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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

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