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Pearson Edexcel
Level 3 GCE
Wednesday 13 May 2020
Paper Reference 8MA0/01
Solved by shahbaz ahmed
November 2024

Mathematics
Advanced Subsidiary
Paper 1: Pure Mathematics
1
...

Write your answer in the form y = mx + c,
1

where m and c are integers to be found
Solution

y = 2x3 − 4x + 5
dy
d
= (2x3 − 4x + 5)
dx dx
d
d
d
dy
= (2x3) − (4x) + (5)
dx dx
dx
dx
dy
d
dx
d
= 2 (x3) − 4 + (5)
dx
dx
dx dx
dy
= 2(3x2) − 4 + 0 = 6x2 − 4
dx
Slope of the line at the point P (2, 13) = P (x, y)

m=

dy
= 6(2)2 − 4 = 20
dx

Where m denotes the slope at the point P (2, 13)
Putting P (2, 13) = P (x, y) and m = 20 in :

y = mx + c
2

13 = 20 × 2 + c
13 = 40 + c
40 + c = 13
c = 13 − 40 = −27
Putting m=20, c=-27 , equation of tangent to the curve
at the point P(2,13)

y = mx + c = 20x − 27
Or

y = 20x − 27

3

Graph of solution


...

To calculate the bearing, we first determine the displacement vector
...


4

Substituting the positions:
d = (−3i − 5j) − (4i − 2j),
d = −3i − 5j − 4i + 2j,
d = −7i − 3j
...
We calculate the angle θ it makes with the
positive i-axis (due east)
...

adjacent component −7 7

Using a calculator:
 
3
θ = tan−1
≈ 23
...

7
This angle is measured counterclockwise from the negative i-axis because both components of the vector are
negative
...
2◦ = 203
...

Final Answer: The bearing on which the boat is
moving is:
203
...

(b) Calculate the speed of the boat
...

Thus:
∥d∥ ≈ 7
...

The time elapsed between 10 : 00 and 12 : 45 is

6

2 hours + 45 minutes, which is:
2+

45
= 2
...

60

The speed of the boat is given by:
Speed =

distance ∥d∥
=

...
75

Substituting the values:
Speed =

7
...
77 km/h
...
75

Final Answer: The speed of the boat is:
2
...


...


Vector Diagram
The vector −3i − 5j can be represented as follows:
−3

O

i (East)

−3i − 5j
−5
j (North)


...


...

203
...

3
...

(ii) Solve the equation
10

1
43x−2 = √
2 2
Solution(i)
√

x 2−

√

18 = x

√
√
x 2− 9×2=x
√

x 2−

√

√

9×

√

2=x

√

x 2−3 2=x
√
x 2−x=3 2
√

√
√
x( 2 − 1) = 3 2
√

√

√ √
x( 2 − 1)( 2 + 1) = 3 2( 2 + 1)
√

√ √
x[( 2) − 1 ] = 3 2( 2 + 1)
2

2

√ √
x[2 − 1] = 3 2( 2 + 1)

11

√ √
x = 3 2( 2 + 1)
Solution(ii)

1
43x−2 = √
2 2
1
43x × 4−2 = √
2 2
1
1
43x × 2 = √
4
2 2
1
1
= √
43x ×
16 2 2
16
43x = √
2 2
8
43x = √
2
2 3x

(2 )

=

23
1

22
1

26x = 23− 2
5

26x = 2 2
=⇒
12

6x =
x=

5
2

5
12


...
In 1997 the average CO2 emissions of new
cars in the UK was 190 g/km
...
Given Ag/km is the
average CO2 emissions of new cars in the UK
n years after 1997 and using a linear model,
(a) form an equation linking A with n
...

(b) Comment on the suitability of your model in light
of this information
...

• In 2005 (n = 8), A = 169 g/km
...


Step 1: Calculate the slope m

The slope m is given by the

rate of change of A with respect to n:
m=

∆A 169 − 190 −21
=
=
= −2
...
625(0) + c =⇒ c = 190
Thus, the equation linking A with n is:
A = −2
...
Using the
model:
A = −2
...
625(19) + 190 = −49
...
125 g/km
However, the actual emissions in 2016 were 120 g/km,
15

which is significantly lower than the predicted 140
...


Suitability of the model

The linear model is reasonable over

short periods, such as from 1997 to 2005, as the data
aligns closely
...
Over longer periods, the model may become less accurate due to factors like technological advancements,
policy changes, and increased environmental awareness
...
The actual decrease in CO2 emissions between 1997
and 2016 is not constant, suggesting a nonlinear trend
...
A better fit for the data may involve a nonlinear
model, such as an exponential or piecewise function,
to account for faster reductions in emissions over
time
...


...


Figure 1 shows the design for a structure
used to support a roof
...

ˆ =
Given AB = 12m, BC = BD = 7m and ∠BAC
27◦
17

(a) find, to one decimal place, the size of
ˆ
∠ACB
The steel beams can only be bought in whole
meter lengths
...

Solution
In the △ABD

18

∠ADB = θ
7
12
=
sin 27 sin θ
7 sin θ = 12(sin 27)
12(sin 27)
7
12(sin 27)
θ = sin−1[
]
7
sin θ =

θ = 51
...
1◦
Since
19

ˆ + ∠BCD
ˆ = 180◦
∠ACB
ˆ + 51
...
1◦
∠ACB
ˆ = 128
...
1◦ + ∠ABD
ˆ = 180◦
78
...
1◦
∠ABD
ˆ = 101
...
9◦ sin 27◦
7
◦
b=
×
sin
101
...
08m = AD
Minimum length of the steel required=AB+BD+AD+BC
=12+7+15
...
08 m

...
(a) Find the first 4 terms, in ascending
powers of x, in the binomial expansion of

(1 + kx)10
where k is a non-zero constant
...


21

Solution
Part (a): Finding the first 4 terms in ascending powers
of x in the binomial expansion of (1 + kx)10

The binomial expansion of (1 + kx)10 is given by:
10  
X
10
(1 + kx)10 =
(1)10−r (kx)r
r
r=0

The first 4 terms in ascending powers of x are:
 
10
(1)1(kx)0 = 1
T0 =
0
 
10
T1 =
(1)9(kx)1 = 10kx
1
 
10
T2 =
(1)8(kx)2 = 45k 2x2
2
 
10
T3 =
(1)7(kx)3 = 120k 3x3
3
Thus, the first 4 terms of the expansion are:
1 + 10kx + 45k 2x2 + 120k 3x3
22

Part (b): Given that the coefficient of x3 is 3 times the
coefficient of x, find the possible values of k
...

7
...

Hence
k ̸= 9
Now putting k= 16
9 in
Z
1

16
9

5
( √ + 3) dx
2 x

16
16 1
) 2 − 5 + 3( ) − 3
9
9
4
16
= 5( ) − 5 + ( ) − 3
3
3

= 5(

29

20
16
)−5+( )−3
3
3
20 + 16
)−8
=(
3

=(

= (12 − 8) = 4
Hence Value of k= 16
9

...
The temperature, θ◦C, of a cup of tea
t minutes after it was placed on a table in a
room, is modeled by the equation

t

θ = 18 + 65e− 8

t≥0

Find, according to the model,
(a) the temperature of the cup of tea when
it was placed on the table
...

(c) Explain why, according to this model,
the temperature of the cup of tea could not
fall to 15◦C
...

31

Figure 2

shows a sketch of µ against t with two data
points that lie on the curve
...

Solution
(a)

t

θ = 18 + 65e− 8

t≥0

Putting t=0

0

θ = 18 + 65e− 8

θ = 18 + 65e0 = 18 + 65(1) = 18 + 65 = 83◦
(b)
32

Again

t

θ = 18 + 65e− 8

t≥0

Putting θ = 35◦C
...
07”
t≥0
17
t

ln e 8 = ln

(c)
Put θ = 15◦C
...
Hence θ = 15◦C is not possible
...
µ(94) = 0, 2
...


35

Step 1: Horizontal Asymptote Analysis
The exponential term e−t/8 approaches 0 as t → ∞
...


t→∞

The horizontal asymptote of µ(t) is therefore given by:

µ(t) = A,

t → ∞
...
At t = 94:
µ(94) = A + Be−94/8 = 0
...
At t = 50:
µ(50) = A + Be−50/8 = 8
...
A+Be−94/8 = 0, or Be−94/8 = −A
...
A+Be−50/8 =
8
...
Substituting x and y
into the equations:
1
...
2
...


37

Factor out B:
B(y − x) = 8
Substitute B =

8
y−x

A=−

=⇒

B=

8

...

y−x
y−x

Simplify x and y:

x = e−94/8 = e−47/4,

y = e−50/8 = e−25/4
...

e
− e−47/4

Step 3: Horizontal Asymptote
From Step 1, the horizontal asymptote is:

µ(t) = A as t → ∞
...

e
− e−47/4

...


Figure 3 shows part of the curve with equation
y = 3 cos x◦
The point P(c, d) is a minimum point on
the curve with c being the smallest negative
value of x at which a minimum occurs
...

(b) State the coordinates of the point to
which P is mapped by the transformation which
transforms the curve with equation y = 3 cos x◦
to the curve with equation
◦

(i) y = 3cos x4

(ii)y=3cos x − 36◦
(c) Solve, for 450◦ ≤ θ < 720◦

3 cos θ = 8 tan θ
giving your solution to one decimal place
...
For cos x◦, the minimum value is −1
...

The smallest negative x is −180◦
...
The x-coordinate of P (−180, −3) be-

41

comes:
c′ = 4(−180) = −720◦
The y-coordinate remains the same
...
The x-coordinate of P (−180, −3)
becomes:
c′ = −180 + 36 = −144◦
The y-coordinate remains the same
...
Rewrite the equation:
3 cos θ = 8 tan θ =⇒ 3 cos θ = 8

sin θ
cos θ

2
...
Expand and rearrange:
3 − 3 sin2 θ = 8 sin θ =⇒ 3 sin2 θ + 8 sin θ − 3 = 0

4
...

The equation becomes:
3u2 + 8u − 3 = 0

43

Solve using the quadratic formula:
u=

u=

−8 ±

p

√

82

− 4(3)(−3) −8 ± 64 + 36 −8 ± 10
=
=
2(3)
6
6

−8 + 10 2 1
= = ,
6
6 3

u=

−8 − 10 −18
=
= −3
6
6

Since sin θ must lie in [−1, 1], only sin θ =

1
3

is

valid
...
Find θ: Using sin θ = 13 , find the angles in
the range 450◦ ≤ θ < 720◦
...
5◦
θ = arcsin
3
In the range 450◦ ≤ θ < 720◦, the solutions

44

are in the 3rd and 4th quadrants:
θ = 540◦+19
...
5◦,

θ = 720◦−19
...
5◦

Final Answers

• (a) c = −180◦, d = −3
• (b)
– (i) (−720, −3)
– (ii) (−144, −3)
• (c) θ = 559
...
5◦

...


g(x) = 2x3 + x2 − 41x − 70
(a) Use the factor theorem to show that g(x)
is divisible by x − 5
...

(c) Find, using algebraic integration, the
exact value of the area of R
...

Using the factor theorem, we evaluate g(5):
g(5) = 2(5)3 + (5)2 − 41(5) − 70
...

Since g(5) = 0, by the factor theorem, (x − 5) is
a factor of g(x)
...


Divide g(x) = 2x3 + x2 − 41x − 70 by (x − 5) :

47

2x3
1
...

x
2
...


3
...
Divide the next term:
= 11x
...
Multiply: (x − 5)(11x) = 11x2 − 55x
...
Subtract: (11x2 − 41x − 70) − (11x2 − 55x) = 14x − 70
...
Divide:

14x
= 14
...
Multiply: (x − 5)(14) = 14x − 70
...
Subtract: (14x − 70) − (14x − 70) = 0
...


48

Next, we factorize 2x2 + 11x + 14:

2x2+11x+14 = 2x2+7x+4x+14 = x(2x+7)+2(2x+7) = (2x+7)(x+2)
...


(c) Find the exact value of the area of the
region R:
The region R is bounded by the curve y =
g(x) and the x-axis, and lies below the x-axis
...


49

The integral for the area is:
Z

5

Z

5

−g(x) dx =

Area =
− 27

−(2x3 +x2 −41x−70) dx
...

2
3
2


Z
1 4 1 3 41 2
x + x − x − 70x + C
...

2
3
2
7
−
2

At x = 5:
1
1
41
1
1
41
− (5)4− (5)3+ (5)2+70(5) = − (625)− (125)+ (25)+350
...
Finally, subtract the value at x =
− 72 from the value at x = 5 to find the area
...


− 72

The integral of −g(x) is:
Z




1 4 1 3 41 2
−g(x) dx = −
x + x − x − 70x
...

Step 1: Evaluate at x = 5:
At x = 5 :



1 4 1 3 41 2
−
(5) + (5) − (5) − 70(5)
...

2
3
2
51

Simplify term by term:
1
(625) = 312
...
67,
3

41
(25) = 512
...


= − (312
...
67 − 512
...

= − (−508
...
33
...

2
2
3
2
2
2
2
Simplify term by term:
 4
1
7
1 2401 2401
−
= ·
=
= 75
...
29,
3
2
3
8
24
 2
41
7
41 49
2009
−
−
=− ·
=−
= −251
...

2
Adding these:
− (75
...
29 − 251
...
61) = −54
...


Step 3: Compute the definite integral:
7
Area = Value at x = 5 − Value at x = −
...
33 − (−54
...
33 + 54
...
94
...
94 square units

...
(i) A circle C1 has equation

x2 + y 2 + 18x − 2y + 30 = 0
53

The line l is the tangent to C1 at the point
P(–5, 7)
Find an equation of l in the form ax + by
+ c = 0, where a, b and c are integers to be
found
...

Given that C2 lies entirely in the 4th quadrant, find the range of possible values for k
...

54

d 2
d
(x + y 2 + 18x − 2y + 30) = 0
dx
dx
d
d
d
d
d 2
x + (y 2) + (18x) − (2y) + (30) = 0
dx
dx
dx
dx
dx
d
dy
dx
dy
2x + (y 2) + 18 − 2 + 0 = 0
dy
dx
dx
dx
dy
dy
+ 18 − 2 = 0
dx
dx
dy
dy
2y − 2 = −2x − 18
dx
dx
dy
2[y − 1] = −2(x + 9)
dx
dy
[y − 1] = −(x + 9)
dx
x+9
dy
=−
dx
y−1

2x + 2y

In particular ,slope at the point P (x1, y1)

dy
x1 + 9
=−
dx
y1 − 1

55

Putting P (–5, 7) = P (x1, y1)
dy
−5 + 9
4
2
=−
=− =− =m
dx
7−1
6
3
Where m is the slope of the line l the tangent
to C1 at the point P(–5, 7)
...

57

Step 1: Rewrite the equation in standard form
The given equation is:
x2 + y 2 − 8x + 12y + k = 0
...

• For y: y 2 + 12y Complete the square:
y 2 + 12y = (y + 6)2 − 36
...


58

Simplify:
(x − 4)2 + (y + 6)2 = 52 − k
...


Step 3: Conditions for the circle to lie entirely
in the fourth quadrant
The circle lies entirely in the fourth quadrant
if:
1
...

2
...

(a) Condition for the circle not touching the x-axis

The distance from the center (4, −6) to the xaxis is 6
...


Squaring both sides:
52 − k < 36

⇒

60

k > 16
...
For the circle to not touch the y-axis:
r<4

√

⇒

52 − k < 4
...


Step 4: Combine conditions
For the circle to lie entirely in the fourth quadrant:
k > 36
...

The given equation is:
x2 + y 2 − 8x + 12y + 36 = 0
...

The circle has center (4, −6) and radius r = 4
...

12
...

The equation

1 ≤ t ≤ 30, t ∈ N

log10 V = 0
...
379

63

is used to model the total number of views
of the advert, V, in the first t days after the
advert went live
...

Give the value of a to the nearest whole
number and give the value of b to 3 significant
figures
...

Using this model, calculate
(c) the total number of views of the advert
in the first 20 days after the advert went live
...
072t + 2
...
072t+2
...
072t102
...
072 = b =⇒ 100
...
379 = a
Hence

abt = V

1 ≤ t ≤ 30, t ∈ N

102
...
072 = 1
...
180 = 282
...
180)20 = V

1 ≤ t ≤ 30, t ∈ N
1 ≤ t ≤ 30, t ∈ N

V = 6, 546
...

13
...
Solution

66

Solution
Part (a): Prove that

4a
b

+

b
a

≥ 4 for all positive a and b
...


ity states that for any two positive numbers x
and y,
x+y √
≥ xy
...

Let x =

4a
b

and y = ab
...

b a

Step 2: Simplify the geometric mean
...

b a
67

Thus,
4a
b

+ ab
≥ 2
...


4a b
+ ≥ 4
...
Hence, the inequality is
true for all positive a and b, with equality when
b = 2a
...


The inequality

4a
b

+

b
a

≥ 4 assumes a and b are

positive
...


68

Counterexample:

Let a = 1 and b = −2 (i
...
, b is

negative):
4a b 4(1) −2
+ =
+
= −2 − 2 = −4,
b
a
−2
1
which is clearly less than 4
...


...
A curve has equation y = g(x)
...

• The curve with equation y = g(x) passes
through the origin
...

(a) find g(x),
(b) prove that the stationary point at (2, 9) is
a maximum

Solution
Part (a) - Find g(x)

We are given:
• g(x) is a cubic expression: g(x) = ax3 + bx2 +
cx + d,
• The coefficient of x3 equals the coefficient
of x: a = c,
• The curve passes through the origin: g(0) =
0, which implies d = 0,
70

• The curve has a stationary point at (2, 9),
so:
– g(2) = 9,
– g ′(2) = 0
...


Step 1: Find the coefficients using g(2) = 9

g(2) = a(2)3 + b(2)2 + a(2) = 8a + 4b + 2a = 10a + 4b
...


71

(1)

Step 2: Use g ′ (2) = 0 to find another equation

The deriva-

tive of g(x) is:
g ′(x) = 3ax2 + 2bx + a
...

Since g ′(2) = 0, we have:
13a + 4b = 0
...


72

(2)

From equations

Subtract equation (2) from equation (1):
(10a + 4b) − (13a + 4b) = 9 − 0,
−3a = 9 =⇒ a = −3
...

4

Part (b) - Prove the stationary point at (2, 9) is a maximum

Step 1: Find the second derivative

The first derivative

is:



39
39
g ′(x) = 3(−3)x2 + 2
x − 3 = −9x2 + x − 3
...

4

The second derivative is:
g ′′(x) = −18x +

39

...

2
2
2
2

Since g ′′(2) < 0, the stationary point at (2, 9)
is a maximum
...


74

Solution
Part (a) - Find g(x)

We are given:
• g(x) is a cubic expression: g(x) = ax3 + bx2 +
cx + d,
• The coefficient of x3 equals the coefficient
of x: a = c,
• The curve passes through the origin: g(0) =
0, which implies d = 0,
• The curve has a stationary point at (2, 9),
so:
– g(2) = 9,
– g ′(2) = 0
...


Step 1: Find the coefficients using g(2) = 9

g(2) = a(2)3 + b(2)2 + a(2) = 8a + 4b + 2a = 10a + 4b
...


Step 2: Use g ′ (2) = 0 to find another equation

(1)

The deriva-

tive of g(x) is:
g ′(x) = 3ax2 + 2bx + a
...


76

Since g ′(2) = 0, we have:
13a + 4b = 0
...

Subtracting equation (2) from equation (1):
(10a + 4b) − (13a + 4b) = 9 − 0,
−3a = 9 =⇒ a = −3
...

4

Thus:
g(x) = −3x3 +

39 2
x − 3x
...

4
2
The second derivative is:
g ′′(x) = −18x +

39

...

2
2
2
2

Since g ′′(2) < 0, the stationary point at (2, 9)
78

is a maximum
...

Solution is shown below graphically
...

Derivation of the Law of Cosines

79

Derivation of the Law of Cosines
Consider a triangle △ABC with sides of lengths
a, b, and c, where c is the side opposite angle
C
...

2bc

Step 1: Place the Triangle in the Coordinate Plane

Place vertex A at the origin, so A = (0, 0), and
place vertex B along the positive x-axis, so
B = (b, 0)
...

Step 2: Distance Formula for Side c

The distance from point A to point C is given
by the distance formula:
c2 = x2C + yC2
80

Step 3: Distance Formula for Side a

The distance from point B to point C is given
by:
a2 = (xC − b)2 + yC2
Expanding this:
a2 = x2C − 2bxC + b2 + yC2
Substitute x2C + yC2 = c2 from Step 2:
a2 = c2 − 2bxC + b2
Rearranging for xC :
2bxC = b2 + c2 − a2
b2 + c2 − a2
xC =
2b

81

Step 4: Relation to Cosine of Angle C

Using the definition of cosine in the coordinate
plane:
cos(C) =

xC
b

Substituting the expression for xC :
cos(C) =

b2 +c2 −a2
2b

b

b2 + c2 − a2
=
2bc

Thus, the Law of Cosines is derived as:
b2 + c 2 − a2
cos(C) =
2bc
Step 5: Graphical Representation of Triangle △ABC

C
c
A

a
b

B

82



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