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Edexcel - AS Pure Mathematics 15th May 2019 Unofficial Markscheme

It's a bit late, but here it is anyway! :tongue:

:tongue:

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

(edited 4 years ago)

Scroll to see replies

OP

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

(edited 4 years ago)

OP

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 ∫²_0.5(4/x³ + kx)dx = 8 (3 marks)
[ -2/x² + kx²/2 ]²_0.5= 8
(-2/4 + 2k) - (-8 + k/8) = 8
15/2 + 15k/8 = 8
15k/8 = 1/2
k = 4/15

(edited 4 years ago)

OP

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.

(edited 4 years ago)

OP

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

(edited 4 years ago)

OP

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

OP

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

OP

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

OP

I'll finish the rest tomorrow :wink:

:wink:

Thank you , God bless

Original post by ~Fractal~

I'll finish the rest tomorrow :wink:

:wink:

OP

9) Quadratic modelling
A company started mining tin on 1^st 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 1^st 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)
1^st January 2023 is 5 years after 1^st 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 1^st 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.

OP

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

OP

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

(edited 4 years ago)

OP

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)

OP

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

OP

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.

OP

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)

OP

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)

OP

Annnd finally done. Hope this helps some of you :smile:

:smile:

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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