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IB Maths AA HL · Paper 1

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2 TZ3

Session: May 2025
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 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 · 56 marks

Question 1

5 marks

The local minimum of a cubic

Number and algebra

5 marks

Differentiate $f (x) = \frac{4 x ^{3}}{3} - 16 x$ .
Set the derivative to zero.
The required point has $p > 0$ , so $p = 2$ .
The local minimum is $(2, - \frac{64}{3})$ .

f^{'} (x) = 4 x^{2} - 16 4 x^{2} - 16 = 0 \Rightarrow x = \pm 2 q = f (2) = \frac{4 ( 8 )}{3} - 32 = - \frac{64}{3}

Question 2

7 marks

Quarterly interest and the binomial expansion of (1 + x)^4

Number and algebra

(a)

1 mark

The quarterly rate is $\frac{4%}{4}$ .

k = 0.01

(b)

2 marks

Use Pascal's triangle for the coefficients 1, 4, 6, 4, 1.

(1 + x)^{4} = 1 + 4 x + 6 x^{2} + 4 x^{3} + x^{4}

(c)

4 marks

Put $x = 0.01$ in the expansion and multiply by 1000.
To the nearest dinar the amount is

1000 (1 + 0.04 + 0.0006 + 0.000004 + 0.00000001) = 1000 (1.04060401) = 1040.60 \dots 1041 dinar

Question 3

4 marks

The area enclosed by e^x, -e^x and two vertical lines

Number and algebra

4 marks

The vertical distance between the curves is $e^{x} - (- e^{x}) = 2 e^{x}$ .

Area = \int_{- 1}^{1} 2 e^{x} d x = 2 [e^{x}]_{- 1}^{1} 2 (e - e^{- 1}) = 2 e - \frac{2}{e}

Question 4

6 marks

Two events: an intersection and independence

Statistics and probability

(a)

3 marks

$P (A^{'}) = \frac{3}{4}$ , so $P (A) = \frac{1}{4}$ . Use $P (A \cap B) = P (B ∣ A) P (A)$ .

P (A \cap B) = \frac{2}{3} \times \frac{1}{4} = \frac{1}{6}

(b)

3 marks

Use $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ with $P (A \cup B) = \frac{3}{4}$ .
Now compare.
Since $P (A \cap B) = P (A) P (B)$ , the events are independent.

\frac{3}{4} = \frac{1}{4} + P (B) - \frac{1}{6} \Rightarrow P (B) = \frac{1}{2} + \frac{1}{6} = \frac{2}{3} P (A) P (B) = \frac{1}{4} \times \frac{2}{3} = \frac{1}{6} = P (A \cap B)

Question 5

8 marks

Picture frames growing by 50% and their mean and median area

Number and algebra

(a)

5 marks

(i) The first area is $4 \times 5 = 20$ and each area is $\frac{9}{4}$ times the one before, so the areas form a geometric sequence.

(ii) The total area is the sum of ten terms.

The mean is the sum over 10.

area of F_{n} = 20 (\frac{9}{4})^{n - 1} S_{10} = \frac{20 ( ( \frac{9}{4} ) ^{10} - 1 )}{\frac{9}{4} - 1} = 16 ((\frac{9}{4})^{10} - 1) mean = \frac{16}{10} ((\frac{9}{4})^{10} - 1) = \frac{8}{5} ((\frac{9}{4})^{10} - 1) p = \frac{8}{5}, a = 10

(b)

3 marks

The median is the mean of $F_{5}$ and $F_{6}$ .

\frac{20 ( \frac{9}{4} ) ^{4} + 20 ( \frac{9}{4} ) ^{5}}{2} = 10 (\frac{9}{4})^{4} (1 + \frac{9}{4}) = 10 (\frac{9}{4})^{4} \times \frac{13}{4} median = \frac{65}{2} (\frac{9}{4})^{4} q = \frac{65}{2}

Question 6

6 marks

Two lines that are perpendicular and intersect

Number and algebra

6 marks

The direction vectors are $0 a 1$ and $123$ . Their scalar product is zero.
At the intersection the components are equal.
From the first, $μ = 3$ . Then $- \frac{3}{2} λ = 6$ , so $λ = - 4$ .
From the third, $- 5 = - b + 9$ .

2 a + 3 = 0 \Rightarrow a = - \frac{3}{2} 4 = 1 + μ, aλ = 2 μ, - 1 + λ = - b + 3 μ b = 14

Question 7

5 marks

z = 3^(i-1): real parts

Number and algebra

(a)

1 mark

Since $e^{l n 3} = 3$ the required form is

3 = e^{l n 3}

(b)

4 marks

(i) $z = 3^{i - 1} = \frac{1}{3} e^{i l n 3} = \frac{1}{3} (cos (ln 3) + i sin (ln 3))$ .

(ii) $\frac{1}{z} = 3^{1 - i} = 3 e^{- i l n 3} = 3 (cos (ln 3) - i sin (ln 3))$ .

Re (z) = \frac{1}{3} cos (ln 3) Re (\frac{1}{z}) = 3 cos (ln 3)

Question 8

7 marks

Induction proof that n > log2 n

Number and algebra

(a)

2 marks

Combine the terms.
For $n \in Z^{+}$ , $2 n \geq n + 1$ , and $lo g_{2}$ is increasing.

1 + lo g_{2} n = lo g_{2} 2 + lo g_{2} n = lo g_{2} (2 n) lo g_{2} (2 n) \geq lo g_{2} (n + 1) \Rightarrow 1 + lo g_{2} n \geq lo g_{2} (n + 1)

(b)

5 marks

Claim: $n > lo g_{2} n$ for all $n \in Z^{+}$ .
Base case: for $n = 1$ , $lo g_{2} 1 = 0$ and $1 > 0$ , so it is true.
Assume it is true for $n = k$ , that is $k > lo g_{2} k$ .
Add 1 to both sides and use part (a).
So $k + 1 > lo g_{2} (k + 1)$ , the statement for $n = k + 1$ .
It is true for $n = 1$ and, if true for $n = k$ , it is true for $n = k + 1$ . By induction it is true for all $n \in Z^{+}$ , so Seema's claim is valid.

k + 1 > 1 + lo g_{2} k \geq lo g_{2} (k + 1)

Question 9

8 marks

A homogeneous differential equation solved with y = vx

Number and algebra

8 marks

The equation is $\frac{d y}{d x} = \frac{x - y}{x + y}$ . Put $y = v x$ .
Move $v$ to the right.
Separate the variables and integrate.
The numerator is $- \frac{1}{2}$ of the derivative of the denominator, $- 2 - 2 v$ .
Combine the logarithms.
Replace $v$ by $\frac{y}{x}$ .
Use $y = 0$ when $x = 2$ .

v + x \frac{d v}{d x} = \frac{1 - v}{1 + v} x \frac{d v}{d x} = \frac{1 - v}{1 + v} - v = \frac{1 - 2 v - v ^{2}}{1 + v} \int \frac{1 + v}{1 - 2 v - v ^{2}} d v = \int \frac{1}{x} d x - \frac{1}{2} ln 1 - 2 v - v^{2} = ln ∣ x ∣ + c ln 1 - 2 v - v^{2} = - 2 ln ∣ x ∣ - 2 c \Rightarrow x^{2} (1 - 2 v - v^{2}) = A x^{2} - 2 x y - y^{2} = A 4 = A x^{2} - 2 x y - y^{2} = 4

Section B · Extended response

Questions 10–12 · 54 marks

Question 10

18 marks

A quadratic, an inequality and composite functions

Functions

(a)

4 marks

The roots are $- 3$ and $- 1$ , so the vertex is at $x = - 2$ . Expand to check: $5 (x + 1) (x + 3) = 5 (x^{2} + 4 x + 3) = 5 (x + 2)^{2} - 5$ .

f (x) = 5 (x + 2)^{2} - 5, a = 5, h = - 2, k = - 5

(b)

4 marks

The graph is a parabola opening upwards with the vertex at $(- 2, - 5)$ , crossing the $x$ -axis at $(- 3, 0)$ and $(- 1, 0)$ , and the $y$ -axis at $f (0) = 15$ .
Sketch: a symmetric U-shaped curve about the line $x = - 2$ , with the minimum point at $(- 2, - 5)$ , $x$ -intercepts $- 3$ and $- 1$ and $y$ -intercept $(0, 15)$ .

(c)

4 marks

Solve $5 (x + 2)^{2} - 5 \leq 40$ .

5 (x + 2)^{2} \leq 45 \Rightarrow (x + 2)^{2} \leq 9 \Rightarrow - 3 \leq x + 2 \leq 3 - 5 \leq x \leq 1

(d)

3 marks

(i) Replace $x$ by $ln x$ .

(ii) Use the result of (c) with $x$ replaced by $ln x$ .

(f \circ g) (x) = 5 (ln x + 1) (ln x + 3) - 5 \leq ln x \leq 1 \Rightarrow e^{- 5} \leq x \leq e

(e)

3 marks

$(g \circ f) (x) = ln (f (x))$ needs $f (x) > 0$ .

5 (x + 1) (x + 3) > 0 \Rightarrow x < - 3 or x > - 1

Question 11

17 marks

Two planes, a third perpendicular plane and a reflected ray of light

Number and algebra

(a)

6 marks

The normals of $Π_{1}$ and $Π_{2}$ are $121$ and $1 - 1 - 2$ .
The acute angle between the planes is $18 0^{\circ} - 12 0^{\circ} = 6 0^{\circ}$ .

cos θ = \frac{1 - 2 - 2}{6 6} = \frac{- 3}{6} = - \frac{1}{2} θ = 12 0^{\circ}

(b)

4 marks

A vector perpendicular to both normals is their vector product.
So $Π_{3}$ has normal $(1, - 1, 1)$ and equation $x - y + z = d$ . Put in the point $R (5, - 5, 5)$ .

121 \times 1 - 1 - 2 = - 4 + 1 1 + 2 - 1 - 2 = - 3 3 - 3 d = 5 + 5 + 5 = 15 x - y + z = 15

(c)

7 marks

(i) Use the compound angle identity with $7 5^{\circ} = 3 0^{\circ} + 4 5^{\circ}$ .

(ii) In triangle $P QR$ , the angle at $R$ is the $6 0^{\circ}$ between the planes and the angle at $P$ is the $4 5^{\circ}$ the reflected ray makes with $Π_{1}$ . So the angle at $Q$ is $18 0^{\circ} - 6 0^{\circ} - 4 5^{\circ} = 7 5^{\circ}$ .

The side $P R = 5$ is opposite the angle at $Q$ , and $QR$ is opposite the angle at $P$ . Use the sine rule.

Rationalise the denominator.

sin 7 5^{\circ} = sin 3 0^{\circ} cos 4 5^{\circ} + cos 3 0^{\circ} sin 4 5^{\circ} = \frac{1}{2} \cdot \frac{2}{2} + \frac{3}{2} \cdot \frac{2}{2} = \frac{2 + 6}{4} \frac{QR}{sin 4 5 ^{\circ}} = \frac{5}{sin 7 5 ^{\circ}} \Rightarrow QR = \frac{5 sin 4 5 ^{\circ}}{sin 7 5 ^{\circ}} = \frac{10 2}{6 + 2} = \frac{10}{3 + 1} QR = \frac{10 ( 3 - 1 )}{2} = 5 (3 - 1) cm, p = 5, q = 3

Question 12

19 marks

A reduction formula for cos^n x, a volume and a Maclaurin limit

Calculus

(a)

4 marks

Integrate by parts with $u = cos^{n - 1} x$ and $d v = cos x d x$ , so $d u = - (n - 1) cos^{n - 2} x sin x d x$ and $v = sin x$ .
Replace $sin^{2} x$ by $1 - cos^{2} x$ .

\int cos^{n} x d x = cos^{n - 1} x sin x + (n - 1) \int cos^{n - 2} x sin^{2} x d x \int cos^{n} x d x = cos^{n - 1} x sin x + (n - 1) \int cos^{n - 2} x d x - (n - 1) \int cos^{n} x d x

(b)

2 marks

Collect the integrals of $cos^{n} x$ on one side.
Divide by $n$ .

n \int cos^{n} x d x = cos^{n - 1} x sin x + (n - 1) \int cos^{n - 2} x d x \int f_{n} (x) d x = \frac{cos ^{n - 1} x sin x}{n} + \frac{n - 1}{n} \int f_{n - 2} (x) d x

(c)

4 marks

Apply the formula with $n = 4$ , then with $n = 2$ .
Substitute.

\int cos^{4} x d x = \frac{1}{4} cos^{3} x sin x + \frac{3}{4} \int cos^{2} x d x \int cos^{2} x d x = \frac{1}{2} cos x sin x + \frac{1}{2} \int 1 d x = \frac{1}{2} cos x sin x + \frac{x}{2} \int cos^{4} x d x = \frac{1}{4} cos^{3} x sin x + \frac{3}{8} cos x sin x + \frac{3}{8} x + c

(d)

4 marks

The volume is $π \int y^{2} d x$ with $y = cos^{2} x$ , so $y^{2} = cos^{4} x$ .
At $x = \pm \frac{π}{2}$ the terms with $cos x$ in the antiderivative are zero, so only the term in $x$ contributes.

V = π \int_{- π /2}^{π /2} cos^{4} x d x V = π [\frac{3}{8} x]_{- π /2}^{π /2} = π \times \frac{3}{8} \times π = \frac{3 π ^{2}}{8}

(e)

5 marks

(i) Use $cos x \approx 1 - \frac{x ^{2}}{2}$ and raise to the power $n$ , keeping terms up to $x^{2}$ .

(ii) Substitute in the limit.

f_{n} (x) = (1 - \frac{x ^{2}}{2} + \dots)^{n} = 1 - \frac{n x ^{2}}{2} + \dots x \to 0 lim \frac{f _{n} ( x ) - 1}{x ^{2}} = x \to 0 lim \frac{- \frac{n x ^{2}}{2} + \dots}{x ^{2}} = - \frac{n}{2}

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 algebra67 marks · 61%
Q1, Q2, Q3, Q5, Q6, Q7, Q8, Q9, Q11
Calculus19 marks · 17%
Q12
Functions18 marks · 16%
Q10
Statistics and probability6 marks · 5%
Q4

Mentor insights

Traps to watch for

01
Geometric sequence · Q5
Each width and height is multiplied by 1.5, so each area is multiplied by $1. 5^{2} = \frac{9}{4}$ .
02
Substitution · Q9
Put $y = v x$ so $\frac{d y}{d x} = v + x \frac{d v}{d x}$ .
03
Complete the square · Q10
The vertex is halfway between the roots.
04
Angle between planes · Q11
The angle between planes is the angle between their normal vectors.
05
Integration by parts · Q12
Write $cos^{n} x = cos^{n - 1} x \cdot cos x$ and use $sin^{2} x = 1 - cos^{2} x$ .

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

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Worked solutions

IB Maths AA HL · Paper 1

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ2 TZ3

Session: May 2025
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 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 · 56 marks

Question 1

5 marks

The local minimum of a cubic

Number and algebra

5 marks

Differentiate $f (x) = \frac{4 x ^{3}}{3} - 16 x$ .
Set the derivative to zero.
The required point has $p > 0$ , so $p = 2$ .
The local minimum is $(2, - \frac{64}{3})$ .

f^{'} (x) = 4 x^{2} - 16 4 x^{2} - 16 = 0 \Rightarrow x = \pm 2 q = f (2) = \frac{4 ( 8 )}{3} - 32 = - \frac{64}{3}

Question 2

7 marks

Quarterly interest and the binomial expansion of (1 + x)^4

Number and algebra

(a)

1 mark

The quarterly rate is $\frac{4%}{4}$ .

k = 0.01

(b)

2 marks

Use Pascal's triangle for the coefficients 1, 4, 6, 4, 1.

(1 + x)^{4} = 1 + 4 x + 6 x^{2} + 4 x^{3} + x^{4}

(c)

4 marks

Put $x = 0.01$ in the expansion and multiply by 1000.
To the nearest dinar the amount is

1000 (1 + 0.04 + 0.0006 + 0.000004 + 0.00000001) = 1000 (1.04060401) = 1040.60 \dots 1041 dinar

Question 3

4 marks

The area enclosed by e^x, -e^x and two vertical lines

Number and algebra

4 marks

The vertical distance between the curves is $e^{x} - (- e^{x}) = 2 e^{x}$ .

Area = \int_{- 1}^{1} 2 e^{x} d x = 2 [e^{x}]_{- 1}^{1} 2 (e - e^{- 1}) = 2 e - \frac{2}{e}

Question 4

6 marks

Two events: an intersection and independence

Statistics and probability

(a)

3 marks

$P (A^{'}) = \frac{3}{4}$ , so $P (A) = \frac{1}{4}$ . Use $P (A \cap B) = P (B ∣ A) P (A)$ .

P (A \cap B) = \frac{2}{3} \times \frac{1}{4} = \frac{1}{6}

(b)

3 marks

Use $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ with $P (A \cup B) = \frac{3}{4}$ .
Now compare.
Since $P (A \cap B) = P (A) P (B)$ , the events are independent.

\frac{3}{4} = \frac{1}{4} + P (B) - \frac{1}{6} \Rightarrow P (B) = \frac{1}{2} + \frac{1}{6} = \frac{2}{3} P (A) P (B) = \frac{1}{4} \times \frac{2}{3} = \frac{1}{6} = P (A \cap B)

Question 5

8 marks

Picture frames growing by 50% and their mean and median area

Number and algebra

(a)

5 marks

(i) The first area is $4 \times 5 = 20$ and each area is $\frac{9}{4}$ times the one before, so the areas form a geometric sequence.

(ii) The total area is the sum of ten terms.

The mean is the sum over 10.

area of F_{n} = 20 (\frac{9}{4})^{n - 1} S_{10} = \frac{20 ( ( \frac{9}{4} ) ^{10} - 1 )}{\frac{9}{4} - 1} = 16 ((\frac{9}{4})^{10} - 1) mean = \frac{16}{10} ((\frac{9}{4})^{10} - 1) = \frac{8}{5} ((\frac{9}{4})^{10} - 1) p = \frac{8}{5}, a = 10

(b)

3 marks

The median is the mean of $F_{5}$ and $F_{6}$ .

\frac{20 ( \frac{9}{4} ) ^{4} + 20 ( \frac{9}{4} ) ^{5}}{2} = 10 (\frac{9}{4})^{4} (1 + \frac{9}{4}) = 10 (\frac{9}{4})^{4} \times \frac{13}{4} median = \frac{65}{2} (\frac{9}{4})^{4} q = \frac{65}{2}

Question 6

6 marks

Two lines that are perpendicular and intersect

Number and algebra

6 marks

The direction vectors are $0 a 1$ and $123$ . Their scalar product is zero.
At the intersection the components are equal.
From the first, $μ = 3$ . Then $- \frac{3}{2} λ = 6$ , so $λ = - 4$ .
From the third, $- 5 = - b + 9$ .

2 a + 3 = 0 \Rightarrow a = - \frac{3}{2} 4 = 1 + μ, aλ = 2 μ, - 1 + λ = - b + 3 μ b = 14

Question 7

5 marks

z = 3^(i-1): real parts

Number and algebra

(a)

1 mark

Since $e^{l n 3} = 3$ the required form is

3 = e^{l n 3}

(b)

4 marks

(i) $z = 3^{i - 1} = \frac{1}{3} e^{i l n 3} = \frac{1}{3} (cos (ln 3) + i sin (ln 3))$ .

(ii) $\frac{1}{z} = 3^{1 - i} = 3 e^{- i l n 3} = 3 (cos (ln 3) - i sin (ln 3))$ .

Re (z) = \frac{1}{3} cos (ln 3) Re (\frac{1}{z}) = 3 cos (ln 3)

Question 8

7 marks

Induction proof that n > log2 n

Number and algebra

(a)

2 marks

Combine the terms.
For $n \in Z^{+}$ , $2 n \geq n + 1$ , and $lo g_{2}$ is increasing.

1 + lo g_{2} n = lo g_{2} 2 + lo g_{2} n = lo g_{2} (2 n) lo g_{2} (2 n) \geq lo g_{2} (n + 1) \Rightarrow 1 + lo g_{2} n \geq lo g_{2} (n + 1)

(b)

5 marks

Claim: $n > lo g_{2} n$ for all $n \in Z^{+}$ .
Base case: for $n = 1$ , $lo g_{2} 1 = 0$ and $1 > 0$ , so it is true.
Assume it is true for $n = k$ , that is $k > lo g_{2} k$ .
Add 1 to both sides and use part (a).
So $k + 1 > lo g_{2} (k + 1)$ , the statement for $n = k + 1$ .
It is true for $n = 1$ and, if true for $n = k$ , it is true for $n = k + 1$ . By induction it is true for all $n \in Z^{+}$ , so Seema's claim is valid.

k + 1 > 1 + lo g_{2} k \geq lo g_{2} (k + 1)

Question 9

8 marks

A homogeneous differential equation solved with y = vx

Number and algebra

8 marks

The equation is $\frac{d y}{d x} = \frac{x - y}{x + y}$ . Put $y = v x$ .
Move $v$ to the right.
Separate the variables and integrate.
The numerator is $- \frac{1}{2}$ of the derivative of the denominator, $- 2 - 2 v$ .
Combine the logarithms.
Replace $v$ by $\frac{y}{x}$ .
Use $y = 0$ when $x = 2$ .

v + x \frac{d v}{d x} = \frac{1 - v}{1 + v} x \frac{d v}{d x} = \frac{1 - v}{1 + v} - v = \frac{1 - 2 v - v ^{2}}{1 + v} \int \frac{1 + v}{1 - 2 v - v ^{2}} d v = \int \frac{1}{x} d x - \frac{1}{2} ln 1 - 2 v - v^{2} = ln ∣ x ∣ + c ln 1 - 2 v - v^{2} = - 2 ln ∣ x ∣ - 2 c \Rightarrow x^{2} (1 - 2 v - v^{2}) = A x^{2} - 2 x y - y^{2} = A 4 = A x^{2} - 2 x y - y^{2} = 4

Section B · Extended response

Questions 10–12 · 54 marks

Question 10

18 marks

A quadratic, an inequality and composite functions

Functions

(a)

4 marks

The roots are $- 3$ and $- 1$ , so the vertex is at $x = - 2$ . Expand to check: $5 (x + 1) (x + 3) = 5 (x^{2} + 4 x + 3) = 5 (x + 2)^{2} - 5$ .

f (x) = 5 (x + 2)^{2} - 5, a = 5, h = - 2, k = - 5

(b)

4 marks

The graph is a parabola opening upwards with the vertex at $(- 2, - 5)$ , crossing the $x$ -axis at $(- 3, 0)$ and $(- 1, 0)$ , and the $y$ -axis at $f (0) = 15$ .
Sketch: a symmetric U-shaped curve about the line $x = - 2$ , with the minimum point at $(- 2, - 5)$ , $x$ -intercepts $- 3$ and $- 1$ and $y$ -intercept $(0, 15)$ .

(c)

4 marks

Solve $5 (x + 2)^{2} - 5 \leq 40$ .

5 (x + 2)^{2} \leq 45 \Rightarrow (x + 2)^{2} \leq 9 \Rightarrow - 3 \leq x + 2 \leq 3 - 5 \leq x \leq 1

(d)

3 marks

(i) Replace $x$ by $ln x$ .

(ii) Use the result of (c) with $x$ replaced by $ln x$ .

(f \circ g) (x) = 5 (ln x + 1) (ln x + 3) - 5 \leq ln x \leq 1 \Rightarrow e^{- 5} \leq x \leq e

(e)

3 marks

$(g \circ f) (x) = ln (f (x))$ needs $f (x) > 0$ .

5 (x + 1) (x + 3) > 0 \Rightarrow x < - 3 or x > - 1

Question 11

17 marks

Two planes, a third perpendicular plane and a reflected ray of light

Number and algebra

(a)

6 marks

The normals of $Π_{1}$ and $Π_{2}$ are $121$ and $1 - 1 - 2$ .
The acute angle between the planes is $18 0^{\circ} - 12 0^{\circ} = 6 0^{\circ}$ .

cos θ = \frac{1 - 2 - 2}{6 6} = \frac{- 3}{6} = - \frac{1}{2} θ = 12 0^{\circ}

(b)

4 marks

A vector perpendicular to both normals is their vector product.
So $Π_{3}$ has normal $(1, - 1, 1)$ and equation $x - y + z = d$ . Put in the point $R (5, - 5, 5)$ .

121 \times 1 - 1 - 2 = - 4 + 1 1 + 2 - 1 - 2 = - 3 3 - 3 d = 5 + 5 + 5 = 15 x - y + z = 15

(c)

7 marks

(i) Use the compound angle identity with $7 5^{\circ} = 3 0^{\circ} + 4 5^{\circ}$ .

The side $P R = 5$ is opposite the angle at $Q$ , and $QR$ is opposite the angle at $P$ . Use the sine rule.

Rationalise the denominator.

sin 7 5^{\circ} = sin 3 0^{\circ} cos 4 5^{\circ} + cos 3 0^{\circ} sin 4 5^{\circ} = \frac{1}{2} \cdot \frac{2}{2} + \frac{3}{2} \cdot \frac{2}{2} = \frac{2 + 6}{4} \frac{QR}{sin 4 5 ^{\circ}} = \frac{5}{sin 7 5 ^{\circ}} \Rightarrow QR = \frac{5 sin 4 5 ^{\circ}}{sin 7 5 ^{\circ}} = \frac{10 2}{6 + 2} = \frac{10}{3 + 1} QR = \frac{10 ( 3 - 1 )}{2} = 5 (3 - 1) cm, p = 5, q = 3

Question 12

19 marks

A reduction formula for cos^n x, a volume and a Maclaurin limit

Calculus

(a)

4 marks

Integrate by parts with $u = cos^{n - 1} x$ and $d v = cos x d x$ , so $d u = - (n - 1) cos^{n - 2} x sin x d x$ and $v = sin x$ .
Replace $sin^{2} x$ by $1 - cos^{2} x$ .

\int cos^{n} x d x = cos^{n - 1} x sin x + (n - 1) \int cos^{n - 2} x sin^{2} x d x \int cos^{n} x d x = cos^{n - 1} x sin x + (n - 1) \int cos^{n - 2} x d x - (n - 1) \int cos^{n} x d x

(b)

2 marks

Collect the integrals of $cos^{n} x$ on one side.
Divide by $n$ .

n \int cos^{n} x d x = cos^{n - 1} x sin x + (n - 1) \int cos^{n - 2} x d x \int f_{n} (x) d x = \frac{cos ^{n - 1} x sin x}{n} + \frac{n - 1}{n} \int f_{n - 2} (x) d x

(c)

4 marks

Apply the formula with $n = 4$ , then with $n = 2$ .
Substitute.

\int cos^{4} x d x = \frac{1}{4} cos^{3} x sin x + \frac{3}{4} \int cos^{2} x d x \int cos^{2} x d x = \frac{1}{2} cos x sin x + \frac{1}{2} \int 1 d x = \frac{1}{2} cos x sin x + \frac{x}{2} \int cos^{4} x d x = \frac{1}{4} cos^{3} x sin x + \frac{3}{8} cos x sin x + \frac{3}{8} x + c

(d)

4 marks

The volume is $π \int y^{2} d x$ with $y = cos^{2} x$ , so $y^{2} = cos^{4} x$ .
At $x = \pm \frac{π}{2}$ the terms with $cos x$ in the antiderivative are zero, so only the term in $x$ contributes.

V = π \int_{- π /2}^{π /2} cos^{4} x d x V = π [\frac{3}{8} x]_{- π /2}^{π /2} = π \times \frac{3}{8} \times π = \frac{3 π ^{2}}{8}

(e)

5 marks

(i) Use $cos x \approx 1 - \frac{x ^{2}}{2}$ and raise to the power $n$ , keeping terms up to $x^{2}$ .

(ii) Substitute in the limit.

f_{n} (x) = (1 - \frac{x ^{2}}{2} + \dots)^{n} = 1 - \frac{n x ^{2}}{2} + \dots x \to 0 lim \frac{f _{n} ( x ) - 1}{x ^{2}} = x \to 0 lim \frac{- \frac{n x ^{2}}{2} + \dots}{x ^{2}} = - \frac{n}{2}

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 algebra67 marks · 61%
Q1, Q2, Q3, Q5, Q6, Q7, Q8, Q9, Q11
Calculus19 marks · 17%
Q12
Functions18 marks · 16%
Q10
Statistics and probability6 marks · 5%
Q4

Mentor insights

Traps to watch for

01
Geometric sequence · Q5
Each width and height is multiplied by 1.5, so each area is multiplied by $1. 5^{2} = \frac{9}{4}$ .
02
Substitution · Q9
Put $y = v x$ so $\frac{d y}{d x} = v + x \frac{d v}{d x}$ .
03
Complete the square · Q10
The vertex is halfway between the roots.
04
Angle between planes · Q11
The angle between planes is the angle between their normal vectors.
05
Integration by parts · Q12
Write $cos^{n} x = cos^{n - 1} x \cdot cos x$ and use $sin^{2} x = 1 - cos^{2} x$ .

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

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