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Worked solutions

IB Maths AA HL · Paper 1

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2022
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 2022 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 · 56 marks

Question 1

5 marks

An arithmetic sequence from its general term

Number and algebra

(a)

1 mark

Put $n = 1$ .

u_{1} = 15 - 3 = 12

(b)

2 marks

Solve $15 - 3 n = - 33$ .

3 n = 48 \Rightarrow n = 16

(c)

2 marks

The difference between consecutive terms.

d = u_{2} - u_{1} = 9 - 12 = - 3

Question 2

6 marks

Sums of consecutive integers and of their squares

Number and algebra

(a)

2 marks

Add the three integers.
This is a multiple of 3, so it is always divisible by 3.

(n - 1) + n + (n + 1) = 3 n

(b)

4 marks

Expand the squares and add.
The term $3 n^{2}$ is divisible by 3, but 2 is not, so the sum leaves remainder 2 on division by 3. It is never divisible by 3.

(n - 1)^{2} + n^{2} + (n + 1)^{2} = n^{2} - 2 n + 1 + n^{2} + n^{2} + 2 n + 1 = 3 n^{2} + 2

Question 3

8 marks

A rational function: asymptotes, a sketch and inequalities with a modulus

Functions

(a)

2 marks

(i) The denominator is zero at

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

x = - 1 y = 2

(b)

3 marks

The $x$ -intercept is where $2 x - 1 = 0$ , and the $y$ -intercept is $f (0)$ .
Draw the asymptotes $x = - 1$ and $y = 2$ . The right branch passes through $(0, - 1)$ and $(\frac{1}{2}, 0)$ and rises towards $y = 2$ from below as $x \to \infty$ , falling to $- \infty$ as $x \to - 1^{+}$ . The left branch is above $y = 2$ , coming down from $+ \infty$ as $x \to - 1^{-}$ towards $y = 2$ as $x \to - \infty$ .

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

(c)

1 mark

From the graph, $f (x)$ is between 0 and 2 only on the right branch after the $x$ -intercept.

x > \frac{1}{2}

(d)

2 marks

Now $f$ is replaced by $\frac{2∣ x ∣ - 1}{∣ x ∣ + 1}$ , which is $f (∣ x ∣)$ . By part (c) the inequality holds when $∣ x ∣ > \frac{1}{2}$ .

x < - \frac{1}{2} or x > \frac{1}{2}

Question 4

5 marks

The least positive solution of a cosine equation

Number and algebra

5 marks

Cosine equals $\frac{1}{2}$ when the angle is $\frac{π}{4}$ (reference angle), or $- \frac{π}{4}$ , then repeating every $2 π$ . We need $\frac{x}{2} + \frac{π}{3} \geq \frac{π}{3}$ , so $\frac{π}{4}$ would give a negative $x$ and is rejected.
The next angle with the same cosine is $2 π - \frac{π}{4} = \frac{7 π}{4}$ .

\frac{x}{2} + \frac{π}{3} = \frac{π}{4} \Rightarrow x < 0 \frac{x}{2} = \frac{7 π}{4} - \frac{π}{3} = \frac{17 π}{12} \Rightarrow x = \frac{17 π}{6}

Question 5

7 marks

A binomial expansion: a coefficient and terms in a mean

Number and algebra

(a)

2 marks

The term in $x^{5}$ in $(x + 1)^{7}$ is $(2 7) x^{5} \cdot 1^{2}$ .

b = (2 7) = \frac{7 \times 6}{2} = 21

(b)

5 marks

The coefficient $a = (1 7) = 7$ . The terms are $7 x^{6}$ , $21 x^{5}$ and $35 x^{4}$ . The third term is the mean of the second and fourth.
Move everything to one side and divide by $7 x^{4}$ (valid since $x \neq = 0$ ).

21 x^{5} = \frac{7 x ^{6} + 35 x ^{4}}{2} \Rightarrow 42 x^{5} = 7 x^{6} + 35 x^{4} x^{2} - 6 x + 5 = 0 \Rightarrow (x - 1) (x - 5) = 0 \Rightarrow x = 1 or x = 5

Question 6

8 marks

An odd function with a square root: proof and range

Functions

(a)

2 marks

Substitute $- x$ .
So $f$ is an odd function.

f (- x) = (- x) 1 - (- x)^{2} = - x 1 - x^{2} = - f (x)

(b)

6 marks

Differentiate with the product and chain rules.
Set $f^{'} (x) = 0$ to find the stationary points.
Evaluate $f$ at these points.
The ends have $f (\pm 1) = 0$ , so the range is

f^{'} (x) = 1 - x^{2} + x \cdot \frac{- 2 x}{2 1 - x ^{2}} = \frac{1 - 2 x ^{2}}{1 - x ^{2}} 1 - 2 x^{2} = 0 \Rightarrow x = \pm \frac{1}{2} f (\frac{1}{2}) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2}, f (- \frac{1}{2}) = - \frac{1}{2} a = - \frac{1}{2}, b = \frac{1}{2}

Question 7

6 marks

Integrating a power of secant times tangent

Number and algebra

6 marks

Write $sec^{n} x tan x = sec^{n - 1} x \cdot sec x tan x$ and let $u = sec x$ , so $d u = sec x tan x d x$ . The limits $0$ and $\frac{π}{3}$ become $u = 1$ and $u = 2$ .

\int_{0}^{π /3} sec^{n} x tan x d x = \int_{1}^{2} u^{n - 1} d u = [\frac{u ^{n}}{n}]_{1}^{2} \frac{2 ^{n} - 1}{n}

Question 8

6 marks

The median of a triangular distribution

Number and algebra

6 marks

The rising part of the density is the straight line from $(a, 0)$ up to $x = c$ . Let $m$ be the median, so that the area from $a$ to $m$ is $\frac{1}{2}$ . This area is a triangle with base $m - a$ and height $\frac{2 ( m - a )}{( b - a ) ( c - a )}$ .
Solve for $m$ , taking the positive root since $m > a$ .

\frac{1}{2} (m - a) \cdot \frac{2 ( m - a )}{( b - a ) ( c - a )} = \frac{( m - a ) ^{2}}{( b - a ) ( c - a )} = \frac{1}{2} (m - a)^{2} = \frac{( b - a ) ( c - a )}{2} \Rightarrow m = a + \frac{( b - a ) ( c - a )}{2}

Question 9

5 marks

Proof by contradiction: no integer root

Number and algebra

5 marks

Assume the equation has an integer root $α$ , so $2 α^{3} + 6 α + 1 = 0$ . Rearrange.
Since $α$ is an integer, $α^{3} + 3 α$ is an integer and the left side is even. The right side, $- 1$ , is odd. This is a contradiction.
So the equation $2 x^{3} + 6 x + 1 = 0$ has no integer roots.

2 α^{3} + 6 α = - 1 \Rightarrow 2 (α^{3} + 3 α) = - 1

Section B · Extended response

Questions 10–12 · 54 marks

Question 10

16 marks

A biased die, a dice game, and two pairs of dice with the same sum distribution

Statistics and probability

(a)

5 marks

The probabilities add to 1, and the mean is 2.
Subtract the first equation from the second: $2 q = 0.4$ .

p + 0.3 + q + 0.1 = 1 \Rightarrow p + q = 0.6 E (X) = p + 0.6 + 3 q + 0.4 = 2 \Rightarrow p + 3 q = 1 q = 0.2, p = 0.4

(b)

2 marks

$X > 2$ means 3 or 4.

P (X > 2) = 0.2 + 0.1 = 0.3

(c)

5 marks

With a score of 4 after three rolls, Nicky needs at least 6 more from her last two rolls, which can total 6, 7 or 8.
Add them.

P (total 6) = 2 (0.3) (0.1) + 0. 2^{2} = 0.1, P (total 7) = 2 (0.2) (0.1) = 0.04, P (total 8) = 0. 1^{2} = 0.01 P (wins) = 0.1 + 0.04 + 0.01 = 0.15

(d)

2 marks

The largest sum for the yellow pair is 8, and for the red pair it is $3 + b$ .

3 + b = 8 \Rightarrow b = 5

(e)

2 marks

Compare the probability of a sum of 5. For the yellow pair $P (S = 5) = \frac{4}{16}$ . For the red pair a sum of 5 comes from the 2 on the first die (probability $\frac{2}{4}$ ) with the $a$ on the second die (probability $\frac{2}{4}$ ) when $a + 2 = 5$ , which gives $\frac{1}{2} \cdot \frac{1}{2} = \frac{4}{16}$ .
Check the other sums: with faces ${1, 2, 2, 3}$ and ${1, 3, 3, 5}$ the sums 2 to 8 have probabilities $\frac{1}{16}, \frac{2}{16}, \frac{3}{16}, \frac{4}{16}, \frac{3}{16}, \frac{2}{16}, \frac{1}{16}$ , as required.

a = 3

Question 11

20 marks

A reciprocal quadratic: its graph, the inverse of a branch, and a composite with arctan

Functions

(a)

6 marks

Since $f (x) = \frac{1}{( x - 3 ) ( x + 1 )}$ , the vertical asymptotes are $x = - 1$ and $x = 3$ , and the horizontal asymptote is $y = 0$ . The $y$ -intercept is $f (0)$ , and there are no $x$ -intercepts.
The denominator is smallest (most negative) halfway between the roots, at $x = 1$ .
The sketch has three branches: for $x < - 1$ it is positive, falling from $+ \infty$ at $x = - 1^{-}$ towards 0 as $x \to - \infty$ . In the middle it is negative: from $- \infty$ at $x = - 1^{+}$ up to the maximum $(1, - \frac{1}{4})$ through $(0, - \frac{1}{3})$ and back to $- \infty$ at $x = 3^{-}$ . For $x > 3$ it is positive, from $+ \infty$ at $x = 3^{+}$ falling towards 0.

f (0) = - \frac{1}{3} local maximum at (1, - \frac{1}{4})

(b)

7 marks

(i) Interchange $x$ and $y$ in $y = \frac{1}{x ^{2} - 2 x - 3}$ to get $x = \frac{1}{y ^{2} - 2 y - 3}$ , and solve for $y$ as a quadratic.

Since the original domain is $x > 3$ , the output $y$ of the inverse is greater than 3, so the minus sign is rejected.

(ii) The domain of $g^{- 1}$ is the range of $g$ . For $x > 3$ the values of $g$ fall from $+ \infty$ towards 0.

x y^{2} - 2 x y - (3 x + 1) = 0 \Rightarrow y = \frac{2 x \pm 4 x ^{2} + 4 x ( 3 x + 1 )}{2 x} = \frac{x \pm 4 x ^{2} + x}{x} g^{- 1} (x) = \frac{x + 4 x ^{2} + x}{x} = 1 + \frac{4 x ^{2} + x}{x} x > 0

(c)

7 marks

$(h \circ g) (a) = arctan \frac{g ( a )}{2} = \frac{π}{4}$ , so $\frac{g ( a )}{2} = 1$ .
Solve the quadratic and keep the root with $a > 3$ .

g (a) = 2 \Rightarrow \frac{1}{a ^{2} - 2 a - 3} = 2 \Rightarrow 2 a^{2} - 4 a - 7 = 0 a = \frac{4 \pm 72}{4} = 1 \pm \frac{3 2}{2} \Rightarrow a = 1 + \frac{3}{2} 2 p = 1, q = 3, r = 2

Question 12

18 marks

Equilateral triangles from complex numbers and the roots of a quadratic

Number and algebra

(a)

2 marks

The conjugate of $z_{2} = r_{2} e^{i θ}$ is $z_{2}^{*} = r_{2} e^{- i θ}$ .

z_{1} z_{2}^{*} = r_{1} e^{i α} \cdot r_{2} e^{- i θ} = r_{1} r_{2} e^{i (α - θ)}

(b)

2 marks

The real part is $r_{1} r_{2} cos (α - θ)$ . Setting it to 0 with $r_{1} r_{2} > 0$ :
So the angle at O is a right angle and the triangle $Z_{1} O Z_{2}$ is right-angled.

cos (α - θ) = 0 \Rightarrow α - θ = \frac{π}{2} (since 0 < α - θ < π)

(c)

6 marks

(i) In an equilateral triangle $r_{1} = r_{2}$ and the angle at O is $\frac{π}{3}$ , so $α = θ + \frac{π}{3}$ .

(ii) Substitute into $z_{1}^{2} + z_{2}^{2}$ .

Since $1 + e^{2 π i /3} = e^{i π /3}$ (the sum of the two vectors has length 1 at angle $\frac{π}{3}$ ):

z_{1} = r_{2} e^{i (θ + π /3)} = z_{2} e^{i π /3} z_{1}^{2} + z_{2}^{2} = z_{2}^{2} e^{2 π i /3} + z_{2}^{2} = z_{2}^{2} (1 + e^{2 π i /3}) z_{1}^{2} + z_{2}^{2} = z_{2}^{2} e^{i π /3} = z_{2} \cdot z_{2} e^{i π /3} = z_{2} z_{1}

(d)

5 marks

For roots $z_{1}$ and $z_{2}$ of $z^{2} + a z + b = 0$ : $z_{1} + z_{2} = - a$ and $z_{1} z_{2} = b$ . Square the sum.
By part (c)(ii), $z_{1}^{2} + z_{2}^{2} = z_{1} z_{2} = b$ .

a^{2} = (z_{1} + z_{2})^{2} = z_{1}^{2} + z_{2}^{2} + 2 z_{1} z_{2} a^{2} = b + 2 b = 3 b \Rightarrow a^{2} - 3 b = 0

(e)

3 marks

With $b = 12$ , $a^{2} = 36$ , so $a = \pm 6$ . The equation is $z^{2} \pm 6 z + 12 = 0$ .
For $a = - 6$ the roots are $3 \pm 3 i$ with arguments $\frac{π}{6}$ and $\frac{11 π}{6}$ , so $α - θ = \pm \frac{5 π}{3}$ , which is not in $(0, π)$ . This labelling is rejected.
For $a = 6$ the roots are $- 3 \pm 3 i$ with arguments $\frac{7 π}{6}$ and $\frac{5 π}{6}$ , and $α - θ = \frac{π}{3}$ , which is in $(0, π)$ .
So exactly one equilateral triangle $Z_{1} O Z_{2}$ satisfies the conditions.

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 algebra58 marks · 53%
Q1, Q2, Q4, Q5, Q7, Q8, Q9, Q12
Functions36 marks · 33%
Q3, Q6, Q11
Statistics and probability16 marks · 15%
Q10

Mentor insights

Traps to watch for

01
Use the graph for inequalities · Q3
In (c) read off where the graph lies between $y = 0$ and the asymptote $y = 2$ . In (d) replace $x$ by $∣ x ∣$ : the graph is reflected, so solutions come in a $\pm$ pair.
02
Odd means $f (- x) = - f (x)$ · Q6
Replace $x$ by $- x$ and compare with $- f (x)$ . For the range, find the stationary points and use the symmetry.
03
Add the ways to reach each total · Q10
In (c) Nicky needs the last two rolls to total at least 6. List each pair of rolls giving 6, 7 and 8, using the probabilities from (a).
04
Choose the branch for the inverse · Q11
The inverse comes from a quadratic with two roots. The condition $x > 3$ on the original function tells you which root to keep.
05
Use the angle at O · Q12
The angle between $O Z_{1}$ and $O Z_{2}$ is $α - θ$ . For an equilateral triangle it is $\frac{π}{3}$ and the two lengths are equal. The condition $0 < α - θ < π$ then decides which labelling is allowed.

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Worked solutions

IB Maths AA HL · Paper 1

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2

Session: May 2022
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 2022 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 · 56 marks

Question 1

5 marks

An arithmetic sequence from its general term

Number and algebra

(a)

1 mark

Put $n = 1$ .

u_{1} = 15 - 3 = 12

(b)

2 marks

Solve $15 - 3 n = - 33$ .

3 n = 48 \Rightarrow n = 16

(c)

2 marks

The difference between consecutive terms.

d = u_{2} - u_{1} = 9 - 12 = - 3

Question 2

6 marks

Sums of consecutive integers and of their squares

Number and algebra

(a)

2 marks

Add the three integers.
This is a multiple of 3, so it is always divisible by 3.

(n - 1) + n + (n + 1) = 3 n

(b)

4 marks

Expand the squares and add.
The term $3 n^{2}$ is divisible by 3, but 2 is not, so the sum leaves remainder 2 on division by 3. It is never divisible by 3.

(n - 1)^{2} + n^{2} + (n + 1)^{2} = n^{2} - 2 n + 1 + n^{2} + n^{2} + 2 n + 1 = 3 n^{2} + 2

Question 3

8 marks

A rational function: asymptotes, a sketch and inequalities with a modulus

Functions

(a)

2 marks

(i) The denominator is zero at

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

x = - 1 y = 2

(b)

3 marks

The $x$ -intercept is where $2 x - 1 = 0$ , and the $y$ -intercept is $f (0)$ .
Draw the asymptotes $x = - 1$ and $y = 2$ . The right branch passes through $(0, - 1)$ and $(\frac{1}{2}, 0)$ and rises towards $y = 2$ from below as $x \to \infty$ , falling to $- \infty$ as $x \to - 1^{+}$ . The left branch is above $y = 2$ , coming down from $+ \infty$ as $x \to - 1^{-}$ towards $y = 2$ as $x \to - \infty$ .

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

(c)

1 mark

From the graph, $f (x)$ is between 0 and 2 only on the right branch after the $x$ -intercept.

x > \frac{1}{2}

(d)

2 marks

Now $f$ is replaced by $\frac{2∣ x ∣ - 1}{∣ x ∣ + 1}$ , which is $f (∣ x ∣)$ . By part (c) the inequality holds when $∣ x ∣ > \frac{1}{2}$ .

x < - \frac{1}{2} or x > \frac{1}{2}

Question 4

5 marks

The least positive solution of a cosine equation

Number and algebra

5 marks

Cosine equals $\frac{1}{2}$ when the angle is $\frac{π}{4}$ (reference angle), or $- \frac{π}{4}$ , then repeating every $2 π$ . We need $\frac{x}{2} + \frac{π}{3} \geq \frac{π}{3}$ , so $\frac{π}{4}$ would give a negative $x$ and is rejected.
The next angle with the same cosine is $2 π - \frac{π}{4} = \frac{7 π}{4}$ .

\frac{x}{2} + \frac{π}{3} = \frac{π}{4} \Rightarrow x < 0 \frac{x}{2} = \frac{7 π}{4} - \frac{π}{3} = \frac{17 π}{12} \Rightarrow x = \frac{17 π}{6}

Question 5

7 marks

A binomial expansion: a coefficient and terms in a mean

Number and algebra

(a)

2 marks

The term in $x^{5}$ in $(x + 1)^{7}$ is $(2 7) x^{5} \cdot 1^{2}$ .

b = (2 7) = \frac{7 \times 6}{2} = 21

(b)

5 marks

The coefficient $a = (1 7) = 7$ . The terms are $7 x^{6}$ , $21 x^{5}$ and $35 x^{4}$ . The third term is the mean of the second and fourth.
Move everything to one side and divide by $7 x^{4}$ (valid since $x \neq = 0$ ).

21 x^{5} = \frac{7 x ^{6} + 35 x ^{4}}{2} \Rightarrow 42 x^{5} = 7 x^{6} + 35 x^{4} x^{2} - 6 x + 5 = 0 \Rightarrow (x - 1) (x - 5) = 0 \Rightarrow x = 1 or x = 5

Question 6

8 marks

An odd function with a square root: proof and range

Functions

(a)

2 marks

Substitute $- x$ .
So $f$ is an odd function.

f (- x) = (- x) 1 - (- x)^{2} = - x 1 - x^{2} = - f (x)

(b)

6 marks

Differentiate with the product and chain rules.
Set $f^{'} (x) = 0$ to find the stationary points.
Evaluate $f$ at these points.
The ends have $f (\pm 1) = 0$ , so the range is

f^{'} (x) = 1 - x^{2} + x \cdot \frac{- 2 x}{2 1 - x ^{2}} = \frac{1 - 2 x ^{2}}{1 - x ^{2}} 1 - 2 x^{2} = 0 \Rightarrow x = \pm \frac{1}{2} f (\frac{1}{2}) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2}, f (- \frac{1}{2}) = - \frac{1}{2} a = - \frac{1}{2}, b = \frac{1}{2}

Question 7

6 marks

Integrating a power of secant times tangent

Number and algebra

6 marks

Write $sec^{n} x tan x = sec^{n - 1} x \cdot sec x tan x$ and let $u = sec x$ , so $d u = sec x tan x d x$ . The limits $0$ and $\frac{π}{3}$ become $u = 1$ and $u = 2$ .

\int_{0}^{π /3} sec^{n} x tan x d x = \int_{1}^{2} u^{n - 1} d u = [\frac{u ^{n}}{n}]_{1}^{2} \frac{2 ^{n} - 1}{n}

Question 8

6 marks

The median of a triangular distribution

Number and algebra

6 marks

The rising part of the density is the straight line from $(a, 0)$ up to $x = c$ . Let $m$ be the median, so that the area from $a$ to $m$ is $\frac{1}{2}$ . This area is a triangle with base $m - a$ and height $\frac{2 ( m - a )}{( b - a ) ( c - a )}$ .
Solve for $m$ , taking the positive root since $m > a$ .

\frac{1}{2} (m - a) \cdot \frac{2 ( m - a )}{( b - a ) ( c - a )} = \frac{( m - a ) ^{2}}{( b - a ) ( c - a )} = \frac{1}{2} (m - a)^{2} = \frac{( b - a ) ( c - a )}{2} \Rightarrow m = a + \frac{( b - a ) ( c - a )}{2}

Question 9

5 marks

Proof by contradiction: no integer root

Number and algebra

5 marks

Assume the equation has an integer root $α$ , so $2 α^{3} + 6 α + 1 = 0$ . Rearrange.
Since $α$ is an integer, $α^{3} + 3 α$ is an integer and the left side is even. The right side, $- 1$ , is odd. This is a contradiction.
So the equation $2 x^{3} + 6 x + 1 = 0$ has no integer roots.

2 α^{3} + 6 α = - 1 \Rightarrow 2 (α^{3} + 3 α) = - 1

Section B · Extended response

Questions 10–12 · 54 marks

Question 10

16 marks

A biased die, a dice game, and two pairs of dice with the same sum distribution

Statistics and probability

(a)

5 marks

The probabilities add to 1, and the mean is 2.
Subtract the first equation from the second: $2 q = 0.4$ .

p + 0.3 + q + 0.1 = 1 \Rightarrow p + q = 0.6 E (X) = p + 0.6 + 3 q + 0.4 = 2 \Rightarrow p + 3 q = 1 q = 0.2, p = 0.4

(b)

2 marks

$X > 2$ means 3 or 4.

P (X > 2) = 0.2 + 0.1 = 0.3

(c)

5 marks

With a score of 4 after three rolls, Nicky needs at least 6 more from her last two rolls, which can total 6, 7 or 8.
Add them.

P (total 6) = 2 (0.3) (0.1) + 0. 2^{2} = 0.1, P (total 7) = 2 (0.2) (0.1) = 0.04, P (total 8) = 0. 1^{2} = 0.01 P (wins) = 0.1 + 0.04 + 0.01 = 0.15

(d)

2 marks

The largest sum for the yellow pair is 8, and for the red pair it is $3 + b$ .

3 + b = 8 \Rightarrow b = 5

(e)

2 marks

Compare the probability of a sum of 5. For the yellow pair $P (S = 5) = \frac{4}{16}$ . For the red pair a sum of 5 comes from the 2 on the first die (probability $\frac{2}{4}$ ) with the $a$ on the second die (probability $\frac{2}{4}$ ) when $a + 2 = 5$ , which gives $\frac{1}{2} \cdot \frac{1}{2} = \frac{4}{16}$ .
Check the other sums: with faces ${1, 2, 2, 3}$ and ${1, 3, 3, 5}$ the sums 2 to 8 have probabilities $\frac{1}{16}, \frac{2}{16}, \frac{3}{16}, \frac{4}{16}, \frac{3}{16}, \frac{2}{16}, \frac{1}{16}$ , as required.

a = 3

Question 11

20 marks

A reciprocal quadratic: its graph, the inverse of a branch, and a composite with arctan

Functions

(a)

6 marks

Since $f (x) = \frac{1}{( x - 3 ) ( x + 1 )}$ , the vertical asymptotes are $x = - 1$ and $x = 3$ , and the horizontal asymptote is $y = 0$ . The $y$ -intercept is $f (0)$ , and there are no $x$ -intercepts.
The denominator is smallest (most negative) halfway between the roots, at $x = 1$ .
The sketch has three branches: for $x < - 1$ it is positive, falling from $+ \infty$ at $x = - 1^{-}$ towards 0 as $x \to - \infty$ . In the middle it is negative: from $- \infty$ at $x = - 1^{+}$ up to the maximum $(1, - \frac{1}{4})$ through $(0, - \frac{1}{3})$ and back to $- \infty$ at $x = 3^{-}$ . For $x > 3$ it is positive, from $+ \infty$ at $x = 3^{+}$ falling towards 0.

f (0) = - \frac{1}{3} local maximum at (1, - \frac{1}{4})

(b)

7 marks

(i) Interchange $x$ and $y$ in $y = \frac{1}{x ^{2} - 2 x - 3}$ to get $x = \frac{1}{y ^{2} - 2 y - 3}$ , and solve for $y$ as a quadratic.

Since the original domain is $x > 3$ , the output $y$ of the inverse is greater than 3, so the minus sign is rejected.

(ii) The domain of $g^{- 1}$ is the range of $g$ . For $x > 3$ the values of $g$ fall from $+ \infty$ towards 0.

x y^{2} - 2 x y - (3 x + 1) = 0 \Rightarrow y = \frac{2 x \pm 4 x ^{2} + 4 x ( 3 x + 1 )}{2 x} = \frac{x \pm 4 x ^{2} + x}{x} g^{- 1} (x) = \frac{x + 4 x ^{2} + x}{x} = 1 + \frac{4 x ^{2} + x}{x} x > 0

(c)

7 marks

$(h \circ g) (a) = arctan \frac{g ( a )}{2} = \frac{π}{4}$ , so $\frac{g ( a )}{2} = 1$ .
Solve the quadratic and keep the root with $a > 3$ .

g (a) = 2 \Rightarrow \frac{1}{a ^{2} - 2 a - 3} = 2 \Rightarrow 2 a^{2} - 4 a - 7 = 0 a = \frac{4 \pm 72}{4} = 1 \pm \frac{3 2}{2} \Rightarrow a = 1 + \frac{3}{2} 2 p = 1, q = 3, r = 2

Question 12

18 marks

Equilateral triangles from complex numbers and the roots of a quadratic

Number and algebra

(a)

2 marks

The conjugate of $z_{2} = r_{2} e^{i θ}$ is $z_{2}^{*} = r_{2} e^{- i θ}$ .

z_{1} z_{2}^{*} = r_{1} e^{i α} \cdot r_{2} e^{- i θ} = r_{1} r_{2} e^{i (α - θ)}

(b)

2 marks

The real part is $r_{1} r_{2} cos (α - θ)$ . Setting it to 0 with $r_{1} r_{2} > 0$ :
So the angle at O is a right angle and the triangle $Z_{1} O Z_{2}$ is right-angled.

cos (α - θ) = 0 \Rightarrow α - θ = \frac{π}{2} (since 0 < α - θ < π)

(c)

6 marks

(i) In an equilateral triangle $r_{1} = r_{2}$ and the angle at O is $\frac{π}{3}$ , so $α = θ + \frac{π}{3}$ .

(ii) Substitute into $z_{1}^{2} + z_{2}^{2}$ .

Since $1 + e^{2 π i /3} = e^{i π /3}$ (the sum of the two vectors has length 1 at angle $\frac{π}{3}$ ):

z_{1} = r_{2} e^{i (θ + π /3)} = z_{2} e^{i π /3} z_{1}^{2} + z_{2}^{2} = z_{2}^{2} e^{2 π i /3} + z_{2}^{2} = z_{2}^{2} (1 + e^{2 π i /3}) z_{1}^{2} + z_{2}^{2} = z_{2}^{2} e^{i π /3} = z_{2} \cdot z_{2} e^{i π /3} = z_{2} z_{1}

(d)

5 marks

For roots $z_{1}$ and $z_{2}$ of $z^{2} + a z + b = 0$ : $z_{1} + z_{2} = - a$ and $z_{1} z_{2} = b$ . Square the sum.
By part (c)(ii), $z_{1}^{2} + z_{2}^{2} = z_{1} z_{2} = b$ .

a^{2} = (z_{1} + z_{2})^{2} = z_{1}^{2} + z_{2}^{2} + 2 z_{1} z_{2} a^{2} = b + 2 b = 3 b \Rightarrow a^{2} - 3 b = 0

(e)

3 marks

With $b = 12$ , $a^{2} = 36$ , so $a = \pm 6$ . The equation is $z^{2} \pm 6 z + 12 = 0$ .
For $a = - 6$ the roots are $3 \pm 3 i$ with arguments $\frac{π}{6}$ and $\frac{11 π}{6}$ , so $α - θ = \pm \frac{5 π}{3}$ , which is not in $(0, π)$ . This labelling is rejected.
For $a = 6$ the roots are $- 3 \pm 3 i$ with arguments $\frac{7 π}{6}$ and $\frac{5 π}{6}$ , and $α - θ = \frac{π}{3}$ , which is in $(0, π)$ .
So exactly one equilateral triangle $Z_{1} O Z_{2}$ satisfies the conditions.

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 algebra58 marks · 53%
Q1, Q2, Q4, Q5, Q7, Q8, Q9, Q12
Functions36 marks · 33%
Q3, Q6, Q11
Statistics and probability16 marks · 15%
Q10

Mentor insights

Traps to watch for

01
Use the graph for inequalities · Q3
In (c) read off where the graph lies between $y = 0$ and the asymptote $y = 2$ . In (d) replace $x$ by $∣ x ∣$ : the graph is reflected, so solutions come in a $\pm$ pair.
02
Odd means $f (- x) = - f (x)$ · Q6
Replace $x$ by $- x$ and compare with $- f (x)$ . For the range, find the stationary points and use the symmetry.
03
Add the ways to reach each total · Q10
In (c) Nicky needs the last two rolls to total at least 6. List each pair of rolls giving 6, 7 and 8, using the probabilities from (a).
04
Choose the branch for the inverse · Q11
The inverse comes from a quadratic with two roots. The condition $x > 3$ on the original function tells you which root to keep.
05
Use the angle at O · Q12
The angle between $O Z_{1}$ and $O Z_{2}$ is $α - θ$ . For an equilateral triangle it is $\frac{π}{3}$ and the two lengths are equal. The condition $0 < α - θ < π$ then decides which labelling is allowed.

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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