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

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Section A · Short response

Questions 1–9 · 57 marks

Question 1

6 marks

A die rolled sixteen times: finding two frequencies from the mean

Statistics and probability

(a)

5 marks

The frequencies add to 16.
The mean is 3, so the total of the scores is $3 \times 16 = 48$ .
Subtract the two equations.

p + q + 4 + 2 + 0 + 3 = 16 \Rightarrow p + q = 7 p + 2 q + 12 + 8 + 0 + 18 = 48 \Rightarrow p + 2 q = 10 q = 3, p = 4

(b)

1 mark

Multiplying each score by 10 multiplies the mean by 10.

3 \times 10 = 30

Question 2

5 marks

Logarithms of 1/a and with base 1000 given log of a

Functions

(a)

2 marks

Use $lo g_{10} \frac{1}{a} = - lo g_{10} a$ with $lo g_{10} a = \frac{1}{3}$ .

lo g_{10} \frac{1}{a} = - \frac{1}{3}

(b)

3 marks

Change the base: $lo g_{1000} a = \frac{lo g _{10} a}{lo g _{10} 1000}$ .

lo g_{1000} a = \frac{1/3}{3} = \frac{1}{9}

Question 3

8 marks

A circular sector with perimeter 10 and area 6.25

Geometry and trigonometry

(a)

4 marks

Write the two conditions.
From the first, $r θ = 10 - 2 r$ . Substitute into the area.

2 r + r θ = 10, \frac{1}{2} r^{2} θ = 6.25 \frac{1}{2} r (10 - 2 r) = 6.25 \Rightarrow 5 r - r^{2} = 6.25 \Rightarrow 4 r^{2} - 20 r + 25 = 0

(b)

4 marks

Factorise: $(2 r - 5)^{2} = 0$ .
Then $r θ = 10 - 5 = 5$ .

r = \frac{5}{2} θ = \frac{5}{2.5} = 2

Question 4

7 marks

Intersections of cos x and sin 2x and the area between them

Functions
Calculus

(a)

3 marks

Set $cos x = sin 2 x = 2 sin x cos x$ .
The point B at $x = \frac{π}{2}$ comes from $cos x = 0$ . The other solutions in $0 \leq x \leq π$ are

cos x (1 - 2 sin x) = 0 \Rightarrow cos x = 0 or sin x = \frac{1}{2} x_{A} = \frac{π}{6}, x_{C} = \frac{5 π}{6}

(b)

4 marks

Between B and C the curve of $f$ lies above $g$ .

Area = \int_{π /2}^{5 π /6} (cos x - sin 2 x) d x = [sin x + \frac{1}{2} cos 2 x]_{π /2}^{5 π /6} (\frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2}) - (1 - \frac{1}{2}) = \frac{3}{4} - \frac{1}{2} = \frac{1}{4}

Question 5

5 marks

Sums of the geometric sequence with ratio 10 and a sum of sums

Number and algebra

(a)

1 mark

Use $S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1}$ with $a = 1$ and $r = 10$ .

S_{n} = \frac{1 0 ^{n} - 1}{9}

(b)

4 marks

Add the first $n$ sums.
The geometric series has first term 10, ratio 10 and $n$ terms.

S_{1} + S_{2} + \dots + S_{n} = \frac{10 - 1}{9} + \frac{1 0 ^{2} - 1}{9} + \dots + \frac{1 0 ^{n} - 1}{9} = \frac{1}{9} [(10 + 1 0^{2} + \dots + 1 0^{n}) - n] 10 + 1 0^{2} + \dots + 1 0^{n} = \frac{10 ( 1 0 ^{n} - 1 )}{9} S_{1} + \dots + S_{n} = \frac{1}{9} [\frac{10 ( 1 0 ^{n} - 1 )}{9} - n] = \frac{1 0 ^{n + 1} - 10 - 9 n}{81}

Question 6

6 marks

Volume of revolution of sqrt(x sin(x^2))

Number and algebra

6 marks

The volume of revolution about the $x$ -axis is
Integrate by inspection (or put $u = x^{2}$ , $d u = 2 x d x$ ).
Substitute $cos \frac{π}{4} = \frac{2}{2}$ .

V = π \int_{0}^{π /2} (x sin (x^{2}))^{2} d x = π \int_{0}^{π /2} x sin (x^{2}) d x V = π [- \frac{1}{2} cos (x^{2})]_{0}^{π /2} = \frac{π}{2} (1 - cos \frac{π}{4}) V = \frac{π}{2} (1 - \frac{2}{2}) = \frac{π ( 2 - 2 )}{4}

Question 7

7 marks

Induction: a sum of binomial coefficients equals another binomial coefficient

Number and algebra

7 marks

Claim: $r = 1 \sum n (1 r) = (2 n + 1)$ for all $n \in Z^{+}$ .
Base case $n = 1$ : the left side is $(1 1) = 1$ and the right side is $(2 2) = 1$ . It holds.
Assume it holds for $n = k$ : $r = 1 \sum k (1 r) = (2 k + 1)$ .
For $n = k + 1$ , split off the last term and use the assumption.
Put both fractions over $2! k!$ .
This is the statement for $n = k + 1$ . The statement is true for $n = 1$ and, if true for $n = k$ , it is true for $n = k + 1$ . By mathematical induction it is true for all $n \in Z^{+}$ .

r = 1 \sum k + 1 (1 r) = (2 k + 1) + (1 k + 1) = \frac{( k + 1 )!}{2 ! ( k - 1 )!} + \frac{( k + 1 )!}{1 ! k !} \frac{( k + 1 )! k}{2 ! k !} + \frac{( k + 1 )! \cdot 2}{2 ! k !} = \frac{( k + 1 )! ( k + 2 )}{2 ! k !} = \frac{( k + 2 )!}{2 ! k !} = (2 k + 2)

Question 8

7 marks

Maclaurin series of sin(x^2), sin^2(x^2) and a derived series

Calculus

(a)

5 marks

Replace $x$ by $x^{2}$ in the series for sine.
Square the series.

sin (x^{2}) = x^{2} - \frac{x ^{6}}{6} + \dots sin^{2} (x^{2}) = (x^{2} - \frac{x ^{6}}{6})^{2} = x^{4} - \frac{x ^{8}}{3} + \dots

(b)

2 marks

Note that $\frac{d}{d x} sin^{2} (x^{2}) = 2 sin (x^{2}) cos (x^{2}) \cdot 2 x = 4 x sin (x^{2}) cos (x^{2})$ . Differentiate the series from (a(ii)).

4 x sin (x^{2}) cos (x^{2}) = 4 x^{3} - \frac{8 x ^{7}}{3} + \dots

Question 9

6 marks

Sketching f(|x|), an odd function and integrals

Functions

(a)

2 marks

Reflect all the parts of the graph below the $x$ -axis in the $x$ -axis.
Sketch: a symmetric curve about the $y$ -axis with sharp points (cusps) at the $x$ -intercepts. The small dips near $x = \pm 1$ become small humps, lower than the maxima at $x = \pm 3.5$ , and the ends, which fell below $- 4$ near $x = \pm 6$ , are reflected to rise steeply above $4$ .

(b)

2 marks

For $x \geq 0$ the graph of $f (x)$ is the same as $f (∣ x ∣)$ . For $x < 0$ use the half turn about the origin (odd symmetry).
Sketch: the right half is unchanged: a dip below the axis near $x = 1$ , a maximum near $x = 3.5$ and a fall through the axis near $x = 5.3$ . The left half is the right half rotated by $18 0^{\circ}$ about the origin.

(c)

2 marks

Since $f$ is an odd function, $\int_{- 4}^{0} f (x) d x = - \int_{0}^{4} f (x) d x$ , and for $x \geq 0$ , $f (∣ x ∣) = f (x)$ so $\int_{0}^{4} f (x) d x = 1.6$ .
Over the symmetric range, the integral of the odd part is 0 and the integral of $f (∣ x ∣)$ is twice $\int_{0}^{4}$ .

\int_{- 4}^{0} f (x) d x = - 1.6 \int_{- 4}^{4} (f (∣ x ∣) + f (x)) d x = 2 (1.6) + 0 = 3.2

Section B · Extended response

Questions 10–12 · 53 marks

Question 10

16 marks

A rational function, a quadratic and the product of solutions of f = g

Functions

(a)

5 marks

$f (x) = \frac{4 x + 2}{x - 2}$ . The $y$ -intercept is $f (0) = - 1$ and the $x$ -intercept is where $4 x + 2 = 0$ .
The vertical asymptote is $x = 2$ and the horizontal asymptote is $y = 4$ .
Sketch: two branches, one in the upper right of the asymptotes (for $x > 2$ , coming down from $+ \infty$ towards $y = 4$ ) and one in the lower left (for $x < 2$ , through $(- \frac{1}{2}, 0)$ and $(0, - 1)$ , tending to $- \infty$ as $x \to 2^{-}$ and to 4 from below as $x \to - \infty$ ).

(- \frac{1}{2}, 0), (0, - 1)

(b)

1 mark

The function takes every value except the horizontal asymptote.

f (x) \in R, f (x) \neq = 4

(c)

1 mark

The axis of symmetry is halfway between the roots.

\frac{- \frac{1}{2} + p}{2} = 2 \Rightarrow p = \frac{9}{2}

(d)

3 marks

Write $g (x) = (x + \frac{1}{2}) (x - \frac{9}{2})$ and expand.

g (x) = x^{2} - 4 x - \frac{9}{4} \Rightarrow b = - 4, c = - \frac{9}{4}

(e)

2 marks

The vertex is on the axis of symmetry, $x = 2$ .

g (2) = 4 - 8 - \frac{9}{4} = - \frac{25}{4}

(f)

4 marks

Set $f (x) = g (x)$ and multiply by $x - 2$ .
The product of the roots of $x^{3} + b x^{2} + c x + d = 0$ is $- d$ . (The root $x = 2$ is not a root of this cubic, since $2^{3} - 24 + 3.5 + 2.5 = - 10 \neq = 0$ .)

4 x + 2 = (x - 2) (x^{2} - 4 x - \frac{9}{4}) = x^{3} - 6 x^{2} + \frac{7}{4} x + \frac{9}{2} x^{3} - 6 x^{2} + \frac{7}{4} x + \frac{5}{2} = 0 product = - \frac{5}{2}

Question 11

17 marks

A cubic with a factor, partial fractions, integration and limits

Number and algebra
Calculus

(a)

2 marks

Evaluate $P (- 1)$ .
So $(x + 1)$ is a factor.

P (- 1) = 3 (- 1)^{3} + 5 (- 1)^{2} + (- 1) - 1 = - 3 + 5 - 1 - 1 = 0

(b)

3 marks

Divide $P (x)$ by $(x + 1)$ .

P (x) = (x + 1) (3 x^{2} + 2 x - 1) = (x + 1) (x + 1) (3 x - 1) P (x) = (x + 1)^{2} (3 x - 1)

(c)

3 marks

Write $\frac{1}{( x + 1 ) ( 2 x + 1 )} = \frac{A}{x + 1} + \frac{B}{2 x + 1}$ , so $1 = A (2 x + 1) + B (x + 1)$ .
Put $x = - 1$ : $1 = - A$ . Put $x = - \frac{1}{2}$ : $1 = \frac{1}{2} B$ .

A = - 1, B = 2 \frac{1}{Q ( x )} = - \frac{1}{x + 1} + \frac{2}{2 x + 1}

(d)

2 marks

Divide the result of (c) by $(x + 1)$ .
Use (c) again for the last fraction.

\frac{1}{( x + 1 ) Q ( x )} = - \frac{1}{( x + 1 ) ^{2}} + \frac{2}{( x + 1 ) ( 2 x + 1 )} \frac{2}{( x + 1 ) ( 2 x + 1 )} = - \frac{2}{x + 1} + \frac{4}{2 x + 1} \frac{1}{( x + 1 ) Q ( x )} = \frac{4}{2 x + 1} - \frac{2}{x + 1} - \frac{1}{( x + 1 ) ^{2}}

(e)

4 marks

Integrate each term.

\int [\frac{4}{2 x + 1} - \frac{2}{x + 1} - \frac{1}{( x + 1 ) ^{2}}] d x = 2 ln ∣2 x + 1∣ - 2 ln ∣ x + 1∣ + \frac{1}{x + 1} + c

(f)

3 marks

Cancel the common factor: for $x \neq = - 1$ , $f (x) = \frac{( x + 1 ) ^{2} ( 3 x - 1 )}{( x + 1 ) ^{2} ( 2 x + 1 )} = \frac{3 x - 1}{2 x + 1}$ .

x \to - 1 lim f (x) = \frac{- 4}{- 1} = 4 x \to \infty lim f (x) = \frac{3}{2}

Question 12

20 marks

Cube roots of -8, a triangle, a rotation and the smallest n

Number and algebra
Geometry and trigonometry

(a)

3 marks

Expand the cube and collect the real and imaginary parts.

(a + bi)^{3} = a^{3} + 3 a^{2} bi - 3 a b^{2} - b^{3} i Re = a^{3} - 3 a b^{2}, Im = 3 a^{2} b - b^{3}

(b)

2 marks

Use $a = 1$ and $b = 3$ in the results of (a).
So $(1 + 3 i)^{3} = - 8$ .

Re = 1 - 3 (1) (3) = - 8, Im = 3 (1) 3 - 33 = 0

(c)

2 marks

Since $u = 1 + 3 i$ is a root of $z^{3} = - 8$ , the other two are the real root and the conjugate of $u$ .

v = - 2, w = 1 - 3 i

(d)

3 marks

The points are $U (1, 3)$ , $V (- 2, 0)$ and $W (1, - 3)$ . Take $[U W]$ as the base, a vertical segment of length $23$ .
The height is the horizontal distance from $V$ to the line $x = 1$ , which is 3.

Area = \frac{1}{2} \times 23 \times 3 = 33

(e)

4 marks

Write each root as $r e^{i θ}$ : $u = 2 e^{iπ /3}$ , $v = 2 e^{iπ}$ and $w = 2 e^{- iπ /3}$ . Multiply each by $e^{iπ /4}$ .
The arguments are reduced to the interval $- π < θ \leq π$ , which is why $v^{'}$ is written with $- \frac{3 π}{4}$ .

u^{'} = 2 e^{i 7 π /12} v^{'} = 2 e^{i 5 π /4} = 2 e^{- i 3 π /4} w^{'} = 2 e^{- i π /12}

(f)

3 marks

Cube one of the new roots: $u^{'3} = 8 e^{i 7 π /4} = 8 e^{- iπ /4}$ .

8 (cos \frac{π}{4} - i sin \frac{π}{4}) = 42 - 42 i c = 42, d = - 42

(g)

3 marks

All six numbers have modulus 2. In degrees, their arguments are $10 5^{\circ}$ , $- 13 5^{\circ}$ , $- 1 5^{\circ}$ and $6 0^{\circ}$ , $18 0^{\circ}$ , $- 6 0^{\circ}$ .
Sorted: $- 13 5^{\circ}$ , $- 6 0^{\circ}$ , $- 1 5^{\circ}$ , $6 0^{\circ}$ , $10 5^{\circ}$ , $18 0^{\circ}$ . The gaps are $7 5^{\circ}$ and $4 5^{\circ}$ , alternately, and each is a multiple of $1 5^{\circ}$ .
The solutions of $z^{n} = a$ sit at equal angles of $\frac{36 0 ^{\circ}}{n}$ . All six points must fit on such a polygon, so $\frac{36 0 ^{\circ}}{n}$ divides both 75 and 45, the largest value being $1 5^{\circ}$ .

n = \frac{360}{15} = 24

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 algebra45 marks · 41%
Q5, Q6, Q7, Q11, Q12
Functions30 marks · 27%
Q2, Q4, Q9, Q10
Calculus18 marks · 16%
Q4, Q8, Q11
Geometry and trigonometry11 marks · 10%
Q3, Q12
Statistics and probability6 marks · 5%
Q1

Mentor insights

Traps to watch for

01
Sector formulas · Q3
Perimeter $2 r + r θ$ and area $\frac{1}{2} r^{2} θ$ , with $θ$ in radians.
02
Double angle · Q4
Replace $sin 2 x$ with $2 sin x cos x$ and factorise rather than dividing by $cos x$ .
03
Intercepts and asymptotes · Q10
Intercepts by setting $x$ or $y$ to 0. Asymptotes from the denominator and the ratio of leading coefficients.
04
Factor theorem · Q11
$(x + 1)$ is a factor if $P (- 1) = 0$ .
05
Expand a cube · Q12
$(a + bi)^{3} = a^{3} + 3 a^{2} (bi) + 3 a (bi)^{2} + (bi)^{3}$ , with $i^{2} = - 1$ and $i^{3} = - i$ .

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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 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 · 57 marks

Question 1

6 marks

A die rolled sixteen times: finding two frequencies from the mean

Statistics and probability

(a)

5 marks

The frequencies add to 16.
The mean is 3, so the total of the scores is $3 \times 16 = 48$ .
Subtract the two equations.

p + q + 4 + 2 + 0 + 3 = 16 \Rightarrow p + q = 7 p + 2 q + 12 + 8 + 0 + 18 = 48 \Rightarrow p + 2 q = 10 q = 3, p = 4

(b)

1 mark

Multiplying each score by 10 multiplies the mean by 10.

3 \times 10 = 30

Question 2

5 marks

Logarithms of 1/a and with base 1000 given log of a

Functions

(a)

2 marks

Use $lo g_{10} \frac{1}{a} = - lo g_{10} a$ with $lo g_{10} a = \frac{1}{3}$ .

lo g_{10} \frac{1}{a} = - \frac{1}{3}

(b)

3 marks

Change the base: $lo g_{1000} a = \frac{lo g _{10} a}{lo g _{10} 1000}$ .

lo g_{1000} a = \frac{1/3}{3} = \frac{1}{9}

Question 3

8 marks

A circular sector with perimeter 10 and area 6.25

Geometry and trigonometry

(a)

4 marks

Write the two conditions.
From the first, $r θ = 10 - 2 r$ . Substitute into the area.

2 r + r θ = 10, \frac{1}{2} r^{2} θ = 6.25 \frac{1}{2} r (10 - 2 r) = 6.25 \Rightarrow 5 r - r^{2} = 6.25 \Rightarrow 4 r^{2} - 20 r + 25 = 0

(b)

4 marks

Factorise: $(2 r - 5)^{2} = 0$ .
Then $r θ = 10 - 5 = 5$ .

r = \frac{5}{2} θ = \frac{5}{2.5} = 2

Question 4

7 marks

Intersections of cos x and sin 2x and the area between them

Functions
Calculus

(a)

3 marks

Set $cos x = sin 2 x = 2 sin x cos x$ .
The point B at $x = \frac{π}{2}$ comes from $cos x = 0$ . The other solutions in $0 \leq x \leq π$ are

cos x (1 - 2 sin x) = 0 \Rightarrow cos x = 0 or sin x = \frac{1}{2} x_{A} = \frac{π}{6}, x_{C} = \frac{5 π}{6}

(b)

4 marks

Between B and C the curve of $f$ lies above $g$ .

Area = \int_{π /2}^{5 π /6} (cos x - sin 2 x) d x = [sin x + \frac{1}{2} cos 2 x]_{π /2}^{5 π /6} (\frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2}) - (1 - \frac{1}{2}) = \frac{3}{4} - \frac{1}{2} = \frac{1}{4}

Question 5

5 marks

Sums of the geometric sequence with ratio 10 and a sum of sums

Number and algebra

(a)

1 mark

Use $S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1}$ with $a = 1$ and $r = 10$ .

S_{n} = \frac{1 0 ^{n} - 1}{9}

(b)

4 marks

Add the first $n$ sums.
The geometric series has first term 10, ratio 10 and $n$ terms.

S_{1} + S_{2} + \dots + S_{n} = \frac{10 - 1}{9} + \frac{1 0 ^{2} - 1}{9} + \dots + \frac{1 0 ^{n} - 1}{9} = \frac{1}{9} [(10 + 1 0^{2} + \dots + 1 0^{n}) - n] 10 + 1 0^{2} + \dots + 1 0^{n} = \frac{10 ( 1 0 ^{n} - 1 )}{9} S_{1} + \dots + S_{n} = \frac{1}{9} [\frac{10 ( 1 0 ^{n} - 1 )}{9} - n] = \frac{1 0 ^{n + 1} - 10 - 9 n}{81}

Question 6

6 marks

Volume of revolution of sqrt(x sin(x^2))

Number and algebra

6 marks

The volume of revolution about the $x$ -axis is
Integrate by inspection (or put $u = x^{2}$ , $d u = 2 x d x$ ).
Substitute $cos \frac{π}{4} = \frac{2}{2}$ .

V = π \int_{0}^{π /2} (x sin (x^{2}))^{2} d x = π \int_{0}^{π /2} x sin (x^{2}) d x V = π [- \frac{1}{2} cos (x^{2})]_{0}^{π /2} = \frac{π}{2} (1 - cos \frac{π}{4}) V = \frac{π}{2} (1 - \frac{2}{2}) = \frac{π ( 2 - 2 )}{4}

Question 7

7 marks

Induction: a sum of binomial coefficients equals another binomial coefficient

Number and algebra

7 marks

Claim: $r = 1 \sum n (1 r) = (2 n + 1)$ for all $n \in Z^{+}$ .
Base case $n = 1$ : the left side is $(1 1) = 1$ and the right side is $(2 2) = 1$ . It holds.
Assume it holds for $n = k$ : $r = 1 \sum k (1 r) = (2 k + 1)$ .
For $n = k + 1$ , split off the last term and use the assumption.
Put both fractions over $2! k!$ .
This is the statement for $n = k + 1$ . The statement is true for $n = 1$ and, if true for $n = k$ , it is true for $n = k + 1$ . By mathematical induction it is true for all $n \in Z^{+}$ .

r = 1 \sum k + 1 (1 r) = (2 k + 1) + (1 k + 1) = \frac{( k + 1 )!}{2 ! ( k - 1 )!} + \frac{( k + 1 )!}{1 ! k !} \frac{( k + 1 )! k}{2 ! k !} + \frac{( k + 1 )! \cdot 2}{2 ! k !} = \frac{( k + 1 )! ( k + 2 )}{2 ! k !} = \frac{( k + 2 )!}{2 ! k !} = (2 k + 2)

Question 8

7 marks

Maclaurin series of sin(x^2), sin^2(x^2) and a derived series

Calculus

(a)

5 marks

Replace $x$ by $x^{2}$ in the series for sine.
Square the series.

sin (x^{2}) = x^{2} - \frac{x ^{6}}{6} + \dots sin^{2} (x^{2}) = (x^{2} - \frac{x ^{6}}{6})^{2} = x^{4} - \frac{x ^{8}}{3} + \dots

(b)

2 marks

Note that $\frac{d}{d x} sin^{2} (x^{2}) = 2 sin (x^{2}) cos (x^{2}) \cdot 2 x = 4 x sin (x^{2}) cos (x^{2})$ . Differentiate the series from (a(ii)).

4 x sin (x^{2}) cos (x^{2}) = 4 x^{3} - \frac{8 x ^{7}}{3} + \dots

Question 9

6 marks

Sketching f(|x|), an odd function and integrals

Functions

(a)

2 marks

Reflect all the parts of the graph below the $x$ -axis in the $x$ -axis.
Sketch: a symmetric curve about the $y$ -axis with sharp points (cusps) at the $x$ -intercepts. The small dips near $x = \pm 1$ become small humps, lower than the maxima at $x = \pm 3.5$ , and the ends, which fell below $- 4$ near $x = \pm 6$ , are reflected to rise steeply above $4$ .

(b)

2 marks

For $x \geq 0$ the graph of $f (x)$ is the same as $f (∣ x ∣)$ . For $x < 0$ use the half turn about the origin (odd symmetry).
Sketch: the right half is unchanged: a dip below the axis near $x = 1$ , a maximum near $x = 3.5$ and a fall through the axis near $x = 5.3$ . The left half is the right half rotated by $18 0^{\circ}$ about the origin.

(c)

2 marks

Since $f$ is an odd function, $\int_{- 4}^{0} f (x) d x = - \int_{0}^{4} f (x) d x$ , and for $x \geq 0$ , $f (∣ x ∣) = f (x)$ so $\int_{0}^{4} f (x) d x = 1.6$ .
Over the symmetric range, the integral of the odd part is 0 and the integral of $f (∣ x ∣)$ is twice $\int_{0}^{4}$ .

\int_{- 4}^{0} f (x) d x = - 1.6 \int_{- 4}^{4} (f (∣ x ∣) + f (x)) d x = 2 (1.6) + 0 = 3.2

Section B · Extended response

Questions 10–12 · 53 marks

Question 10

16 marks

A rational function, a quadratic and the product of solutions of f = g

Functions

(a)

5 marks

$f (x) = \frac{4 x + 2}{x - 2}$ . The $y$ -intercept is $f (0) = - 1$ and the $x$ -intercept is where $4 x + 2 = 0$ .
The vertical asymptote is $x = 2$ and the horizontal asymptote is $y = 4$ .
Sketch: two branches, one in the upper right of the asymptotes (for $x > 2$ , coming down from $+ \infty$ towards $y = 4$ ) and one in the lower left (for $x < 2$ , through $(- \frac{1}{2}, 0)$ and $(0, - 1)$ , tending to $- \infty$ as $x \to 2^{-}$ and to 4 from below as $x \to - \infty$ ).

(- \frac{1}{2}, 0), (0, - 1)

(b)

1 mark

The function takes every value except the horizontal asymptote.

f (x) \in R, f (x) \neq = 4

(c)

1 mark

The axis of symmetry is halfway between the roots.

\frac{- \frac{1}{2} + p}{2} = 2 \Rightarrow p = \frac{9}{2}

(d)

3 marks

Write $g (x) = (x + \frac{1}{2}) (x - \frac{9}{2})$ and expand.

g (x) = x^{2} - 4 x - \frac{9}{4} \Rightarrow b = - 4, c = - \frac{9}{4}

(e)

2 marks

The vertex is on the axis of symmetry, $x = 2$ .

g (2) = 4 - 8 - \frac{9}{4} = - \frac{25}{4}

(f)

4 marks

Set $f (x) = g (x)$ and multiply by $x - 2$ .
The product of the roots of $x^{3} + b x^{2} + c x + d = 0$ is $- d$ . (The root $x = 2$ is not a root of this cubic, since $2^{3} - 24 + 3.5 + 2.5 = - 10 \neq = 0$ .)

4 x + 2 = (x - 2) (x^{2} - 4 x - \frac{9}{4}) = x^{3} - 6 x^{2} + \frac{7}{4} x + \frac{9}{2} x^{3} - 6 x^{2} + \frac{7}{4} x + \frac{5}{2} = 0 product = - \frac{5}{2}

Question 11

17 marks

A cubic with a factor, partial fractions, integration and limits

Number and algebra
Calculus

(a)

2 marks

Evaluate $P (- 1)$ .
So $(x + 1)$ is a factor.

P (- 1) = 3 (- 1)^{3} + 5 (- 1)^{2} + (- 1) - 1 = - 3 + 5 - 1 - 1 = 0

(b)

3 marks

Divide $P (x)$ by $(x + 1)$ .

P (x) = (x + 1) (3 x^{2} + 2 x - 1) = (x + 1) (x + 1) (3 x - 1) P (x) = (x + 1)^{2} (3 x - 1)

(c)

3 marks

Write $\frac{1}{( x + 1 ) ( 2 x + 1 )} = \frac{A}{x + 1} + \frac{B}{2 x + 1}$ , so $1 = A (2 x + 1) + B (x + 1)$ .
Put $x = - 1$ : $1 = - A$ . Put $x = - \frac{1}{2}$ : $1 = \frac{1}{2} B$ .

A = - 1, B = 2 \frac{1}{Q ( x )} = - \frac{1}{x + 1} + \frac{2}{2 x + 1}

(d)

2 marks

Divide the result of (c) by $(x + 1)$ .
Use (c) again for the last fraction.

\frac{1}{( x + 1 ) Q ( x )} = - \frac{1}{( x + 1 ) ^{2}} + \frac{2}{( x + 1 ) ( 2 x + 1 )} \frac{2}{( x + 1 ) ( 2 x + 1 )} = - \frac{2}{x + 1} + \frac{4}{2 x + 1} \frac{1}{( x + 1 ) Q ( x )} = \frac{4}{2 x + 1} - \frac{2}{x + 1} - \frac{1}{( x + 1 ) ^{2}}

(e)

4 marks

Integrate each term.

\int [\frac{4}{2 x + 1} - \frac{2}{x + 1} - \frac{1}{( x + 1 ) ^{2}}] d x = 2 ln ∣2 x + 1∣ - 2 ln ∣ x + 1∣ + \frac{1}{x + 1} + c

(f)

3 marks

Cancel the common factor: for $x \neq = - 1$ , $f (x) = \frac{( x + 1 ) ^{2} ( 3 x - 1 )}{( x + 1 ) ^{2} ( 2 x + 1 )} = \frac{3 x - 1}{2 x + 1}$ .

x \to - 1 lim f (x) = \frac{- 4}{- 1} = 4 x \to \infty lim f (x) = \frac{3}{2}

Question 12

20 marks

Cube roots of -8, a triangle, a rotation and the smallest n

Number and algebra
Geometry and trigonometry

(a)

3 marks

Expand the cube and collect the real and imaginary parts.

(a + bi)^{3} = a^{3} + 3 a^{2} bi - 3 a b^{2} - b^{3} i Re = a^{3} - 3 a b^{2}, Im = 3 a^{2} b - b^{3}

(b)

2 marks

Use $a = 1$ and $b = 3$ in the results of (a).
So $(1 + 3 i)^{3} = - 8$ .

Re = 1 - 3 (1) (3) = - 8, Im = 3 (1) 3 - 33 = 0

(c)

2 marks

Since $u = 1 + 3 i$ is a root of $z^{3} = - 8$ , the other two are the real root and the conjugate of $u$ .

v = - 2, w = 1 - 3 i

(d)

3 marks

The points are $U (1, 3)$ , $V (- 2, 0)$ and $W (1, - 3)$ . Take $[U W]$ as the base, a vertical segment of length $23$ .
The height is the horizontal distance from $V$ to the line $x = 1$ , which is 3.

Area = \frac{1}{2} \times 23 \times 3 = 33

(e)

4 marks

Write each root as $r e^{i θ}$ : $u = 2 e^{iπ /3}$ , $v = 2 e^{iπ}$ and $w = 2 e^{- iπ /3}$ . Multiply each by $e^{iπ /4}$ .
The arguments are reduced to the interval $- π < θ \leq π$ , which is why $v^{'}$ is written with $- \frac{3 π}{4}$ .

u^{'} = 2 e^{i 7 π /12} v^{'} = 2 e^{i 5 π /4} = 2 e^{- i 3 π /4} w^{'} = 2 e^{- i π /12}

(f)

3 marks

Cube one of the new roots: $u^{'3} = 8 e^{i 7 π /4} = 8 e^{- iπ /4}$ .

8 (cos \frac{π}{4} - i sin \frac{π}{4}) = 42 - 42 i c = 42, d = - 42

(g)

3 marks

All six numbers have modulus 2. In degrees, their arguments are $10 5^{\circ}$ , $- 13 5^{\circ}$ , $- 1 5^{\circ}$ and $6 0^{\circ}$ , $18 0^{\circ}$ , $- 6 0^{\circ}$ .
Sorted: $- 13 5^{\circ}$ , $- 6 0^{\circ}$ , $- 1 5^{\circ}$ , $6 0^{\circ}$ , $10 5^{\circ}$ , $18 0^{\circ}$ . The gaps are $7 5^{\circ}$ and $4 5^{\circ}$ , alternately, and each is a multiple of $1 5^{\circ}$ .
The solutions of $z^{n} = a$ sit at equal angles of $\frac{36 0 ^{\circ}}{n}$ . All six points must fit on such a polygon, so $\frac{36 0 ^{\circ}}{n}$ divides both 75 and 45, the largest value being $1 5^{\circ}$ .

n = \frac{360}{15} = 24

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 algebra45 marks · 41%
Q5, Q6, Q7, Q11, Q12
Functions30 marks · 27%
Q2, Q4, Q9, Q10
Calculus18 marks · 16%
Q4, Q8, Q11
Geometry and trigonometry11 marks · 10%
Q3, Q12
Statistics and probability6 marks · 5%
Q1

Mentor insights

Traps to watch for

01
Sector formulas · Q3
Perimeter $2 r + r θ$ and area $\frac{1}{2} r^{2} θ$ , with $θ$ in radians.
02
Double angle · Q4
Replace $sin 2 x$ with $2 sin x cos x$ and factorise rather than dividing by $cos x$ .
03
Intercepts and asymptotes · Q10
Intercepts by setting $x$ or $y$ to 0. Asymptotes from the denominator and the ratio of leading coefficients.
04
Factor theorem · Q11
$(x + 1)$ is a factor if $P (- 1) = 0$ .
05
Expand a cube · Q12
$(a + bi)^{3} = a^{3} + 3 a^{2} (bi) + 3 a (bi)^{2} + (bi)^{3}$ , with $i^{2} = - 1$ and $i^{3} = - i$ .

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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