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IB Maths AA SL · Paper 1

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2 TZ3

Session: May 2026
Marks: 80
Duration: 1 hour 30 min
Calculator: Not allowed
Questions: 9

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2026 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 1–6 · 38 marks

Question 1

6 marks

Differentiation and a tangent line

Calculus

(a)

2 marks

Differentiate term by term: $2 x$ gives $2$ , and $sin x$ gives $cos x$ .

f^{'} (x) = 2 + cos x

(b)

4 marks

Find the point: substitute $x = π$ into $f$ .
Find the gradient: substitute $x = π$ into $f^{'} (x)$ from part (a).
Put both into $y - y_{1} = m (x - x_{1})$ and tidy up.

f (π) = 2 π + sin π = 2 π m = f^{'} (π) = 2 + cos π = 1 y - 2 π = 1 (x - π) y = x + π

Mentor tip

When a part says “hence”, it wants you to use the part before it. Reusing your derivative from (a) is the quick route.

Question 2

5 marks

A quadratic: symmetry, vertex and range

Functions

(a)

2 marks

The graph crosses the $x$ -axis at $x = 2$ and $x = q$ . The axis of symmetry sits exactly halfway between them, so set the midpoint equal to $4$ .

\frac{2 + q}{2} = 4 ⟹ q = 6

Another way. Expand to $3 x^{2} - 3 (2 + q) x + 6 q$ and set $- \frac{b}{2 a} = 4$ . It gives the same result.

Mentor tip

On a “show that” you already know the answer, so the working is what counts. Write the midpoint equation out rather than just stating $q = 6$ .

(b)

2 marks

The vertex lies on the axis of symmetry, so $x = 4$ . Substitute it (with $q = 6$ ) for the $y$ -coordinate.

f (4) = 3 (4 - 2) (4 - 6) = - 12 vertex (4, - 12)

Mentor tip

Completing the square, $3 (x - 4)^{2} - 12$ , is a quick way to check.

(c)

1 mark

The parabola opens upwards (the leading coefficient $3$ is positive), so its lowest value is at the vertex.

f (x) \geq - 12

Mentor tip

A range describes outputs, so write it with $f (x)$ or $y$ . Writing $x \geq - 12$ describes the wrong thing.

Question 3

5 marks

Kinematics: from velocity to displacement

Calculus

(a)

2 marks

Integrate each term: $\frac{1}{t + 1}$ integrates to $ln (t + 1)$ , and $e^{- t}$ integrates to $- e^{- t}$ . Add the constant.

\int (\frac{1}{t + 1} + e^{- t}) d t = ln (t + 1) - e^{- t} + C

Mentor tip

Watch the sign on the exponential. Integrating $e^{- t}$ brings out a factor of $- 1$ , so a minus sign appears even though the integrand has none.

(b)

3 marks

Displacement is the integral of velocity, so start from your answer to (a).
Starting at $O$ means $s = 0$ when $t = 0$ . Substitute to find $C$ .

0 = ln 1 - e^{0} + C = - 1 + C C = 1 s (t) = ln (t + 1) - e^{- t} + 1

Another way. Integrate the velocity from $0$ to $t$ as a definite integral. The constant takes care of itself.

Mentor tip

The trap is assuming $C = 0$ because $s$ starts at $0$ . But $e^{0} = 1$ , not $0$ , so $C = 1$ .

Question 4

6 marks

A geometric sequence

Number and algebra

(a)

3 marks

Two terms apart means multiplying by $r$ twice, so $u_{8} = u_{6} r^{2}$ .
Divide to get $r^{2}$ , take the square root, and use the fact that every term is positive to choose the sign.

r^{2} = \frac{u _{8}}{u _{6}} = \frac{6.4 \times 1 0 ^{5}}{1.6 \times 1 0 ^{5}} = 4 r = 2 (not - 2 : every term is positive)

Mentor tip

Say why $r = - 2$ goes. Leaving $\pm 2$ on the page makes it look as if you never used that condition.

(b)

3 marks

Use $u_{n} = u_{1} r^{n - 1}$ with $n = 6$ , solve for $u_{1}$ , then write it in the form asked for.

1.6 \times 1 0^{5} = u_{1} \cdot 2^{5} = 32 u_{1} u_{1} = \frac{160 000}{32} = 5000 = 5 \times 1 0^{3}

Mentor tip

Finish in the requested form, $5 \times 1 0^{3}$ , rather than stopping at $5000$ .

Question 5

7 marks

Logarithm laws

Number and algebra

(a)

4 marks

Change the base from $25$ to $5$ , using $lo g_{5} 25 = 2$ .
Write $81 x^{2}$ as $(9 x)^{2}$ , then use the power law and the product law.

lo g_{25} (81 x^{2}) = \frac{lo g _{5} ( 9 x ) ^{2}}{lo g _{5} 25} = \frac{2 lo g _{5} ( 9 x )}{2} = lo g_{5} (9 x) = lo g_{5} 9 + lo g_{5} x

Another way. Let $lo g_{25} (81 x^{2}) = y$ , so $2 5^{y} = 81 x^{2}$ . That is $(5^{y})^{2} = (9 x)^{2}$ , so $5^{y} = 9 x$ and $y = lo g_{5} (9 x)$ .

Mentor tip

Put each log law on its own line. On a “show that”, the reasoning is the whole answer.

(b)

3 marks

Replace $lo g_{25} (81 x^{2})$ using part (a).
Gather the log terms in $x$ on one side and the numbers on the other, then drop the logs.
Undo the power $\frac{2}{3}$ by raising both sides to $\frac{3}{2}$ .

lo g_{5} x^{1/3} = lo g_{5} 9 + lo g_{5} x - lo g_{5} 4 lo g_{5} x^{- 2/3} = lo g_{5} \frac{9}{4} ⟹ x^{2/3} = \frac{4}{9} x = (\frac{4}{9})^{3/2} = (\frac{2}{3})^{3} = \frac{8}{27}

Mentor tip

Check the answer against the domain: $x = \frac{8}{27}$ is positive, so it stands.

Question 6

9 marks

Probability and independence

Statistics and probability

(a)

3 marks

$A$ splits into two pieces that do not overlap, $A \cap B^{'}$ and $A \cap B$ . Add their probabilities over a common denominator of $24$ .

P (A) = P (A \cap B^{'}) + P (A \cap B) = \frac{x + 5}{24} + \frac{2 ( 7 - x )}{24} = \frac{19 - x}{24}

Mentor tip

Show the common denominator and the expansion of $2 (7 - x)$ . That is the working a “show that” asks for.

(b)

6 marks

Find $P (B)$ the same way: $B$ splits into $A \cap B$ and $A^{'} \cap B$ .
Independence means $P (A \cap B) = P (A) \times P (B)$ . Set that up.
Clear the fractions, collect into a quadratic and factorise.

P (B) = \frac{7 - x}{12} + \frac{1}{6} = \frac{9 - x}{12} \frac{7 - x}{12} = \frac{19 - x}{24} \cdot \frac{9 - x}{12} 24 (7 - x) = (19 - x) (9 - x) ⟹ x^{2} - 4 x + 3 = 0 (x - 1) (x - 3) = 0 ⟹ x = 1 or x = 3

Mentor tip

Check both roots against the allowed range $0 \leq x \leq 7$ before keeping them. Here both work.

Section B · Extended response

Questions 7–9 · 42 marks

Question 7

12 marks

A rational function and a transformation

Functions

(a)

3 marks

Vertical asymptote: where the denominator is zero.
Horizontal asymptote: what $f (x)$ tends to as $x \to \pm \infty$ . Compare the leading terms.

3 - x = 0 ⟹ x = 3 \frac{2 x - 3}{3 - x} \to \frac{2}{- 1} = - 2 ⟹ y = - 2

Another way. Rewrite $f (x) = - 2 + \frac{3}{3 - x}$ . Both asymptotes can be read straight off it.

Mentor tip

An asymptote is a line, so give its equation: $x = 3$ and $y = - 2$ , not just $3$ and $- 2$ .

(b)

4 marks

$x$ -axis: set $f (x) = 0$ , which means the numerator is zero.
$y$ -axis: substitute $x = 0$ .

2 x - 3 = 0 ⟹ (\frac{3}{2}, 0) f (0) = \frac{- 3}{3} = - 1 ⟹ (0, - 1)

Mentor tip

Give each crossing as a point with both coordinates.

(c)

5 marks

$g \circ f$ means apply $f$ , then $g$ : $g (f (x)) = f (x) + 4$ . Adding $4$ outside the function shifts the whole graph up by $4$ .
The horizontal asymptote moves up with it, from $y = - 2$ to $y = 2$ , so $g \circ f$ takes every value except $2$ .

(g \circ f) (x) = f (x) + 4 translation by (04) (g \circ f) (x) \neq = 2

Mentor tip

Name the transformation precisely: a translation, 4 units up (or by the vector $(4 0)$ ). “Move” is too vague to describe it.

Question 8

15 marks

A cubic: inflexion and tangents

Calculus

(a)

6 marks

Differentiate twice.
At a point of inflexion the second derivative is zero: $f^{''} (- 2) = 0$ gives $b = 6 a$ .
The point lies on the curve: substituting $x = - 2$ into $f$ gives $- 2 a + b = 6$ .
Solve the two equations together.

f^{''} (x) = 6 a x + 2 b ⟹ - 12 a + 2 b = 0 ⟹ b = 6 a - 8 a + 4 b - 2 c + d = - 2 c + d + 24 ⟹ - 2 a + b = 6 4 a = 6 ⟹ a = \frac{3}{2}, b = 9

Mentor tip

Label your two equations. It keeps the simultaneous solve easy to follow.

(b)

7 marks

(i) A horizontal tangent at $x = - 3$ means $f^{'} (- 3) = 0$ .

(ii) Solve $f^{'} (k) = 0$ . One root is $- 3$ , which you already know, so the other is $k$ .

(iii) Use the second derivative at $x = k$ : positive means concave up, a minimum.

f^{'} (- 3) = \frac{81}{2} - 54 + c = 0 ⟹ c = \frac{27}{2} \frac{9}{2} k^{2} + 18 k + \frac{27}{2} = 0 ⟹ (k + 3) (k + 1) = 0 ⟹ k = - 1 f^{''} (- 1) = - 9 + 18 = 9 > 0 ⟹ local minimum

Mentor tip

A quick check for (ii): a cubic’s point of inflexion sits halfway between its stationary points, so $\frac{- 3 + k}{2} = - 2$ gives $k = - 1$ .

(c)

2 marks

$P$ is where the graph meets the $y$ -axis, so $P = (0, d)$ .
The tangent at $x = - 3$ is horizontal, so it is the line $y = f (- 3)$ . Show that $f (- 3) = d$ .

f (- 3) = - \frac{81}{2} + 81 - \frac{81}{2} + d = d tangent: y = d, which passes through P (0, d)

Mentor tip

You need both halves: $P$ is $(0, d)$ , and the tangent is the line $y = d$ .

Question 9

15 marks

Optimisation: the smallest pulling force

Geometry and trigonometry
Calculus

(a)

2 marks

A horizontal handle means $θ = 0$ . Substitute it, with the given force.

300 = \frac{A}{cos 0 + μ sin 0} = A

(b)

4 marks

Use the exact values at $θ = \frac{π}{6}$ : $cos = \frac{3}{2}$ , $sin = \frac{1}{2}$ .
Substitute with $A = 300$ and the given force, then solve for $μ$ .

1003 = \frac{300}{\frac{3}{2} + \frac{μ}{2}} = \frac{600}{3 + μ} 3 + μ = \frac{6}{3} = 23 ⟹ μ = 3

Mentor tip

Keep everything exact, and rationalise $\frac{6}{3}$ to $23$ .

(c)

9 marks

(i) Quotient rule with $u = 300$ (so $u^{'} = 0$ ) and $v = cos θ + 3 sin θ$ .

(ii) The force is smallest where $\frac{d F}{d θ} = 0$ . Only the numerator can be zero, which gives $tan α = 3$ , so $α = \frac{π}{3}$ in the physical range.

Substitute $α$ back into $F$ for the force itself.

\frac{d F}{d θ} = \frac{300 ( sin θ - 3 cos θ )}{( cos θ + 3 sin θ ) ^{2}} sin α = 3 cos α ⟹ tan α = 3 ⟹ α = \frac{π}{3} F (\frac{π}{3}) = \frac{300}{\frac{1}{2} + \frac{3}{2}} = 150 N

Mentor tip

Finish both halves of (ii): the angle, $\frac{π}{3}$ , and the force, $150$ N.

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Calculus carries the most marks on this paper.

Calculus35 marks · 44%
Q1, Q3, Q8, Q9
Functions17 marks · 21%
Q2, Q7
Number and algebra13 marks · 16%
Q4, Q5
Statistics and probability9 marks · 11%
Q6
Geometry and trigonometry6 marks · 8%
Q9

Mentor insights

Traps to watch for

01
The constant of integration · Q3
Starting at the origin tempts you to set C = 0. But e⁰ = 1, so C = 1. Substitute the starting values; never assume.
02
Choosing the common ratio · Q4
r² = 4 gives r = ±2. Use the condition that every term is positive, and write down why −2 goes.
03
Asymptotes are lines · Q7
Give each one as an equation, x = 3 and y = −2, not just the numbers.
04
Naming a transformation · Q7
Say exactly what happens: a translation, 4 units up. “Move” with no direction leaves the reader guessing.

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 SL · Paper 1

Mathematics: Analysis & Approaches, Standard Level

Timezone

TZ1TZ2 TZ3

Session: May 2026
Marks: 80
Duration: 1 hour 30 min
Calculator: Not allowed
Questions: 9

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2026 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 1–6 · 38 marks

Question 1

6 marks

Differentiation and a tangent line

Calculus

(a)

2 marks

Differentiate term by term: $2 x$ gives $2$ , and $sin x$ gives $cos x$ .

f^{'} (x) = 2 + cos x

(b)

4 marks

Find the point: substitute $x = π$ into $f$ .
Find the gradient: substitute $x = π$ into $f^{'} (x)$ from part (a).
Put both into $y - y_{1} = m (x - x_{1})$ and tidy up.

f (π) = 2 π + sin π = 2 π m = f^{'} (π) = 2 + cos π = 1 y - 2 π = 1 (x - π) y = x + π

Mentor tip

When a part says “hence”, it wants you to use the part before it. Reusing your derivative from (a) is the quick route.

Question 2

5 marks

A quadratic: symmetry, vertex and range

Functions

(a)

2 marks

The graph crosses the $x$ -axis at $x = 2$ and $x = q$ . The axis of symmetry sits exactly halfway between them, so set the midpoint equal to $4$ .

\frac{2 + q}{2} = 4 ⟹ q = 6

Another way. Expand to $3 x^{2} - 3 (2 + q) x + 6 q$ and set $- \frac{b}{2 a} = 4$ . It gives the same result.

Mentor tip

On a “show that” you already know the answer, so the working is what counts. Write the midpoint equation out rather than just stating $q = 6$ .

(b)

2 marks

The vertex lies on the axis of symmetry, so $x = 4$ . Substitute it (with $q = 6$ ) for the $y$ -coordinate.

f (4) = 3 (4 - 2) (4 - 6) = - 12 vertex (4, - 12)

Mentor tip

Completing the square, $3 (x - 4)^{2} - 12$ , is a quick way to check.

(c)

1 mark

The parabola opens upwards (the leading coefficient $3$ is positive), so its lowest value is at the vertex.

f (x) \geq - 12

Mentor tip

A range describes outputs, so write it with $f (x)$ or $y$ . Writing $x \geq - 12$ describes the wrong thing.

Question 3

5 marks

Kinematics: from velocity to displacement

Calculus

(a)

2 marks

Integrate each term: $\frac{1}{t + 1}$ integrates to $ln (t + 1)$ , and $e^{- t}$ integrates to $- e^{- t}$ . Add the constant.

\int (\frac{1}{t + 1} + e^{- t}) d t = ln (t + 1) - e^{- t} + C

Mentor tip

Watch the sign on the exponential. Integrating $e^{- t}$ brings out a factor of $- 1$ , so a minus sign appears even though the integrand has none.

(b)

3 marks

Displacement is the integral of velocity, so start from your answer to (a).
Starting at $O$ means $s = 0$ when $t = 0$ . Substitute to find $C$ .

0 = ln 1 - e^{0} + C = - 1 + C C = 1 s (t) = ln (t + 1) - e^{- t} + 1

Another way. Integrate the velocity from $0$ to $t$ as a definite integral. The constant takes care of itself.

Mentor tip

The trap is assuming $C = 0$ because $s$ starts at $0$ . But $e^{0} = 1$ , not $0$ , so $C = 1$ .

Question 4

6 marks

A geometric sequence

Number and algebra

(a)

3 marks

Two terms apart means multiplying by $r$ twice, so $u_{8} = u_{6} r^{2}$ .
Divide to get $r^{2}$ , take the square root, and use the fact that every term is positive to choose the sign.

r^{2} = \frac{u _{8}}{u _{6}} = \frac{6.4 \times 1 0 ^{5}}{1.6 \times 1 0 ^{5}} = 4 r = 2 (not - 2 : every term is positive)

Mentor tip

Say why $r = - 2$ goes. Leaving $\pm 2$ on the page makes it look as if you never used that condition.

(b)

3 marks

Use $u_{n} = u_{1} r^{n - 1}$ with $n = 6$ , solve for $u_{1}$ , then write it in the form asked for.

1.6 \times 1 0^{5} = u_{1} \cdot 2^{5} = 32 u_{1} u_{1} = \frac{160 000}{32} = 5000 = 5 \times 1 0^{3}

Mentor tip

Finish in the requested form, $5 \times 1 0^{3}$ , rather than stopping at $5000$ .

Question 5

7 marks

Logarithm laws

Number and algebra

(a)

4 marks

Change the base from $25$ to $5$ , using $lo g_{5} 25 = 2$ .
Write $81 x^{2}$ as $(9 x)^{2}$ , then use the power law and the product law.

lo g_{25} (81 x^{2}) = \frac{lo g _{5} ( 9 x ) ^{2}}{lo g _{5} 25} = \frac{2 lo g _{5} ( 9 x )}{2} = lo g_{5} (9 x) = lo g_{5} 9 + lo g_{5} x

Another way. Let $lo g_{25} (81 x^{2}) = y$ , so $2 5^{y} = 81 x^{2}$ . That is $(5^{y})^{2} = (9 x)^{2}$ , so $5^{y} = 9 x$ and $y = lo g_{5} (9 x)$ .

Mentor tip

Put each log law on its own line. On a “show that”, the reasoning is the whole answer.

(b)

3 marks

Replace $lo g_{25} (81 x^{2})$ using part (a).
Gather the log terms in $x$ on one side and the numbers on the other, then drop the logs.
Undo the power $\frac{2}{3}$ by raising both sides to $\frac{3}{2}$ .

lo g_{5} x^{1/3} = lo g_{5} 9 + lo g_{5} x - lo g_{5} 4 lo g_{5} x^{- 2/3} = lo g_{5} \frac{9}{4} ⟹ x^{2/3} = \frac{4}{9} x = (\frac{4}{9})^{3/2} = (\frac{2}{3})^{3} = \frac{8}{27}

Mentor tip

Check the answer against the domain: $x = \frac{8}{27}$ is positive, so it stands.

Question 6

9 marks

Probability and independence

Statistics and probability

(a)

3 marks

$A$ splits into two pieces that do not overlap, $A \cap B^{'}$ and $A \cap B$ . Add their probabilities over a common denominator of $24$ .

P (A) = P (A \cap B^{'}) + P (A \cap B) = \frac{x + 5}{24} + \frac{2 ( 7 - x )}{24} = \frac{19 - x}{24}

Mentor tip

Show the common denominator and the expansion of $2 (7 - x)$ . That is the working a “show that” asks for.

(b)

6 marks

Find $P (B)$ the same way: $B$ splits into $A \cap B$ and $A^{'} \cap B$ .
Independence means $P (A \cap B) = P (A) \times P (B)$ . Set that up.
Clear the fractions, collect into a quadratic and factorise.

P (B) = \frac{7 - x}{12} + \frac{1}{6} = \frac{9 - x}{12} \frac{7 - x}{12} = \frac{19 - x}{24} \cdot \frac{9 - x}{12} 24 (7 - x) = (19 - x) (9 - x) ⟹ x^{2} - 4 x + 3 = 0 (x - 1) (x - 3) = 0 ⟹ x = 1 or x = 3

Mentor tip

Check both roots against the allowed range $0 \leq x \leq 7$ before keeping them. Here both work.

Section B · Extended response

Questions 7–9 · 42 marks

Question 7

12 marks

A rational function and a transformation

Functions

(a)

3 marks

Vertical asymptote: where the denominator is zero.
Horizontal asymptote: what $f (x)$ tends to as $x \to \pm \infty$ . Compare the leading terms.

3 - x = 0 ⟹ x = 3 \frac{2 x - 3}{3 - x} \to \frac{2}{- 1} = - 2 ⟹ y = - 2

Another way. Rewrite $f (x) = - 2 + \frac{3}{3 - x}$ . Both asymptotes can be read straight off it.

Mentor tip

An asymptote is a line, so give its equation: $x = 3$ and $y = - 2$ , not just $3$ and $- 2$ .

(b)

4 marks

$x$ -axis: set $f (x) = 0$ , which means the numerator is zero.
$y$ -axis: substitute $x = 0$ .

2 x - 3 = 0 ⟹ (\frac{3}{2}, 0) f (0) = \frac{- 3}{3} = - 1 ⟹ (0, - 1)

Mentor tip

Give each crossing as a point with both coordinates.

(c)

5 marks

$g \circ f$ means apply $f$ , then $g$ : $g (f (x)) = f (x) + 4$ . Adding $4$ outside the function shifts the whole graph up by $4$ .
The horizontal asymptote moves up with it, from $y = - 2$ to $y = 2$ , so $g \circ f$ takes every value except $2$ .

(g \circ f) (x) = f (x) + 4 translation by (04) (g \circ f) (x) \neq = 2

Mentor tip

Name the transformation precisely: a translation, 4 units up (or by the vector $(4 0)$ ). “Move” is too vague to describe it.

Question 8

15 marks

A cubic: inflexion and tangents

Calculus

(a)

6 marks

Differentiate twice.
At a point of inflexion the second derivative is zero: $f^{''} (- 2) = 0$ gives $b = 6 a$ .
The point lies on the curve: substituting $x = - 2$ into $f$ gives $- 2 a + b = 6$ .
Solve the two equations together.

f^{''} (x) = 6 a x + 2 b ⟹ - 12 a + 2 b = 0 ⟹ b = 6 a - 8 a + 4 b - 2 c + d = - 2 c + d + 24 ⟹ - 2 a + b = 6 4 a = 6 ⟹ a = \frac{3}{2}, b = 9

Mentor tip

Label your two equations. It keeps the simultaneous solve easy to follow.

(b)

7 marks

(i) A horizontal tangent at $x = - 3$ means $f^{'} (- 3) = 0$ .

(ii) Solve $f^{'} (k) = 0$ . One root is $- 3$ , which you already know, so the other is $k$ .

(iii) Use the second derivative at $x = k$ : positive means concave up, a minimum.

f^{'} (- 3) = \frac{81}{2} - 54 + c = 0 ⟹ c = \frac{27}{2} \frac{9}{2} k^{2} + 18 k + \frac{27}{2} = 0 ⟹ (k + 3) (k + 1) = 0 ⟹ k = - 1 f^{''} (- 1) = - 9 + 18 = 9 > 0 ⟹ local minimum

Mentor tip

A quick check for (ii): a cubic’s point of inflexion sits halfway between its stationary points, so $\frac{- 3 + k}{2} = - 2$ gives $k = - 1$ .

(c)

2 marks

$P$ is where the graph meets the $y$ -axis, so $P = (0, d)$ .
The tangent at $x = - 3$ is horizontal, so it is the line $y = f (- 3)$ . Show that $f (- 3) = d$ .

f (- 3) = - \frac{81}{2} + 81 - \frac{81}{2} + d = d tangent: y = d, which passes through P (0, d)

Mentor tip

You need both halves: $P$ is $(0, d)$ , and the tangent is the line $y = d$ .

Question 9

15 marks

Optimisation: the smallest pulling force

Geometry and trigonometry
Calculus

(a)

2 marks

A horizontal handle means $θ = 0$ . Substitute it, with the given force.

300 = \frac{A}{cos 0 + μ sin 0} = A

(b)

4 marks

Use the exact values at $θ = \frac{π}{6}$ : $cos = \frac{3}{2}$ , $sin = \frac{1}{2}$ .
Substitute with $A = 300$ and the given force, then solve for $μ$ .

1003 = \frac{300}{\frac{3}{2} + \frac{μ}{2}} = \frac{600}{3 + μ} 3 + μ = \frac{6}{3} = 23 ⟹ μ = 3

Mentor tip

Keep everything exact, and rationalise $\frac{6}{3}$ to $23$ .

(c)

9 marks

(i) Quotient rule with $u = 300$ (so $u^{'} = 0$ ) and $v = cos θ + 3 sin θ$ .

(ii) The force is smallest where $\frac{d F}{d θ} = 0$ . Only the numerator can be zero, which gives $tan α = 3$ , so $α = \frac{π}{3}$ in the physical range.

Substitute $α$ back into $F$ for the force itself.

\frac{d F}{d θ} = \frac{300 ( sin θ - 3 cos θ )}{( cos θ + 3 sin θ ) ^{2}} sin α = 3 cos α ⟹ tan α = 3 ⟹ α = \frac{π}{3} F (\frac{π}{3}) = \frac{300}{\frac{1}{2} + \frac{3}{2}} = 150 N

Mentor tip

Finish both halves of (ii): the angle, $\frac{π}{3}$ , and the force, $150$ N.

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Calculus carries the most marks on this paper.

Calculus35 marks · 44%
Q1, Q3, Q8, Q9
Functions17 marks · 21%
Q2, Q7
Number and algebra13 marks · 16%
Q4, Q5
Statistics and probability9 marks · 11%
Q6
Geometry and trigonometry6 marks · 8%
Q9

Mentor insights

Traps to watch for

01
The constant of integration · Q3
Starting at the origin tempts you to set C = 0. But e⁰ = 1, so C = 1. Substitute the starting values; never assume.
02
Choosing the common ratio · Q4
r² = 4 gives r = ±2. Use the condition that every term is positive, and write down why −2 goes.
03
Asymptotes are lines · Q7
Give each one as an equation, x = 3 and y = −2, not just the numbers.
04
Naming a transformation · Q7
Say exactly what happens: a translation, 4 units up. “Move” with no direction leaves the reader guessing.

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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