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

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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

l

l

a) Find m: (2 marks)

Rewrite l

2x + 4y -3 = 0

so y = -1/2(x) + 3/4 where gradient of l

Since l

m = 2

b) l

Set equations for l

2x + 7 = -1/2(x) + 3/4

Solve for x:

x = -5/2

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

l

_{1}: 2x + 4y -3 = 0l

_{2}: y = mx + 7a) Find m: (2 marks)

Rewrite l

_{1}in the form y = mx + c to find the gradient of l_{1}:2x + 4y -3 = 0

so y = -1/2(x) + 3/4 where gradient of l

_{1}is -1/2Since l

_{2}is perpendicular to l_{1}, m is the negative reciprocal of the gradient of l_{1}.m = 2

b) l

_{1}and l_{2}meet at P. Find the x coordinate of P (2 marks)Set equations for l

_{1}and l_{2}equal to each other:2x + 7 = -1/2(x) + 3/4

Solve for x:

x = -5/2

Last edited by ~Fractal~; 2 years ago

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2) Find all real solutions for:

i) 16a

256a

64a

a(64a

a = 0 or a = 1/4

ii) b

Treat like a quadratic and factorise:

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

But original equation is a quartic:

(b

so b

and b

i) 16a

^{2}= 2√a (4 marks)256a

^{4}= 4a (square both sides)64a

^{4}- a = 0a(64a

^{3}- 1) = 0a = 0 or a = 1/4

ii) b

^{4}+ 7b^{2 }- 18 = 0 (4 marks)Treat like a quadratic and factorise:

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

But original equation is a quartic:

(b

^{2}+9)(b^{2}-2) = 0so b

^{2}+9 = 0 --> no real solutionsand b

^{2}-2 = 0 --> b = +√a or b = -√a
Last edited by ~Fractal~; 2 years ago

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3) Find: (3 marks)

a) ∫(4/x

= ∫(4x

Then integrate using the power rule for integration:

= -2/x

b) Hence find the value of k such that ∫

[ -2/x

(-2/4 + 2k) - (-8 + k/8) = 8

15/2 + 15k/8 = 8

15k/8 = 1/2

k = 4/15

a) ∫(4/x

^{3}+ kx)dx= ∫(4x

^{-3}+ kx)dxThen integrate using the power rule for integration:

= -2/x

^{2}+ kx^{2}/2 + cb) Hence find the value of k such that ∫

^{2}_{0.5}(4/x^{3}+ kx)dx = 8 (3 marks)[ -2/x

^{2}+ kx^{2}/2 ]^{2}_{0.5 }= 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

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 - y

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.

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 - y

_{1}= m(x - x_{1}) 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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5) Differentiation and increasing functions

y = 3x

a) Find dy/dx (3 marks)

Rewrite equation as powers of x:

y = 3x

Differentiate using power rule for differentiation:

dy/dx = 6x - 24x

b) Hence find the range of values of x for which the curve is increasing (2 marks)

Increasing when dy/dx > 0

6x - 24/x

6x > 24/x

6x

x

x > ∛4

y = 3x

^{2}+ 24/x + 2a) Find dy/dx (3 marks)

Rewrite equation as powers of x:

y = 3x

^{2}+ 24x^{-1}+ 2Differentiate 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/x

^{2}> 06x > 24/x

^{2}6x

^{3}> 24x

^{3}> 4x > ∛4

Last edited by ~Fractal~; 2 years ago

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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 cm

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

3x

3x

3x

x

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:

BC

BC = √(4√3)

BC = 2√21 cm

In triangle ABC, AC = 2x cm, AB = 3x cm and angle CAB = 60°

Given the area of the triangle is 18√3 cm

^{2}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

3x

^{2}(√3/2) = 18√33x

^{2}√3 = 36√33x

^{2}= 36x

^{2}= 12x = 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:

BC

^{2}= AC^{2}+ AB^{2}- 2(AC)(AB)cosABC = √(4√3)

^{2}+ (6√3)^{2}- 2(4√3)(6√3)cos(60)BC = 2√21 cm

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7) Reciprocal graph curve skecting and intersections

Curve has the equation y = k

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 2x

Intersection occurs when -2x + 5 = k

-2x

Rearranging: 2x

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

b

Using 2x

16 - 4(2)(k

16 - 8k

k

k = +√2 or k = -√2

Curve has the equation y = k

^{2}/x + 1a) 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 2x

^{2}- 4x + k^{2}= 0 (2 marks)Intersection occurs when -2x + 5 = k

^{2}/x + 1-2x

^{2}+ 5x = k^{2}+ xRearranging: 2x

^{2}- 4x + k^{2}= 0c) 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

b

^{2}- 4ac = 0Using 2x

^{2}- 4x + k^{2}= 0 a = 2, b = -4, c = k^{2}16 - 4(2)(k

^{2}) = 016 - 8k

^{2}= 0k

^{2}= 2k = +√2 or k = -√2

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8) Binomial expansion

a) Find the first 3 terms in ascending powers of x of the binomial expansion of (2 + 3x/4)

Apply the standard method:

= 64 + 144x + 135x

b) Explain how you could use your expansion to estimate the value of 1.925

Set 2 + 3x/4 = 1.925

so x = -0.1

Substitue x = -0.1 in the expansion to estimate 1.925

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:

^{6}C_{0}(2)^{6}(3x/4)^{0}+^{6}C_{1}(2)^{5}(3x/4)^{1}+^{6}C_{2}(2)^{4}(3x/4)^{2}= 64 + 144x + 135x

^{2}+...b) Explain how you could use your expansion to estimate the value of 1.925

^{6}.*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 1.925

^{6}
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9) Quadratic modelling

A company started mining tin on 1

A model to find the total mass of tin is given by T = 1200 - 3(n - 20)

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

1 year has passed, so n = 1

T = 1200 - 3(1 - 20)

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

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)

d) State, giving reasons, the limitation on the values of n (2 marks)

0 ≤ n ≤ 20

n cannot be negative since the company cannot mine before 1

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.

A company started mining tin on 1

^{st}January 2019A 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 tonnesb) 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 tonnesd) 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.

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10) Equation of a circle

x

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:

x

(x - 2)

(x - 2)

Centre is (2, -4)

ii) The exact radius of the circle

r

r = 2√7

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

x

^{2}+ y^{2}- 4x + 8y - 8 = 0a) 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:

x

^{2}- 4x + y^{2 }+ 8y - 8 = 0(x - 2)

^{2}- 4 + (y + 4)^{2}- 16 - 8 = 0(x - 2)

^{2}+ (y + 4)^{2}= 28Centre is (2, -4)

ii) The exact radius of the circle

r

^{2}= 28r = 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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11) Factor theorem and algebraic division

f(x) = 2x

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)

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)(2x

Factorising the quadratic: f(x) = (x - 4)

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 2x

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

f(x) = 2x

^{3}- 13x^{2}+ 8x + 48a) 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 = 0b) 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)(2x

^{2}- 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 2x

^{3}- 13x^{2}+ 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

12) Using trigonometric identities

a) Show that

Use sin

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°

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)

a) Show that

^{≡ 4 - 5cosθ }(4 marks)Use sin

^{2}θ + cos^{2}θ ≡ 1 to write sin^{2}θ in terms of cos^{2}θ.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)

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13) Integration to find an area (7 marks)

Curve has equation y = 2x

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 = 6x

Stationary points occur when dy/dx = 0

Solve 6x

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:

∫

= [x

= (4)

Curve has equation y = 2x

^{3}- 17x^{2}+ 40xThe 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 = 6x

^{2}- 34x + 40Stationary points occur when dy/dx = 0

Solve 6x

^{2}- 34x + 40 = 0x = 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:

∫

^{4}_{0}(2x^{3}- 17x^{2}+ 40x)dx= [x

^{4}/2 - 17x^{3}/3 + 20x^{2}]^{4}_{0}= (4)

^{4}/2 -17(4)^{3}/3 + 20(4)^{2}= 256/3
0

14) Exponential model

The value of a car is modelled by the equation V = 15700e

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

Since rate is mentioned, dV/dt will need to be found:

dV/dt = -3925e

when t = T,

dV/dt = -500 (since value is decreasing by £500 per year)

so -3925e

Therefore, 3925e

ii) Hence find the age of the car at this instant, giving answer in years and months to the nearest month.

e

-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)

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.

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}= 500Since 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}= -500Therefore, 3925e

^{-0.25T}= 500ii) 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

15) Proof by deduction

Given that n ∈ ℕ, prove that n

First prove conjecture is true for even numbers,

Let n = 2m

(2m)

which is a number two greater than a multiple of 8, so if n is even, n

Now prove conjecture is true for odd numbers,

Let n = 2r + 1

consider (2r + 1)

= 8r

= 2(4r

a multiple of two add 1 is always an odd number, so n

If n

Odd numbers are indivisible by 8.

Therefore, n

Given that n ∈ ℕ, prove that n

^{3}+ 2 is not divisible by 8 (4 marks)First prove conjecture is true for even numbers,

Let n = 2m

(2m)

^{3}+ 2 = 8m^{3}+ 2which is a number two greater than a multiple of 8, so if n is even, n

^{3}+ 2 is not divisible by 8.Now prove conjecture is true for odd numbers,

Let n = 2r + 1

consider (2r + 1)

^{3}= 8r

^{3}+ 12r^{2}+ 6r + 1 (using binomal expansion)= 2(4r

^{3}+ 6r^{2}+ 3r) + 1a multiple of two add 1 is always an odd number, so n

^{3}is odd.If n

^{3}is odd, n^{3}+ 2 is also odd.Odd numbers are indivisible by 8.

Therefore, n

^{3}+ 2 is indivisible by 8 for all natural numbers (i.e. the set of positive integers)
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16) Vectors

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

ii) Two different vectors,

The angle between

Find the angle between vector

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

Use the sine rule to determine angle between vector

sin(30)/6 = sinθ/3

sinθ = 1/4

θ = 14.477...°

Since the vectors form a triangle, the remaining angle between

**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)
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#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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