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Edexcel IGCSE 4MA1/1H, Thursday 16 May 2024: Worked Solutions, Questions 15 to 25

Sir Faraz Hassan

6 Aug 2026

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Edexcel International GCSE Mathematics A (4MA1)4MA1/1H - Higher Tier - Thursday 16 May 2024100 marks · 2 hours · Calculator allowed

Back to questions 1 to 14

This is the rest of the paper. Questions 1 to 14, the paper's overview and the frequently asked questions are on the first page.

Original worked solutions for Edexcel International GCSE Mathematics A, Paper 4MA1/1H (Higher Tier), June 2024 series, sat Thursday 16 May 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).

Try each question yourself first, then open the worked solution to check your method and see exactly where each method mark (M1) and accuracy mark (A1) is earned. The questions follow the same order as the original paper and carry the same marks.

Download printable PDF

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

Worked solutions, questions 15 to 25 of 25

The function $\mathrm{f}$ is defined as

$\mathrm{f} : x \mapsto \dfrac{3x+1}{x-2}$

(a) Write down the value of $x$ that cannot belong to any domain of the function $\mathrm{f}$
[1 mark]
(b) Find the inverse function $\mathrm{f}^{-1}$ , giving your answer in the form $\mathrm{f}^{-1}(x) = \ldots$ [3 marks]

(a)(b)

[Total 4 marks]

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Question 15 - Exam Solution

Understanding the Question

Given

The function

\mathrm{f}

sends an input

x

\dfrac{3x+1}{x-2}

The rule is one fraction, so its denominator decides which inputs the function can handle.

No domain is stated, no figure is given and there is no data to read off.

Find

(a) The one value of

x

that no domain of

\mathrm{f}

may contain. (b) A formula for

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

, written in terms of

x

Plan the Solution

For part (a), set the denominator equal to $0$ and solve it. That input is the only one the rule cannot handle.
For part (b), write $y$ for the output, so that $y = \mathrm{f}(x)$ , and then rearrange to make $x$ the subject.
Multiply both sides by $x-2$ straight away, so no fraction is left to carry through the algebra.
Gather every term containing $x$ on one side, factorise, then divide by the bracket.
Finish by renaming $y$ as $x$ , because the answer is asked for in terms of $x$ .

Worked Solution [4 marks]

Rule - a fraction has no value when its denominator is

0

, and a function is inverted by putting

y = \mathrm{f}(x)

, rearranging until

x

is the subject, then writing

x

in place of

y

Step 1: (a) find where the rule breaks down

x - 2 = 0

x = 2

(Reason: The rule divides by

x-2

, and division by

0

gives no number at all. Every other input leaves a denominator that is not

0

, so

2

is the single value that must be kept out of every domain.)

Step 2: (b) write y for the output and clear the fraction

y = \dfrac{3x+1}{x-2}

y(x - 2) = 3x + 1

(Reason: Calling the output

y

turns the mapping into an equation linking

x

and

y

. Multiplying both sides by

x-2

clears the fraction, which is what earns the first method mark.)

Step 3: (b) expand, then collect the x terms on one side

yx - 2y = 3x + 1

yx - 3x = 1 + 2y

(Reason: Expanding the bracket puts

x

in two separate places. Moving

3x

to the left and the term in

y

to the right leaves every

x

on one side and nothing else there.)

Step 4: (b) factorise and make x the subject

x(y - 3) = 1 + 2y

x = \dfrac{1 + 2y}{y - 3}

(Reason: Both terms on the left share a factor of

x

, so taking it outside a bracket leaves a single

x

multiplied by

y-3

. Dividing by that bracket makes

x

the subject, and this factorising step is the second method mark.)

Step 5: (b) write the inverse in terms of x

\mathrm{f}^{-1}(x) = \dfrac{1 + 2x}{x - 3}

(Reason: The rearranged equation says what input

\mathrm{f}

needed in order to produce

y

, which is exactly what

\mathrm{f}^{-1}

does. The letter is only a label, so

y

is renamed

x

to match the form the question asks for.)

(a)

x = 2

(b)

\mathrm{f}^{-1}(x) = \dfrac{1 + 2x}{x - 3}

Verification

Check 1: Put

x = 2

into the top and the bottom of the rule separately, and see what the fraction would ask for.

3 \times 2 + 1 = 7

and

2 - 2 = 0

, so

\mathrm{f}(2)

asks for

7

divided by

0

, which is not a number.

Check 2: Send a number through

\mathrm{f}

and then feed the output back through

\mathrm{f}^{-1}

\mathrm{f}(3) = \dfrac{3 \times 3 + 1}{3 - 2} = 10

and

\mathrm{f}^{-1}(10) = \dfrac{1 + 2 \times 10}{10 - 3} = 3

, the number we started with.

Check 3: Do it the other way round, starting with

\mathrm{f}^{-1}

and finishing with

\mathrm{f}

\mathrm{f}^{-1}(4) = \dfrac{1 + 2 \times 4}{4 - 3} = 9

and

\mathrm{f}(9) = \dfrac{3 \times 9 + 1}{9 - 2} = 4

, so the two rules really do undo each other.

Check 4: Split the original rule to see which outputs

\mathrm{f}

can never produce, and compare that with the inverse.

\dfrac{3x+1}{x-2} = 3 + \dfrac{7}{x-2}

, and

\dfrac{7}{x-2}

is never

0

, so

\mathrm{f}

never outputs

3

. The inverse must leave that value out, and its denominator

x-3

does exactly that.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) The value left out of every domain	B1	$(x =) 2$ . Accept $x = 2$ and $x \neq 2$ , and accept it stated in words as the value $x$ cannot take. Any response that contains $2$ is also acceptable.	✓
(a) Forms that are not accepted	Note	Do not accept the answer written with an inequality sign: $x > 2$ , $x < 2$ , $x \geq 2$ or $x \leq 2$ . Do not accept $2$ given alongside another number, for example $2$ and $3$ .	✓
(b) Clear the fraction	M1	$y(x - 2) = 3x + 1$ or equivalent, or $yx - 2y = 3x + 1$ or equivalent. The same working with the two letters interchanged, $x(y - 2) = 3y + 1$ or $yx - 2x = 3y + 1$ , earns the mark just as well.	✓
(b) Factorise correctly	M1	$x(y - 3) = 1 + 2y$ or equivalent, or $y(x - 3) = 1 + 2x$ or equivalent from the interchanged working. The mark is for the factorising, not for the tidying that follows.	✓
(b) The inverse function	A1	$\dfrac{1 + 2x}{x - 3}$ or equivalent, for example $\dfrac{-1 - 2x}{3 - x}$ . The answer must be in terms of $x$ .	✓
Answer with no working	Note	Working is not required in part (b), so a correct answer scores full marks unless it comes from obviously incorrect working.	✓

Full marks: 4/4

There are $20$ beads in a jar.
$15$ of the beads are green
$5$ of the beads are white

Owen takes at random $3$ beads from the jar.

Work out the probability that Owen takes at least one bead of each colour from the jar. [4 marks]

[Total 4 marks]

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Question 16 - Exam Solution

Understanding the Question

Given

A jar holding

20

beads:

15

green and

5

white.

3

beads are taken at random, and none of them is put back.

Find

The probability that the

3

beads include at least one green bead and at least one white bead.

Plan the Solution

At least one of each colour means anything except all one colour, and there are only two ways to get all one colour, so the complement is much quicker than listing every mixed order.
Work out the probability of $3$ greens, work out the probability of $3$ whites, add them, then take the total away from $1$ .
Nothing is put back, so the count of that colour and the total both drop by one at every pick.

Worked Solution [4 marks]

For picks without replacement, multiply the probabilities of the successive picks, taking one off the count of that colour and one off the total each time. Then take the probability that all

3

beads are the same colour away from

1

Step 1: the probability that all three beads are green

\dfrac{15}{20} \times \dfrac{14}{19} \times \dfrac{13}{18} = \dfrac{2730}{6840} = \dfrac{91}{228}

(Reason: There are

15

green beads out of

20

. With one green gone there are

14

green out of

19

, and then

13

green out of

18

Step 2: the probability that all three beads are white

\dfrac{5}{20} \times \dfrac{4}{19} \times \dfrac{3}{18} = \dfrac{60}{6840} = \dfrac{1}{114}

(Reason: The same reducing pattern, starting from the

5

white beads:

5

out of

20

, then

4

out of

19

, then

3

out of

18

. Only one white bead is used up at each pick, and the jar holds just

5

of them, so all three whites can still be drawn.)

Step 3: add the two all-one-colour probabilities

\dfrac{91}{228} + \dfrac{1}{114} = \dfrac{91}{228} + \dfrac{2}{228} = \dfrac{93}{228}

(Reason: All green and all white cannot both happen, so the two probabilities simply add. Doubling

114

gives the common denominator

228

Step 4: take the all-one-colour probability away from

1

1 - \dfrac{93}{228} = \dfrac{135}{228} = \dfrac{45}{76}

(Reason: Every draw that is not all green and not all white has at least one bead of each colour, so the complement is the answer. Dividing top and bottom by

3

turns

\dfrac{135}{228}

into

\dfrac{45}{76}

\dfrac{45}{76}

Verification

Check 1 - count the favourable orders instead: Two greens and a white can come out in

3

orders, each worth

\dfrac{15}{20} \times \dfrac{14}{19} \times \dfrac{5}{18}

, and two whites and a green in

3

orders, each worth

\dfrac{5}{20} \times \dfrac{4}{19} \times \dfrac{15}{18}

\dfrac{3150}{6840} + \dfrac{900}{6840} = \dfrac{4050}{6840} = \dfrac{45}{76}

Check 2 - count selections rather than orders: There are

1140

ways of choosing

3

beads from

20

when the order is ignored. Of those,

455

are all green and

10

are all white, so the rest are mixed.

\dfrac{1140 - 455 - 10}{1140} = \dfrac{675}{1140} = \dfrac{45}{76}

Check 3 - the three outcomes must account for everything: All green, all white and a mixture are the only possibilities, so their probabilities have to total

1

. The all-one-colour probability was

\dfrac{93}{228}

, which cancels to

\dfrac{31}{76}

\dfrac{45}{76} + \dfrac{31}{76} = \dfrac{76}{76} = 1

Mark Scheme Breakdown

Step	Mark	Description	Got it?
One correct product of three fractions	M1	For one mixed order, $\dfrac{15}{20} \times \dfrac{14}{19} \times \dfrac{5}{18}$ or $\dfrac{5}{20} \times \dfrac{4}{19} \times \dfrac{15}{18}$ in any order, or for one all-one-colour product, $\dfrac{15}{20} \times \dfrac{14}{19} \times \dfrac{13}{18}$ or $\dfrac{5}{20} \times \dfrac{4}{19} \times \dfrac{3}{18}$ . The products must be correct but need not be evaluated, and equivalent decimals to $2$ decimal places, truncated or rounded, are accepted.	✓
All the products one route needs	M1	For $3 \times \dfrac{15}{20} \times \dfrac{14}{19} \times \dfrac{5}{18}$ or $3 \times \dfrac{5}{20} \times \dfrac{4}{19} \times \dfrac{15}{18}$ , or for both mixed products in any order, or for both all-one-colour products.	✓
A complete method using correct products	M1	For adding all six mixed orders, or for the complement $1 - \left( \dfrac{15}{20} \times \dfrac{14}{19} \times \dfrac{13}{18} + \dfrac{5}{20} \times \dfrac{4}{19} \times \dfrac{3}{18} \right)$ .	✓
The probability	A1	$\dfrac{45}{76}$ or an equivalent fraction. Accept $0.59$ to $2$ decimal places, truncated or rounded, or $59.2\%$ to $2$ significant figures. Working is not required, so a correct answer scores full marks unless it comes from obviously incorrect working.	✓
Alternative route, using the first two beads only	Note	A candidate who works with a green and a white among the first two picks earns the first two method marks together, for $\dfrac{15}{20} \times \dfrac{5}{19} = \dfrac{15}{76}$ and $\dfrac{5}{20} \times \dfrac{15}{19} = \dfrac{15}{76}$ . This row awards nothing on its own.	✓

Full marks: 4/4

Show that $\dfrac{1 + \sqrt{5}}{3 - \sqrt{5}}$ can be written in the form $a + \sqrt{b}$ , where $a$ and $b$ are integers.
You must show each stage of your working clearly. [3 marks]

[Total 3 marks]

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Question 17 - Exam Solution

Understanding the Question

Given

One fraction with a surd in its denominator:

\dfrac{1 + \sqrt{5}}{3 - \sqrt{5}}

The useful fact behind the whole method:

\sqrt{5} \times \sqrt{5} = 5

, a whole number.

Find

Integers

a

and

b

with

\dfrac{1 + \sqrt{5}}{3 - \sqrt{5}} = a + \sqrt{b}

. This is a show-that question, so the marks are for the stages of working, not for the final line on its own.

Plan the Solution

Multiply the numerator and the denominator by the conjugate of the denominator, $3 + \sqrt{5}$ . That is multiplying by $1$ , so the value of the fraction does not change.
Expand the numerator with all four products, then collect the two $\sqrt{5}$ terms.
Expand the denominator. The two middle terms cancel, so a whole number is left.
Divide every term of the numerator by that whole number, then read off $a$ and $b$ .

Worked Solution [3 marks]

Rule - Rationalising a denominator: multiply the top and the bottom by the conjugate, using

(p - \sqrt{q})(p + \sqrt{q}) = p^2 - q

, which is a difference of two squares and always leaves a whole number underneath.

Step 1: Multiply top and bottom by the conjugate

3 + \sqrt{5}

\dfrac{1 + \sqrt{5}}{3 - \sqrt{5}} = \dfrac{1 + \sqrt{5}}{3 - \sqrt{5}} \times \dfrac{3 + \sqrt{5}}{3 + \sqrt{5}}

(Reason: (Reason:

\dfrac{3 + \sqrt{5}}{3 + \sqrt{5}}

is equal to

1

, so the value is untouched. The conjugate is chosen because it is the one multiplier that clears the surd from the bottom.))

Step 2: Expand the numerator

(1 + \sqrt{5})(3 + \sqrt{5}) = 3 + \sqrt{5} + 3\sqrt{5} + 5

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

(Reason: (Reason: all four products are written out. The last one is

\sqrt{5} \times \sqrt{5} = 5

, so

3 + 5 = 8

and the two surd terms collect as

1\sqrt{5} + 3\sqrt{5} = 4\sqrt{5}

.))

Step 3: Expand the denominator

(3 - \sqrt{5})(3 + \sqrt{5}) = 9 + 3\sqrt{5} - 3\sqrt{5} - 5

(3 - \sqrt{5})(3 + \sqrt{5}) = 9 - 5 = 4

(Reason: (Reason: the two middle terms are equal and opposite, so they cancel. What is left is

3^2

with

\sqrt{5} \times \sqrt{5}

taken away from it, and no surd survives.))

Step 4: Divide every term by the denominator

\dfrac{8 + 4\sqrt{5}}{4} = \dfrac{8}{4} + \dfrac{4\sqrt{5}}{4}

\dfrac{8}{4} + \dfrac{4\sqrt{5}}{4} = 2 + \sqrt{5}

(Reason: (Reason: the fraction bar applies to the whole numerator, so both terms are divided.

\dfrac{4\sqrt{5}}{4}

leaves

1\sqrt{5}

, which is written

\sqrt{5}

.))

Step 5: Read off

a

and

b

2 + \sqrt{5} = a + \sqrt{b} \text{ gives } a = 2 \text{ and } b = 5

(Reason: (Reason: the whole-number part is

a

and the number under the root is

b

. Both are integers, which is exactly what the question asked to be shown.))

\dfrac{1 + \sqrt{5}}{3 - \sqrt{5}} = 2 + \sqrt{5}

a = 2

and

b = 5

, both integers

Verification

Check 1: Multiply the answer back by the original denominator. If

2 + \sqrt{5}

is right,

(2 + \sqrt{5})(3 - \sqrt{5})

must return the original numerator

1 + \sqrt{5}

(2 + \sqrt{5})(3 - \sqrt{5}) = 6 - 2\sqrt{5} + 3\sqrt{5} - 5 = 1 + \sqrt{5}

Check 2: A calculator is allowed here, so work out the original fraction and the answer as decimals and compare them.

\dfrac{1 + \sqrt{5}}{3 - \sqrt{5}} \approx 4.236 \text{ and } 2 + \sqrt{5} \approx 4.236

Check 3: Test the form itself. The question demands integers, so

4

must divide both terms of

8 + 4\sqrt{5}

exactly, and the surd must be left with a coefficient of

1

\dfrac{8}{4} = 2

and

\dfrac{4}{4} = 1

, so

a = 2

and

b = 5

are integers and the answer is genuinely in the form

a + \sqrt{b}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Multiply the numerator and the denominator by $3 + \sqrt{5}$	M1	For rationalising the denominator by multiplying the numerator and the denominator by $3 + \sqrt{5}$ , or by $-3 - \sqrt{5}$ , or equivalent.	✓
Expand both brackets	M1	Numerator correctly expanded, and may be simplified to at least $2$ terms, such as $8 + 4\sqrt{5}$ ; and denominator correctly expanded, and may be simplified to $1$ term, such as $4$ .	✓
Simplify to the required form	A1 dep on M2	For $2 + \sqrt{5}$ from correct working. Dependent on both method marks.	✓
Working required	Note	This row earns nothing. The mark scheme states that working is required, so the answer $2 + \sqrt{5}$ written down with no rationalising stage shown scores no marks at all.	✓

Full marks: 3/3

The curve $C$ has equation $y = x^3 - 40x + 1$
Work out the coordinates of the two points on $C$ at which the gradient of the curve is $8$ [5 marks]

First pointSecond point

[Total 5 marks]

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Question 18 - Exam Solution

Understanding the Question

Given

The curve

C

has equation

y = x^3 - 40x + 1

The gradient at each of the points wanted is

8

Find

The coordinates of both points on

C

where the gradient is

8

The gradient function of a cubic is a quadratic, so expect

2

points, one with a positive

x

and one with a negative

x

Plan the Solution

Differentiate the equation of the curve to get the gradient function $\dfrac{dy}{dx}$
Set that gradient function equal to $8$ and solve the quadratic for $x$
Substitute each value of $x$ back into the equation of the curve to get its $y$ , then pair the two coordinates up

Worked Solution [5 marks]

Rule - Gradient of a curve: the gradient at any point is

\dfrac{dy}{dx}

, so a question that gives you a gradient is telling you to solve

\dfrac{dy}{dx} = 8

. Differentiate term by term, using

\dfrac{d}{dx}(ax^n) = anx^{n-1}

Step 1: differentiate the equation of the curve

y = x^3 - 40x + 1

\dfrac{dy}{dx} = 3x^2 - 40

(Reason: Each term drops one power and is multiplied by the old power:

x^3

becomes

3x^2

, the

-40x

becomes

-40

because

x

has power

1

, and the constant

1

differentiates to nothing at all.)

Step 2: set the gradient function equal to 8 and solve

3x^2 - 40 = 8

3x^2 = 48

x^2 = 16

x = 4 \text{ or } x = -4

(Reason: The gradient function is a quadratic, so

x^2 = 16

has two square roots and both of them count. Losing the negative root here is what turns a five-mark answer into a three-mark one.)

Step 3: find the y-coordinate at x = 4

y = 4^3 - 40 \times 4 + 1 = 64 - 160 + 1 = -95

(Reason: The

x

values come from the gradient function, but the

y

values must come from the curve itself, so substitute back into

y = x^3 - 40x + 1

and not into the derivative.)

Step 4: find the y-coordinate at x = -4

y = (-4)^3 - 40 \times (-4) + 1 = -64 + 160 + 1 = 97

(Reason: Watch both signs: cubing a negative keeps it negative, so

(-4)^3 = -64

, while

-40 \times (-4)

is a minus times a minus and so comes out positive.)

First point

(4, -95)

Second point

(-4, 97)

Verification

Check 1: Put both answers back into the gradient function. Squaring either

4

-4

gives

16

, so both points must give the same gradient.

3 \times 16 - 40 = 8

at each point, as required

Check 2: Use symmetry. The part

x^3 - 40x

is an odd function, so the two points sit symmetrically about the point of the curve where

x = 0

. The midpoint of the two answers should therefore have height

1

, which is what the curve gives at

x = 0

\dfrac{-95 + 97}{2} = 1

, matching the curve at

x = 0

Check 3: Measure the gradient at

x = 4

from the curve alone, with no differentiation: take the chord from

x = 3.999

x = 4.001

and divide the change in

y

by the change in

x

. the chord gradient is

8.000001

, which is

8

to every figure a calculator would show

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$3x^2$ or $-40$	M1	for differentiating one of the first two terms correctly	✓
$3x^2 - 40$	A1	for both terms correct and no additions	✓
$3x^2 - 40 = 8$	M1ft	dep on M1, for equating their quadratic derivative with $8$ . The derivative must be of the form $ax^2 - 40$ or $3x^2 - b$ with $a \neq 0$ and $b \neq 0$	✓
$y = 4^3 - 40 \times 4 + 1$ or $y = (-4)^3 - 40 \times (-4) + 1$	M1ft	dep on the previous M1, for substituting at least one $x$ value into $y$	✓
$(4, -95) \text{ and } (-4, 97)$	A1	both coordinates must be paired correctly	✓
Working not required, so a correct answer scores full marks unless it comes from obviously incorrect working.	Note	Following through from $ax^2 - 40 = 8$ or $3x^2 - b = 8$ , their $x$ values must be correct.	✓

Full marks: 5/5

$ABCD$ is a quadrilateral.

Work out the value of $x$ .
Give your answer correct to $3$ significant figures. [5 marks]

x =

[Total 5 marks]

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Question 19 - Exam Solution

Understanding the Question

Given

Quadrilateral

ABCD

, with

AB = 15.2

AD = 13.4

m and

BC = 12.8

Angle

BDC = 40^\circ

and angle

BCD = 62^\circ

The diagonal

BD

splits the shape into triangle

ABD

and triangle

BCD

. It is the only length that belongs to both.

Find

The value of

x

, which is the size of angle

DAB

in degrees, correct to

3

significant figures.

Plan the Solution

Triangle $BCD$ is the only one with enough information to start: it has two angles and the side $BC = 12.8$ m opposite one of them, which is the sine rule situation.
Use the sine rule there to find the diagonal $BD$ .
Triangle $ABD$ then has all three sides and no angle, which is the cosine rule situation. Rearrange it for $\cos x$ .
Keep the unrounded $BD$ in the calculator throughout and round only at the very end.

Worked Solution [5 marks]

Rule - Sine rule:

\dfrac{a}{\sin A} = \dfrac{b}{\sin B}

, each side paired with the angle opposite it. Cosine rule, rearranged for an angle:

\cos A = \dfrac{b^2 + c^2 - a^2}{2bc}

Step 1: pair each side with its opposite angle in triangle

BCD

\dfrac{BD}{\sin 62^\circ} = \dfrac{12.8}{\sin 40^\circ}

(Reason:

BD

is opposite the

62^\circ

angle at

C

, and the

12.8

m side is opposite the

40^\circ

angle at

D

, so those are the two pairs the sine rule needs)

Step 2: work out the diagonal

BD

BD = \dfrac{12.8 \times \sin 62^\circ}{\sin 40^\circ} = 17.58\ldots \text{ m}

(Reason: multiplying both sides by

\sin 62^\circ

leaves

BD

on its own; the full unrounded value stays in the calculator for the next step)

Step 3: put the cosine rule into triangle

ABD

BD^2 = 13.4^2 + 15.2^2 - 2 \times 13.4 \times 15.2 \times \cos x

309.13\ldots = 179.56 + 231.04 - 407.36\cos x

(Reason:

x

sits between the two known sides

AD

and

AB

, and

BD = 17.58\ldots

m is the side opposite it, so all three sides are known and only

\cos x

is not)

Step 4: rearrange for the cosine

\cos x = \dfrac{13.4^2 + 15.2^2 - 17.58\ldots^2}{2 \times 13.4 \times 15.2}

\cos x = \dfrac{410.6 - 309.13\ldots}{407.36} = 0.2490\ldots

(Reason: the cosine rule is one equation in one unknown here, so the two squares go to one side and the product

2 \times 13.4 \times 15.2 = 407.36

divides through)

Step 5: undo the cosine

x = \cos^{-1}(0.2490\ldots) = 75.577\ldots

x = 75.6 \text{ (3 s.f.)}

(Reason: the cosine came out positive, so

x

is acute; rounding only at this last line keeps every figure of

BD

in play)

x = 75.6

Verification

Check 1 - rebuild the diagonal from the answer: Run the cosine rule forwards instead of backwards: put

x = 75.6

into

BD = \sqrt{410.6 - 407.36\cos x}

\sqrt{410.6 - 407.36 \times \cos 75.6^\circ} = 17.586\ldots

, which is

17.6

m to

3

significant figures - the same as the

17.58\ldots

m the sine rule gave.

Check 2 - close the angles of triangle ABD: The cosine rule also gives the other two angles of triangle

ABD

: angle

ADB = 56.85^\circ

and angle

ABD = 47.57^\circ

. A triangle that has been solved correctly must close.

75.58^\circ + 56.85^\circ + 47.57^\circ = 180^\circ

Check 3 - is the size sensible? In triangle

ABD

the longest side is

BD = 17.58\ldots

m, so the largest angle must be the one facing it, which is

x

. And

13.4^2 + 15.2^2 = 410.6

is bigger than

BD^2 = 309.13\ldots

, so that angle is acute.

75.6^\circ

is the largest of the three angles and is under

90^\circ

, as both tests require.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Sine rule set up in triangle $BCD$ : $\dfrac{BD}{\sin 62^\circ} = \dfrac{12.8}{\sin 40^\circ}$ oe	M1	for correct use of the sine rule for BD	✓
$BD = \dfrac{12.8}{\sin 40^\circ} \times \sin 62^\circ = 17.5(82\ldots)$	M1	for finding $BD$ , truncated or rounded	✓
$17.5(82\ldots)^2 = 13.4^2 + 15.2^2 - 2 \times 13.4 \times 15.2 \times \cos x$	M1	for correct use of the cosine rule	✓
$\cos x = \dfrac{13.4^2 + 15.2^2 - 17.5(82\ldots)^2}{2 \times 13.4 \times 15.2}$ oe, or $\cos x = 0.247$ to $0.256$ oe	M1	for a correct rearrangement of $\cos x$	✓
$x = 75.6$	A1	accept $75.1$ to $75.7$ . Working is not required, so a correct answer scores full marks unless it comes from obviously incorrect working.	✓
Alternative route to $BD$ , through $CD$ : $CD = \dfrac{12.8}{\sin 40^\circ} \times \sin 78^\circ = 19.4(781\ldots)$ and $BD = \dfrac{19.4(781\ldots)}{\sin 78^\circ} \times \sin 62^\circ = 17.5(82\ldots)$	M2	this pair of lines replaces the first two rows and is worth the same two marks; the third angle of triangle $BCD$ is $180^\circ - 40^\circ - 62^\circ = 78^\circ$	✓

Full marks: 5/5

In the diagram, $OABC$ is a sector of a circle with centre $O$ .

Angle $AOC = 60^\circ$
The shaded segment $ABC$ has an area of $38$ cm²

Work out the perimeter of the shaded segment $ABC$ .
Give your answer correct to one decimal place. [4 marks]

[Total 4 marks]

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Question 20 - Exam Solution

Understanding the Question

Given

A sector

OABC

of a circle with centre

O

Angle

AOC = 60^\circ

The area of the shaded segment

ABC

38

cm²

The radius is not given, and every length in the question depends on it.

Find

The perimeter of the shaded segment

ABC

, correct to one decimal place The perimeter of a segment is its arc plus its chord, so the radius has to be found first.

Plan the Solution

The segment is what is left when triangle $OAC$ is cut away from the sector, so its area is the sector's area minus the triangle's area.
Both of those are a multiple of $r^2$ , so the given area becomes one equation in $r^2$ alone. One division and one square root then give the radius.
The boundary of the segment is its curved edge plus its straight edge: the arc $ABC$ plus the chord $AC$ . The two radii lie inside the sector and are not part of that boundary.
Because $OA = OC$ and the angle between them is $60^\circ$ , triangle $OAC$ is equilateral, so the chord is the same length as the radius and needs no separate calculation.

Worked Solution [4 marks]

Rule - Segment of a circle: its area is

\dfrac{60}{360} \times \pi r^2 - \dfrac{1}{2} r^2 \sin 60^\circ

and its perimeter is the arc

\dfrac{60}{360} \times 2 \pi r

plus the chord

AC

Step 1: write the area of the segment in terms of

r

\dfrac{60}{360} \times \pi r^2 - \dfrac{1}{2} \times r^2 \times \sin 60^\circ = 38

(Reason: the sector is

\dfrac{60}{360}

of the whole circle, and triangle

OAC

has two sides of length

r

with

60^\circ

between them, so its area is

\dfrac{1}{2} r^2 \sin 60^\circ

. Taking the triangle away from the sector leaves exactly the shaded segment.)

Step 2: collect the two terms into a single number

\dfrac{\pi}{6} - \dfrac{\sqrt{3}}{4} = 0.5235988 - 0.4330127

0.5235988 - 0.4330127 = 0.0905861

(Reason: both terms carry a factor of

r^2

, so the equation becomes

r^2 \times 0.0905861 = 38

. Keeping seven decimal places here is deliberate: this number is a small difference between two much larger ones, so rounding it early would move the final answer.)

Step 3: divide to find

r^2

, then take the square root

\dfrac{38}{0.0905861} = 419.49

r = \sqrt{419.49} = 20.4815

(Reason: dividing both sides by

0.0905861

leaves

r^2

on its own, and the radius is its positive square root, because a length cannot be negative.)

Step 4: the length of the arc

ABC

\dfrac{60}{360} \times 2 \times \pi \times 20.4815 = 21.4481

(Reason: the arc is the same fraction of the circumference,

\dfrac{60}{360}

, that the sector is of the whole circle.)

Step 5: the length of the chord

AC

OA = OC = 20.4815

AC = 20.4815

(Reason: two sides of triangle

OAC

are radii, so they are equal, and the angle between them is

60^\circ

. The other two angles are equal to each other and together make

120^\circ

, so each of them is

60^\circ

as well and the triangle is equilateral. The chord is therefore the radius itself.)

Step 6: add the two edges of the segment

21.4481 + 20.4815 = 41.9296

(Reason: the boundary of the shaded segment is made of exactly two pieces, the curved arc

ABC

and the straight chord

AC

. Nothing else lies on its edge.)

Step 7: round to one decimal place

41.9296 \to 41.9 \text{ cm}

(Reason: the digit after the first decimal place is a

2

, which is less than

5

, so the first decimal place is left as it is.)

41.9

Verification

Check 1: Put the unrounded

r^2 = 419.4905

back into the two areas the segment is built from: the sector,

\dfrac{\pi}{6} \times 419.4905

, and the triangle,

\dfrac{\sqrt{3}}{4} \times 419.4905

. Their difference must return the area the question gives.

219.6447 - 181.6447 = 38

Check 2: Divide the perimeter by the radius. Every

60^\circ

segment is the same shape, so this ratio is fixed at

\dfrac{\pi}{3} + 1 = 2.0472

whatever the radius happens to be.

\dfrac{41.9296}{20.4815} = 2.0472

Check 3: Scale a segment up instead. A

60^\circ

segment of a circle of radius

1

has perimeter

2.0472

, and multiplying every length by

20.4815

must give the perimeter found above.

2.0472 \times 20.4815 = 41.93

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A correct expression for the area of the segment, e.g. $\dfrac{60}{360} \times \pi r^2 - \dfrac{1}{2} r^2 \sin 60^\circ = 38$	M1	for a correct expression for the area of the segment, or equivalent. The expression may be embedded in an equation, e.g. $\dfrac{60}{360} \times \pi r^2 = 38 + \dfrac{1}{2} r^2 \sin 60^\circ$ , or written with the two coefficients already simplified as $\dfrac{\pi r^2}{6} - \dfrac{\sqrt{3}}{4} r^2$ .	✓
A correct expression for $r^2$ or $r$ , e.g. $r^2 = \dfrac{38}{\dfrac{\pi}{6} - \dfrac{\sqrt{3}}{4}} = 419.49$	M1	dependent on the first M1, for a correct expression for $r^2$ or $r$ . The values these come to are $419.490$ and $20.4815$ , and a value rounded or truncated from either is accepted.	✓
Using the radius to find the arc, e.g. $\dfrac{60}{360} \times 2 \times \pi \times 20.4815 = 21.4481$	M1	for using their value of $r$ to find the arc length, e.g. $\dfrac{\pi}{6} \times 20.4815 \times 2$ , or equivalent. Follow through on their radius.	✓
$41.9$	A1	for $41.9$ cm. Allow anything from $41$ to $42$ , which covers a radius that was rounded or truncated on the way. Working is not required, so a correct answer scores full marks unless it follows obviously incorrect working.	✓

Full marks: 4/4

A curve has equation $y = \text{f}(x)$ .
The curve has exactly one minimum point, and the coordinates of that minimum point are $(5, -4)$ .

Write down the coordinates of the minimum point on the curve with equation

(i) $y = \text{f}(x + 7)$ [1 mark]

(ii) $y = \text{f}(x) - 6$ [1 mark]

(i)(ii)

[Total 2 marks]

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Question 21 - Exam Solution

Understanding the Question

Given

A curve with equation

y = \text{f}(x)

, and nothing else about

\text{f}

at all.

Its one and only minimum point is at

(5, -4)

Find

(i) the minimum point of

y = \text{f}(x + 7)

(ii) the minimum point of

y = \text{f}(x) - 6

Plan the Solution

Decide, for each equation, whether the change is INSIDE the bracket or OUTSIDE it. Inside means a horizontal translation; outside means a vertical one.
A translation slides the whole curve without turning or stretching it, so the minimum point stays a minimum point. Only its coordinates move.
Apply the translation to the single point $(5, -4)$ . A horizontal translation changes only the $x$ -coordinate; a vertical one changes only the $y$ -coordinate.
Watch the sign inside the bracket: $+7$ inside moves the curve in the negative $x$ direction, not the positive one.

Worked Solution [2 marks]

Rule - Translating a curve:

y = \text{f}(x + a)

moves every point

a

units in the negative

x

direction, and

y = \text{f}(x) + b

moves every point

b

units in the positive

y

direction.

Step 1: read what

y = \text{f}(x + 7)

does

y = \text{f}(x + 7)

(Reason: The

+7

is inside the bracket, so it acts on

x

before

\text{f}

does anything: the translation is horizontal. The curve reaches its lowest point when the input to

\text{f}

5

, which now happens when

x + 7 = 5

, so the whole curve has slid

7

units in the negative

x

direction.)

Step 2: translate the minimum point for part (i)

(5 - 7, -4) = (-2, -4)

(Reason: Only the

x

-coordinate is affected. A horizontal translation cannot change how high a point sits, so the

y

-coordinate is still

-4

Step 3: read what

y = \text{f}(x) - 6

does

y = \text{f}(x) - 6

(Reason: The

-6

is outside the bracket, so it is applied after

\text{f}

has done its work: every output is

6

smaller, so the whole curve drops

6

units.)

Step 4: translate the minimum point for part (ii)

(5, -4 - 6) = (5, -10)

(Reason: Only the

y

-coordinate is affected. A vertical translation cannot move a point sideways, so the

x

-coordinate is still

5

(i)

(-2, -4)

(ii)

(5, -10)

Verification

Check 1: Ask where the new curve repeats the old low point.

y = \text{f}(x + 7)

takes the value

\text{f}(5)

when

x + 7 = 5

, that is when

x = -2

, and the height there is the same

\text{f}(5) = -4

. the minimum is at

(-2, -4)

Check 2: Test both parts on a curve that really does have its minimum at

(5, -4)

. Take

\text{f}(x) = (x - 5)^2 - 4

. Then

\text{f}(x + 7) = (x + 2)^2 - 4

, whose minimum is where

x + 2 = 0

, and

\text{f}(x) - 6 = (x - 5)^2 - 10

(-2, -4)

for (i) and

(5, -10)

for (ii)

Check 3: Count which coordinate is allowed to move. Part (i) is a horizontal translation, so the

y

-coordinate must survive unchanged as

-4

; part (ii) is a vertical translation, so the

x

-coordinate must survive unchanged as

5

. each answer changes exactly one coordinate, and it is the expected one

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(i) minimum point of $y = \text{f}(x + 7)$	B1	cao $(-2, -4)$ . Accept $x = -2$ and $y = -4$ written separately, or the pair labelled on a sketch.	✓
(ii) minimum point of $y = \text{f}(x) - 6$	B1	cao $(5, -10)$ . Accept $x = 5$ and $y = -10$ written separately.	✓
Guidance on the two common slips	Note	A translation applied the wrong way round - moving the curve to the right for $y = \text{f}(x + 7)$ , or upwards for $y = \text{f}(x) - 6$ - scores nothing. There is no method mark in either part: the answer is written down, so each mark is all or nothing.	✓

Full marks: 2/2

The incomplete histogram gives information about the distances, in kilometres, cycled by $100$ members of a cycling club last Saturday.

Every member cycled at least $5$ kilometres.
No member cycled more than $55$ kilometres.
$14$ members cycled between $15$ kilometres and $20$ kilometres.

Complete the histogram. [3 marks]

[Total 3 marks]

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Question 22 - Exam Solution

Understanding the Question

Given

A histogram of the distances cycled by

100

members, with three bars already drawn.

Every distance is at least

5

kilometres and at most

55

kilometres, so the bars must cover

5

55

with no gaps.

14

members are in the

15

20

class - the only frequency the question states.

The vertical axis carries no numbers, so the scale has to be worked out first.

Find

The one bar that is missing: the distances it covers, and how tall to draw it. Expect a frequency density, not a frequency - the height of a histogram bar is

\dfrac{\text{frequency}}{\text{class width}}

Plan the Solution

Use the one stated frequency, $14$ over the $15$ to $20$ class, to fix what one small square is worth.
Read the other two bars off that scale and turn each height back into a frequency.
Subtract the three frequencies from $100$ to get the missing frequency.
The missing class runs from $5$ to $15$ , so divide by that width to get the height to draw.

Worked Solution [3 marks]

Rule - Histograms: the AREA of a bar is its frequency, so

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

Step 1: use the stated frequency to fix the scale

\dfrac{14}{20 - 15} = \dfrac{14}{5} = 2.8

\dfrac{2.8}{28} = 0.1

(Reason: (Reason: the

15

20

bar is

28

small squares tall and stands for a frequency density of

2.8

, so one small square is

0.1

on the vertical axis. Nothing else on the graph can be read until this is known.))

Step 2: read the other two bars off that scale

34 \times 0.1 = 3.4

3.4 \times (35 - 20) = 51

4 \times 0.1 = 0.4

0.4 \times (55 - 35) = 8

(Reason: (Reason: the middle bar is

34

small squares tall and the flat one is

4

. Multiplying each frequency density by its class width turns the area of the bar back into a number of members.))

Step 3: take the known frequencies away from the total

14 + 51 + 8 = 73

100 - 73 = 27

(Reason: (Reason: every member is somewhere on the histogram, so the four bars must account for all

100

. The three drawn bars hold

73

of them, leaving

27

for the missing one.))

Step 4: turn that frequency into a height

15 - 5 = 10

\dfrac{27}{10} = 2.7

(Reason: (Reason: the shortest distance is

5

kilometres and the next bar starts at

15

, so the missing class is

5

15

. Dividing the frequency by that class width gives the frequency density, which is what the vertical axis measures.))

Step 5: draw the bar

2.7 \times 10 = 27

(Reason: (Reason: a bar from

5

15

with its top at

2.7

has area

27

, which is exactly the

27

members it has to represent. On the printed grid that top edge is

27

small squares above the horizontal axis.))

A bar from

5

15

kilometresdrawn at a frequency density of

2.7

Verification

Check 1 - do the four bars hold 100 members? Work out the area of every bar and add them, using frequency density times class width.

2.7 \times 10 + 14 + 51 + 8 = 100

Check 2 - count small squares instead of using the scale: One small square is

1

kilometre across and

0.1

up, so ten small squares stand for one member. The missing bar is

10

squares across and

27

up.

\dfrac{10 \times 27}{10} = 27

Check 3 - count large squares: The heavy grid lines fall every

5

small squares, so one large square is

25

small squares, which is

2.5

members. The completed histogram covers

40

large squares.

40 \times 2.5 = 100

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Fix the frequency density scale	M1	for finding the frequency density $\dfrac{14}{5} = 2.8$ , or a correct value on the frequency density scale, or ten small squares standing for one member oe, or one large square standing for $2.5$ members oe, or $51$ and $8$ assigned to the correct bars	✓
Find the missing frequency	M1	for a method to find the total frequency of the bars given, or a method to find the missing frequency: $14 + 51 + 8 = 73$ oe, or $100 - (14 + 51 + 8) = 27$ oe	✓
Draw the bar	A1	for the correct bar, frequency $27$ : drawn from $5$ to $15$ with its top at $2.7$	✓
Special case - right height, wrong left edge	SC B2	for a bar of height $2.7$ drawn from $0$ to $15$ : the frequency density is right, but the bar has been started at the axis instead of at $5$	✓
Special case - divided by the wrong width	SC B2	for a bar of height $1.8$ drawn from $0$ to $15$ : the missing frequency of $27$ has been spread over a width of $15$ instead of $10$	✓
Working is not required	Note	A correct histogram scores full marks unless it comes from obviously incorrect working.	✓

Full marks: 3/3

An ornament is made by removing a hemisphere, shown shaded, from a solid cone, as shown in the diagram.

The radius of the hemisphere is $2x$ cm
The radius of the base of the cone is $5x$ cm
The vertical height of the cone is $6x$ cm

The volume of the ornament is $6948\pi$ cm³

Work out the total surface area of the solid hemisphere that has been removed from the cone.
Give your answer correct to the nearest integer. [5 marks]

cm²

[Total 5 marks]

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Question 23 - Exam Solution

Understanding the Question

Given

A cone of base radius

5x

cm and vertical height

6x

A hemisphere of radius

2x

cm removed from the cone's top face

The volume of what is left is

6948\pi

cm³

Find

The total surface area of the solid hemisphere that was removed, to the nearest integer

Plan the Solution

What is left is the cone minus the hemisphere, so write both volumes in terms of $x$ .
Every length is a multiple of $x$ , so both volumes come out as a multiple of $\pi x^{3}$ . Subtract, set the result equal to $6948\pi$ and solve for $x^{3}$ .
Take the cube root to get $x$ , then the hemisphere's own radius $r = 2x$ .
The removed hemisphere is SOLID, so its total surface area is the curved half-sphere plus the flat circular face it was cut on.

Worked Solution [5 marks]

Cone:

V = \dfrac{1}{3}\pi r^{2} h

. Hemisphere:

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

. Total surface area of a solid hemisphere:

2\pi r^{2} + \pi r^{2} = 3\pi r^{2}

Step 1: Write the volume of the cone in terms of

x

V_{\text{cone}} = \dfrac{1}{3}\pi (5x)^{2}(6x)

V_{\text{cone}} = \dfrac{1}{3}\pi \times 25x^{2} \times 6x = 50\pi x^{3}

(Reason: The base radius is

5x

and the vertical height is

6x

. Squaring the radius gives

25x^{2}

, and

\dfrac{1}{3}\times 25 \times 6 = 50

Step 2: Write the volume of the hemisphere that is removed

V_{\text{hemisphere}} = \dfrac{1}{2} \times \dfrac{4}{3}\pi (2x)^{3}

V_{\text{hemisphere}} = \dfrac{2}{3}\pi \times 8x^{3} = \dfrac{16}{3}\pi x^{3}

(Reason: A hemisphere is half a sphere, and the sphere here has radius

2x

. Cubing the whole bracket gives

8x^{3}

. This is where a missing bracket costs the mark, since

2x^{3}

is not

(2x)^{3}

Step 3: Form an equation for the volume of the ornament and solve for

x^{3}

50\pi x^{3} - \dfrac{16}{3}\pi x^{3} = 6948\pi

\dfrac{134}{3}\pi x^{3} = 6948\pi

x^{3} = \dfrac{6948 \times 3}{134} = 155.552\ldots

(Reason: What is left is the cone minus the hemisphere, and

50 - \dfrac{16}{3} = \dfrac{134}{3}

. Every term carries a factor of

\pi

, so the

\pi

cancels from both sides before

x^{3}

is found.)

Step 4: Take the cube root to find

x

x = \sqrt[3]{155.552\ldots} = 5.378\ldots

(Reason: Keep the unrounded value in the calculator. Rounding

x

5.4

here would drag the final area off by several square centimetres.)

Step 5: Work out the total surface area of the solid hemisphere

r = 2x = 2 \times 5.378\ldots = 10.756\ldots

S = 3\pi r^{2} = 3\pi \times (10.756\ldots)^{2}

S = 1090.39\ldots

(Reason: The hemisphere is solid, so its surface is the curved half-sphere

2\pi r^{2}

together with the flat circular face

\pi r^{2}

, which is

3\pi r^{2}

altogether. To the nearest integer

1090.39\ldots

1090

1090

cm²

Verification

Check 1: Put the value back into the volume. The cone is

50\pi x^{3}

and the hemisphere is

\dfrac{16}{3}\pi x^{3}

, so what is left is

\dfrac{134}{3}\pi x^{3}

\dfrac{134}{3} \times 155.552\ldots = 6948

, so the volume really is

6948\pi

cm³

Check 2: Add the two surfaces separately instead of using

3\pi r^{2}

. With

r = 10.756\ldots

cm the curved part is

2\pi r^{2}

and the flat circle is

\pi r^{2}

726.93\ldots + 363.46\ldots = 1090.39\ldots

, the same total

Check 3: A whole sphere of radius

10.756\ldots

cm has surface area

4\pi r^{2} = 1453.85\ldots

. A solid hemisphere is three quarters of that: half the curved surface plus one flat circle.

\dfrac{3}{4} \times 1453.85\ldots = 1090.39\ldots

, which agrees

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Volume of the cone, or of the hemisphere, in terms of $x$	M1	For $\dfrac{1}{3}\pi \times (5x)^{2} \times 6x$ oe or $50\pi x^{3}$ oe, or for $\dfrac{1}{2} \times \dfrac{4}{3} \times \pi \times (2x)^{3}$ oe or $\dfrac{16}{3}\pi x^{3}$ oe, or for the whole sphere $\dfrac{4}{3} \times \pi \times (2x)^{3}$ oe or $\dfrac{32}{3}\pi x^{3}$ oe. Ignore missing brackets around $5x$ and $2x$ for this mark.	✓
A correct equation for the volume of the shape	M1	For $\dfrac{1}{3}\pi \times (5x)^{2} \times 6x - \dfrac{1}{2} \times \dfrac{4}{3} \times \pi \times (2x)^{3} = 6948\pi$ oe, or $50\pi x^{3} - \dfrac{16}{3}\pi x^{3} = 6948\pi$ oe, or $\dfrac{134}{3}\pi x^{3} = 6948\pi$ oe. If it is not expanded at this stage then the brackets must be seen.	✓
Rearranging the correct equation to find $x^{3}$ or $x$	M1	For $x^{3} = \dfrac{6948\pi \times 3}{134\pi} = 155.552\ldots$ oe, or $x = \sqrt[3]{155.552\ldots} = 5.37(8\ldots)$ oe. Accept $5.4$ or better.	✓
Surface area of the hemisphere	M1	For $3 \times \pi \times (2 \times 5.37\ldots)^{2}$ oe or $12 \times \pi \times (5.37\ldots)^{2}$ oe.	✓
The answer	A1	$1090$ , allow $1086$ to $1100$ . Working is not required, so a correct answer scores full marks unless it comes from obviously incorrect working.	✓
Special case: using $6948$ without the $\pi$	SC	SC B3 for $x^{3} = 49.5(138\ldots)$ or $x = 3.67(205\ldots)$ , and SC B4 for a final answer of awrt $508$ . This is the one wrong route the scheme still rewards, and naming it is the point: the $\pi$ must be cancelled from both sides, not divided out of one.	✓

Full marks: 5/5

A polygon has $n$ sides, where $n > 5$
Written in order of size, the interior angles of this polygon form an arithmetic sequence.
The first term of the sequence is $84^{\circ}$
The common difference of the sequence is $4^{\circ}$

Find the total of all the interior angles of the polygon.
You must show clear algebraic working. [6 marks]

[Total 6 marks]

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Question 24 - Exam Solution

Understanding the Question

Given

A polygon with

n

sides, where

n > 5

Its interior angles, in order of size, are an arithmetic sequence with first term

a = 84

and common difference

d = 4

Standard fact: the interior angles of any

n

-sided polygon add up to

(n - 2) \times 180^{\circ}

Find

The total of all

n

interior angles, in degrees Expect a quadratic in

n

, so expect two roots and only one that satisfies

n > 5

Plan the Solution

Write that same total twice: once as an arithmetic series in $n$ , once with the polygon angle rule.
Set the two expressions equal, since they count the same angles, and tidy the result into a quadratic.
Solve the quadratic, then use the condition $n > 5$ to decide which root is this polygon.
Substitute that value of $n$ back into $(n - 2) \times 180$ to get the sum.

Worked Solution [6 marks]

Rule: an arithmetic series with first term

a

and common difference

d

has

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

, and the interior angles of an

n

-sided polygon sum to

(n - 2) \times 180^{\circ}

Step 1: write the angle total as an arithmetic series

S_n = \dfrac{n}{2}\left[2(84) + (n - 1)(4)\right]

S_n = \dfrac{n}{2}\left[168 + 4n - 4\right] = \dfrac{n}{2}\left[164 + 4n\right]

S_n = 82n + 2n^{2}

(Reason: The

n

angles are an arithmetic sequence with

a = 84

and

d = 4

, so their total is the sum of the first

n

terms. Halving the bracket now keeps the algebra tidy.)

Step 2: write the same total using the polygon rule

(n - 2) \times 180 = 180n - 360

(Reason: Every polygon with

n

sides has interior angles adding to

(n - 2) \times 180

degrees, so this is the same total written a second way.)

Step 3: equate the two totals and form a quadratic

82n + 2n^{2} = 180n - 360

2n^{2} - 98n + 360 = 0

n^{2} - 49n + 180 = 0

(Reason: The two expressions count the same angles, so they are equal. Collecting every term on one side gives a three term quadratic, and every coefficient has a factor of

2

, so divide right through by

2

Step 4: solve the quadratic

(n - 4)(n - 45) = 0

n = 4 \text{ or } n = 45

(Reason: Two numbers multiplying to

180

and adding to

49

are

4

and

45

, so it factorises. The quadratic formula gives the same pair:

\sqrt{49^{2} - 4(180)} = \sqrt{1681} = 41

Step 5: use the condition on n to choose the root

n > 5 \implies n = 45

(Reason: The root

n = 4

really does solve the equation, since a quadrilateral with angles

84^{\circ}

88^{\circ}

92^{\circ}

and

96^{\circ}

has an angle sum of

360^{\circ}

. The question rules it out with

n > 5

, which is exactly why that condition is given.)

Step 6: work out the sum of the interior angles

(45 - 2) \times 180 = 43 \times 180 = 7740

(Reason: With

n = 45

the polygon angle rule gives the total straight away, in degrees.)

7740^{\circ}

Verification

Check 1: Total the

45

angles as a series instead of using the polygon rule:

\dfrac{45}{2}\left[2(84) + 44(4)\right]

\dfrac{45}{2} \times 344 = 45 \times 172 = 7740

Check 2: The largest angle is

84 + 44(4) = 260

, and an arithmetic sequence has mean equal to the average of its first and last terms, so the mean angle is

\dfrac{84 + 260}{2}

\dfrac{84 + 260}{2} = 172 \implies 45 \times 172 = 7740

Check 3: Each exterior angle is

180

minus its interior angle, and exterior angles always total

360

, so

45 \times 180

minus the answer must come to

360

8100 - 7740 = 360

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Substitute $a = 84$ and $d = 4$ into $S_n = \dfrac{n}{2}\left[2a + (n - 1)d\right]$	M1	For $\dfrac{n}{2}\left[2(84) + (n - 1)(4)\right]$ or $\dfrac{n}{2}\left[168 + 4n - 4\right]$ or $\dfrac{n}{2}\left[164 + 4n\right]$ or $82n + 2n^{2}$ oe	✓
Equate that sum with $(n - 2) \times 180$	M1	For $\dfrac{n}{2}\left[164 + 4n\right] = (n - 2) \times 180$ or $82n + 2n^{2} = (n - 2) \times 180$ oe. The sum must come from a correct substitution of $a$ and $d$ into $\dfrac{n}{2}\left[2a + (n - 1)d\right]$	✓
Multiply out and collect terms to form a three term quadratic	M1	For $n^{2} - 49n + 180 = 0$ oe, or any form of $an^{2} + bn + c = 0$ with at least two of the coefficients correct. Allow $n^{2} - 49n = -180$	✓
Solve their three term quadratic	M1ft	Any correct method, allowing one sign error and some simplification, for example $(n - 45)(n - 4) = 0$ or $n = \dfrac{49 \pm \sqrt{2401 - 720}}{2}$ or $\left(n - \dfrac{49}{2}\right)^{2} - \left(\dfrac{49}{2}\right)^{2} = -180$ . If factorising, allow brackets which expand to give two of the three terms correct. Or the correct value $n = 45$ (ignore $n = 4$ )	✓
Substitute their $n$ to find the angle sum	M1	For $(45 - 2) \times 180$ or $\dfrac{45}{2}\left[2(84) + (45 - 1)(4)\right]$ oe, or the same with $44$ in place of $45$ . Note $n > 5$	✓
State the sum of the interior angles	A1	$7740$ . Working required. Accept $7560$ or $7480$	✓

Full marks: 6/6

The function $f$ is given by
$f(x) = 17 - 3x^2 + 12x$

Express $f(x)$ in the form $a - b(x - c)^2$ , where $a$ , $b$ and $c$ are constants. [4 marks]

f(x) =

[Total 4 marks]

Show solution & mark schemeHide solution & mark scheme

Question 25 - Exam Solution

Understanding the Question

Given

f(x) = 17 - 3x^2 + 12x

The coefficient of

x^2

-3

, so the curve is a parabola that opens downwards.

Find

The constants

a

b

and

c

that write

f(x)

a - b(x - c)^2

. The target form has a minus in front of the square, so

b

is the positive number

3

, not

-3

Plan the Solution

Rewrite the terms in the usual order, then take the factor $-3$ out of the two $x$ terms only, leaving the $17$ outside.
Complete the square on the simple quadratic left inside the bracket.
Multiply the whole bracket by $-3$ again and collect the constants, so the answer ends up in the form the question asks for.

Worked Solution [4 marks]

Rule - Completing the square: take the

x^2

coefficient out as a factor first, then use

x^2 + px = (x + \dfrac{p}{2})^2 - (\dfrac{p}{2})^2

on what is left inside the bracket.

Step 1: take the factor

-3

out of the

x

terms

f(x) = 17 - 3x^2 + 12x

f(x) = 17 - 3(x^2 - 4x)

(Reason: Only the two

x

terms go inside the bracket. Dividing

-3x^2

-3

gives

x^2

, and dividing

12x

-3

gives

-4x

Step 2: complete the square inside the bracket

x^2 - 4x = (x - 2)^2 - 4

(Reason: Half of

-4

-2

, so the square is

(x - 2)^2

. Expanding that gives

x^2 - 4x + 4

, which is

4

too big, so

4

is taken off again.)

Step 3: put that back in and multiply through by

-3

f(x) = 17 - 3[(x - 2)^2 - 4]

f(x) = 17 - 3(x - 2)^2 + 12

(Reason: Every term inside the square bracket is multiplied by the factor

-3

, including the constant.)

Step 4: collect the two constants

f(x) = 17 + 12 - 3(x - 2)^2

f(x) = 29 - 3(x - 2)^2

(Reason: The question wants the constant written first, so reading off the form

a - b(x - c)^2

gives

a = 29

b = 3

and

c = 2

f(x) = 29 - 3(x - 2)^2

Verification

Check 1: Expand the answer again and compare it, term by term, with the

f(x)

the question prints.

29 - 3(x^2 - 4x + 4) = 29 - 3x^2 + 12x - 12 = 17 - 3x^2 + 12x

Check 2: Put

x = 5

into both forms. They are the same function, so they must give the same value.

17 - 3(5)^2 + 12(5) = 2

and

29 - 3(5 - 2)^2 = 2

Check 3: The coefficient of

x^2

is negative, so

f

has a maximum, and the completed square says that maximum is

a

x = c

. Substituting

x = 2

tests both constants at once.

f(2) = 17 - 3(2)^2 + 12(2) = 29

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Factorise the $x$ terms, e.g. $-3(x^2 - 4x)$ , or state $b = 3$ .	M1	For factorising $-3x^2 + 12x$ , or for the correct value of $b$ (that is $b = 3$ ) embedded in an incorrect final answer of the form $a - 3(x - c)^2$ .	✓
$-3[(x - 2)^2 \ldots]$ or $-3(x - 2)^2 \ldots$	M1	For a correct first step to complete the square.	✓
$-3[(x - 2)^2 - (2)^2] \ldots$ or $-3(x - 2)^2 + 12 \ldots$	M1	For a correct second step to complete the square, or equivalent.	✓
$f(x) = 29 - 3(x - 2)^2$	A1	Or equivalent, e.g. $-3(x - 2)^2 + 29$ . Working is not required, so a correct answer scores full marks unless it follows obviously incorrect working.	✓
Alternative method: multiply out $a - b(x - c)^2$ to get $-bx^2 + 2bcx - bc^2 + a$ , then equate coefficients.	Note	The mark scheme allows this route for the same $4$ marks: M1 for multiplying out, M1 for equating coefficients ( $2bc = 12$ or $a - bc^2 = 17$ ), M1 for finding at least two of $a$ , $b$ , $c$ , then A1 for the answer. This row carries no mark of its own.	✓

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 4MA1/1H, Thursday 16 May 2024: Worked Solutions, Questions 15 to 25

Worked solutions, questions 15 to 25 of 25

Question 15, Calculator allowed

Question 15 - Exam Solution

Question 16, Calculator allowed

Question 16 - Exam Solution

Question 17, Calculator allowed

Question 17 - Exam Solution

Question 18, Calculator allowed

Question 18 - Exam Solution

Question 19, Calculator allowed

Question 19 - Exam Solution

Question 20, Calculator allowed

Question 20 - Exam Solution

Question 21, Calculator allowed

Question 21 - Exam Solution

Question 22, Calculator allowed

Question 22 - Exam Solution

Question 23, Calculator allowed

Question 23 - Exam Solution

Question 24, Calculator allowed

Question 24 - Exam Solution

Question 25, Calculator allowed

Question 25 - Exam Solution

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