A Level Mathematics
Edexcel – AQA – OCR – WJEC - CCEA
Paper 1 – Pure Mathematics
Set 1
Name
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Date
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2 hours allowed
Calculator Paper
Maximum Mark: 100
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Question 1
The diagram below shows the graph of
(i)
𝑦 = 𝑓(𝑥)
On the diagram draw the graph of
1
𝑦 = 𝑓( 𝑥)
2
(1)
(ii)
On the diagram draw the graph of
𝑦 = 2𝑓(𝑥) − 2
(2)
(Total 3 Marks)
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Question 2
The equation
(i)
𝑥 3 − 2𝑥 + 1 = 0
has three real roots
Show that one of the roots lies between
−1 and −2
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(2)
(ii)
Taking
𝑥1 = −2 as the first approximation to one of the roots, use the
Newton-Raphson method to find 𝑥2 , the second approximation.
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(3)
(iii)
Explain what happens with the Newton-Raphson method in the case when
the first approximation is
𝑥1 = 1.
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(1)
(Total 6 Marks)
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Question 3
A circle C has equation
(i)
𝑥 2 + 𝑦 2 + 10𝑥 − 6𝑦 + 14 = 0
Find the centre and radius of C
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(3)
(ii)
Find the coordinates of any points where the line
𝑦 = 2𝑥 + 3 meets circle C.
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(4)
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(iii)
State what can be deduced about the line and circle.
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(4)
(Total 8 Marks)
Question 4
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Business A also made a £4000 profit during its first year. In each subsequent year the
profit was 95% of the previous year’s profit.
Business B made £4000 profit during its first year. In each subsequent year the profit
increased by £750.
(i)
Find an expression for the total profit made by Business A during the first n
years.
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(3)
(ii)
Find an expression for the total profit made by Business B during the first n
years.
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(2)
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(iii)
Find how many years it will take for the total profit of Business B to reach
£325,000.
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(3)
(iv)
Comment on the profits made by each business in the long term.
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(2)
(Total 10 Marks)
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Question 5
The cross section of a modern building is modelled by the equation
𝑥 2 + 𝑥𝑦 + 4𝑦 2 = 60
Find the maximum vertical height above the ground the building reaches if
𝑥
and
𝑦
are measured in 10 metre units.
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(Total 8 Marks)
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Question 6
An athlete starts gaining muscle mass, assuming that the mass, m kg after t days,
increases at a rate that is inversely proportional to the cube root of the mass.
(i)
Construct a differential equation involving m, t and a positive constant k to
model this situation.
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(3)
(ii)
Explain why these assumptions may not be appropriate.
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(1)
(Total 4 Marks)
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Question 7
(i)
Differentiate
𝑒 4𝑡𝑎𝑛 𝑥
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(2)
(ii)
Differentiate
𝑠𝑖𝑛 3𝑥
𝑥3
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(3)
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(iii)
A function is defined implicitly by
2𝑥 2 𝑦 + 3𝑦 2 − 4𝑥 = 12
Find the equation of the normal at the point (1, 2)
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(6)
(Total 11 Marks)
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Question 8
(i)
Use binomial expansions to show that
√
1+3𝑥
1−𝑥
≈ 1 + 2𝑥 + 2𝑥 3
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(6)
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(ii)
A student substitutes 𝑥 = 2 into both sides of√
1+3𝑥
1−𝑥
≈ 1 + 2𝑥 + 2𝑥 3
as an attempt to find an approximation of √7.
Give a reason why the student should not use 𝑥 = 2
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(1)
(iii)
1
Substitute 𝑥 = 2 into
√
1+3𝑥
1−𝑥
≈ 1 + 2𝑥 + 2𝑥 3
to gain an approximation of
√5. Give your answer as a fraction in its simplest form.
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(3)
(Total 10 Marks)
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Question 9
𝑓(𝑥) = 18𝑥 3 − 15𝑥 2 − 𝑥 + 2
(i)
Prove that
(3𝑥 + 1) is a factor of 𝑓(𝑥)
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(2)
(ii)
Factorise
𝑓(𝑥)
completely.
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(3)
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(iii)
Hence prove that there are no real solutions to the equation
18𝑠𝑒𝑐 2 𝑥 + 2𝑐𝑜𝑠 𝑥 − 1
= 𝑠𝑒𝑐 𝑥
15
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(5)
(Total 10 Marks)
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Question 10
The diagram shows curve C with equation
𝑥 2 𝑙𝑛𝑥
𝑦=
− 2𝑥 + 4
2
𝑥>0
The finite region, S, shown on the diagram, is bounded by the curve C, the lines
𝑥=3
and the
𝑥 -axis.
The table shows corresponding values of
𝑥
𝑦
(i)
𝑥 = 1,
1
2
𝑥
1.5
1.45615
and
𝑦
2
1.38629
Use the trapezium rule with all of the values of
estimate for the area of S
2.5
1.86341
3
2.94376
𝑦 in the table to obtain an
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(3)
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(ii)
Explain how this trapezium rule method could be adapted to obtain a more
accurate estimate for the area of S
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(1)
(iii)
Show that the exact area of S can be written in the form
and
𝑎
𝑏
ln𝑐 + 𝑑
where
𝑎, 𝑏
𝑐 are integers
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(6)
(Total 10 Marks)
Question 11
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The curve C with equation
𝑦=
where
𝑝 − 2𝑥
(3𝑥 − 𝑞)(𝑥 + 2)
𝑝 and 𝑞 are constants, passes through point (5, 0) and has two vertical
asymptotes with equations
(i)
𝑥 ∈ ℝ, 𝑥 ≠ −2, 𝑥 ≠ 2
Show that
𝑥 = 2 and 𝑥 = −2
𝑝 = 10 and 𝑞 = 6
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(3)
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The diagram shows a sketch of part of the curve C. The shaded region is bounded by the
curve C, the
(ii)
𝑥 -axis and the line 𝑥 = 3
Find the area of the shaded region, giving your answer to 4 decimal places.
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(8)
(Total 11 Marks)
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Question 12
A company decides to manufacture a cardboard tube which contains crisps with a capacity
of 3500 ml.
The company models the tube in the shape of a cylinder with radius
𝑟
cm and height
ℎ
cm.
(i)
Prove that the total surface area,
𝑆 𝑐𝑚2 , of the tube is given by
𝑆 = 2𝜋𝑟 2 +
7000
𝑟
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(3)
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(ii)
Given that
𝑟
can vary, find the dimensions of a tube that has the minimum
surface area. You do not need to calculate the resultant surface area.
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(5)
(iii)
With reference to the shape of the tube, suggest a reason why the company
may choose not to manufacture a tube with the minimum surface area.
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(1)
(Total 9 Marks)
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Answers
1 (i)
A1M for correct stretch (factor 2 in y-direction)
1 (ii)
A2M for correct stretch and translation (A1M for each)
This diagram shows the stretch. Then move the graph down by 2.
2 (i)
A1M for substitution of
f(−1)>0
−1 and −2 to show that f(−2)<0 and
A1M for stating that f(𝑥 ) = 0 when there is a root so one must lie
between those values
2 (ii)
A1M for using Newton-Raphson method
𝑥𝑛+1
A1M for substitution of
A1M for
(𝑥𝑛 )3 − 2(𝑥𝑛 ) + 1
= 𝑥𝑛 −
3(𝑥𝑛 )2 − 2
−2 into the formula
−1.7
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2 (iii)
A1M for stating that zero appears in the formula, however you are not
diving by 0 which would cause the method to fail, you are dividing 0
by 1 which is 0. The method repeats which tells us that 1 is actually a
root.
3 (i)
A1M for attempt to complete the square
A2M for Centre (−5, 3) and Radius √20
3 (ii)
A1M for substitution of the line equation into the circle equation
𝑥 2 + (2𝑥 + 3 )2 + 10𝑥 − 6(2𝑥 + 3) + 14 = 0
A1M for rearrangement
𝑥 2 + 2𝑥 + 1 = 0
A1M for solving quadratic using any method to find
𝑥 = −1
A1M for substitution into line equation to find (-1, 1)
3 (iii)
A1M for stating that the line is a tangent to the circle at the point x =
-1 as they only intersect once.
4 (i)
A1M for using a geometric progression with a = 4000 and r = 0.95
A1M for
A1M for
4 (ii)
4000(1 − (0.95𝑛 ))
1 − 0.95
80000 − 80000(0.95)𝑛
A1M for using an arithmetic progression with a = 4000 and d = 750
𝑛
(2(4000) + (𝑛 − 1)750)
2
A1M for
4 (iii)
375𝑛2 + 3625𝑛
A1M for 375𝑛2
A1M for
+ 3625𝑛 = 325000
𝑛 = 25
and
𝑛 = −
104
3
A1M for stating 25 years
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4 (iv)
A1M for Business A’s profits will eventually plateau as (0.95)n tends to
zero
A1M for Business B’s profits will continue to grow
5
A1M for attempting differentiation of
A1M for
2𝑥 + 𝑦 + 𝑥
A1M for stating that
𝑑𝑦
𝑑𝑥
𝑑𝑦
2𝑥 + 𝑦 = 0
A1M for
𝑥 = −
𝑑𝑦
𝑑𝑥
= 0 at a stationary point
𝑑𝑥
A1M for
+ 8𝑦
𝑥 2 + 𝑥𝑦 + 4𝑦 2 − 60
𝑦
2
A1M for substitution
𝑦
𝑦
2
2
(− )2 + (− )𝑦 + 4𝑦 2 = 60
A1M for solving to find
𝑦 = ±4
A1M for 80m (40m also gets the mark)
6 (i)
A1M for using
𝑑𝑚
𝑑𝑡
for rate of change
𝑘
A1M for using 3
as the proportional increase
√𝑚
A1M for complete correct equation
𝑑𝑚
𝑑𝑡
=
𝑘
3
√𝑚
6 (ii)
A1M for stating that mass cannot increase indefinitely.
7 (i)
A1M for using the chain rule
A1M for
7 (ii)
𝑑𝑦
𝑑𝑥
= 4𝑠𝑒𝑐 2 𝑥 𝑒 4𝑡𝑎𝑛 𝑥
A1M for using the quotient rule
A1M for
𝑑𝑦
A1M for
𝑑𝑦
𝑑𝑥
𝑑𝑥
=
=
𝑥 2 𝑓(𝑥) − 𝑠𝑖𝑛 3𝑥 𝑔(𝑥)
𝑥6
3𝑥 𝑐𝑜𝑠(3𝑥) − 3𝑠𝑖𝑛(3𝑥)
𝑥4
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7 (iii)
A3M for
2𝑥 2
𝑑𝑦
𝑑𝑥
A1M for rearrangement
A1M for substitution
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
=
𝑑𝑥
=
− 4 = 0 (A1M for first three terms)
4−4𝑥𝑦
2𝑥 2 +6𝑦
4−4(1)(2)
2(1)2 +6(2)
A1M for equation of normal
8 (i)
𝑑𝑦
+ 4𝑥𝑦 + 6𝑦
=−
2
7
=
tangent gradient
7
7
3
2
2
2
𝑦 − 2 = (𝑥 − 1) and 𝑦 = 𝑥 −
1+3𝑥
A1M for √ 1−𝑥 = (1 + 3𝑥)0.5 × (1 − 𝑥)−0.5
A1M for (1 + 3𝑥)0.5 = 1 + 0.5(3𝑥) +
3
0.5 × −0.5
(3𝑥)2
2
9
+
0.5 × −0.5 × −1.5
(3𝑥)3 +…
6
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A1M for (1 + 3𝑥)0.5 = 1 + 2 𝑥 − 8 𝑥 2 + 16 𝑥 3
A1M for (1 − 𝑥)−0.5 = 1 + −0.5(−𝑥) +
−0.5 × −1.5 × −2.5
(−𝑥)3 +…
6
1
3
−0.5 × −1.5
(−𝑥)2
2
+
5
A1M for (1 − 𝑥)−0.5 = 1 + 2 𝑥 + 8 𝑥 2 + 16 𝑥 3
A1M for showing that (1 + 3𝑥)0.5 × (1 − 𝑥)−0.5 = 1 + 2𝑥 + 2𝑥 3
8 (ii)
A1M for stating that
1+3(2)
√ 1−(2) = √−7 which is not going to give a
solution. You cannot root a negative number and get a real answer.
8 (iii)
A1M for substitution
√
1
2
1
1−( )
2
1+3( )
= √5
1
2
1
2
A1M for substitution 1 + 2( ) + 2( )3
9
A1M for √5 ≈ 4
9 (i)
A1M for stating that
𝑓(𝑥) = 0
A1M for showing that
if
𝑥=−
1
3
is a factor
1
1
1
3
3
3
18(− )3 − 15(− )2 − (− ) + 2 = 0
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9 (ii)
A1M for obtaining any quadratic factor
A1M for obtaining a second linear factor
A1M for
9 (iii)
𝑓(𝑥) = (3𝑥 + 1)(2𝑥 − 1)(3𝑥 − 2)
A1M for rearrangement
18𝑠𝑒𝑐 3 𝑥 − 15𝑠𝑒𝑐 2 𝑥 − 𝑠𝑒𝑐 𝑥 + 2
A1M for
(3𝑠𝑒𝑐 𝑥 + 1)(2𝑠𝑒𝑐 𝑥 − 1)(3𝑠𝑒𝑐 𝑥 − 2) = 0
A1M for
𝑠𝑒𝑐 𝑥 = − , ,
1
1 2
3
2 3
A1M for stating that the range of
𝑠𝑒𝑐 𝑥 is (−∞,−1]∪[1,∞)
A1M for stating that the solutions found are between -1 and 1,
therefore outside the possible range of 𝑠𝑒𝑐 𝑥, so there are no
solutions
10 (i)
A1M for using h=0.5
A1M for correct form of the trapezium rule
A1M for area = 3.58887 to 5 decimal places.
10 (ii)
10 (iii)
A1M for stating that more strips would be needed
1
A4M for integration by parts
6
𝑥 3 𝑙𝑛𝑥 −
1
18
𝑥 3 − 𝑥 2 + 4𝑥
A1M for
1
1
1
1
[6 (3)3 𝑙𝑛3 − 18 (3)3 − (3)2 + 4(3)] − [6 (1)3 𝑙𝑛1 − 18 (1)3 − (1)2 + 4(1)]
A1M for
11 (i)
A1M for
9
2
𝑙𝑛3 −
13
9
; hence a = 9, b = 2, c = 3, d = -13/9.
(3𝑥 − 𝑞) = 0
(3(2) − 𝑞) = 0
and
therefore
A1M for substitution
0=
A1M for solving to find
(𝑥 + 2) = 0
at the asymptotes
𝑞=6
𝑝−2(5)
(3(5)−6)((5)+2)
𝑝 = 10
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11 (ii)
A1M for attempting to rewrite
A2M for
𝑦=
1
2(𝑥−2)
−
𝑦=
𝑝−2𝑥
(3𝑥−𝑞)(𝑥+2)
as partial fractions
7
6(𝑥+2)
A1M for attempting integration, with an upper limit of 5
A2M for integration
1
2
7
𝑙𝑜𝑔(𝑥 − 2) − 𝑙𝑜𝑔(𝑥 + 2)
6
1
7
1
7
2
6
2
6
A1M for [ 𝑙𝑜𝑔(5 − 2) − 𝑙𝑜𝑔(5 + 2)] − [ 𝑙𝑜𝑔(3 − 2) − 𝑙𝑜𝑔(3 + 2)]
A1M for 0.1568
12 (i)
A1M for volume equation
3500 = 𝜋𝑟 2 ℎ
A1M for surface area equation
and
3500
𝜋𝑟 2
= ℎ
𝑆 = 2𝜋𝑟 2 + 2𝜋𝑟ℎ
A1M for substitution
3500
𝜋𝑟 2
7000
𝑆 = 2𝜋𝑟 2 +
𝑟
𝑆 = 2𝜋𝑟 2 + 2𝜋𝑟
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12 (ii)
A2M for correct differentiation (A1M for each term)
𝑑𝑆
7000
= 4𝜋𝑟 −
𝑑𝑟
𝑟2
A1M for setting
𝑑𝑆
A1M solving for
𝑟
𝑑𝑟
= 0
4𝜋𝑟 =
𝑟3 =
7000
𝑟2
1750
𝜋
𝑟 = 8.23 𝑐𝑚
A1M for substitution to find
ℎ
3500
= ℎ
𝜋(8.23. . . )2
ℎ = 16.46 𝑐𝑚
12 (iii)
A1M for stating that the tube may be too large to hold at 16 cm wide,
not stack well with similar products on shelves or that the crisps may
be difficult to make and eat if they are also 16 cm wide.
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