# Edexcel - AS Pure Mathematics 15th May 2019 Unofficial Markscheme

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Thread starter 2 years ago
#1
It's a bit late, but here it is anyway! I will add more to the markscheme when I get the time. If you want any clarification or think a mistake was made, feel free to leave a message.

1) Straight line equations
l1: 2x + 4y -3 = 0
l2: y = mx + 7

a) Find m: (2 marks)

Rewrite l1 in the form y = mx + c to find the gradient of l1:
2x + 4y -3 = 0
so y = -1/2(x) + 3/4 where gradient of l1 is -1/2
Since l2 is perpendicular to l1, m is the negative reciprocal of the gradient of l1.

m = 2

b) l1 and l2 meet at P. Find the x coordinate of P (2 marks)

Set equations for l1 and l2 equal to each other:
2x + 7 = -1/2(x) + 3/4
Solve for x:
x = -5/2
Last edited by ~Fractal~; 2 years ago
0
Thread starter 2 years ago
#2
2) Find all real solutions for:

i) 16a2 = 2√a (4 marks)

256a4 = 4a (square both sides)

64a4 - a = 0

a(64a3 - 1) = 0

a = 0 or a = 1/4

ii) b4 + 7b2 - 18 = 0 (4 marks)

Treat like a quadratic and factorise:

(b+9)(b-2) = 0

But original equation is a quartic:

(b2+9)(b2-2) = 0

so b2+9 = 0 --> no real solutions

and b2-2 = 0 --> b = +√a or b = -√a
Last edited by ~Fractal~; 2 years ago
0
Thread starter 2 years ago
#3
3) Find: (3 marks)
a) ∫(4/x3 + kx)dx
= ∫(4x-3 + kx)dx
Then integrate using the power rule for integration:
= -2/x2 + kx2/2 + c

b) Hence find the value of k such that ∫20.5(4/x3 + kx)dx = 8 (3 marks)
[ -2/x2 + kx2/2 ]20.5 = 8
(-2/4 + 2k) - (-8 + k/8) = 8
15/2 + 15k/8 = 8
15k/8 = 1/2
k = 4/15
Last edited by ~Fractal~; 2 years ago
1
Thread starter 2 years ago
#4
4) Linear modelling:
A tree with height H metres was measured t years after planting.
3 years after planting, H = 2.35
6 years after planting, H = 3.28

a) Find an equation linking H with t (3 marks)
(3, 2.35) and (6, 3.28) must both lie on the straight line for the linear model.

Find the gradient of the line:
m = ∆y/∆x
m = (3.28 - 2.35)/(6-3) = 0.31
Use y - y1 = m(x - x1) using either (3, 2.35) or (6, 3.28)
H - 2.35 = 0.31(t - 3)
H = 0.31t + 1.42

The height of the tree was approximately 140cm when planted
b) Explain whether or not this supports the use of the linear model (2 marks)

The y intercept in the model predicts the initial height to be 1.42m or 142cm (as this occurs when t = 0).
This is very similar to 140cm, so use of the model is supported.
Last edited by ~Fractal~; 2 years ago
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Thread starter 2 years ago
#5
5) Differentiation and increasing functions
y = 3x2 + 24/x + 2

a) Find dy/dx (3 marks)
Rewrite equation as powers of x:
y = 3x2 + 24x-1 + 2
Differentiate using power rule for differentiation:
dy/dx = 6x - 24x-2

b) Hence find the range of values of x for which the curve is increasing (2 marks)
Increasing when dy/dx > 0
6x - 24/x2 > 0
6x > 24/x2
6x3 > 24
x3 > 4
x > ∛4
Last edited by ~Fractal~; 2 years ago
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Thread starter 2 years ago
#6
6) Non-right angled trigonometry
In triangle ABC, AC = 2x cm, AB = 3x cm and angle CAB = 60°

Given the area of the triangle is 18√3 cm2
a) show that x = 2√3 (3 marks)
Area of triangle given by 1/2(a)(b)sinC
1/2(2x)(3x)sin(60) = 18√3
3x2(√3/2) = 18√3
3x2√3 = 36√3
3x2 = 36
x2 = 12
x = 2√3

b) Hence find the exact length of BC (3 marks)
AC = 2(2√3) = 4√3 cm
AB = 3(2√3) = 6√3 cm

Use cosine rule:
BC2 = AC2 + AB2 - 2(AC)(AB)cosA
BC = √(4√3)2 + (6√3)2 - 2(4√3)(6√3)cos(60)
BC = 2√21 cm
1
Thread starter 2 years ago
#7
7) Reciprocal graph curve skecting and intersections
Curve has the equation y = k2/x + 1

a) Skecth the curve, indicating the horizontal asymptote (3 marks)

Horizontal asymptote at y = 1

Line l has equation y = -2x + 5
b) Show that the x coordinate of any point of intersection of l with the curve is given by 2x2 - 4x + k2 = 0 (2 marks)
Intersection occurs when -2x + 5 = k2/x + 1
-2x2 + 5x = k2 + x
Rearranging: 2x2 - 4x + k2 = 0

c) Hence find the exact values of k for which l is a tangent to the curve (3 marks)
The line is a tangent when the discriminant is equal to zero as there is only one point of intersection
b2 - 4ac = 0
Using 2x2 - 4x + k2 = 0 a = 2, b = -4, c = k2
16 - 4(2)(k2) = 0
16 - 8k2 = 0
k2 = 2
k = +√2 or k = -√2
0
Thread starter 2 years ago
#8
8) Binomial expansion
a) Find the first 3 terms in ascending powers of x of the binomial expansion of (2 + 3x/4)6 (4 marks)
Apply the standard method:
6C0(2)6(3x/4)0 + 6C1(2)5(3x/4)1 + 6C2(2)4(3x/4)2
= 64 + 144x + 135x2 +...

b) Explain how you could use your expansion to estimate the value of 1.9256. You do not need to perform the calculation. (1 mark)
Set 2 + 3x/4 = 1.925
so x = -0.1
Substitue x = -0.1 in the expansion to estimate 1.9256
1
Thread starter 2 years ago
#9
I'll finish the rest tomorrow 1
2 years ago
#10
Thank you , God bless
(Original post by ~Fractal~)
I'll finish the rest tomorrow 0
Thread starter 2 years ago
#11
9) Quadratic modelling
A company started mining tin on 1st January 2019
A model to find the total mass of tin is given by T = 1200 - 3(n - 20)2
where T is the total mass of tin mined in the n years after the start of mining

a) Calculate the mass of tin that will be mined up to 1st January 2020 (1 mark)
1 year has passed, so n = 1
T = 1200 - 3(1 - 20)2 = 117 tonnes

b) Deduce the maximum total mass of tin that can be mined (1 mark)
Maximum amount is given when n = 0 (before mining has begun)
= 1200 tonnes

c) Calculate the mass of tin that will be mined in 2023 (2 marks)
1st January 2023 is 5 years after 1st January 2019 (so n = 5).
To find the mass mined during this year, find the total mass mined from 2019 to 2022 and substract from the total mass mined from 2019 to 2023:
[1200 - 3(5 - 20)2] - [1200 - 3(4 - 20)2] = 93 tonnes

d) State, giving reasons, the limitation on the values of n (2 marks)
This is my best guess at what would be on the official markscheme
0 ≤ n ≤ 20
n cannot be negative since the company cannot mine before 1st January 2019.
n must be less than or equal to 20. After 20 years, all 1200 tonnes will have been extracted, so values of n above 20 are meaningless.
0
Thread starter 2 years ago
#12
10) Equation of a circle

x2 + y2 - 4x + 8y - 8 = 0
a) Find: (3 marks)
i) The coordinates of the centre of the circle

Rearrange original equation and complete the sqaure separately on the x and y terms:
x2 - 4x + y2 + 8y - 8 = 0
(x - 2)2 - 4 + (y + 4)2 - 16 - 8 = 0
(x - 2)2 + (y + 4)2 = 28

Centre is (2, -4)

ii) The exact radius of the circle
r2 = 28
r = 2√7

1 mark is likely to be given for putting the equation into the general form for the equation of a circle and the other 2 for correctly finding the centre and radius.

The straight line with equation x = k is a tangent to the circle.

b) Find the possible values for k (2 marks)

Two possible tangents.
The x coordinates of the tangent lines are the distance of the radius away from the centre of the circle, so k = 2 + 2√7 or k = 2 - 2√7
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Thread starter 2 years ago
#13
11) Factor theorem and algebraic division
f(x) = 2x3 - 13x2 + 8x + 48

a) Prove that (x - 4) is a factor of f(x) (2 marks)
If (x - 4) is a factor, f(4) = 0 by the factor theorem.
f(4) = 2(4)3 - 13(4)2 + 8(4) + 48 = 0

b) Hence, using algebra, show that f(x) = 0 has only two distinct roots (4 marks)
Use algebraic division or inspection to take out the factor of (x - 4) from f(x)

f(x) = (x - 4)(2x2 - 5x - 12)
Factorising the quadratic: f(x) = (x - 4)2(2x + 3)
So roots are x = 4 (repeated root) and x = -3/2
Therefore, only two distinct roots.

c) Deduce the number of real roots of the equation 2x3 - 13x2 + 8x + 46 = 0 (2 marks)
This curve is translated two units downwards, so the curve will now cross the x axis at 3 places, so will have 3 roots.

Given that k is constant and the curve y = f(x + k) passes through the origin
d) find two possible values of k (2 marks)
f(x + k) represent a translation of the graph by vector (-k, 0)
k = -3/2 or k = 4
Last edited by ~Fractal~; 2 years ago
1
Thread starter 2 years ago
#14
12) Using trigonometric identities

a) Show that
≡ 4 - 5cosθ (4 marks)

Use sin2θ + cos2θ ≡ 1 to write sin2θ in terms of cos2θ.
Treat numerator as a quadratic and factorise. Cancel out the common factors to obtain expression identical to RHS.

b) Hence or otherwise, solve for 0 ≤ θ < 360°
= 4 + 3sinθ (3 marks)

Rewrite as 4 - 5cosθ = 4 + 3sinθ
3sinθ + 5cosθ = 0
3tanθ + 5 = 0 (using tanθ = sinθ/cosθ identity)
tanθ = -5/3
θ = -59.03...° (not in range)
Use symmetry of y = tanθ graph or the cast diagram to determine all values of θ in the range 0 ≤ θ < 360°
θ = 180 - 59.03...° = 121.0° (1dp)
or θ = 360 - 59.03...° = 301.0° (1dp)
0
Thread starter 2 years ago
#15
13) Integration to find an area (7 marks)

Curve has equation y = 2x3 - 17x2 + 40x
The curve has a minimum turning point at x = k.
The region R is bounded by the curve, the x axis and line x = k.
Show that the area of R is 256/3

Differentiate
dy/dx = 6x2 - 34x + 40
Stationary points occur when dy/dx = 0
Solve 6x2 - 34x + 40 = 0
x = 5/3 or x = 4
The minimum turning point is further along the x axis than the maximum turning point, so the x coordinate at the minimum is 4.
k = 4
Area is given by definite integral with limits 4 and 0:
40(2x3 - 17x2 + 40x)dx
= [x4/2 - 17x3/3 + 20x2]40
= (4)4/2 -17(4)3/3 + 20(4)2 = 256/3
0
Thread starter 2 years ago
#16
14) Exponential model

The value of a car is modelled by the equation V = 15700e-0.25t + 2300 where t is the age in years.
a) Find the initial value of the car (1 mark)
t = 0,
V = 15700 + 2300 = £18000

Given the model predicts the value of the car is decreasing at a rate of £500 per year when t = T,
b) i) Show that 3925e-0.25T = 500

Since rate is mentioned, dV/dt will need to be found:
dV/dt = -3925e-0.25t
when t = T,
dV/dt = -500 (since value is decreasing by £500 per year)
so -3925e-0.25T = -500
Therefore, 3925e-0.25T = 500

ii) Hence find the age of the car at this instant, giving answer in years and months to the nearest month.
e-0.25T = 20/157
-0.25T = ln(20/157)
T = - 4(ln(20/157)) = 8.242...
12 months in 1 year, so 0.242... years is 0.242... x 12 = 2.904... months
So age of car is 8 years and 3 months (nearest month)

(6 marks) for b)
Most likely 3 for i) and 3 for ii)

The model predicts the value of the car approaches, but does not fall below £A.
c) State the value of A (1 mark)
As T gets large, V approaches £2300
So A = 2300

d) State a limitation of the model (1 mark)
Other factors will affect the price of the car such as mileage or condition which the model does not account for.
1
Thread starter 2 years ago
#17
15) Proof by deduction
Given that n ∈ ℕ, prove that n3 + 2 is not divisible by 8 (4 marks)

First prove conjecture is true for even numbers,
Let n = 2m
(2m)3 + 2 = 8m3 + 2
which is a number two greater than a multiple of 8, so if n is even, n3 + 2 is not divisible by 8.

Now prove conjecture is true for odd numbers,
Let n = 2r + 1
consider (2r + 1)3
= 8r3 + 12r2 + 6r + 1 (using binomal expansion)
= 2(4r3 + 6r2 + 3r) + 1
a multiple of two add 1 is always an odd number, so n3 is odd.
If n3 is odd, n3 + 2 is also odd.
Odd numbers are indivisible by 8.
Therefore, n3 + 2 is indivisible by 8 for all natural numbers (i.e. the set of positive integers)
1
Thread starter 2 years ago
#18
16) Vectors
a and b are non-zero vectors such that the magnitude of a + b = the magnitude of a + the magnitude of b
i) Explain, geometrically, the significance of this statement (1 mark)
The vectors a and b lie in the same direction.
You should get the mark for saying that a and b are parallel too.

ii) Two different vectors, m and n, are such that the magnitude of m = 3 and the magnitude of m - n = 6.
The angle between m and n is 30°.
Find the angle between vector m and vector m - n, in degrees to 1dp. (4 marks)

Drawing a vector diagram helps to understand what you're working with:

Use the sine rule to determine angle between vector n and m - n.
sin(30)/6 = sinθ/3
sinθ = 1/4
θ = 14.477...°
Since the vectors form a triangle, the remaining angle between m and m - n is given by 180 - 30 - 14.477... = 135.5° (1dp)
0
Thread starter 2 years ago
#19
Annnd finally done. Hope this helps some of you 4
2 years ago
#20
Thanks. I got my post moderated today because I placed a link (that I was sent by someone else) to the exam paper
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