Loading…

Worked solutions

IB Maths AA HL · Paper 1

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2024
Marks: 110
Duration: 2 hours
Calculator: Not 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 · 55 marks

Question 1

4 marks

Solving tan(2x - 5 degrees) = 1

Number and algebra

4 marks

The reference angle is $tan^{- 1} 1 = 4 5^{\circ}$ . In the interval for $2 x - 5^{\circ}$ the solutions are $4 5^{\circ}$ and $22 5^{\circ}$ .

2 x - 5 = 45 \Rightarrow x = 2 5^{\circ} 2 x - 5 = 225 \Rightarrow x = 11 5^{\circ}

Question 2

5 marks

An exponential equation that is quadratic in 3^x

Number and algebra

5 marks

Put $t = 3^{x}$ .
Since $3^{x} > 0$ , reject $t = - 2$ and use $t = \frac{1}{3}$ .

3 t^{2} + 5 t - 2 = 0 \Rightarrow (3 t - 1) (t + 2) = 0 3^{x} = \frac{1}{3} \Rightarrow x = - 1

Question 3

7 marks

A symmetrical quadrilateral OABC: midpoint, line OB and area

Geometry and trigonometry

(a)

4 marks

The midpoint of $[A C]$ with $A (6, 0)$ and $C (3, 33)$ .
The line through $O$ and $M$ has gradient

M = (\frac{9}{2}, \frac{3 3}{2}) m = \frac{3 3 /2}{9/2} = \frac{3}{3} y = \frac{3}{3} x

(b)

3 marks

$[O A]$ is perpendicular to $[A B]$ , so B lies directly above A, at $x = 6$ .
Triangle OAB has base 6 and height $23$ . The quadrilateral is two such triangles (OAB and OCB, by symmetry).

y_{B} = \frac{3}{3} \times 6 = 23 Area = 2 \times \frac{1}{2} \times 6 \times 23 = 123

Question 4

6 marks

Bird nesting in two seasons: a tree diagram and a quadratic in k

Statistics and probability

(a)

2 marks

The probability of not nesting is one minus the probability of nesting.

Spring: 1 - k, Summer (on both branches): 1 - \frac{k}{2}

(b)

4 marks

The seasons are multiplied along the branches for not nesting in both.
Expand and clear fractions.
The probability of nesting in Spring is $k$ , which cannot exceed 1. Since $\frac{8}{3} > 1$ , only $k = \frac{1}{3}$ is valid.

(1 - k) (1 - \frac{k}{2}) = \frac{5}{9} 1 - \frac{3 k}{2} + \frac{k ^{2}}{2} = \frac{5}{9} \Rightarrow 18 - 27 k + 9 k^{2} = 10 \Rightarrow 9 k^{2} - 27 k + 8 = 0

Question 5

8 marks

A rational function against a line: the number of intersections

Functions

(a)

1 mark

As $x \to \pm \infty$ , $f (x) = \frac{2 ( x + 3 )}{3 ( x + 2 )} \to \frac{2}{3}$ .

y = \frac{2}{3}

(b)

7 marks

Set $f (x) = g (x)$ and multiply by $3 (x + 2)$ .
The roots are $x = 0$ and $x = - \frac{6 m + 1}{3 m}$ .
(i) For $m > 0$ the second root is negative and not equal to $- 2$ , so there are two different solutions.
(ii) One solution needs the two roots to coincide.
This agrees with $f^{'} (0)$ : since $f^{'} (x) = \frac{2}{3} \cdot \frac{( x + 2 ) - ( x + 3 )}{( x + 2 ) ^{2}} = - \frac{2}{3 ( x + 2 ) ^{2}}$ , we get $f^{'} (0) = - \frac{1}{6}$ .
(iii) Two solutions with $x \geq 0$ need the second root to be positive.

2 (x + 3) = 3 (x + 2) (m x + 1) \Rightarrow 3 m x^{2} + (6 m + 1) x = 0 \Rightarrow x (3 m x + 6 m + 1) = 0 2 solutions - \frac{6 m + 1}{3 m} = 0 \Rightarrow m = - \frac{1}{6} - \frac{6 m + 1}{3 m} > 0 \Rightarrow - \frac{1}{6} < m < 0

Question 6

5 marks

Eating and cooking apples: the conditional probability of an eating apple

Statistics and probability

(a)

1 mark

For eating apples, 140 g is the mean plus $\frac{40}{20} = 2$ standard deviations, so 2.5% lie above it.

2.5%

(b)

4 marks

For cooking apples, 140 g is the mean, so half of them are above 140 g.
Weigh each type by its share of the apples.

P (> 140 ∣ cooking) = 0.5 P (eating ∣> 140) = \frac{0.8 \times 0.025}{0.8 \times 0.025 + 0.2 \times 0.5} = \frac{0.02}{0.12} = \frac{1}{6}

Question 7

7 marks

Roots of a cubic and a quintic with a complex root

Number and algebra

(a)

1 mark

For $g (x) = 2 x^{3} - 7 x^{2} + d x - e$ , the roots add to the negative of the $x^{2}$ coefficient over the leading coefficient.

α + β + γ = \frac{7}{2}

(b)

3 marks

The coefficients of $h$ are real, so $p - 3 i$ is also a root. There are five roots: $α$ , $β$ , $γ$ , $p + 3 i$ and $p - 3 i$ .
The sum of the roots of $h$ is $\frac{11}{2}$ .

\frac{7}{2} + (p + 3 i) + (p - 3 i) = \frac{11}{2} \Rightarrow 2 p = 2 \Rightarrow p = 1

(c)

3 marks

The product of the five roots is $- \frac{- 20}{2} = 10$ , and $(1 + 3 i) (1 - 3 i) = 10$ .
Since $h (\frac{1}{2}) = 0$ and $α$ and $β$ are integers, $γ = \frac{1}{2}$ . Then
The positive integers with product 2 and $α < β$ are

α β γ \times 10 = 10 \Rightarrow α β γ = 1 α β = 2 α = 1, β = 2

Question 8

6 marks

l'Hopital's rule twice on a limit at 0

Number and algebra

6 marks

The limit has the form $\frac{0}{0}$ , so differentiate the top and bottom.
At $x = 0$ this is again $\frac{0}{0}$ , so differentiate once more. The derivative of $sec^{4} x tan x$ is $4 sec^{4} x tan^{2} x + sec^{6} x$ , and the derivative of $2 sin x cos x$ is $2 cos 2 x$ .

x \to 0 lim \frac{4 sec ^{4} x tan x + 2 sin x cos x}{4 x ^{3} - 2 x} x \to 0 lim \frac{16 sec ^{4} x tan ^{2} x + 4 sec ^{6} x + 2 cos 2 x}{12 x ^{2} - 2} = \frac{0 + 4 + 2}{- 2} = - 3

Question 9

7 marks

Assigning students to groups, with two kept apart

Number and algebra

(a)

1 mark

Choose the 3 students for the first group.

(3 n)

(b)

6 marks

With the two students apart, exactly one of them is in the first group.
This is half of the total.
Divide by $n - 2$ and simplify.
The second group needs at least 3 students, so $n \geq 6$ and $n = 4$ is rejected.

2 (2 n - 2) 2 (2 n - 2) = \frac{1}{2} (3 n) \Rightarrow (n - 2) (n - 3) = \frac{n ( n - 1 ) ( n - 2 )}{12} 12 (n - 3) = n (n - 1) \Rightarrow n^{2} - 13 n + 36 = 0 \Rightarrow (n - 4) (n - 9) = 0 n = 9

Section B · Extended response

Questions 10–12 · 55 marks

Question 10

16 marks

An arithmetic and a geometric sequence combined into a new arithmetic sequence

Number and algebra

(a)

2 marks

The common difference gives $p - a = q - p$ .

2 p - q = a

(b)

2 marks

The common ratio gives $\frac{s}{a} = \frac{t}{s}$ .

s^{2} = a t

(c)

2 marks

With $q = t = 1$ , part (a) gives $a = 2 p - 1$ and part (b) gives $s^{2} = a$ .
Since $s \neq = 0$ , $s^{2} > 0$ .

2 p - 1 = s^{2} > 0 p > \frac{1}{2}

(d)

4 marks

With $a = 9$ and $q = 1$ , $p = \frac{a + q}{2} = 5$ , and the common difference is $- 4$ .
With $s^{2} = 9 \times 1$ and $s > 0$ , $s = 3$ and the ratio is $\frac{1}{3}$ .

9, 5, 1, - 3 9, 3, 1, \frac{1}{3}

(e)

6 marks

The common difference of $u_{n}$ .
Check against the next pair: $u_{3} - u_{2} = (1 + 0) - (5 + ln 3) = - 4 - ln 3$ .
Sum the first ten terms with $u_{1} = 9 + 2 ln 3$ .

d = u_{2} - u_{1} = (5 + ln 3) - (9 + ln 9) = - 4 + ln 3 - 2 ln 3 = - 4 - ln 3 S_{10} = \frac{10}{2} [2 (9 + 2 ln 3) + 9 (- 4 - ln 3)] = 5 (18 + 4 ln 3 - 36 - 9 ln 3) i = 1 \sum 10 u_{i} = 5 (- 18 - 5 ln 3) = - 90 - 25 ln 3

Question 11

19 marks

Two planes: perpendicular, then parallel, and the distance between them

Number and algebra

(a)

1 mark

Substitute $A = (2, \frac{1}{2}, 1)$ into $2 x + 6 y - 2 z$ .

2 (2) + 6 (\frac{1}{2}) - 2 (1) = 4 + 3 - 2 = 5

(b)

5 marks

The normals are $(2, 6, - 2)$ and $(k^{2} - 6, 2 k + 3, - 6)$ . Set their scalar product to 0.
Then $P_{2}$ is $3 x - 3 y - 6 z = q$ . Substitute $A$ .

2 (k^{2} - 6) + 6 (2 k + 3) + 12 = 0 \Rightarrow 2 k^{2} + 12 k + 18 = 0 \Rightarrow (k + 3)^{2} = 0 k = - 3 q = 3 (2) - 3 (\frac{1}{2}) - 6 (1) = - \frac{3}{2}

(c)

2 marks

With $k = 3$ , the normal of $P_{2}$ is $(3, 9, p)$ , which is parallel to $(2, 6, - 2)$ . The multiplier is $\frac{3}{2}$ .

p = \frac{3}{2} \times (- 2) = - 3

(d)

7 marks

The line through $A$ perpendicular to $P_{1}$ has direction $(2, 6, - 2)$ .
$P_{2}$ is $3 x + 9 y - 3 z = - \frac{51}{2}$ . Substitute the line.
The distance is $∣ A B ∣ = ∣ λ ∣ 2^{2} + 6^{2} + 2^{2}$ .

r = 2 \frac{1}{2} 1 + λ 26 - 2 3 (2 + 2 λ) + 9 (\frac{1}{2} + 6 λ) - 3 (1 - 2 λ) = - \frac{51}{2} \frac{15}{2} + 66 λ = - \frac{51}{2} \Rightarrow λ = - \frac{1}{2} B = (1, - \frac{5}{2}, 2) ∣ A B ∣ = \frac{1}{2} 44 = 11

(e)

4 marks

$P_{3}$ is on the other side of $P_{1}$ , at $λ = + \frac{1}{2}$ from $A$ along the same line.
$P_{3}$ is parallel to $P_{1}$ , so it is $2 x + 6 y - 2 z = d$ . Substitute $C$ .

C = (2 + 1, \frac{1}{2} + 3, 1 - 1) = (3, \frac{7}{2}, 0) d = 6 + 21 - 0 = 27 P_{3} : 2 x + 6 y - 2 z = 27

Question 12

20 marks

Induction on derivatives of (1 - ax)^(-1/2), a Maclaurin series and an estimate of root 3

Number and algebra
Calculus

(a)

8 marks

Claim: $f^{(n)} (x) = \frac{a ^{n} ( 2 n - 1 )! ( 1 - a x ) ^{- \frac{2 n + 1}{2}}}{2 ^{2 n - 1} ( n - 1 )!}$ for all $n \in Z^{+}$ .
Base case $n = 1$ : $f^{'} (x) = - \frac{1}{2} (1 - a x)^{- 3/2} (- a) = \frac{a}{2} (1 - a x)^{- 3/2}$ . The formula gives $\frac{a \cdot 1 ! ( 1 - a x ) ^{- 3/2}}{2 \cdot 0 !}$ , the same. It holds.
Assume it holds for $n = k$ .
Differentiate, using $\frac{d}{d x} (1 - a x)^{- \frac{2 k + 1}{2}} = \frac{( 2 k + 1 ) a}{2} (1 - a x)^{- \frac{2 k + 3}{2}}$ .
Multiply the top and bottom by $2 k$ . Since $(2 k + 1) (2 k) (2 k - 1)! = (2 k + 1)!$ and $k (k - 1)! = k!$ ,
This is the formula with $n = k + 1$ . The statement is true for $n = 1$ and, if true for $n = k$ , true for $n = k + 1$ . By induction it holds for all $n \in Z^{+}$ .

f^{(k)} (x) = \frac{a ^{k} ( 2 k - 1 )! ( 1 - a x ) ^{- \frac{2 k + 1}{2}}}{2 ^{2 k - 1} ( k - 1 )!} f^{(k + 1)} (x) = \frac{a ^{k + 1} ( 2 k + 1 ) ( 2 k - 1 )! ( 1 - a x ) ^{- \frac{2 k + 3}{2}}}{2 ^{2 k} ( k - 1 )!} f^{(k + 1)} (x) = \frac{a ^{k + 1} ( 2 k + 1 )! ( 1 - a x ) ^{- \frac{2 ( k + 1 ) + 1}{2}}}{2 ^{2 (k + 1) - 1} k !}

(b)

2 marks

Use $f (0) = 1$ and the formula at $x = 0$ .
The Maclaurin series up to $x^{2}$ is $f (0) + f^{'} (0) x + \frac{f ^{''} ( 0 )}{2} x^{2}$ .

f^{'} (0) = \frac{a}{2}, f^{''} (0) = \frac{a ^{2} \cdot 3 !}{2 ^{3} \cdot 1 !} = \frac{3 a ^{2}}{4} f (x) = 1 + \frac{1}{2} a x + \frac{3}{8} a^{2} x^{2} + \dots

(c)

4 marks

Put $a = 2$ and $a = 4$ in the series of (b).
Multiply and keep the terms up to $x^{2}$ .

(1 - 2 x)^{- 1/2} = 1 + x + \frac{3}{2} x^{2} + \dots (1 - 4 x)^{- 1/2} = 1 + 2 x + 6 x^{2} + \dots (1 + x + \frac{3}{2} x^{2}) (1 + 2 x + 6 x^{2}) = 1 + 3 x + (6 + 2 + \frac{3}{2}) x^{2} = 1 + 3 x + \frac{19}{2} x^{2} (1 - 2 x)^{- 1/2} (1 - 4 x)^{- 1/2} \approx \frac{2 + 6 x + 19 x ^{2}}{2}

(d)

1 mark

Each series converges when $∣ a x ∣ < 1$ . For $a = 4$ this is the stricter condition, which also covers $a = 2$ .

∣ x ∣ < \frac{1}{4}

(e)

5 marks

With $x = \frac{1}{10}$ the approximation gives
The exact product is
Equate and solve for $3$ .

\frac{2 + 0.6 + 0.19}{2} = \frac{279}{200} (1 - 0.2)^{- 1/2} (1 - 0.4)^{- 1/2} = \frac{1}{0.48} = \frac{10}{4 3} = \frac{5}{2 3} \frac{5}{2 3} \approx \frac{279}{200} \Rightarrow 3 \approx \frac{500}{279}

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 algebra72 marks · 65%
Q1, Q2, Q7, Q8, Q9, Q10, Q11, Q12
Calculus12 marks · 11%
Q12
Statistics and probability11 marks · 10%
Q4, Q6
Functions8 marks · 7%
Q5
Geometry and trigonometry7 marks · 6%
Q3

Mentor insights

Traps to watch for

01
Symmetry about a diagonal · Q3
The diagonal OB is the perpendicular bisector of $[A C]$ , so it passes through the midpoint of $[A C]$ .
02
Intersections as a quadratic · Q5
Setting $f (x) = g (x)$ and clearing the fraction gives $3 m x^{2} + (6 m + 1) x = 0$ , which has the root $x = 0$ .
03
Equal differences · Q10
In $a, p, q$ we have $p - a = q - p$ . In $a, s, t$ the ratios are equal: $\frac{s}{a} = \frac{t}{s}$ .
04
Perpendicular planes · Q11
The normal vectors have zero scalar product.
05
Induction layout · Q12
Check $n = 1$ , assume $n = k$ , then differentiate the formula to get $n = k + 1$ .

Solutions and tips are GeorgeTheTutor's own; the questions are not reproduced here. Not affiliated with or endorsed by the International Baccalaureate Organization.

Want someone to go through your paper with you?

Our maths mentors took these exams recently. One-to-one, they'll find where your marks are going and fix it before May.

Find your maths mentor More past papers

Worked solutions

IB Maths AA HL · Paper 1

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2024
Marks: 110
Duration: 2 hours
Calculator: Not 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 · 55 marks

Question 1

4 marks

Solving tan(2x - 5 degrees) = 1

Number and algebra

4 marks

The reference angle is $tan^{- 1} 1 = 4 5^{\circ}$ . In the interval for $2 x - 5^{\circ}$ the solutions are $4 5^{\circ}$ and $22 5^{\circ}$ .

2 x - 5 = 45 \Rightarrow x = 2 5^{\circ} 2 x - 5 = 225 \Rightarrow x = 11 5^{\circ}

Question 2

5 marks

An exponential equation that is quadratic in 3^x

Number and algebra

5 marks

Put $t = 3^{x}$ .
Since $3^{x} > 0$ , reject $t = - 2$ and use $t = \frac{1}{3}$ .

3 t^{2} + 5 t - 2 = 0 \Rightarrow (3 t - 1) (t + 2) = 0 3^{x} = \frac{1}{3} \Rightarrow x = - 1

Question 3

7 marks

A symmetrical quadrilateral OABC: midpoint, line OB and area

Geometry and trigonometry

(a)

4 marks

The midpoint of $[A C]$ with $A (6, 0)$ and $C (3, 33)$ .
The line through $O$ and $M$ has gradient

M = (\frac{9}{2}, \frac{3 3}{2}) m = \frac{3 3 /2}{9/2} = \frac{3}{3} y = \frac{3}{3} x

(b)

3 marks

$[O A]$ is perpendicular to $[A B]$ , so B lies directly above A, at $x = 6$ .
Triangle OAB has base 6 and height $23$ . The quadrilateral is two such triangles (OAB and OCB, by symmetry).

y_{B} = \frac{3}{3} \times 6 = 23 Area = 2 \times \frac{1}{2} \times 6 \times 23 = 123

Question 4

6 marks

Bird nesting in two seasons: a tree diagram and a quadratic in k

Statistics and probability

(a)

2 marks

The probability of not nesting is one minus the probability of nesting.

Spring: 1 - k, Summer (on both branches): 1 - \frac{k}{2}

(b)

4 marks

The seasons are multiplied along the branches for not nesting in both.
Expand and clear fractions.
The probability of nesting in Spring is $k$ , which cannot exceed 1. Since $\frac{8}{3} > 1$ , only $k = \frac{1}{3}$ is valid.

(1 - k) (1 - \frac{k}{2}) = \frac{5}{9} 1 - \frac{3 k}{2} + \frac{k ^{2}}{2} = \frac{5}{9} \Rightarrow 18 - 27 k + 9 k^{2} = 10 \Rightarrow 9 k^{2} - 27 k + 8 = 0

Question 5

8 marks

A rational function against a line: the number of intersections

Functions

(a)

1 mark

As $x \to \pm \infty$ , $f (x) = \frac{2 ( x + 3 )}{3 ( x + 2 )} \to \frac{2}{3}$ .

y = \frac{2}{3}

(b)

7 marks

Set $f (x) = g (x)$ and multiply by $3 (x + 2)$ .
The roots are $x = 0$ and $x = - \frac{6 m + 1}{3 m}$ .
(i) For $m > 0$ the second root is negative and not equal to $- 2$ , so there are two different solutions.
(ii) One solution needs the two roots to coincide.
This agrees with $f^{'} (0)$ : since $f^{'} (x) = \frac{2}{3} \cdot \frac{( x + 2 ) - ( x + 3 )}{( x + 2 ) ^{2}} = - \frac{2}{3 ( x + 2 ) ^{2}}$ , we get $f^{'} (0) = - \frac{1}{6}$ .
(iii) Two solutions with $x \geq 0$ need the second root to be positive.

2 (x + 3) = 3 (x + 2) (m x + 1) \Rightarrow 3 m x^{2} + (6 m + 1) x = 0 \Rightarrow x (3 m x + 6 m + 1) = 0 2 solutions - \frac{6 m + 1}{3 m} = 0 \Rightarrow m = - \frac{1}{6} - \frac{6 m + 1}{3 m} > 0 \Rightarrow - \frac{1}{6} < m < 0

Question 6

5 marks

Eating and cooking apples: the conditional probability of an eating apple

Statistics and probability

(a)

1 mark

For eating apples, 140 g is the mean plus $\frac{40}{20} = 2$ standard deviations, so 2.5% lie above it.

2.5%

(b)

4 marks

For cooking apples, 140 g is the mean, so half of them are above 140 g.
Weigh each type by its share of the apples.

P (> 140 ∣ cooking) = 0.5 P (eating ∣> 140) = \frac{0.8 \times 0.025}{0.8 \times 0.025 + 0.2 \times 0.5} = \frac{0.02}{0.12} = \frac{1}{6}

Question 7

7 marks

Roots of a cubic and a quintic with a complex root

Number and algebra

(a)

1 mark

For $g (x) = 2 x^{3} - 7 x^{2} + d x - e$ , the roots add to the negative of the $x^{2}$ coefficient over the leading coefficient.

α + β + γ = \frac{7}{2}

(b)

3 marks

The coefficients of $h$ are real, so $p - 3 i$ is also a root. There are five roots: $α$ , $β$ , $γ$ , $p + 3 i$ and $p - 3 i$ .
The sum of the roots of $h$ is $\frac{11}{2}$ .

\frac{7}{2} + (p + 3 i) + (p - 3 i) = \frac{11}{2} \Rightarrow 2 p = 2 \Rightarrow p = 1

(c)

3 marks

The product of the five roots is $- \frac{- 20}{2} = 10$ , and $(1 + 3 i) (1 - 3 i) = 10$ .
Since $h (\frac{1}{2}) = 0$ and $α$ and $β$ are integers, $γ = \frac{1}{2}$ . Then
The positive integers with product 2 and $α < β$ are

α β γ \times 10 = 10 \Rightarrow α β γ = 1 α β = 2 α = 1, β = 2

Question 8

6 marks

l'Hopital's rule twice on a limit at 0

Number and algebra

6 marks

The limit has the form $\frac{0}{0}$ , so differentiate the top and bottom.
At $x = 0$ this is again $\frac{0}{0}$ , so differentiate once more. The derivative of $sec^{4} x tan x$ is $4 sec^{4} x tan^{2} x + sec^{6} x$ , and the derivative of $2 sin x cos x$ is $2 cos 2 x$ .

x \to 0 lim \frac{4 sec ^{4} x tan x + 2 sin x cos x}{4 x ^{3} - 2 x} x \to 0 lim \frac{16 sec ^{4} x tan ^{2} x + 4 sec ^{6} x + 2 cos 2 x}{12 x ^{2} - 2} = \frac{0 + 4 + 2}{- 2} = - 3

Question 9

7 marks

Assigning students to groups, with two kept apart

Number and algebra

(a)

1 mark

Choose the 3 students for the first group.

(3 n)

(b)

6 marks

With the two students apart, exactly one of them is in the first group.
This is half of the total.
Divide by $n - 2$ and simplify.
The second group needs at least 3 students, so $n \geq 6$ and $n = 4$ is rejected.

2 (2 n - 2) 2 (2 n - 2) = \frac{1}{2} (3 n) \Rightarrow (n - 2) (n - 3) = \frac{n ( n - 1 ) ( n - 2 )}{12} 12 (n - 3) = n (n - 1) \Rightarrow n^{2} - 13 n + 36 = 0 \Rightarrow (n - 4) (n - 9) = 0 n = 9

Section B · Extended response

Questions 10–12 · 55 marks

Question 10

16 marks

An arithmetic and a geometric sequence combined into a new arithmetic sequence

Number and algebra

(a)

2 marks

The common difference gives $p - a = q - p$ .

2 p - q = a

(b)

2 marks

The common ratio gives $\frac{s}{a} = \frac{t}{s}$ .

s^{2} = a t

(c)

2 marks

With $q = t = 1$ , part (a) gives $a = 2 p - 1$ and part (b) gives $s^{2} = a$ .
Since $s \neq = 0$ , $s^{2} > 0$ .

2 p - 1 = s^{2} > 0 p > \frac{1}{2}

(d)

4 marks

With $a = 9$ and $q = 1$ , $p = \frac{a + q}{2} = 5$ , and the common difference is $- 4$ .
With $s^{2} = 9 \times 1$ and $s > 0$ , $s = 3$ and the ratio is $\frac{1}{3}$ .

9, 5, 1, - 3 9, 3, 1, \frac{1}{3}

(e)

6 marks

The common difference of $u_{n}$ .
Check against the next pair: $u_{3} - u_{2} = (1 + 0) - (5 + ln 3) = - 4 - ln 3$ .
Sum the first ten terms with $u_{1} = 9 + 2 ln 3$ .

d = u_{2} - u_{1} = (5 + ln 3) - (9 + ln 9) = - 4 + ln 3 - 2 ln 3 = - 4 - ln 3 S_{10} = \frac{10}{2} [2 (9 + 2 ln 3) + 9 (- 4 - ln 3)] = 5 (18 + 4 ln 3 - 36 - 9 ln 3) i = 1 \sum 10 u_{i} = 5 (- 18 - 5 ln 3) = - 90 - 25 ln 3

Question 11

19 marks

Two planes: perpendicular, then parallel, and the distance between them

Number and algebra

(a)

1 mark

Substitute $A = (2, \frac{1}{2}, 1)$ into $2 x + 6 y - 2 z$ .

2 (2) + 6 (\frac{1}{2}) - 2 (1) = 4 + 3 - 2 = 5

(b)

5 marks

The normals are $(2, 6, - 2)$ and $(k^{2} - 6, 2 k + 3, - 6)$ . Set their scalar product to 0.
Then $P_{2}$ is $3 x - 3 y - 6 z = q$ . Substitute $A$ .

2 (k^{2} - 6) + 6 (2 k + 3) + 12 = 0 \Rightarrow 2 k^{2} + 12 k + 18 = 0 \Rightarrow (k + 3)^{2} = 0 k = - 3 q = 3 (2) - 3 (\frac{1}{2}) - 6 (1) = - \frac{3}{2}

(c)

2 marks

With $k = 3$ , the normal of $P_{2}$ is $(3, 9, p)$ , which is parallel to $(2, 6, - 2)$ . The multiplier is $\frac{3}{2}$ .

p = \frac{3}{2} \times (- 2) = - 3

(d)

7 marks

The line through $A$ perpendicular to $P_{1}$ has direction $(2, 6, - 2)$ .
$P_{2}$ is $3 x + 9 y - 3 z = - \frac{51}{2}$ . Substitute the line.
The distance is $∣ A B ∣ = ∣ λ ∣ 2^{2} + 6^{2} + 2^{2}$ .

r = 2 \frac{1}{2} 1 + λ 26 - 2 3 (2 + 2 λ) + 9 (\frac{1}{2} + 6 λ) - 3 (1 - 2 λ) = - \frac{51}{2} \frac{15}{2} + 66 λ = - \frac{51}{2} \Rightarrow λ = - \frac{1}{2} B = (1, - \frac{5}{2}, 2) ∣ A B ∣ = \frac{1}{2} 44 = 11

(e)

4 marks

$P_{3}$ is on the other side of $P_{1}$ , at $λ = + \frac{1}{2}$ from $A$ along the same line.
$P_{3}$ is parallel to $P_{1}$ , so it is $2 x + 6 y - 2 z = d$ . Substitute $C$ .

C = (2 + 1, \frac{1}{2} + 3, 1 - 1) = (3, \frac{7}{2}, 0) d = 6 + 21 - 0 = 27 P_{3} : 2 x + 6 y - 2 z = 27

Question 12

20 marks

Induction on derivatives of (1 - ax)^(-1/2), a Maclaurin series and an estimate of root 3

Number and algebra
Calculus

(a)

8 marks

Claim: $f^{(n)} (x) = \frac{a ^{n} ( 2 n - 1 )! ( 1 - a x ) ^{- \frac{2 n + 1}{2}}}{2 ^{2 n - 1} ( n - 1 )!}$ for all $n \in Z^{+}$ .
Base case $n = 1$ : $f^{'} (x) = - \frac{1}{2} (1 - a x)^{- 3/2} (- a) = \frac{a}{2} (1 - a x)^{- 3/2}$ . The formula gives $\frac{a \cdot 1 ! ( 1 - a x ) ^{- 3/2}}{2 \cdot 0 !}$ , the same. It holds.
Assume it holds for $n = k$ .
Differentiate, using $\frac{d}{d x} (1 - a x)^{- \frac{2 k + 1}{2}} = \frac{( 2 k + 1 ) a}{2} (1 - a x)^{- \frac{2 k + 3}{2}}$ .
Multiply the top and bottom by $2 k$ . Since $(2 k + 1) (2 k) (2 k - 1)! = (2 k + 1)!$ and $k (k - 1)! = k!$ ,
This is the formula with $n = k + 1$ . The statement is true for $n = 1$ and, if true for $n = k$ , true for $n = k + 1$ . By induction it holds for all $n \in Z^{+}$ .

f^{(k)} (x) = \frac{a ^{k} ( 2 k - 1 )! ( 1 - a x ) ^{- \frac{2 k + 1}{2}}}{2 ^{2 k - 1} ( k - 1 )!} f^{(k + 1)} (x) = \frac{a ^{k + 1} ( 2 k + 1 ) ( 2 k - 1 )! ( 1 - a x ) ^{- \frac{2 k + 3}{2}}}{2 ^{2 k} ( k - 1 )!} f^{(k + 1)} (x) = \frac{a ^{k + 1} ( 2 k + 1 )! ( 1 - a x ) ^{- \frac{2 ( k + 1 ) + 1}{2}}}{2 ^{2 (k + 1) - 1} k !}

(b)

2 marks

Use $f (0) = 1$ and the formula at $x = 0$ .
The Maclaurin series up to $x^{2}$ is $f (0) + f^{'} (0) x + \frac{f ^{''} ( 0 )}{2} x^{2}$ .

f^{'} (0) = \frac{a}{2}, f^{''} (0) = \frac{a ^{2} \cdot 3 !}{2 ^{3} \cdot 1 !} = \frac{3 a ^{2}}{4} f (x) = 1 + \frac{1}{2} a x + \frac{3}{8} a^{2} x^{2} + \dots

(c)

4 marks

Put $a = 2$ and $a = 4$ in the series of (b).
Multiply and keep the terms up to $x^{2}$ .

(1 - 2 x)^{- 1/2} = 1 + x + \frac{3}{2} x^{2} + \dots (1 - 4 x)^{- 1/2} = 1 + 2 x + 6 x^{2} + \dots (1 + x + \frac{3}{2} x^{2}) (1 + 2 x + 6 x^{2}) = 1 + 3 x + (6 + 2 + \frac{3}{2}) x^{2} = 1 + 3 x + \frac{19}{2} x^{2} (1 - 2 x)^{- 1/2} (1 - 4 x)^{- 1/2} \approx \frac{2 + 6 x + 19 x ^{2}}{2}

(d)

1 mark

Each series converges when $∣ a x ∣ < 1$ . For $a = 4$ this is the stricter condition, which also covers $a = 2$ .

∣ x ∣ < \frac{1}{4}

(e)

5 marks

With $x = \frac{1}{10}$ the approximation gives
The exact product is
Equate and solve for $3$ .

\frac{2 + 0.6 + 0.19}{2} = \frac{279}{200} (1 - 0.2)^{- 1/2} (1 - 0.4)^{- 1/2} = \frac{1}{0.48} = \frac{10}{4 3} = \frac{5}{2 3} \frac{5}{2 3} \approx \frac{279}{200} \Rightarrow 3 \approx \frac{500}{279}

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 algebra72 marks · 65%
Q1, Q2, Q7, Q8, Q9, Q10, Q11, Q12
Calculus12 marks · 11%
Q12
Statistics and probability11 marks · 10%
Q4, Q6
Functions8 marks · 7%
Q5
Geometry and trigonometry7 marks · 6%
Q3

Mentor insights

Traps to watch for

01
Symmetry about a diagonal · Q3
The diagonal OB is the perpendicular bisector of $[A C]$ , so it passes through the midpoint of $[A C]$ .
02
Intersections as a quadratic · Q5
Setting $f (x) = g (x)$ and clearing the fraction gives $3 m x^{2} + (6 m + 1) x = 0$ , which has the root $x = 0$ .
03
Equal differences · Q10
In $a, p, q$ we have $p - a = q - p$ . In $a, s, t$ the ratios are equal: $\frac{s}{a} = \frac{t}{s}$ .
04
Perpendicular planes · Q11
The normal vectors have zero scalar product.
05
Induction layout · Q12
Check $n = 1$ , assume $n = k$ , then differentiate the formula to get $n = k + 1$ .

Solutions and tips are GeorgeTheTutor's own; the questions are not reproduced here. Not affiliated with or endorsed by the International Baccalaureate Organization.

Want someone to go through your paper with you?

Our maths mentors took these exams recently. One-to-one, they'll find where your marks are going and fix it before May.

Find your maths mentor More past papers