The regression line of $y$ on $x$ always passes through $(\overset{x}{ˉ}, \overset{y}{ˉ})$ .
02
Amplitude · Q4
Half the difference between the maximum and the minimum.
03
Displacement · Q10
Integrate each velocity. The plane starts 100 m ahead of the car.
04
Integration by parts · Q11
Take $u = arccos x$ and $d v = d x$ .
05
Intersection · Q12
Equate the three components and check that all three agree.

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

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 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Section A · Short response

Questions 1–9 · 54 marks

Question 1

6 marks

A regular pentagon inside a circle: arc, radius and the area outside the pentagon

Geometry and trigonometry

(a)

3 marks

(i) The five equal angles at the centre make a full turn.

(ii) Use arc length.

θ = \frac{2 π}{5} \approx 1.26 12 = r θ \Rightarrow r = \frac{12}{2 π /5} = \frac{30}{π} \approx 9.55 cm

(b)

3 marks

The shaded region is the five equal segments between the pentagon and the circle.
Multiply by 5.

Segment = \frac{1}{2} (9.549)^{2} (1.2566 - sin 1.2566) = 57.30 - 43.36 \approx 13.93 5 \times 13.93 \approx 69.7 cm^{2}

Question 2

4 marks

Choosing a committee with three named roles

Number and algebra

4 marks

Choose the chairperson, vice-chairperson and treasurer in order from 10 people.
Choose the four additional members from the remaining 7.
Multiply.

^{10} P_{3} = 10 \times 9 \times 8 = 720 (4 7) = 35 720 \times 35 = 25200

Question 3

7 marks

A regression line through the mean point and a missing data value

Statistics and probability

(a)

2 marks

Add the $x$ -values and divide by 8.

\overset{x}{ˉ} = \frac{3.8 + 6.3 + 2.5 + 2.4 + 8.9 + 10.7 + 6.7 + 10.3}{8} = \frac{51.6}{8} = 6.45

(b)

4 marks

(i) Substitute $\overset{x}{ˉ}$ into the regression line.

(ii) The eight $y$ -values add up to $8 \overset{y}{ˉ}$ .

\overset{y}{ˉ} = 0.014375 + 2.1625 \times 6.45 = 13.9625 \approx 14.0 11.8 + 9.7 + 6.1 + 4.1 + p + 23.3 + 13.7 + 24.5 = 93.2 + p = 8 \times 13.9625 = 111.7 p = 18.5

(c)

1 mark

Enter the eight pairs on the calculator, with $p = 18.5$ .

r \approx 0.954

Question 4

7 marks

A cosine graph from its extreme points, its period and a sine form

Functions

(a)

3 marks

(i) The maximum is 2 and the minimum is $- 8$ . The graph starts at a minimum, so $a$ is negative.

(ii) The minimum is at $x = 0$ and the maximum at $x = 16$ , which is half a period, so the period is 32.

∣ a ∣ = \frac{2 - ( - 8 )}{2} = 5 \Rightarrow a = - 5 b = \frac{2 π}{32} = \frac{π}{16} \approx 0.196

(b)

1 mark

The graph repeats after one period.

k = 32

(c)

3 marks

(i) The sine form must start its rising midline crossing where the cosine graph does. The graph of $g$ rises through the midline $y = - 3$ at $x = 8$ , so the sine curve is shifted 8 to the right.

(ii) The sine graph has amplitude 5, and a positive $p$ gives a rising start from the midline.

q = 8 p = 5, g (x) = 5 sin (\frac{π}{16} (x - 8)) - 3

Question 5

5 marks

Grouped data: mean, standard deviation and an interquartile range from a normal model

Statistics and probability

(a)

2 marks

Use the midpoints 17.5, 22.5, 27.5, 32.5 and 37.5 with the frequencies 31, 42, 61, 46 and 29 (209 values in all).
(i) The mean.
(ii) The standard deviation, from the calculator.

\overset{y}{ˉ} = 27.5 σ \approx 6.26

(b)

3 marks

The quartiles of $N (27.5, 6.263 7^{2})$ .
Check using $IQR = 2 \times 0.6745 σ$ .

Q_{3} \approx 31.72, Q_{1} \approx 23.28 IQR \approx 31.72 - 23.28 \approx 8.4 2 \times 0.6745 \times 6.2637 \approx 8.4

Question 6

5 marks

A repeating decimal as an infinite geometric series

Number and algebra

(a)

3 marks

The first term is $\frac{53}{1000}$ (when $k = 0$ ) and the common ratio is $\frac{1}{100}$ .

S_{\infty} = \frac{0.053}{1 - 0.01} = \frac{0.053}{0.99} = \frac{53}{990}

(b)

2 marks

Split $0.2 \overline{53}$ as $0.2 + 0.0535353 \dots$ and notice the second part is the series from part (a).
The numbers 251 (prime) and 990 have no common factor, so this is in lowest terms.

0.2 + \frac{53}{990} = \frac{198}{990} + \frac{53}{990} = \frac{251}{990}

Question 7

7 marks

A defective semiconductor from one of two machines: a binomial and Bayes

Statistics and probability

(a)

4 marks

Given the machine, the number of defective items is binomial with $n = 10$ . For machine A, $p = 0.1$ ; for machine B, $p = 0.05$ .
Each machine is chosen with probability $\frac{1}{2}$ .

P (2 ∣ A) = (2 10) (0.1)^{2} (0.9)^{8} \approx 0.19371, P (2 ∣ B) = (2 10) (0.05)^{2} (0.95)^{8} \approx 0.07463 P (2) = 0.5 (0.19371) + 0.5 (0.07463) \approx 0.134

(b)

3 marks

Use the conditional probability formula.

P (A ∣ 2) = \frac{0.5 \times 0.19371}{0.13417} \approx 0.722

Question 8

6 marks

Volume of revolution of a region between two curves about the y-axis

Number and algebra

6 marks

Write both curves with $x^{2}$ as a function of $y$ .
Rotating about the $y$ -axis, the region for $0 \leq y \leq 3$ reaches out to the curve $f$ . For $2 \leq y \leq 3$ there is also a hollow part inside the curve $g$ , so subtract it.
Evaluate the first integral.
Evaluate the second.

y = x^{2} - 16 \Rightarrow x^{2} = y^{2} + 16, y = 2 + \frac{x ^{2}}{25} \Rightarrow x^{2} = 25 y - 50 V = π \int_{0}^{3} (y^{2} + 16) d y - π \int_{2}^{3} (25 y - 50) d y π [\frac{y ^{3}}{3} + 16 y]_{0}^{3} = π (9 + 48) = 57 π π [\frac{25 y ^{2}}{2} - 50 y]_{2}^{3} = π [(112.5 - 150) - (50 - 100)] = \frac{25 π}{2} V = 57 π - \frac{25 π}{2} = \frac{89 π}{2} \approx 140 cubic units

Question 9

7 marks

Related rates in a triangle with a shrinking angle

Number and algebra

7 marks

The area is $\frac{1}{2} (15) (20) sin θ = 140$ .
Since $θ$ is acute, $cos θ = 1 - \frac{196}{225} = \frac{29}{15}$ and $θ \approx 1.2036$ .
Differentiate $x^{2} = 625 - 600 cos θ$ with respect to $t$ .
Find $x$ .
Substitute $\frac{d θ}{d t} = - \frac{π}{60}$ .
So $A C$ is decreasing at about $0.724$ cm per minute.

150 sin θ = 140 \Rightarrow sin θ = \frac{14}{15} 2 x \frac{d x}{d t} = 600 sin θ \frac{d θ}{d t} \Rightarrow \frac{d x}{d t} = \frac{300 sin θ}{x} \frac{d θ}{d t} x = 625 - 4029 \approx 20.24 \frac{d x}{d t} = \frac{300 \times \frac{14}{15}}{20.24} \times (- \frac{π}{60}) \approx - 0.724

Section B · Extended response

Questions 10–12 · 56 marks

Question 10

17 marks

An airplane and a car: velocities, displacement and the gap between them

Calculus

(a)

2 marks

Put $t = 0$ .
(i) The airplane: $v = 60 e^{0} = 60$ m/s.
(ii) The car: $v = 5 (0) = 0$ m/s.

(b)

3 marks

(i) Equal speeds.

(ii) Substitute.

60 e^{- 0.1 t} = 5 t \Rightarrow t \approx 6.36 s v = 5 (6.3556) \approx 31.8 m/s

(c)

7 marks

Integrate each velocity to get a displacement.
The gap is the plane's position minus the car's.
Use $d (0) = 100$ .

s_{ai r} (t) = \int 60 e^{- 0.1 t} d t = - 600 e^{- 0.1 t} + C, s_{c a r} (t) = \int 5 t d t = \frac{5}{2} t^{2} d (t) = - 600 e^{- 0.1 t} - \frac{5}{2} t^{2} + C 100 = - 600 + C \Rightarrow C = 700 d (t) = 700 - 600 e^{- 0.1 t} - \frac{5}{2} t^{2}

(d)

2 marks

The car reaches the back of the airplane when $d (t) = 0$ .

700 - 600 e^{- 0.1 t} - 2.5 t^{2} = 0 \Rightarrow t \approx 15.1 s

(e)

3 marks

The distance the car has travelled is its displacement.

\frac{5}{2} (15.0586)^{2} = 2.5 \times 226.76 \approx 567 m

Question 11

18 marks

A density with arccos: integration by parts, finding k, mean, variance and a probability

Calculus

(a)

4 marks

Let $u = arccos x$ and $d v = d x$ , so $d u = - \frac{d x}{1 - x ^{2}}$ and $v = x$ .
The last integral is $- 1 - x^{2}$ , which can be found by inspection or by the substitution $w = 1 - x^{2}$ .

\int arccos x d x = x arccos x + \int \frac{x}{1 - x ^{2}} d x \int arccos x d x = x arccos x - 1 - x^{2} + C

(b)

6 marks

(i) The total probability is 1. Use $u = x^{2}$ with $d u = 2 x d x$ and then part (a).

Rearrange.

(ii) Solve on the calculator for $0 < k < 1$ .

\int_{0}^{k} 3 x arccos (x^{2}) d x = \frac{3}{2} \int_{0}^{k^{2}} arccos u d u = \frac{3}{2} [u arccos u - 1 - u^{2}]_{0}^{k^{2}} \frac{3}{2} [k^{2} arccos (k^{2}) - 1 - k^{4} - (0 - 1)] = 1 k^{2} arccos (k^{2}) - 1 - k^{4} + 1 = \frac{2}{3} \Rightarrow k^{2} arccos (k^{2}) - 1 - k^{4} + \frac{1}{3} = 0 k \approx 0.713250

(c)

5 marks

(i) The mean.

(ii) The mean of the squares.

The variance.

E (X) = \int_{0}^{0.71325} x \cdot 3 x arccos (x^{2}) d x \approx 0.456 E (X^{2}) = \int_{0}^{0.71325} x^{2} \cdot 3 x arccos (x^{2}) d x \approx 0.2372 Var (X) = 0.2372 - 0.456 3^{2} \approx 0.0290

(d)

3 marks

The standard deviation.
Integrate the density over $(μ - σ, μ + σ) = (0.28607, 0.62655)$ .

σ = 0.028980 \approx 0.17024 P (μ - σ < X < μ + σ) = \int_{0.28607}^{0.62655} 3 x arccos (x^{2}) d x \approx 0.620

Question 12

21 marks

Two lines, the plane containing them, a triangle and a pyramid

Number and algebra

(a)

3 marks

Write $L_{2}$ in parametric form using the parameter $t$ : $(x, y, z) = (- 1 + 2 t, 7 + t, - 5 + 3 t)$ . Equate with $L_{1}$ , which is $(5 + s, 4 - s, 2 + s)$ .
From the second equation $s = - 3 - t$ . Substitute in the first.
Check the third: $2 - 4 = - 2$ and $- 5 + 3 = - 2$ . They agree. Substitute $s = - 4$ in $L_{1}$ .

5 + s = - 1 + 2 t, 4 - s = 7 + t, 2 + s = - 5 + 3 t 5 - 3 - t = - 1 + 2 t \Rightarrow t = 1, s = - 4 5 - 4 4 + 4 2 - 4 = 18 - 2

(b)

3 marks

(i) The plane contains the intersection point and the two direction vectors.

(ii) The normal is $b \times c = - 4 - 1 3$ , so the plane is $- 4 x - y + 3 z = d$ . Substitute the point $(1, 8, - 2)$ .

r = 18 - 2 + λ 1 - 1 1 + μ 213 - 4 - 8 - 6 = - 18 \Rightarrow - 4 x - y + 3 z = - 18 \Rightarrow 4 x + y - 3 z = 18

(c)

2 marks

Put the other two variables to zero in $4 x + y - 3 z = 18$ .
(i) On the $y$ -axis: $y = 18$ , so $q = 18$ .
(ii) On the $z$ -axis: $- 3 z = 18$ , so $r = - 6$ .

(d)

5 marks

The vertices are $P (4.5, 0, 0)$ , $Q (0, 18, 0)$ and $R (0, 0, - 6)$ .
Take the cross product.
The area is half its magnitude.

P Q = - 4.5 180, P R = - 4.5 0 - 6 P Q \times P R = - 108 - 27 81 Area = \frac{1}{2} 10 8^{2} + 2 7^{2} + 8 1^{2} = \frac{1}{2} 18954 \approx 68.8

(e)

2 marks

$L_{3}$ passes through $(1, 8, - 2)$ and has the normal $(4, 1, - 3)$ as its direction.

r_{3} = 18 - 2 + γ 41 - 3

(f)

2 marks

Equate with $S (- 11, 5, 7)$ .
Check: $8 + (- 3) = 5$ and $- 2 - 3 (- 3) = 7$ .

1 + 4 γ = - 11 \Rightarrow γ = - 3

(g)

4 marks

The point $(1, 8, - 2)$ is in the plane, and $L_{3}$ is perpendicular to the plane there. So the height of the pyramid is the distance from it to $S$ .
Use the volume formula with the base area from part (d).

N S = - 12 - 3 9, h = 144 + 9 + 81 = 234 = 326 V = \frac{1}{3} \times \frac{1}{2} 18954 \times 234 = \frac{1}{3} \times \frac{1}{2} \times 2726 \times 326 = 351

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 algebra43 marks · 39%
Q2, Q6, Q8, Q9, Q12
Calculus35 marks · 32%
Q10, Q11
Statistics and probability19 marks · 17%
Q3, Q5, Q7
Functions7 marks · 6%
Q4
Geometry and trigonometry6 marks · 5%
Q1

Mentor insights

Traps to watch for

01
Mean point · Q3
The regression line of $y$ on $x$ always passes through $(\overset{x}{ˉ}, \overset{y}{ˉ})$ .
02
Amplitude · Q4
Half the difference between the maximum and the minimum.
03
Displacement · Q10
Integrate each velocity. The plane starts 100 m ahead of the car.
04
Integration by parts · Q11
Take $u = arccos x$ and $d v = d x$ .
05
Intersection · Q12
Equate the three components and check that all three agree.

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