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IB Maths AA HL · Paper 1

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ3

Session: November 2025
Marks: 110
Duration: 2 hours
Calculator: Not allowed
Questions: 11

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2025 TZ1 paper, which your school or the IB can provide.

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Section A · Short response

Questions 1–8 · 56 marks

Question 1

6 marks

An arithmetic sequence from its first and fifth terms

Number and algebra

(a)

4 marks

Use $u_{5} = u_{1} + 4 d$ .
The 13th term.

12 = 36 + 4 d \Rightarrow d = - 6 u_{13} = 36 + 12 (- 6) = - 36

(b)

2 marks

Set $S_{n} = \frac{n}{2} [2 (36) + (n - 1) (- 6)] = 0$ and divide by $n$ (as $n \neq = 0$ ).

72 - 6 (n - 1) = 0 \Rightarrow n - 1 = 12 n = 13

Question 2

6 marks

Simplifying a rational expression and solving a logarithmic equation

Number and algebra

(a)

2 marks

Factorise the quadratics, then cancel the common factors $(x - 1)$ and $(x + 1)$ , which are not zero since $x > 7$ .

\frac{x}{x ^{2} - 8 x + 7} \times \frac{x ^{2} - 1}{x + 1} = \frac{x}{( x - 1 ) ( x - 7 )} \times \frac{( x - 1 ) ( x + 1 )}{x + 1} = \frac{x}{x - 7}

(b)

4 marks

Use $lo g_{2} 2 = 1$ and combine the logarithms on the left.
The two arguments are equal.
Use part (a): the left side is $\frac{x}{x - 7}$ .

lo g_{2} [x (x^{2} - 1)] - lo g_{2} 2 = lo g_{2} [(x^{2} - 8 x + 7) (x + 1)] lo g_{2} \frac{x ( x ^{2} - 1 )}{2} = lo g_{2} [(x^{2} - 8 x + 7) (x + 1)] \frac{x ( x ^{2} - 1 )}{2} = (x^{2} - 8 x + 7) (x + 1) \Rightarrow \frac{x ( x ^{2} - 1 )}{( x ^{2} - 8 x + 7 ) ( x + 1 )} = 2 \frac{x}{x - 7} = 2 \Rightarrow x = 2 x - 14 \Rightarrow x = 14

Question 3

7 marks

A line and a parabola through the same two points

Calculus

(a)

3 marks

(i) The parabola $y = a x^{2} + c$ passes through $P (0, 9)$ .

(ii) It also passes through $Q (3, 0)$ .

c = 9 0 = 9 a + 9 \Rightarrow a = - 1

(b)

4 marks

The parabola $y = 9 - x^{2}$ is above the line $y = 9 - 3 x$ between $x = 0$ and $x = 3$ .

Area = \int_{0}^{3} [(9 - x^{2}) - (9 - 3 x)] d x = \int_{0}^{3} (3 x - x^{2}) d x [\frac{3 x ^{2}}{2} - \frac{x ^{3}}{3}]_{0}^{3} = \frac{27}{2} - 9 = \frac{9}{2}

Question 4

8 marks

Two parabolas whose vertices lie on a line

Calculus

(a)

3 marks

For $f (x) = - x^{2} + 4 x + p$ the vertex $A$ is at $x = 2$ . It lies on $y = 2 x - 1$ , so its height is $2 (2) - 1 = 3$ .

f (2) = - 4 + 8 + p = 3 \Rightarrow p = - 1

(b)

5 marks

For $g (x) = x^{2} + q x - 1$ the vertex $B$ is at $x = - \frac{q}{2}$ . On the line, its height is $y = 2 (- \frac{q}{2}) - 1 = - q - 1$ .
On the parabola, its height is
Equate the two heights.
Since $q \neq = 0$ , $q = 4$ .

g (- \frac{q}{2}) = \frac{q ^{2}}{4} - \frac{q ^{2}}{2} - 1 = - \frac{q ^{2}}{4} - 1 - \frac{q ^{2}}{4} - 1 = - q - 1 \Rightarrow q^{2} = 4 q \Rightarrow q (q - 4) = 0

Question 5

7 marks

A Maclaurin series of (1 - x)^(-4) and a depreciating car

Calculus

(a)

5 marks

(i) Differentiate $f (x) = (1 - x)^{- 4}$ .

(ii) Differentiate again.

f^{'} (x) = 4 (1 - x)^{- 5} \Rightarrow f^{'} (0) = 4 \Rightarrow a = 4 f^{''} (x) = 20 (1 - x)^{- 6} \Rightarrow f^{''} (0) = 20 b = \frac{20}{2 !} = 10

(b)

2 marks

The value four years ago, $V$ , satisfies $1000 = V (0.9)^{4}$ , so $V = 1000 (1 - 0.1)^{- 4}$ . Use the series with $x = 0.1$ .

V \approx 1000 (1 + 4 (0.1) + 10 (0.1)^{2} + 20 (0.1)^{3}) = 1000 (1.52) = $1520

Question 6

6 marks

An infinite number of solutions to a linear system

Number and algebra

6 marks

The equations are $x - z = 4$ , $2 x + 2 y + z = 14$ and $x + 2 y + α z = β$ . Subtract the first equation from the second.
The third equation $x + 2 y + α z = β$ must be the same as this one for infinitely many solutions.

(2 x + 2 y + z) - (x - z) = 14 - 4 \Rightarrow x + 2 y + 2 z = 10 α = 2, β = 10

Question 7

8 marks

A separable differential equation with cos squared x

Calculus

(a)

6 marks

Separate the variables in $\frac{d y}{d x} = y^{2} cos^{2} x$ .
Use the double angle identity.
Rearrange for $y$ .

\int \frac{1}{y ^{2}} d y = \int cos^{2} x d x - \frac{1}{y} = \int \frac{1}{2} (1 + cos 2 x) d x = \frac{x}{2} + \frac{1}{4} sin 2 x + C y = \frac{- 1}{\frac{x}{2} + \frac{1}{4} sin 2 x + C} = \frac{- 4}{2 x + sin 2 x + 4 C} a = - 4

(b)

2 marks

The solution is $y = \frac{- 4}{2 x + sin 2 x + b}$ . Use $y = 1$ when $x = \frac{π}{2}$ .

1 = \frac{- 4}{π + 0 + b} \Rightarrow π + b = - 4 b = - 4 - π

Question 8

8 marks

The triple angle formula and a transformation of the sine graph

Number and algebra

(a)

4 marks

Use the angle sum formula.
Use $sin 2 θ = 2 sin θ cos θ$ and $cos 2 θ = 1 - 2 sin^{2} θ$ .
Replace $cos^{2} θ$ by $1 - sin^{2} θ$ .

sin 3 θ = sin 2 θ cos θ + cos 2 θ sin θ sin 3 θ = 2 sin θ cos^{2} θ + (1 - 2 sin^{2} θ) sin θ sin 3 θ = 2 sin θ (1 - sin^{2} θ) + sin θ - 2 sin^{3} θ = 3 sin θ - 4 sin^{3} θ

(b)

4 marks

The expression is $6 sin (θ + \frac{π}{6}) - 8 sin^{3} (θ + \frac{π}{6}) = 2 [3 sin u - 4 sin^{3} u]$ with $u = θ + \frac{π}{6}$ . By part (a):
To go from $y = sin θ$ : apply a vertical stretch with scale factor 2, a horizontal stretch with scale factor $\frac{1}{3}$ , and then a horizontal translation of $\frac{π}{6}$ to the left.

y = 2 sin 3 u = 2 sin (3 θ + \frac{π}{2}) = 2 sin (3 (θ + \frac{π}{6}))

Section B · Extended response

Questions 9–11 · 54 marks

Question 9

14 marks

A function from its derivative: turning points, inflexion and the function itself

Calculus

(a)

3 marks

Horizontal tangents mean $f^{'} (x) = 0$ .

3 x^{2} + 12 x - 15 = 3 (x + 5) (x - 1) = 0 a = - 5, b = 1

(b)

2 marks

At $x = a$ the sign diagram shows $f^{'}$ going from positive to negative. So there is a local maximum, since the gradient changes from positive to negative.

(c)

3 marks

Differentiate again.

f^{''} (x) = 6 x + 12 = 0 \Rightarrow c = - 2

(d)

2 marks

Yes, there is a point of inflexion at $x = c$ , because $f^{''} (x)$ changes sign there (from negative to positive, so the concavity changes).

(e)

4 marks

Integrate $f^{'} (x) = 3 x^{2} + 12 x - 15$ .
Use $f (- 2) = 36$ .

f (x) = x^{3} + 6 x^{2} - 15 x + C - 8 + 24 + 30 + C = 36 \Rightarrow C = - 10 f (x) = x^{3} + 6 x^{2} - 15 x - 10

Question 10

18 marks

Binomial series for 1/(1 + x^2) and 1/sqrt(1 - x^2) and an approximation of pi

Number and algebra

(a)(i)

3 marks

Write $\frac{1}{1 + x ^{2}} = (1 + x^{2})^{- 1}$ and use $n = - 1$ with $y = x^{2}$ .

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

(a)(ii)

2 marks

Integrate term by term.

\int \frac{1}{1 + x ^{2}} d x \approx x - \frac{x ^{3}}{3} + \frac{x ^{5}}{5} - \frac{x ^{7}}{7} + c

(b)

5 marks

(i) Write $\frac{1}{1 - x ^{2}} = (1 - x^{2})^{- 1/2}$ and use $n = - \frac{1}{2}$ with $y = - x^{2}$ .

(ii) Integrate term by term.

1 + (- \frac{1}{2}) (- x^{2}) + \frac{( - \frac{1}{2} ) ( - \frac{3}{2} )}{2} (- x^{2})^{2} = 1 + \frac{1}{2} x^{2} + \frac{3}{8} x^{4} \int \frac{1}{1 - x ^{2}} d x \approx x + \frac{x ^{3}}{6} + \frac{3 x ^{5}}{40} + c

(c)

4 marks

The antiderivative of $\frac{1}{1 - x ^{2}}$ is $arcsin x$ , which is 0 when $x = 0$ , so $c = 0$ . Put $x = \frac{1}{2}$ .

arcsin \frac{1}{2} \approx \frac{1}{2} + \frac{( 1/2 ) ^{3}}{6} + \frac{3 ( 1/2 ) ^{5}}{40} = \frac{1}{2} + \frac{1}{48} + \frac{3}{1280} \frac{1}{2} + \frac{1}{48} = \frac{25}{48} \Rightarrow arcsin \frac{1}{2} \approx \frac{25}{48} + \frac{3}{1280} k = 3

(d)

4 marks

Since $arcsin \frac{1}{2} = \frac{π}{6}$ .

\frac{π}{6} \approx \frac{25}{48} + \frac{3}{1280} = \frac{25 \times 80 + 3 \times 3}{3840} \Rightarrow π \approx 6 (\frac{25}{48} + \frac{3}{1280}) = \frac{25}{8} + \frac{9}{640} π \approx \frac{2000 + 9}{640} = \frac{2009}{640}

Question 11

22 marks

A complex number, its square roots and tan of pi over 12

Number and algebra

(a)

2 marks

Multiply the numerator and denominator of $w = \frac{4 i}{1 + i 3}$ by $1 - i 3$ .

w = \frac{4 i ( 1 - i 3 )}{1 + 3} = i (1 - i 3) = i + 3 = 3 + i

(b)

3 marks

(i) The modulus.

(ii) The argument, with $w$ in the first quadrant.

∣ w ∣ = 3 + 1 = 2 ar g w = tan^{- 1} \frac{1}{3} = \frac{π}{6}

(c)

4 marks

Write $w = 2 e^{iπ /6}$ . By de Moivre's theorem, the square roots have modulus $2$ and argument $\frac{π}{12}$ , or that plus $π$ .

2 e^{iπ /12} and 2 e^{i (\frac{π}{12} - π)} = 2 e^{- 11 π i /12}

(d)

8 marks

(i) Expand $(a + bi)^{2} = a^{2} - b^{2} + 2 abi = 3 + i$ and equate the parts.

Substitute for $b$ .

Solve this quadratic in $a^{2}$ .

Since $a^{2} > 0$ we take the plus sign.

(ii) Use $b^{2} = a^{2} - 3$ .

a^{2} - b^{2} = 3, 2 ab = 1 \Rightarrow b = \frac{1}{2 a} a^{2} - \frac{1}{4 a ^{2}} = 3 \Rightarrow 4 a^{4} - 43 a^{2} - 1 = 0 a^{2} = \frac{4 3 \pm 48 + 16}{8} = \frac{3 \pm 2}{2} a^{2} = \frac{3 + 2}{2} b^{2} = \frac{3 + 2}{2} - 3 = \frac{2 - 3}{2}

(e)

5 marks

The argument of $a + bi$ is half the argument of $w$ , so $tan \frac{π}{12} = \frac{b}{a}$ .
Since $tan \frac{π}{12} > 0$ and $2 - 3 > 0$ ,

tan^{2} \frac{π}{12} = \frac{b ^{2}}{a ^{2}} = \frac{2 - 3}{2 + 3} = \frac{( 2 - 3 ) ^{2}}{4 - 3} = 7 - 43 = (2 - 3)^{2} tan \frac{π}{12} = 2 - 3, p = 2, q = - 1

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 algebra66 marks · 60%
Q1, Q2, Q6, Q8, Q10, Q11
Calculus44 marks · 40%
Q3, Q4, Q5, Q7, Q9

Mentor insights

Traps to watch for

01
Vertex · Q4
The vertex of $a x^{2} + b x + c$ is at $x = - \frac{b}{2 a}$ .
02
Separate the variables · Q7
$\frac{d y}{y ^{2}} = cos^{2} x d x$ , and use $cos^{2} x = \frac{1}{2} (1 + cos 2 x)$ .
03
First derivative sign · Q9
If $f^{'}$ changes from positive to negative, the point is a local maximum.
04
Binomial series · Q10
$(1 + y)^{n} = 1 + n y + \frac{n ( n - 1 )}{2} y^{2} + \dots$
05
Conjugate · Q11
Multiply the numerator and denominator by the conjugate of the denominator.

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Worked solutions

IB Maths AA HL · Paper 1

Mathematics: Analysis & Approaches, Higher Level

Timezone

TZ1TZ3

Session: November 2025
Marks: 110
Duration: 2 hours
Calculator: Not allowed
Questions: 11

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2025 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Section A · Short response

Questions 1–8 · 56 marks

Question 1

6 marks

An arithmetic sequence from its first and fifth terms

Number and algebra

(a)

4 marks

Use $u_{5} = u_{1} + 4 d$ .
The 13th term.

12 = 36 + 4 d \Rightarrow d = - 6 u_{13} = 36 + 12 (- 6) = - 36

(b)

2 marks

Set $S_{n} = \frac{n}{2} [2 (36) + (n - 1) (- 6)] = 0$ and divide by $n$ (as $n \neq = 0$ ).

72 - 6 (n - 1) = 0 \Rightarrow n - 1 = 12 n = 13

Question 2

6 marks

Simplifying a rational expression and solving a logarithmic equation

Number and algebra

(a)

2 marks

Factorise the quadratics, then cancel the common factors $(x - 1)$ and $(x + 1)$ , which are not zero since $x > 7$ .

\frac{x}{x ^{2} - 8 x + 7} \times \frac{x ^{2} - 1}{x + 1} = \frac{x}{( x - 1 ) ( x - 7 )} \times \frac{( x - 1 ) ( x + 1 )}{x + 1} = \frac{x}{x - 7}

(b)

4 marks

Use $lo g_{2} 2 = 1$ and combine the logarithms on the left.
The two arguments are equal.
Use part (a): the left side is $\frac{x}{x - 7}$ .

lo g_{2} [x (x^{2} - 1)] - lo g_{2} 2 = lo g_{2} [(x^{2} - 8 x + 7) (x + 1)] lo g_{2} \frac{x ( x ^{2} - 1 )}{2} = lo g_{2} [(x^{2} - 8 x + 7) (x + 1)] \frac{x ( x ^{2} - 1 )}{2} = (x^{2} - 8 x + 7) (x + 1) \Rightarrow \frac{x ( x ^{2} - 1 )}{( x ^{2} - 8 x + 7 ) ( x + 1 )} = 2 \frac{x}{x - 7} = 2 \Rightarrow x = 2 x - 14 \Rightarrow x = 14

Question 3

7 marks

A line and a parabola through the same two points

Calculus

(a)

3 marks

(i) The parabola $y = a x^{2} + c$ passes through $P (0, 9)$ .

(ii) It also passes through $Q (3, 0)$ .

c = 9 0 = 9 a + 9 \Rightarrow a = - 1

(b)

4 marks

The parabola $y = 9 - x^{2}$ is above the line $y = 9 - 3 x$ between $x = 0$ and $x = 3$ .

Area = \int_{0}^{3} [(9 - x^{2}) - (9 - 3 x)] d x = \int_{0}^{3} (3 x - x^{2}) d x [\frac{3 x ^{2}}{2} - \frac{x ^{3}}{3}]_{0}^{3} = \frac{27}{2} - 9 = \frac{9}{2}

Question 4

8 marks

Two parabolas whose vertices lie on a line

Calculus

(a)

3 marks

For $f (x) = - x^{2} + 4 x + p$ the vertex $A$ is at $x = 2$ . It lies on $y = 2 x - 1$ , so its height is $2 (2) - 1 = 3$ .

f (2) = - 4 + 8 + p = 3 \Rightarrow p = - 1

(b)

5 marks

For $g (x) = x^{2} + q x - 1$ the vertex $B$ is at $x = - \frac{q}{2}$ . On the line, its height is $y = 2 (- \frac{q}{2}) - 1 = - q - 1$ .
On the parabola, its height is
Equate the two heights.
Since $q \neq = 0$ , $q = 4$ .

g (- \frac{q}{2}) = \frac{q ^{2}}{4} - \frac{q ^{2}}{2} - 1 = - \frac{q ^{2}}{4} - 1 - \frac{q ^{2}}{4} - 1 = - q - 1 \Rightarrow q^{2} = 4 q \Rightarrow q (q - 4) = 0

Question 5

7 marks

A Maclaurin series of (1 - x)^(-4) and a depreciating car

Calculus

(a)

5 marks

(i) Differentiate $f (x) = (1 - x)^{- 4}$ .

(ii) Differentiate again.

f^{'} (x) = 4 (1 - x)^{- 5} \Rightarrow f^{'} (0) = 4 \Rightarrow a = 4 f^{''} (x) = 20 (1 - x)^{- 6} \Rightarrow f^{''} (0) = 20 b = \frac{20}{2 !} = 10

(b)

2 marks

The value four years ago, $V$ , satisfies $1000 = V (0.9)^{4}$ , so $V = 1000 (1 - 0.1)^{- 4}$ . Use the series with $x = 0.1$ .

V \approx 1000 (1 + 4 (0.1) + 10 (0.1)^{2} + 20 (0.1)^{3}) = 1000 (1.52) = $1520

Question 6

6 marks

An infinite number of solutions to a linear system

Number and algebra

6 marks

The equations are $x - z = 4$ , $2 x + 2 y + z = 14$ and $x + 2 y + α z = β$ . Subtract the first equation from the second.
The third equation $x + 2 y + α z = β$ must be the same as this one for infinitely many solutions.

(2 x + 2 y + z) - (x - z) = 14 - 4 \Rightarrow x + 2 y + 2 z = 10 α = 2, β = 10

Question 7

8 marks

A separable differential equation with cos squared x

Calculus

(a)

6 marks

Separate the variables in $\frac{d y}{d x} = y^{2} cos^{2} x$ .
Use the double angle identity.
Rearrange for $y$ .

\int \frac{1}{y ^{2}} d y = \int cos^{2} x d x - \frac{1}{y} = \int \frac{1}{2} (1 + cos 2 x) d x = \frac{x}{2} + \frac{1}{4} sin 2 x + C y = \frac{- 1}{\frac{x}{2} + \frac{1}{4} sin 2 x + C} = \frac{- 4}{2 x + sin 2 x + 4 C} a = - 4

(b)

2 marks

The solution is $y = \frac{- 4}{2 x + sin 2 x + b}$ . Use $y = 1$ when $x = \frac{π}{2}$ .

1 = \frac{- 4}{π + 0 + b} \Rightarrow π + b = - 4 b = - 4 - π

Question 8

8 marks

The triple angle formula and a transformation of the sine graph

Number and algebra

(a)

4 marks

Use the angle sum formula.
Use $sin 2 θ = 2 sin θ cos θ$ and $cos 2 θ = 1 - 2 sin^{2} θ$ .
Replace $cos^{2} θ$ by $1 - sin^{2} θ$ .

sin 3 θ = sin 2 θ cos θ + cos 2 θ sin θ sin 3 θ = 2 sin θ cos^{2} θ + (1 - 2 sin^{2} θ) sin θ sin 3 θ = 2 sin θ (1 - sin^{2} θ) + sin θ - 2 sin^{3} θ = 3 sin θ - 4 sin^{3} θ

(b)

4 marks

The expression is $6 sin (θ + \frac{π}{6}) - 8 sin^{3} (θ + \frac{π}{6}) = 2 [3 sin u - 4 sin^{3} u]$ with $u = θ + \frac{π}{6}$ . By part (a):
To go from $y = sin θ$ : apply a vertical stretch with scale factor 2, a horizontal stretch with scale factor $\frac{1}{3}$ , and then a horizontal translation of $\frac{π}{6}$ to the left.

y = 2 sin 3 u = 2 sin (3 θ + \frac{π}{2}) = 2 sin (3 (θ + \frac{π}{6}))

Section B · Extended response

Questions 9–11 · 54 marks

Question 9

14 marks

A function from its derivative: turning points, inflexion and the function itself

Calculus

(a)

3 marks

Horizontal tangents mean $f^{'} (x) = 0$ .

3 x^{2} + 12 x - 15 = 3 (x + 5) (x - 1) = 0 a = - 5, b = 1

(b)

2 marks

At $x = a$ the sign diagram shows $f^{'}$ going from positive to negative. So there is a local maximum, since the gradient changes from positive to negative.

(c)

3 marks

Differentiate again.

f^{''} (x) = 6 x + 12 = 0 \Rightarrow c = - 2

(d)

2 marks

Yes, there is a point of inflexion at $x = c$ , because $f^{''} (x)$ changes sign there (from negative to positive, so the concavity changes).

(e)

4 marks

Integrate $f^{'} (x) = 3 x^{2} + 12 x - 15$ .
Use $f (- 2) = 36$ .

f (x) = x^{3} + 6 x^{2} - 15 x + C - 8 + 24 + 30 + C = 36 \Rightarrow C = - 10 f (x) = x^{3} + 6 x^{2} - 15 x - 10

Question 10

18 marks

Binomial series for 1/(1 + x^2) and 1/sqrt(1 - x^2) and an approximation of pi

Number and algebra

(a)(i)

3 marks

Write $\frac{1}{1 + x ^{2}} = (1 + x^{2})^{- 1}$ and use $n = - 1$ with $y = x^{2}$ .

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

(a)(ii)

2 marks

Integrate term by term.

\int \frac{1}{1 + x ^{2}} d x \approx x - \frac{x ^{3}}{3} + \frac{x ^{5}}{5} - \frac{x ^{7}}{7} + c

(b)

5 marks

(i) Write $\frac{1}{1 - x ^{2}} = (1 - x^{2})^{- 1/2}$ and use $n = - \frac{1}{2}$ with $y = - x^{2}$ .

(ii) Integrate term by term.

1 + (- \frac{1}{2}) (- x^{2}) + \frac{( - \frac{1}{2} ) ( - \frac{3}{2} )}{2} (- x^{2})^{2} = 1 + \frac{1}{2} x^{2} + \frac{3}{8} x^{4} \int \frac{1}{1 - x ^{2}} d x \approx x + \frac{x ^{3}}{6} + \frac{3 x ^{5}}{40} + c

(c)

4 marks

The antiderivative of $\frac{1}{1 - x ^{2}}$ is $arcsin x$ , which is 0 when $x = 0$ , so $c = 0$ . Put $x = \frac{1}{2}$ .

arcsin \frac{1}{2} \approx \frac{1}{2} + \frac{( 1/2 ) ^{3}}{6} + \frac{3 ( 1/2 ) ^{5}}{40} = \frac{1}{2} + \frac{1}{48} + \frac{3}{1280} \frac{1}{2} + \frac{1}{48} = \frac{25}{48} \Rightarrow arcsin \frac{1}{2} \approx \frac{25}{48} + \frac{3}{1280} k = 3

(d)

4 marks

Since $arcsin \frac{1}{2} = \frac{π}{6}$ .

\frac{π}{6} \approx \frac{25}{48} + \frac{3}{1280} = \frac{25 \times 80 + 3 \times 3}{3840} \Rightarrow π \approx 6 (\frac{25}{48} + \frac{3}{1280}) = \frac{25}{8} + \frac{9}{640} π \approx \frac{2000 + 9}{640} = \frac{2009}{640}

Question 11

22 marks

A complex number, its square roots and tan of pi over 12

Number and algebra

(a)

2 marks

Multiply the numerator and denominator of $w = \frac{4 i}{1 + i 3}$ by $1 - i 3$ .

w = \frac{4 i ( 1 - i 3 )}{1 + 3} = i (1 - i 3) = i + 3 = 3 + i

(b)

3 marks

(i) The modulus.

(ii) The argument, with $w$ in the first quadrant.

∣ w ∣ = 3 + 1 = 2 ar g w = tan^{- 1} \frac{1}{3} = \frac{π}{6}

(c)

4 marks

Write $w = 2 e^{iπ /6}$ . By de Moivre's theorem, the square roots have modulus $2$ and argument $\frac{π}{12}$ , or that plus $π$ .

2 e^{iπ /12} and 2 e^{i (\frac{π}{12} - π)} = 2 e^{- 11 π i /12}

(d)

8 marks

(i) Expand $(a + bi)^{2} = a^{2} - b^{2} + 2 abi = 3 + i$ and equate the parts.

Substitute for $b$ .

Solve this quadratic in $a^{2}$ .

Since $a^{2} > 0$ we take the plus sign.

(ii) Use $b^{2} = a^{2} - 3$ .

a^{2} - b^{2} = 3, 2 ab = 1 \Rightarrow b = \frac{1}{2 a} a^{2} - \frac{1}{4 a ^{2}} = 3 \Rightarrow 4 a^{4} - 43 a^{2} - 1 = 0 a^{2} = \frac{4 3 \pm 48 + 16}{8} = \frac{3 \pm 2}{2} a^{2} = \frac{3 + 2}{2} b^{2} = \frac{3 + 2}{2} - 3 = \frac{2 - 3}{2}

(e)

5 marks

The argument of $a + bi$ is half the argument of $w$ , so $tan \frac{π}{12} = \frac{b}{a}$ .
Since $tan \frac{π}{12} > 0$ and $2 - 3 > 0$ ,

tan^{2} \frac{π}{12} = \frac{b ^{2}}{a ^{2}} = \frac{2 - 3}{2 + 3} = \frac{( 2 - 3 ) ^{2}}{4 - 3} = 7 - 43 = (2 - 3)^{2} tan \frac{π}{12} = 2 - 3, p = 2, q = - 1

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 algebra66 marks · 60%
Q1, Q2, Q6, Q8, Q10, Q11
Calculus44 marks · 40%
Q3, Q4, Q5, Q7, Q9

Mentor insights

Traps to watch for

01
Vertex · Q4
The vertex of $a x^{2} + b x + c$ is at $x = - \frac{b}{2 a}$ .
02
Separate the variables · Q7
$\frac{d y}{y ^{2}} = cos^{2} x d x$ , and use $cos^{2} x = \frac{1}{2} (1 + cos 2 x)$ .
03
First derivative sign · Q9
If $f^{'}$ changes from positive to negative, the point is a local maximum.
04
Binomial series · Q10
$(1 + y)^{n} = 1 + n y + \frac{n ( n - 1 )}{2} y^{2} + \dots$
05
Conjugate · Q11
Multiply the numerator and denominator by the conjugate of the denominator.

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