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Edexcel IGCSE 4MA1 Paper 1H, November 2024: Worked Solutions, Questions 16 to 25

Sir Faraz Hassan

29 Jul 2026

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Edexcel International GCSE Mathematics A (4MA1)Paper 1H - Higher Tier - November 2024100 marks · 2 hours · Calculator allowed

Back to questions 1 to 15

This is the second half of the paper. Questions 1 to 15, the paper's overview and the frequently asked questions are on the first page.

Original worked solutions for Edexcel International GCSE Mathematics A (4MA1), Paper 1H (Higher Tier), November 2024 – 100 marks, 2 hours, calculator allowed. The questions have been reworded; all numerical values match the original paper. The official question paper and mark scheme are published by Pearson Edexcel. This resource reproduces neither the exam paper nor the official mark scheme.

Both are PDF files hosted by Pearson: official question paper (PDF) and official mark scheme (PDF).

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All 25 questions with a full worked solution and mark scheme - free PDF

Worked solutions, questions 16 to 25 of 25

$OKLM$ is a sector of a circle, centre $O$ .

Angle $KOM$ is $50^\circ$

The area of triangle $OKM$ is $120 \text{ cm}^2$

Work out the area of the sector $OKLM$ .

Give your answer correct to $3$ significant figures. [4 marks]

cm²

[Total 4 marks]

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Question 16 - Exam Solution

Understanding the Question

Given

OKLM

is a sector of a circle, centre

O

Angle

KOM

50^\circ

The area of triangle

OKM

120 \text{ cm}^2

Find

The area of the sector

OKLM

, correct to

3

significant figures

Plan the Solution

The two straight edges of a sector are radii of the same circle, so the two sides of triangle $OKM$ that meet at $O$ are the same length. Call that length $r$ .
Use the given area of the triangle to work out $r^2$ , then square root it to get the radius.
The sector is $50$ parts out of $360$ of the whole circle, so take that fraction of the circle area.
Keep every decimal on the calculator to the very last line, then round once, to $3$ significant figures.

Worked Solution [4 marks]

Rule - Area of a triangle from two sides and the angle between them:

\dfrac{1}{2} \times \text{side} \times \text{side} \times \sin(\text{angle})

. Area of a sector:

\dfrac{\text{angle}}{360} \times \pi \times \text{radius}^2

Turn the triangle into an equation

120 = \dfrac{1}{2} \times OK \times OM \times \sin 50^\circ

120 = \dfrac{1}{2} \times r^2 \times \sin 50^\circ

(Reason:

OK

and

OM

are radii of the same circle, so they are equal. Writing both of them as

r

leaves

r^2

as the only unknown, which is exactly what the sector formula needs.)

Make

r^2

the subject

r^2 = \dfrac{2 \times 120}{\sin 50^\circ} = 313.2977\ldots

(Reason: Rearranging isolates

r^2

. Nothing is rounded here - the full figure stays on the calculator.)

Square root for the radius

r = \sqrt{313.2977\ldots} = 17.7002\ldots \text{ cm}

(Reason: This is the length of each radius, and the mark scheme awards a method mark for reaching it. Carry the full accuracy forward rather than working on from

17.7

Put the radius into the sector formula

\text{Area of sector } OKLM = \dfrac{50}{360} \times \pi \times r^2

\dfrac{50}{360} \times \pi \times 313.2977\ldots = 136.7019\ldots

(Reason: A whole circle has area

\pi

times

r^2

, and this sector is

50

parts out of

360

of it. Squaring the radius simply gives

r^2

back, so the value from step 2 can go straight in.)

Round to

3

significant figures

136.7019\ldots \approx 137 \text{ cm}^2

(Reason: The first three significant figures are

1

3

and

6

. The next digit is

7

, so the

6

rounds up to

7

137

cm²

Verification

Check 1: Square the radius again. If step 3 undid step 2 correctly, multiplying

17.7002

by itself has to land back on

r^2

, give or take the four decimal places it was written to.

17.7002 \times 17.7002 = 313.2971

Check 2: Work all the way back to the question. If

r^2

is right, putting it through the triangle formula must return the area the question gives.

\dfrac{1}{2} \times 313.2977\ldots \times \sin 50^\circ = 120 \text{ cm}^2

Check 3: Reach the answer without the radius at all. Dividing the sector formula by the triangle formula cancels

r^2

, so the sector is the same multiple of the triangle whatever the radius happens to be. To six decimal places that multiple is

1.139183

120 \times 1.139183 = 136.702

Check 4: Check the size against the whole circle. Since

360

divided by

50

7.2

, a

50

degree sector is one part in

7.2

of the circle, and this circle has area

984.2539

cm² to four decimal places.

\dfrac{984.2539}{7.2} = 136.7019

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A correct equation or expression for the square of the radius	M1	for a correct equation or expression for the square of the radius, such as $120 = \dfrac{1}{2} \times \text{OK} \times \text{OM} \times \sin 50^\circ$ or $\dfrac{2 \times 120}{\sin 50^\circ}$ , or for $313(.2977494)$ . Any pair of letters may be used for the two sides.	✓
Rearrange to reach the radius	M1	for a correct rearrangement to find the radius, or for square rooting $313(.2977494)$ , or for $17.7(0021891)$	✓
Substitute into the area of a sector	M1	for the area of sector $OKLM$ written as $\dfrac{50}{360} \times \pi \times 17.7^2$ or an equivalent	✓
The answer, to $3$ significant figures	A1	awrt $137$ . A correct answer scores full marks unless it comes from obviously incorrect working.	✓

Full marks: 4/4

(a) Write $\sqrt{675}$ in the form $n\sqrt{27}$ , where $n$ is a positive integer.
[1 mark]
(b) Show that $\dfrac{5 - \sqrt{2}}{\sqrt{2} - 1}$ can be expressed in the form $a + b\sqrt{2}$ , where $a$ and $b$ are integers. [3 marks]

(a)

[Total 4 marks]

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Question 17 - Exam Solution

Understanding the Question

Given

\sqrt{675}

, to be written in the form

n\sqrt{27}

with

n

a positive integer

\dfrac{5 - \sqrt{2}}{\sqrt{2} - 1}

, to be written in the form

a + b\sqrt{2}

with

a

and

b

integers

Find

the positive integer

n

in part (a) the integers

a

and

b

in part (b), with the working shown

Plan the Solution

Part (a): the target form keeps $27$ under the root, so divide $675$ by $27$ and check that what is left is a perfect square.
Part (b): the denominator $\sqrt{2} - 1$ is irrational, so rationalise it by multiplying the top and the bottom by the conjugate $\sqrt{2} + 1$ .
Expand the numerator to four terms and turn the denominator into a whole number, then collect the rational part and the $\sqrt{2}$ part to read off $a$ and $b$ .

Worked Solution [4 marks]

Rule - Surds:

\sqrt{ab} = \sqrt{a} \times \sqrt{b}

, and a denominator is rationalised by multiplying the top and the bottom by its conjugate, because

(\sqrt{k} - 1)(\sqrt{k} + 1) = k - 1

Part (a) - find the square factor that leaves

27

behind

\dfrac{675}{27} = 25

675 = 25 \times 27

(Reason: The answer must keep

27

under the root, so

27

has to be one of the factors. Dividing shows the other factor is

25

, and

25

is a perfect square, which is exactly what is needed for

n

to be a whole number.)

Part (a) - split the root and take the square root of

25

\sqrt{675} = \sqrt{25 \times 27} = \sqrt{25} \times \sqrt{27}

\sqrt{675} = 5\sqrt{27}

(Reason: The root of a product is the product of the roots, so the square factor comes outside as the whole number

5

while

27

stays inside. That gives

n = 5

Part (b) - multiply the top and the bottom by the conjugate

\dfrac{5 - \sqrt{2}}{\sqrt{2} - 1} = \dfrac{5 - \sqrt{2}}{\sqrt{2} - 1} \times \dfrac{\sqrt{2} + 1}{\sqrt{2} + 1}

(Reason: The conjugate of

\sqrt{2} - 1

\sqrt{2} + 1

. Multiplying by

\dfrac{\sqrt{2} + 1}{\sqrt{2} + 1}

is multiplying by

1

, so the value of the fraction is unchanged and only its appearance changes.)

Part (b) - expand the denominator

(\sqrt{2} - 1)(\sqrt{2} + 1) = 2 - 1 = 1

(Reason: The two middle terms,

+\sqrt{2}

and

-\sqrt{2}

, cancel each other, so only the difference of the two squares survives. Clearing the surd out of the denominator is the whole purpose of the conjugate.)

Part (b) - expand the numerator to four terms

(5 - \sqrt{2})(\sqrt{2} + 1) = 5\sqrt{2} + 5 - 2 - \sqrt{2}

(Reason: Every term in the first bracket multiplies every term in the second, which is why four terms appear. Note that

-\sqrt{2} \times \sqrt{2} = -2

, because squaring a square root removes the root and leaves the number underneath it.)

Part (b) - collect the like terms

5\sqrt{2} - \sqrt{2} = 4\sqrt{2}

5 - 2 = 3

\dfrac{3 + 4\sqrt{2}}{1} = 3 + 4\sqrt{2}

(Reason: Whole numbers collect with whole numbers and

\sqrt{2}

terms with

\sqrt{2}

terms. The denominator has come out as

1

, so the fraction is simply its numerator, and comparing with

a + b\sqrt{2}

gives

a = 3

and

b = 4

(a)

\sqrt{675} = 5\sqrt{27}

, so

n = 5

.(b)

3 + 4\sqrt{2}

, so

a = 3

and

b = 4

Verification

Check 1: Reverse part (a) by squaring the whole number and putting it back under the root: work out

5^2 \times 27

5^2 \times 27 = 25 \times 27 = 675

, which is the number the question starts with

Check 2: Reduce both forms in part (a) to the simplest surd, using

675 = 225 \times 3

and

27 = 9 \times 3

\sqrt{675} = 15\sqrt{3}

and

5\sqrt{27} = 5 \times 3\sqrt{3} = 15\sqrt{3}

, so the two forms agree

Check 3: Reverse part (b) by multiplying the answer back by the original denominator: expand

(3 + 4\sqrt{2})(\sqrt{2} - 1)

3\sqrt{2} - 3 + 8 - 4\sqrt{2} = 5 - \sqrt{2}

, which is the original numerator, so the denominator really was

1

Check 4: Evaluate the original fraction and the answer to part (b) as decimals and compare

\dfrac{5 - \sqrt{2}}{\sqrt{2} - 1} \approx 8.65685

and

3 + 4\sqrt{2} \approx 8.65685

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $5\sqrt{27}$	B1	for $5\sqrt{27}$ . Allow $n = 5$ . Do not accept $5$ by itself.	✓
(b) Multiply the numerator and the denominator by the conjugate	M1	for rationalising the denominator by multiplying numerator and denominator by $\sqrt{2} + 1$ or $-\sqrt{2} - 1$	✓
(b) Expand, eg $\dfrac{5\sqrt{2} + 5 - 2 - \sqrt{2}}{2 - 1}$	M1	The numerator must be expanded to $4$ terms. The denominator may be shown as $4$ terms, which then need to be all correct, eg $\dfrac{5\sqrt{2} + 5 - 2 - \sqrt{2}}{\sqrt{4} + \sqrt{2} - \sqrt{2} - 1}$ oe, or the numerator $5\sqrt{2} + 5 - 2 - \sqrt{2}$ alone. Accept $1$ in the denominator without working.	✓
(b) $3 + 4\sqrt{2}$	A1	for $3 + 4\sqrt{2}$ , or for stating $a = 3$ and $b = 4$ . Working required.	✓

Full marks: 4/4

The histogram gives information about the times taken by some visitors to complete a maze at a country park.

Work out an estimate for the fraction of these visitors who took between $20$ minutes and $60$ minutes to complete the maze. [4 marks]

[Total 4 marks]

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Question 18 - Exam Solution

Understanding the Question

Given

A histogram of the times, in minutes, that some visitors took to complete a maze

The five bars have frequency densities

1.5

0.8

4

2.5

and

4.6

Their class boundaries are at

0

10

25

30

60

and

65

minutes

No total is printed anywhere, so the number of visitors has to be built from the histogram

Find

An estimate for the fraction of these visitors whose time was between

20

minutes and

60

minutes

Plan the Solution

On a histogram the frequency is the AREA of a bar, not its height, so multiply each frequency density by its own class width.
Add the five frequencies. That total is the denominator of the fraction, and it is nowhere in the question.
$20$ minutes is not a class boundary. It cuts the bar running from $10$ to $25$ , so take only the $5$ minute piece from $20$ to $25$ , at the same frequency density.
Add the three pieces that make up $20$ to $60$ minutes, write that over the total, and leave the answer as a fraction.

Worked Solution [4 marks]

Rule - Frequency from a histogram:

\text{frequency} = \text{frequency density} \times \text{class width}

, which is the area of the bar.

Turn every bar into a frequency

1.5 \times 10 = 15

0.8 \times 15 = 12

4 \times 5 = 20

2.5 \times 30 = 75

4.6 \times 5 = 23

(Reason: The height of a histogram bar is frequency density, not frequency, so every bar has to be multiplied by its own class width - which is the same as reading off the area of the bar. The five widths come from the horizontal axis:

10

15

5

30

and

5

minutes.)

Add the five frequencies for the total

15 + 12 + 20 + 75 + 23 = 145

(Reason: Every visitor is counted in exactly one bar, so the five frequencies add up to the whole group. This total is the denominator of the fraction, and the question never states it.)

Cut the second bar at

20

minutes

0.8 \times 5 = 4

(Reason:

20

minutes is not a class boundary. It falls inside the bar that runs from

10

25

, and a histogram assumes the times are spread evenly across a bar, so this piece keeps the frequency density

0.8

and only the width changes, from

15

minutes down to

5

Add the three pieces that make up

20

60

minutes

4 + 20 + 75 = 99

(Reason: The interval from

20

60

minutes is made of three pieces: the part of the

10

25

bar beyond

20

minutes, the whole

25

30

bar and the whole

30

60

bar. The last bar is left out because it starts exactly where the interval ends.)

Write the estimate as a fraction

\dfrac{4 + 20 + 75}{15 + 12 + 20 + 75 + 23} = \dfrac{99}{145}

(Reason: The question asks for a fraction, so it is left as one rather than turned into a decimal.

99

9 \times 11

and

145

5 \times 29

, so the two share no factor and the fraction is already in its simplest form.)

\dfrac{99}{145}

Verification

Check 1 - the visitors left out: Count everybody OUTSIDE

20

60

minutes instead and subtract. That is the whole first bar,

15

, then the

10

20

piece of the second bar at

0.8 \times 10

, then the whole last bar,

23

145 - (15 + 8 + 23) = 99

Check 2 - counting blocks instead of using the formula: Take one block of the grid to be

5

minutes wide and

0.5

of a unit of frequency density tall, so one block stands for

2.5

visitors. The five bars are then

6

4.8

8

30

and

9.2

blocks, and the

20

60

region is

1.6 + 8 + 30

blocks.

\dfrac{39.6}{58} = \dfrac{99}{145}

Check 3 - working in tenths: Multiply every frequency by

10

so that nothing is a decimal. The bars become

150

120

200

750

and

230

, a total of

1450

, and the

20

60

region becomes

40 + 200 + 750

. Cancelling the

10

has to return the same fraction.

\dfrac{990}{1450} = \dfrac{99}{145}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
At least two correct frequencies	M1	for at least two correct frequencies, for example $1.5 \times 10 (= 15)$ or $0.8 \times 15 (= 12)$ or $4 \times 5 (= 20)$ or $2.5 \times 30 (= 75)$ or $4.6 \times 5 (= 23)$ , or for any two of $15$ , $12$ , $20$ , $75$ and $23$ . Counting squares or blocks earns it just as well, so any two of $150$ , $120$ , $200$ , $750$ and $230$ , or any two of $6$ , $4.8$ , $8$ , $30$ and $9.2$ , also score.	✓
A method to find the number of visitors in each interval, with the intention to add	M1	for $1.5 \times 10 + 0.8 \times 15 + 4 \times 5 + 2.5 \times 30 + 4.6 \times 5 (= 145)$ , or for $15 + 12 + 20 + 75 + 23 (= 145)$ , with one error allowed. The block totals $150 + 120 + 200 + 750 + 230 (= 1450)$ and $6 + 4.8 + 8 + 30 + 9.2 (= 58)$ score the same mark, and so does a correct method for the $20$ to $60$ minute frequency such as $4 + 20 + 75 (= 99)$ .	✓
A complete method for the fraction	M1	for a complete method, for example $\dfrac{0.8 \times 5 + 4 \times 5 + 2.5 \times 30}{15 + 12 + 20 + 75 + 23}$ , or the block form $\dfrac{1.6 + 8 + 30}{58}$ . The one error already allowed in the previous method mark may follow through into this one.	✓
The fraction	A1	for $\dfrac{99}{145}$ , or for any equivalent fraction such as $\dfrac{990}{1450}$ . The answer must be left as a fraction. A correct answer scores full marks unless it has come from obviously incorrect working.	✓
Note - reaching $99$ without listing the frequencies	note	A value of $99$ for the $20$ to $60$ minute interval scores both of the first two method marks (M2), even when the individual frequencies are never written down.	✓

Full marks: 4/4

An arithmetic series has first term $a$ and common difference $d$ .

Adding the first $30$ terms of this series gives $4\,395$

The $10\text{th}$ term and the $20\text{th}$ term have a sum of $284$

Find the sum of the first $45$ terms of the series. You must show clear algebraic working. [5 marks]

[Total 5 marks]

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Question 19 - Exam Solution

Understanding the Question

Given

An arithmetic series with first term

a

and common difference

d

The sum of the first

30

terms is

4\,395

The

10\text{th}

term added to the

20\text{th}

term is

284

Find

the sum of the first

45

terms of the same series

Plan the Solution

Put $n = 30$ into the sum formula, so the first fact becomes an equation in $a$ and $d$ .
Put $n = 10$ and $n = 20$ into the $n\text{th}$ term formula, add the two expressions, and the second fact becomes a second equation in $a$ and $d$ .
Both equations contain $2a$ , so subtracting one from the other removes $a$ and leaves $d$ on its own.
Substitute $d$ back to get $a$ , then put $a$ , $d$ and $n = 45$ into the sum formula.

Worked Solution [5 marks]

Rule - Arithmetic series: the

n\text{th}

term is

a + (n - 1)d

, and the sum of the first

n

terms is

\dfrac{n}{2}\left[2a + (n - 1)d\right]

Turn the sum of the first

30

terms into an equation

S_{30} = \dfrac{30}{2}\left[2a + (30 - 1)d\right] = 4\,395

15(2a + 29d) = 4\,395

\dfrac{4\,395}{15} = 293

2a + 29d = 293

(Reason: A total made from

30

terms can only come from the sum formula, so putting

n = 30

into it turns the first fact into an equation in the two unknowns. Dividing both sides by

15

straight away keeps the numbers small, which makes the elimination later much easier.)

Turn the

10\text{th}

and

20\text{th}

terms into a second equation

u_{10} = a + (10 - 1)d = a + 9d

u_{20} = a + (20 - 1)d = a + 19d

(a + 9d) + (a + 19d) = 284

2a + 28d = 284

(Reason: Every term is the first term plus a number of common differences, and that number is always one less than the position, so the

10\text{th}

term carries

9

lots of

d

and the

20\text{th}

carries

19

. Adding the two expressions collects them into a second equation in the same two unknowns.)

Subtract the two equations to find the common difference

(2a + 29d) - (2a + 28d) = 293 - 284

293 - 284 = 9

d = 9

(Reason: Both equations contain exactly

2a

, so subtracting one from the other cancels the first term completely. On the left,

29

lots of

d

take away

28

lots of

d

leaves a single

d

, so the right hand side is the value of

d

itself.)

Substitute the common difference to find the first term

2a + 28 \times 9 = 284

28 \times 9 = 252

2a = 284 - 252 = 32

a = 16

(Reason: The second equation is the simpler of the two, so put

d = 9

into that one. Taking

252

from both sides leaves

2a

on its own, and halving

32

gives the first term.)

Put the first term, the common difference and

45

terms into the sum formula

S_{45} = \dfrac{45}{2}\left[2 \times 16 + (45 - 1) \times 9\right]

\dfrac{45}{2} \times \left(32 + 396\right) = \dfrac{45}{2} \times 428

\dfrac{45}{2} \times 428 = 45 \times 214 = 9\,630

(Reason: The sum formula needs the number of terms, the first term and the common difference, and all three are now known. Watch the multiplier of the common difference: with

45

terms it is

45

minus

1

, which is

44

, not

45

9630

Verification

Check 1: Rebuild the first given fact from the values found, by working out the sum of the first

30

terms with

a = 16

and

d = 9

\dfrac{30}{2} \times \left(2 \times 16 + 29 \times 9\right) = 15 \times 293 = 4\,395

Check 2: Rebuild the second given fact by adding the

10\text{th}

term to the

20\text{th}

term, again using

a = 16

and

d = 9

(16 + 9 \times 9) + (16 + 19 \times 9) = 97 + 187 = 284

Check 3: Sum the

45

terms a completely different way, as half the number of terms times the first term plus the last term, where the

45\text{th}

term is

412

\dfrac{45}{2} \times \left(16 + 412\right) = 22.5 \times 428 = 9\,630

Check 4: Split the sum instead: add the given total for the first

30

terms to the

15

terms that run from the

31\text{st}

term,

286

, to the

45\text{th}

term,

412

4\,395 + \dfrac{15}{2} \times \left(286 + 412\right) = 4\,395 + 5\,235 = 9\,630

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Use the sum formula with $n = 30$	M1	for using the sum formula $\dfrac{n}{2}\left[2a + (n - 1)d\right]$ with $n = 30$ . Accept $\dfrac{30}{2}\left[2a + (30 - 1)d\right] = 4\,395$ , or the expanded $30a + 435d = 4\,395$ , or the simplified $2a + 29d = 293$	✓
Use the $n\text{th}$ term formula to form a second equation	M1	for using the $n\text{th}$ term formula $a + (n - 1)d$ correctly to form an equation. Accept $a + (10 - 1)d + a + (20 - 1)d = 284$ , or $a + 9d + a + 19d = 284$ , or $2a + 28d = 284$ , or $a + 14d = 142$	✓
A correct method to eliminate $a$ or $d$	dM1	dependent on both earlier method marks, for a correct method to eliminate $a$ or $d$ : the coefficients of $a$ or of $d$ made equal and the correct operator used on the chosen variable, condoning any one arithmetic error. Accept $2a + 29d = 293$ minus $2a + 28d = 284$ giving $d = 9$ , or $28a + 406d = 4\,102$ minus $29a + 406d = 4\,118$ giving $a = 16$ . Writing $a$ or $d$ in terms of the other variable and substituting correctly also earns it	✓
Use the sum formula again, now with $n = 45$	dM1	dependent on the previous method mark, for using the sum formula correctly with $a = 16$ and $d = 9$ , eg $\dfrac{45}{2}\left[2(16) + (45 - 1)9\right]$	✓
The sum of the first $45$ terms, with working shown	A1	for $9\,630$ . Working is required, so a bare answer scores nothing	✓

Full marks: 5/5

A curve $C$ has equation $y = 2x^4 - 64x$
The curve $C$ has a minimum point.
Work out an equation of the tangent to $C$ at this minimum point.
You must show clear algebraic working. [4 marks]

[Total 4 marks]

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Question 20 - Exam Solution

Understanding the Question

Given

The curve

C

has equation

y = 2x^4 - 64x

C

has a minimum point

Find

An equation of the tangent to

C

at that minimum point

Plan the Solution

Differentiate the curve to get an expression for the gradient, $\dfrac{dy}{dx}$
At a minimum point the gradient is zero, so solve $\dfrac{dy}{dx} = 0$ to find the $x$ -coordinate
Substitute that $x$ back into the equation of the curve to get the $y$ -coordinate
A tangent at a minimum point is horizontal, so its equation has the form $y = k$ with no $x$ in it

Worked Solution [4 marks]

Rule - Stationary points: solve

\dfrac{dy}{dx} = 0

for

x

, substitute back into the curve for

y

, and note that a tangent at a minimum has gradient

0

, so it is a horizontal line

y = k

Differentiate the equation of the curve

y = 2x^4 - 64x

\dfrac{dy}{dx} = 8x^3 - 64

(Reason: Multiply by the power and then reduce the power by one. The term

2x^4

gives

4 \times 2x^3 = 8x^3

, and the term

-64x

gives

-64

, because

x

differentiates to

1

. One correct term is already worth the first method mark.)

Set the gradient equal to zero

8x^3 - 64 = 0

(Reason: At a minimum point the curve is momentarily flat, so the gradient there is

0

. Solving

\dfrac{dy}{dx} = 0

is what locates the stationary point.)

Solve the equation for

x

8x^3 = 64

x^3 = \dfrac{64}{8} = 8

x = \sqrt[3]{8} = 2

(Reason: Divide both sides by

8

, then take the cube root. A cube root has just one real value, so

x = 2

is the only stationary point on this curve, and it must therefore be the minimum point the question describes.)

Work out the

y

-coordinate of the minimum point

y = 2 \times 2^4 - 64 \times 2 = 32 - 128 = -96

(Reason: Substitute

x = 2

back into the equation of the curve,

y = 2x^4 - 64x

, to get the height of the minimum point. The derivative has done its job already and is not the expression to substitute into here.)

Write down the equation of the tangent

\text{gradient of the tangent} = 0

y = -96

(Reason: A tangent at a minimum point is horizontal, so its gradient is

0

. Every point on a horizontal line through

(2, -96)

has

y

-coordinate

-96

, so the tangent is

y = -96

. The question asks for an equation, so

-96

on its own, or the coordinates

(2, -96)

, would score nothing.)

y = -96

Verification

Check 1: Differentiate a second time and test the sign at

x = 2

, using

\dfrac{d^2y}{dx^2} = 24x^2

24 \times 2^2 = 96

, which is positive, so the stationary point really is a minimum and not a maximum

Check 2: Work out the height of the curve just either side of

x = 2

and compare it with

-96

y = -95.5358

x = 1.9

and

y = -95.5038

x = 2.1

, both above

-96

, so

-96

is the lowest value the curve reaches

Check 3: Subtract the tangent from the curve and factorise the result,

2x^4 - 64x + 96

2(x - 2)^2(x^2 + 4x + 12)

, and

x^2 + 4x + 12 = (x + 2)^2 + 8

is always positive, so the difference is never negative and is zero only at

x = 2

- the line

y = -96

touches the curve there and never crosses it, which is exactly what a tangent does

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Differentiate one term of the curve correctly	M1	$4 \times 2x^3$ or $8x^3$ or $64$ with either sign	✓
Form the equation that locates the stationary point	dM1	$8x^3 - 64 = 0$ or equivalent. The equation must be of the form $ax^3 - 64 = 0$ with $a$ not equal to $0$ , or $8x^3 + b = 0$ with $b$ not equal to $0$ , where $a$ and $b$ are constants	✓
Solve the equation for $x$	dM1	$x = \sqrt[3]{\dfrac{64}{8}}$ , leading to $x = 2$ . Again the equation being solved must be of the form $ax^3 - 64 = 0$ with $a$ not equal to $0$ , or $8x^3 + b = 0$ with $b$ not equal to $0$	✓
State an equation for the tangent	A1	$y = -96$ or equivalent, eg $y + 96 = 0$ or $y = 0x - 96$ or $-y = 96$ . Working is required, and the answer must be an equation in terms of $y$ : $-96$ on its own, or the coordinates $(2, -96)$ , scores nothing	✓

Full marks: 4/4

$T = \dfrac{x^2 + y^2}{w}$
Each of the three values below has been rounded.
$x = 28.4$ correct to $1$ decimal place
$y = 17$ correct to $2$ significant figures
$w = 90$ correct to the nearest $5$
Work out the upper bound for the value of $T$ .
Give your answer correct to $3$ significant figures.
You must show your working clearly. [3 marks]

[Total 3 marks]

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Question 21 - Exam Solution

Understanding the Question

Given

T = \dfrac{x^2 + y^2}{w}

x = 28.4

correct to

1

decimal place

y = 17

correct to

2

significant figures

w = 90

correct to the nearest

5

Find

The upper bound for the value of

T

, correct to

3

significant figures

Plan the Solution

Turn each rounded value into an interval: a value rounded to a unit lies half a unit either side of the figure given
Decide which end of each interval makes $T$ biggest. $x$ and $y$ are on the top, so both take their UPPER bounds; $w$ is on the bottom, so it takes its LOWER bound
Substitute those three bounds into the formula and work out the numerator first
Divide, then round the result to $3$ significant figures

Worked Solution [3 marks]

Rule - Bounds of a calculation: a value rounded to a unit

u

lies within

\dfrac{u}{2}

of the figure given, and a fraction is biggest when its numerator is biggest and its denominator is smallest, so the upper bound of

\dfrac{x^2 + y^2}{w}

comes from the upper bounds of

x

and

y

over the LOWER bound of

w

Write down the interval each letter lies in

28.35 \le x < 28.45

16.5 \le y < 17.5

87.5 \le w < 92.5

(Reason: Each rounding has its own unit. One decimal place is a unit of

0.1

, so

x

is within

0.05

28.4

. On

17

the second significant figure is the units digit, so

2

significant figures here means the nearest

1

, and

y

is within

0.5

17

. The nearest

5

is a unit of

5

, so

w

is within

2.5

90

. Half the unit each way is what produces all six numbers, and the first mark is for writing down any one of them.)

Choose the bound of each letter that makes

T

biggest

\text{upper bound of } x = 28.45

\text{upper bound of } y = 17.5

\text{lower bound of } w = 87.5

(Reason:

x^2

and

y^2

sit on the top of the fraction, so the bigger they are the bigger

T

is. But

w

sits on the bottom, and dividing by a smaller number gives a bigger answer, so

w

must take its LOWEST value. Reaching for

92.5

because the question says 'upper bound' is the single most common way this mark is lost. The true

x

never actually reaches

28.45

, which is why the mark scheme also accepts the upper bound written as the recurring decimal

28.449

recurring.)

Substitute the three bounds into the formula

T = \dfrac{28.45^2 + 17.5^2}{87.5}

(Reason: This substitution is what the method mark is for, before any arithmetic is done. Every figure in it came out of step 1, and the mark scheme wants to see the upper bounds of

x

and

y

over the lower bound of

w

Work out the numerator

28.45^2 = 809.4025

17.5^2 = 306.25

809.4025 + 306.25 = 1115.6525

(Reason: Square each upper bound, then add. Squaring keeps the order for positive numbers, so the biggest

x

and the biggest

y

really do give the biggest numerator - there is no need to test any other combination.)

Divide by the lower bound of

w

\dfrac{1115.6525}{87.5} \approx 12.75031429

(Reason: The lower bound of

w

87.5

, so

87.5

is the number to divide by. The division is written as a fraction rather than with a division sign, and the mark scheme shows this value as

12.75031429

Round to

3

significant figures

12.75031429 \approx 12.8

(Reason: The first three significant figures are

1

2

and

7

, and the digit after them is

5

, so the

7

rounds up to

8

. The mark scheme asks for awrt

12.8

and insists the working is shown, so an answer written down with no substitution scores nothing here.)

Upper bound of

T = 12.8

Verification

Check 1: Take values strictly inside all three intervals,

x = 28.4499

y = 17.4999

and

w = 87.5001

, and work out

T

T = 12.75019469

, which is below

12.75031429

, so a value the three letters are actually allowed to take falls short of the bound, exactly as an upper bound requires

Check 2: Reverse the division: multiply the unrounded answer back up by the lower bound of

w

and compare it with the numerator

\dfrac{446261}{35000} \times 87.5 = 1115.6525

, which is

809.4025 + 306.25

, so the division was carried out on the right numbers

Check 3: Test the other bound of

w

, and work out where the LOWER bound of

T

sits, to see that

12.8

lands at the top of the range using

92.5

instead gives

12.1

3

significant figures, not

12.8

, and the lower bound

\dfrac{28.35^2 + 16.5^2}{92.5}

11.6

3

significant figures, so

T

lies between

11.6

and

12.8

and

12.8

is the top of that range

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Write down a correct bound for $x$ , $y$ or $w$	B1	$28.35$ or $28.45$ or $16.5$ or $17.5$ or $87.5$ or $92.5$ . Accept $28.449$ recurring for $28.45$ , and $17.49$ recurring for $17.5$	✓
Substitute the correct bounds into the formula for $T$	M1	$T = \dfrac{(UB_x)^2 + (UB_y)^2}{LB_w}$ , for example $\dfrac{28.45^2 + 17.5^2}{87.5}$ , where $28.4 < UB_x \le 28.45$ and $17 < UB_y \le 17.5$ and $87.5 \le LB_w < 90$	✓
Complete the calculation and round it	A1	awrt $12.8$ . The answer must come from the correct figures $28.45$ , $17.5$ and $87.5$ , and working is required, so an unsupported $12.8$ scores nothing	✓

Full marks: 3/3

The diagram shows a garden bird bath in the shape of a hemisphere. The bird bath is cast from bronze.

The outer radius of the bird bath is $12$ cm

The inner radius of the bird bath is $9$ cm

The thickness of the bird bath is uniform.

Work out the volume of the bronze.

Give your answer correct to the nearest whole number. [3 marks]

cm³

[Total 3 marks]

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Question 22 - Exam Solution

Understanding the Question

Given

A bird bath in the shape of a hemisphere, cast in bronze to a uniform thickness

The outer radius is

12

cm and the inner radius is

9

The volume of a sphere is

\dfrac{4\pi r^3}{3}

Find

The volume of the bronze, correct to the nearest whole number

Plan the Solution

The bronze is what is left when the hollow is taken out of the solid outer hemisphere.
Work out the outer hemisphere, radius $12$ cm, then the hollow inner hemisphere, radius $9$ cm.
Subtract one from the other, keep the result as a multiple of $\pi$ , and round only on the last line.

Worked Solution [3 marks]

Volume of a hemisphere (half a sphere):

\dfrac{2\pi r^3}{3}

Volume of the outer hemisphere

12^3 = 1728

V_{\text{outer}} = \dfrac{2}{3} \times \pi \times 1728 = 1152\pi \approx 3619.115

(Reason: A hemisphere is half a sphere, so its volume is two thirds of

\pi

times the radius cubed. The outside of the bird bath has radius

12

cm.)

Volume of the hollow inside

9^3 = 729

V_{\text{inner}} = \dfrac{2}{3} \times \pi \times 729 = 486\pi \approx 1526.814

(Reason: The hollow is a hemisphere as well, of radius

9

cm, and it shares its centre with the outside because the thickness is uniform.)

Subtract the hollow from the outer hemisphere

V = 1152\pi - 486\pi = 666\pi

666\pi \approx 2092.3007

(Reason: The bronze is the shell left between the two hemispheres. Working in multiples of

\pi

keeps the value exact until the very last line.)

Round to the nearest whole number

2092.3007 \approx 2092

(Reason: The first figure after the decimal point is

3

, which is below

5

, so the whole-number part stays as it is.)

2092 \text{ cm}^3

Verification

Check 1: Take the cubes apart first:

1728 - 729 = 999

, then

\dfrac{2}{3} \times 999 = 666

666\pi \approx 2092.3

, the same volume as before

Check 2: Use whole spheres instead:

\dfrac{4}{3} \times \pi \times 1728 = 2304\pi

and

\dfrac{4}{3} \times \pi \times 729 = 972\pi

, so the shell between the two spheres is

1332\pi

The bird bath is half of that shell, and half of

1332\pi

666\pi

, which is

2092

to the nearest whole number

Check 3: Is the size sensible? The bronze must come to less than the whole outer hemisphere, which is about

3619 \text{ cm}^3

2092 \text{ cm}^3

is about

58

per cent of it, which suits a shell whose hollow is three quarters as wide as the outside

Mark Scheme Breakdown

Step	Mark	Description	Got it?
The volume of one sphere or one hemisphere	M1	For the volume of a sphere or of a hemisphere, for example $\dfrac{2}{3} \times \pi \times 12^3 (= 1152\pi = 3619.114)$ or $\dfrac{2}{3} \times \pi \times 9^3 (= 486\pi = 1526.814)$ , or a full sphere $\dfrac{4}{3} \times \pi \times 12^3 (= 2304\pi = 7238.229)$ or $\dfrac{4}{3} \times \pi \times 9^3 (= 972\pi = 3053.628)$	✓
A complete method for the metal	M1	For a complete method, for example $1152\pi - 486\pi (= 666\pi)$ , or $\dfrac{2304\pi - 972\pi}{2} (= 666\pi)$	✓
The rounded answer	A1	For $2092$ ; the scheme accepts anything from $2091$ to awrt $2093$	✓
Note	note	A correct answer scores full marks unless it comes from obviously incorrect working	✓

Full marks: 3/3

The curve $C$ has equation $y = x^2 - 8x - 9$

The straight line $L$ has equation $y = k$ , where $k$ is an integer.

$C$ and $L$ meet at the points $A$ and $B$ .

Point $A$ has coordinates $(p, k)$ and point $B$ has coordinates $(q, k)$ .

Given that $p - q = 14$ , work out the value of $k$ .

You must show clear algebraic working. [5 marks]

k =

[Total 5 marks]

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Question 23 - Exam Solution

Understanding the Question

Given

The curve

C

has equation

y = x^2 - 8x - 9

The straight line

L

has equation

y = k

, where

k

is an integer

C

and

L

meet at

A(p, k)

and

B(q, k)

, so the two points share one

y

value

p - q = 14

Find

The value of

k

Plan the Solution

Where $L$ meets $C$ the two $y$ values are equal, so $p$ and $q$ are the two roots of $x^2 - 8x - 9 = k$ .
A horizontal line cuts a parabola at two points that sit the same distance either side of its line of symmetry, so find that line of symmetry first.
The line of symmetry is $x = 4$ , so the gap of $14$ splits into $7$ on each side, which fixes $p$ and $q$ .
Put either root back into the equation of $C$ , because the $y$ value it returns is $k$ itself.

Worked Solution [5 marks]

Rule - Symmetry of a parabola:

y = ax^2 + bx + c

is symmetrical about the vertical line

x = \dfrac{-b}{2a}

, so the two points where a horizontal line cuts it lie the same distance either side of that line.

Set the equation of the curve equal to the equation of the line

x^2 - 8x - 9 = k

x^2 - 8x - 9 - k = 0

(Reason: At

A

and at

B

a point sits on the curve and on the line at the same moment, so its

y

value is

x^2 - 8x - 9

and it is also

k

. That makes

p

and

q

the two roots of this one quadratic, and it is why both points carry the same second coordinate.)

Find the line of symmetry of the curve

x^2 - 8x - 9 = (x - 4)^2 - 16 - 9 = (x - 4)^2 - 25

x = 4

(Reason: Completing the square writes the curve as

(x - 4)^2 - 25

, so its lowest point is

(4, -25)

and the curve is symmetrical about the vertical line

x = 4

. Differentiating gives the gradient function

2x - 8

, which is zero at that same

x = 4

, and the roots of

x^2 - 8x - 9 = 0

, namely

-1

and

9

, have that same midpoint. Any one of the three earns the first mark.)

Place

p

and

q

either side of the line of symmetry

\dfrac{p + q}{2} = 4

\dfrac{p - q}{2} = \dfrac{14}{2} = 7

p = 4 + 7 = 11

q = 4 - 7 = -3

(Reason: A horizontal line meets the curve at two points the same distance from

x = 4

, so the midpoint of

p

and

q

4

and the gap of

14

between them is

7

on each side.

p

is the larger of the two, because the question gives

p - q

as positive.)

Substitute a root back into the equation of the curve

k = 11^2 - 8 \times 11 - 9

121 - 88 - 9 = 24

(Reason:

A

lies on

C

, so putting

x = 11

into

y = x^2 - 8x - 9

returns the

y

value that

A

and

B

share, and that value is

k

. The question says

k

is an integer, and

24

is one, which is a small sign that nothing has gone wrong.)

k = 24

Verification

Check 1: Use the other root. Substituting

q = -3

into the curve gives

(-3)^2 - 8(-3) - 9

, and it must land on the same

k

A

and

B

really do lie on one horizontal line

9 + 24 - 9 = 24

Check 2: Work backwards. Put

k = 24

into

x^2 - 8x - 9 = k

, so

x^2 - 8x - 33 = 0

, which factorises as

(x - 11)(x + 3) = 0

x = 11

and

x = -3

, so

p - q = 11 - (-3) = 14

, which is exactly the gap the question gives

Check 3: Avoid the roots altogether. From

k = (x - 4)^2 - 25

the two solutions are

x = 4 + \sqrt{k + 25}

and

x = 4 - \sqrt{k + 25}

, so the gap between them is

2\sqrt{k + 25}

whatever

k

2\sqrt{k + 25} = 14

gives

\sqrt{k + 25} = 7

, so

k + 25 = 49

and

k = 24

again, reached without ever naming

p

q

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A correct start on the line of symmetry	M1	For using differentiation, completing the square or symmetry: $\dfrac{dy}{dx} = 2x - 8$ or $(x - 4)^2$ ..... or the roots of $x^2 - 8x - 9 = 0$ combined as $\dfrac{-1 + 9}{2}$	✓
The line of symmetry stated	M1	$x = 4$ or $(4, -25)$ or $(4, .....)$ or $\dfrac{p + q}{2} = 4$ oe	✓
One of the two roots	A1	$q = -3$ or $p = 11$ . Either one alone is enough for this mark	✓
A root substituted into the equation of $C$	M1	$(k =) (-3)^2 - 8(-3) - 9$ or $(k =) (11)^2 - 8(11) - 9$	✓
The value of $k$	A1 dep on M2	$24$ . The paper asks for clear algebraic working, and the scheme prints Working required against this row, so $24$ on its own scores nothing here	✓
Alternative routes	note	The official scheme prints three alternatives, each carrying the same five marks in the same places: substitute $p = q + 14$ and form one equation in one variable; or set $(q + 14)^2 - 8(q + 14) - 9$ equal to $k$ directly; or solve $x^2 - 8x - 9 - k = 0$ by formula to reach $x = 4 + \sqrt{k + 25}$ and $x = 4 - \sqrt{k + 25}$ , then $25 + k = \left(\dfrac{14}{2}\right)^2$	✓

Full marks: 5/5

The diagram shows the triangle $OAB$ .

$\overrightarrow{OA} = 10\mathbf{a}$ and $\overrightarrow{OB} = 10\mathbf{b}$

$P$ is the point on $AB$ for which $\overrightarrow{AP} = \dfrac{1}{4}\overrightarrow{AB}$ , and $Q$ is the point on $OB$ for which $\overrightarrow{OQ} = \dfrac{1}{5}\overrightarrow{OB}$ .

$R$ is the point where the straight line $ARQ$ crosses the straight line $ORP$ .

Write each of these vectors in terms of $\mathbf{a}$ and $\mathbf{b}$ , giving each answer in its simplest form.

(i) $\overrightarrow{AQ}$ [1 mark]

(ii) $\overrightarrow{OP}$ [1 mark]

(iii) $\overrightarrow{OR}$ [4 marks]

(i)(ii)(iii)

[Total 6 marks]

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Question 24 - Exam Solution

Understanding the Question

Given

\overrightarrow{OA} = 10\mathbf{a}

and

\overrightarrow{OB} = 10\mathbf{b}

, so

\mathbf{a}

and

\mathbf{b}

are the two directions everything else is built from.

P

lies on

AB

with

\overrightarrow{AP} = \dfrac{1}{4}\overrightarrow{AB}

Q

lies on

OB

with

\overrightarrow{OQ} = \dfrac{1}{5}\overrightarrow{OB}

ARQ

and

ORP

are straight lines, so

R

lies on both of them.

Find

\overrightarrow{AQ}

\overrightarrow{OP}

and

\overrightarrow{OR}

, each in terms of

\mathbf{a}

and

\mathbf{b}

. Parts (i) and (ii) are single journeys; part (iii) needs the two straight lines to be used together.

Plan the Solution

Travel from one point to another along legs that are already known: to reach $Q$ from $A$ , go back along $AO$ first and then forward along $OQ$ .
For $OP$ , start at $O$ , go to $A$ , then take a quarter of $AB$ .
$R$ is the only point lying on both straight lines, so write $OR$ twice - once as a multiple of $OP$ , once as $OA$ plus a multiple of $AQ$ .
Two expressions for the same vector must have equal $\mathbf{a}$ -parts and equal $\mathbf{b}$ -parts. That gives two equations in the two unknown multipliers.
Solve the pair, then substitute back into either expression.

Worked Solution [6 marks]

Rule - Two routes to one point: if

R

lies on both straight lines then

\overrightarrow{OR}

can be written along either line, and comparing the

\mathbf{a}

parts and the

\mathbf{b}

parts of the two expressions gives a pair of simultaneous equations.

Part (i) - travel from

A

Q

through

O

\overrightarrow{OQ} = \dfrac{1}{5} \times 10\mathbf{b} = 2\mathbf{b}

\overrightarrow{AQ} = \overrightarrow{AO} + \overrightarrow{OQ}

\overrightarrow{AQ} = -10\mathbf{a} + 2\mathbf{b}

(Reason: There is no single arrow from

A

Q

in the given information, so the journey is broken into legs that are given. Going from

A

O

is going along

OA

backwards, which reverses its sign.)

Part (ii) - travel from

O

P

through

A

\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -10\mathbf{a} + 10\mathbf{b}

\overrightarrow{AP} = \dfrac{1}{4}(-10\mathbf{a} + 10\mathbf{b}) = -\dfrac{5}{2}\mathbf{a} + \dfrac{5}{2}\mathbf{b}

\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP} = 10\mathbf{a} - \dfrac{5}{2}\mathbf{a} + \dfrac{5}{2}\mathbf{b}

\overrightarrow{OP} = \dfrac{15}{2}\mathbf{a} + \dfrac{5}{2}\mathbf{b}

(Reason:

AB

has to be found first, because

AP

is given as a fraction of it. Collecting the

\mathbf{a}

terms at the end is what the word simplify is asking for.)

Part (iii), first expression -

R

is on the straight line

ORP

\overrightarrow{OR} = k\overrightarrow{OP}

\overrightarrow{OR} = \dfrac{15k}{2}\mathbf{a} + \dfrac{5k}{2}\mathbf{b}

(Reason:

O

R

and

P

lie on one straight line, so

OR

points the same way as

OP

and is some multiple of it. That multiple is called

k

, and finding it is what the rest of the question is for.)

Part (iii), second expression -

R

is also on the straight line

ARQ

\overrightarrow{OR} = \overrightarrow{OA} + \lambda\overrightarrow{AQ}

\overrightarrow{OR} = 10\mathbf{a} + \lambda(-10\mathbf{a} + 2\mathbf{b})

\overrightarrow{OR} = (10 - 10\lambda)\mathbf{a} + 2\lambda\mathbf{b}

(Reason:

A

R

and

Q

lie on one straight line, so

AR

is some fraction of

AQ

. Reaching

R

this way needs a second unknown, and part (i) has already supplied

AQ

Part (iii) - compare the

\mathbf{a}

parts and the

\mathbf{b}

parts, then solve

\dfrac{15k}{2} = 10 - 10\lambda

\dfrac{5k}{2} = 2\lambda

\lambda = \dfrac{5k}{4}

\dfrac{15k}{2} = 10 - \dfrac{25k}{2}

20k = 10

k = \dfrac{1}{2}

(Reason: Both expressions describe the same vector

OR

, and

\mathbf{a}

and

\mathbf{b}

point in different directions, so the amount of

\mathbf{a}

in each must match and the amount of

\mathbf{b}

in each must match. The

\mathbf{b}

comparison gives

\lambda

in terms of

k

, and substituting it into the

\mathbf{a}

comparison leaves one equation in one unknown.)

Part (iii) - substitute

k

back into the first expression

\overrightarrow{OR} = \dfrac{1}{2}\left(\dfrac{15}{2}\mathbf{a} + \dfrac{5}{2}\mathbf{b}\right)

\overrightarrow{OR} = \dfrac{15}{4}\mathbf{a} + \dfrac{5}{4}\mathbf{b}

(Reason:

k

is a half, so

R

sits exactly halfway along

OP

. Substituting into the

OP

route is quicker than the

AQ

route, and either one gives the same vector.)

(i)

AQ = -10\mathbf{a} + 2\mathbf{b}

; (ii)

OP = 7.5\mathbf{a} + 2.5\mathbf{b}

; (iii)

OR = 3.75\mathbf{a} + 1.25\mathbf{b}

Verification

Check 1: The two expressions for

\overrightarrow{OR}

must agree. Put

k = \dfrac{1}{2}

into the

ORP

route, and

\lambda = \dfrac{5}{8}

into the

ARQ

route. The first gives

\dfrac{15}{4}\mathbf{a} + \dfrac{5}{4}\mathbf{b}

and the second gives

(10 - 6.25)\mathbf{a} + 1.25\mathbf{b}

, the same vector.

Check 2: Reach

R

along a third path, through

Q

instead of through

A

, using

\overrightarrow{OR} = \overrightarrow{OQ} + \mu\overrightarrow{QA}

with

\mu = \dfrac{3}{8}

2\mathbf{b} + \dfrac{3}{8}(10\mathbf{a} - 2\mathbf{b}) = \dfrac{15}{4}\mathbf{a} + \dfrac{5}{4}\mathbf{b}

Check 3: Leave the vectors behind and use coordinates. With

\mathbf{a}

and

\mathbf{b}

as the two unit vectors the points are

O(0, 0)

A(10, 0)

B(0, 10)

P(7.5, 2.5)

and

Q(0, 2)

. Work out where the line through

O

and

P

meets the line through

A

and

Q

. They meet at

(3.75, 1.25)

, which reads back as

\dfrac{15}{4}\mathbf{a} + \dfrac{5}{4}\mathbf{b}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(i) $AQ$	B1	for $-10\mathbf{a} + 2\mathbf{b}$ or $2\mathbf{b} - 10\mathbf{a}$ . The answer must be simplified.	✓
(ii) $OP$	B1	for $\dfrac{15}{2}\mathbf{a} + \dfrac{5}{2}\mathbf{b}$ oe, eg $7.5\mathbf{a} + 2.5\mathbf{b}$ . The answer must be simplified.	✓
(iii) first path to $R$ , along $ORP$	M1ft	for $\overrightarrow{OR} = k\overrightarrow{OP} = k\left(\dfrac{15}{2}\mathbf{a} + \dfrac{5}{2}\mathbf{b}\right)$ oe. Here $\overrightarrow{OP}$ must be in terms of $\mathbf{a}$ and $\mathbf{b}$ (lower case).	✓
(iii) second path to $R$ , along $ARQ$ or through $Q$	M1ft	for $\overrightarrow{OR} = \overrightarrow{OA} + \lambda\overrightarrow{AQ} = 10\mathbf{a} + \lambda(-10\mathbf{a} + 2\mathbf{b}) = (10 - 10\lambda)\mathbf{a} + 2\lambda\mathbf{b}$ oe, or $\overrightarrow{OR} = \overrightarrow{OQ} + \overrightarrow{QR} = 2\mathbf{b} - \mu(-10\mathbf{a} + 2\mathbf{b}) = 10\mu\mathbf{a} + (2 - 2\mu)\mathbf{b}$ oe. Here $\overrightarrow{AQ}$ must be in terms of $\mathbf{a}$ and $\mathbf{b}$ (lower case).	✓
(iii) both equations correct	M1	for $\dfrac{15}{2}k = 10 - 10\lambda$ and $\dfrac{5}{2}k = 2\lambda$ oe, or $\lambda = \dfrac{5}{8}$ oe; or $\dfrac{5}{2}k = 2 - 2\mu$ and $\dfrac{15}{2}k = 10\mu$ oe, or $k = 0.5$ oe.	✓
(iii) $OR$	A1	for $\dfrac{15}{4}\mathbf{a} + \dfrac{5}{4}\mathbf{b}$ oe, eg $3.75\mathbf{a} + 1.25\mathbf{b}$ .	✓
Note	note	A correct answer scores full marks, unless it comes from obviously incorrect working.	✓

Full marks: 6/6

A function $\mathrm{f}$ has domain $x \geq 6$ and is defined by

$\mathrm{f}(x) = 2x^2 - 24x + 7$

Work out the inverse function $\mathrm{f}^{-1}(x)$ [4 marks]

f⁻¹(x) =

[Total 4 marks]

Show solution & mark schemeHide solution & mark scheme

Question 25 - Exam Solution

Understanding the Question

Given

\mathrm{f}(x) = 2x^2 - 24x + 7

The domain is

x \geq 6

, and

6

is exactly where this parabola turns.

On that domain

\mathrm{f}

only increases, so no two inputs share an output and the inverse really is a function.

Find

\mathrm{f}^{-1}(x)

, the inverse function. The mark scheme requires the answer in terms of

x

, so the working must end by swapping the letters back.

Plan the Solution

An inverse undoes the function, so start from $y = \mathrm{f}(x)$ and rearrange until $x$ is the subject.
$x$ appears twice in $2x^2 - 24x + 7$ , once squared and once not, so it cannot be peeled off one operation at a time. Completing the square rewrites the rule with $x$ in only one place.
Once it reads $y = 2(x - 6)^2 - 65$ , undo the operations in reverse order: add $65$ , halve, square root, add $6$ .
Square rooting offers a plus and a minus. The domain $x \geq 6$ is what decides which one is kept.
Finish by writing the answer in terms of $x$ , because that is the letter an inverse function is written in.

Worked Solution [4 marks]

Rule - Inverse by rearranging: write

y = \mathrm{f}(x)

, make

x

the subject, then rewrite the result in terms of

x

. Where

x

appears twice, complete the square first so that it appears only once.

Write

y = \mathrm{f}(x)

, then take the

2

out of the first two terms

y = 2x^2 - 24x + 7

y = 2(x^2 - 12x) + 7

(Reason: Making

x

the subject is the whole job, but

x

sits in the squared term and again in the

-24x

term, so it cannot be undone one operation at a time. Taking the

2

out of the first two terms is the opening move of completing the square, and it is where the mark scheme awards its first method mark. The

+ 7

is deliberately left outside the bracket, because only the

x

terms are being reshaped.)

Complete the square inside the bracket

x^2 - 12x = (x - 6)^2 - 6^2

y = 2((x - 6)^2 - 36) + 7

y = 2(x - 6)^2 - 72 + 7 = 2(x - 6)^2 - 65

(Reason: Half of

12

6

, so

x^2 - 12x

(x - 6)^2

short of

36

. Multiplying the bracket out and collecting the constants leaves

x

in exactly one place, which is what makes the rearrangement possible at all. Watch the constant as it leaves the bracket: it is inside a bracket that is being multiplied by

2

, so it must be doubled on the way out. This line is the mark scheme's second method mark, and it depends on the first.)

Make the squared bracket the subject

y + 65 = 2(x - 6)^2

(x - 6)^2 = \dfrac{y + 65}{2}

(Reason: Undo the operations in the reverse of the order they were applied. The last thing done to the bracket was subtracting

65

, so add

65

first, then undo the multiplication by

2

by dividing. This is the third method mark, and the scheme accepts the equivalent form with

(y - 7)

over

2

, plus

6^2

, on the right instead.)

Take the square root, choose the sign, then swap the letters

x - 6 = \pm\sqrt{\dfrac{y + 65}{2}}

x = 6 + \sqrt{\dfrac{y + 65}{2}}

\mathrm{f}^{-1}(x) = 6 + \sqrt{\dfrac{x + 65}{2}}

(Reason: Square rooting always offers two possibilities. The domain is

x \geq 6

, so

x - 6

can never be negative and only the positive root can be the inverse. That single decision is worth the accuracy mark on its own: the scheme awards M3A0, all three method marks and no accuracy mark, to an answer left as

6

plus-or-minus the root. The last line renames

y

x

, because an inverse function is written in terms of

x

\mathrm{f}^{-1}(x) = 6 + \sqrt{\dfrac{x + 65}{2}}

Verification

Check 1: Take

x = 10

, which is inside the domain. Work out the output, then feed that output into the inverse and see whether

10

comes back.

2 \times 100 - 240 + 7 = -33

, and

6 + \sqrt{\dfrac{-33 + 65}{2}} = 6 + \sqrt{16} = 10

Check 2: Repeat with a different input,

x = 9

, so that the first check is not being confirmed by the value that produced it.

2 \times 81 - 216 + 7 = -47

, and

6 + \sqrt{\dfrac{-47 + 65}{2}} = 6 + \sqrt{9} = 9

Check 3: Test the edge of the domain. The smallest allowed input is

6

, so the smallest possible output is the value at

6

, and the inverse must send it straight back to

6

2 \times 36 - 144 + 7 = -65

, and

6 + \sqrt{\dfrac{-65 + 65}{2}} = 6 + 0 = 6

Check 4: Confirm that the rejected root really is rejected. The equation has a second solution when the output is

-33

; work out what the minus sign gives.

6 - \sqrt{16} = 2

, and

2 \times 4 - 48 + 7 = -33

, so

2

does solve the equation but lies outside the domain

x \geq 6

, which is why only the plus sign survives

Check 5: Show that the other form the mark scheme accepts is the same expression, by putting

\dfrac{x - 7}{2} + 36

over a common denominator.

\dfrac{x - 7}{2} + 36 = \dfrac{x - 7 + 72}{2} = \dfrac{x + 65}{2}

, so

6 + \sqrt{\dfrac{x - 7}{2} + 36}

is the same function written a different way

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A correct first step towards completing the square	M1	for eg $(y =) 2(x^2 - 12x) + 7$ or $(y =) 2(x^2 - 12x + \dfrac{7}{2})$ or $\dfrac{y - 7}{2} = x^2 - 12x$ . For each method mark the function must be correct.	✓
The square completed	M1 dep	for eg $(y =) 2((x - 6)^2 - 6^2) + 7$ or $(y =) 2((x - 6)^2 - 6^2 + \dfrac{7}{2})$ or $(y =) 2(x - 6)^2 - 65$ oe or $\dfrac{y - 7}{2} = (x - 6)^2 - 6^2$ oe	✓
The squared bracket made the subject	M1	for $(x - 6)^2 = \dfrac{y + 65}{2}$ oe, or $(x - 6)^2 = \dfrac{y - 7}{2} + 6^2$ oe	✓
The inverse function	A1	for $6 + \sqrt{\dfrac{x + 65}{2}}$ oe, eg $6 + \sqrt{\dfrac{x - 7}{2} + 36}$ . The answer must be in terms of $x$ . Award M3A0 for $6 \pm \sqrt{\dfrac{x + 65}{2}}$ , for $6 \pm \sqrt{\dfrac{y + 65}{2}}$ and for $6 + \sqrt{\dfrac{y + 65}{2}}$ .	✓
Note	note	A correct answer scores full marks, unless it comes from obviously incorrect working. Candidates are allowed to swap $x$ and $y$ while finding the inverse, provided the final answer is in terms of $x$ .	✓
Alternative method	note	The scheme prints a full alternative. Rearranging to $2x^2 - 24x + 7 - y = 0$ earns the first M1; $(x =) \dfrac{24 \pm \sqrt{576 - 8(7 - y)}}{4}$ earns the dependent M1; $(x =) 6 \pm \sqrt{\dfrac{y + 65}{2}}$ earns the third M1; and the same A1 is awarded for the final answer.	✓

Full marks: 4/4

Keep revising

That is the whole paper. Read what the IGCSE is and how it is graded, or compare Edexcel 4MA1 with Cambridge 0580 if you are still choosing a board. Check the IGCSE grade boundaries to set your target, and if the exam is close, the four-week IGCSE Maths revision plan sets out what to do week by week.

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Edexcel IGCSE 4MA1 Paper 1H, November 2024: Worked Solutions, Questions 16 to 25

Worked solutions, questions 16 to 25 of 25

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