AS Level / Year 1Edexcel Further Maths / CP1

2018 © crashMATHS Limited

March 2018 Mocks

27/03/18 v4final

Question Scheme AO Marks

1

(a) k(2k 3

k=1

n

) = 2 k2

k=1

n

3 kk=1

n

Use linearity AO1.1a M1

k(2k 3k=1

n

) = 2 n6

(n +1)(2n +1) 3 n2

(n +1)

Uses standard results AO1.1a

dM1

= 16

n(n +1) 2(2n +1) 9[ ] = 16

n(n +1)(4n 7) Extracts a factor of 16

n(n +1)

Complete and convincing proof

AO1.1a

AO2.1

dM1

A1 [4]

(b) k(2k 3)

k=n

3n

= k(2k 3)k=1

3n

k(2k 3)k=1

n 1

Correct partitioning

AO1.1a M1

= 16

(3n)(3n +1)(4(3n) 7) 16

(n 1)(n 1+1)(4(n 1)– 7) Uses their (a) to find an expression for

the sum. No need to simplify. ISW after any correct form has been

reached

AO1.1b A1ft [2]

= n2

(3n +1)(12n 7) n6

(n 1)(4n –11) [For illustration only, simplification not

necessary]

6

27/03/18 v4final

Question Scheme AO Marks

2

(a)

Enlargement Description AO1.2 B1

of scale factor 3 about the origin / (0,0) Description AO1.2 B1 [2]

(b) Q =

−1 0 0 1

⎛⎝⎜

⎞⎠⎟

Correct matrix Q AO1.2 B1

R =−1 0 0 1

⎛⎝⎜

⎞⎠⎟

3 00 3

⎛⎝⎜

⎞⎠⎟=

−3 0 0 3

⎛⎝⎜

⎞⎠⎟

Correct R ft their Q AO1.1b B1ft

[2]

(c) Method 1

Area of T = 12(8)(6) = 24 (units2)

Correct area of T AO1.1a M1

detR = –9 Attempts to find the determinant of their R

AO1.1a M1

So Area of T * = 24 × −9 = 216 (units2) Correct area of T* AO1.1b A1 [3]

(c) Method 2

D = (0,0), E = (–12, 18), F = (–24,0)

Attempts to find the coordinates of D, E and F under their R

AO1.1a M1

So area of T * = 12(24)(18) = 216 (units2)

Attempts to find area of T* Correct area

AO1.1a AO1.1b

M1 A1

[3]

7

27/03/18 v4final

Question Scheme AO Marks

3

Method 1

(α+ 2),(β + 2) and (γ + 2) satisfy

(w − 2)3 + p(w − 2)2 + 6(w − 2)+1= 0

Substitutes x = w − 2 AO3.1a M1

⇒w3 − 6w2 +12w − 8 + pw2 − 4 pw + 4 p + 6w −12 +1= 0

⇒w2 + (p − 6)w2 + (18 − 4 p)w + (4 p −19) = 0

Expands the brackets Correct simplified expression

AO1.1a AO1.1b

dM1 A1

Comparing coefficients gives p − 6 = −8⇒ p = −2

q = 18 − 4 p = 18 + 8 = 26

Compares coefficients Correct values of p and q

AO2.2 AO1.1b

M1 A1

[5]

Method 2 α+ β + γ = − p , aβ + βγ +αγ = 6 , αβγ = 1 (*) Seen or implied AO2.2 M1

−8 = (α+ 2)+ (β + 2)+ (γ + 2) = (α+ β + γ)+ 6⇒ 8 = − p + 6⇒ p = −2

Considers sum of the roots and forms

equation in p ft their (*) Correct value of p

AO3.1a

AO1.1b

dM1

A1

q = (α+ 2)(β + 2)+ (α+ 2)(γ + 2)+ (β + 2)(γ + 2)= αβ +αγ + βγ + 4(α+ β + γ)+12

⇒ q = 6 + 4(− − 2)+12⇒ q = 26

Considers pair sum and forms equation in terms of q ft their (*) and

their p Correct value of q

AO1.1a

AO1.1b

dM1

A1 [5]

5

27/03/18 v4final

Question Scheme AO Marks

4

(a)

62 + 3i

= 6(2 − 3i)(2 + 3i)(2 − 3i)

= 12 −18i13

Multiplies top and bottom by 2 – 3i

Correct denominator of 13 AO1.1a AO1.1b

M1 A1

= 1213

− 1813i

Correct simplification

AO1.1b A1

x + iy + 1213

− 1813i = (x − iy)(4 − i)

⇒ x + 1213

⎛⎝⎜

⎞⎠⎟ + i y −

1813

⎛⎝⎜

⎞⎠⎟ = (4x − y)− i(x + 4y)

Uses z*= x − iy and good attempt to combine real and imaginary parts

AO1.2 M1

So x + 1213

= 4x − y and y − 1813

= −x − 4y Compares coefficients correctly ft their 6

2 + 3i and attempts to solve for x or y

AO1.1a dM1

∴ x = 38, y = 21

104

Correct values of x and y AO1.1b A1 [6]

(b) r = 0.4259... Correct value of r ft their (a). See notes

AO1.2 B1

θ = tan−1 21 /104

3 / 8⎛⎝⎜

⎞⎠⎟ = 28.300...°

Method to find the argument. See notes

AO1.1a M1

So z = 0.43(cos(28.3)+ isin(28.3)) Correct expression AO1.1b A1 [3]

27/03/18 v4final

(c)

One point plotted correctly on Argand diagram

All points plotted correctly

(Points should ft their (a))

AO1.2 AO1.2

B1 B1

[2]

11

Question 4 Notes

(b) B1 – for the correct value of r ft their (a). Note that you can accept any degree of accuracy, exact answers or approximated answers

(these should be seen to at least two significant figures), but there must be some computation, so r = 38

⎛⎝⎜

⎞⎠⎟2

+ 21104

⎛⎝⎜

⎞⎠⎟2

alone is B0.

M1 – correct method to find the argument ft their (a). If their complex number is in another quadrant, then this does not matter, we only

need to see tan−1 yx

⎛⎝⎜

⎞⎠⎟

here.

A1 – correct expression. Note the argument must be given to at least two significant figures and you should accept other valid arguments. (c) B1 – one point plotted correctly on the Argand diagram, coordinates ft their (a). Accept complex numbers plotted as points or vectors. B1 – all three points plotted correctly on Argand diagram, coordinates ft their (a). Accept complex numbers plotted as points or vectors. The positions of all points should be relatively correct.

27/03/18 v4final

Question Scheme AO Marks

5

(a)

Let the two roots be α and β , then

α+ β = − b

a, αβ = c

a, αβ= mn

Correctly constrains the problem AO3.1a B1

⇒α2 = cm

an and

β2 = cn

am Finds expressions for α

2 and β2 AO2.1 M1

α2 + β2 = (α+ β)2 − 2αβ⇒ cm

an+ cnam

= b2

a2− 2ca

Attempts to use relation

α2 + β2 = (α+ β)2 − 2αβ

AO2.1 dM1

⇒ acm2 + acn2

a2mn= b

2 − 2aca2

⇒ acm2 + acn2 = mnb2 − 2acmn⇒ acm2 + 2acmn + acn2 = mnb2

⇒ ac(m2 + 2mn + n2 ) = mnb2

⇒ ac(m + n)2 = mnb2 *

Complete and convincing proof AO2.1 A1

[4]

(b)

k(−10)(3+ 2)2 = 3(2)82 ⇒ k = −192125

Uses condition in (a) with correct

values Correct value of k

AO2.2

AO1.1b

M1 A1

[2]

6

27/03/18 v4final

Question 5 Notes

(a) Note there may be alternatives to this question. In general, you should apply the following scheme: B1 – correctly constrains the problem, M1 – correctly expresses the roots in terms of a, b, c, m and n, dM1 – uses their expressions and a suitable relation to obtain an expression only in terms of a, b, c, m and n, A1 – complete and convincing proof

27/03/18 v4final

Question Scheme AO Marks

6

(a)

a = 3 Cao AO2.2 B1 [1]

(b)

Volume = (x 3

3

7)dx = x2

23x

3

7

Correct expression for the volume (condone omission of here and

ignore limits) Correct indefinite integration (condone

omission of here)

AO1.1a

AO1.1b

M1

A1

= 72

23(7) 32

2+ 3(3)

Substitutes limits into the expression the correct way around

AO1.1a dM1

= 8 = 25.1 327...( ) Correct answer oe

AO1.1b A1 [4]

(c/i)

28 – 12 = 16 cm Cao AO3.4

B1 [1]

(c/ii) 2 x 6 = 12 cm Answer is 12 cm, but condone 6 cm as q not clear

AO3.4

B1 [1]

27/03/18 v4final

(c/iii) 3f( 14 x) = 3 14 x − 3 Correct expression, seen or implied AO1.2 B1

Volume = 9π 1

4 x − 3( )12

28

∫ dx Sets up correct integral with correct limits ft their 3f(1/4x) and (c/i)

AO3.1b B1ft

= 9π 1

8(28)2 − 3(28)− 1

8(12)2 + 3(12)⎛

⎝⎜⎞⎠⎟

Correct method to integrate their volume expression definitely

AO1.1a M1

= 288π = 904.77 86...( ) Correct volume oe AO3.4 A1 [4]

(d)

idea that Bowl of the glass curves outwards and then inwards, while solid of revolution doesn’t change direction of curvature / dy/dx is never zero on the curve

Any reasonable description of the curvature issue

AO3.5b B1 [1]

12

27/03/18 v4final

Question Scheme AO Marks

7

(a)

0 2 −13 − 2 13 2 −1

= −23 13 −1

−13 − 23 2

Considers correct 3 x 3 determinant Expands the determinant

AO3.1a AO1.1a

M1 dM1

= −2(−3− 3)−1(6 + 6)= 12 −12 = 0

therefore the system has no solutions

Expands at least one 2 x 2 determinant correctly

Convincingly obtains 0 and gives a conclusion

AO1.1b

AO2.1

dM1 A1

[4]

(b) 3x − 4 = 1⇒ 3x = 5

3x + 4 = 5⇒ 3x = 1

so the system is not consistent

Attempts to eliminate one or two variables to show inconsistency

Convincing proof and conclusion

AO2.1 AO2.1

M1 A1

[2]

(c) (The planes defined by the equations) form a prism Description AO1.2 B1 [1]

7

Question 7 Notes

(b) 1st M1 – this is for an attempt to form two equations which are inconsistent, i.e. replacing 2y – z in the 2nd and 3rd equation (shown in scheme), subtracting 2nd and 3rd equations (to obtain 2y – z = 1), etc.

27/03/18 v4final

Question Scheme AO Marks

8

(a)

r =−420

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ + t

2−16

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

Correct expression oe. Accept row vectors in particular

AO1.2 B1 [1]

(b) e.g. Let t = –1, then

r =−420

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ −1

2−16

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ =

−63−6

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ ,so (–6, 3, –6) lies on the line

Convincing proof AO2.1 B1 [1]

(c) Method 1

120

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ ⋅

2−16

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ = 2 − 2 = 0

Considers dot product of v with their direction vector of the line

AO1.1a M1

120

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ ⋅

2−14

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ −

−63−6

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥=120

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ ⋅

8−410

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ = 8 − 8 = 0

so v is perpendicular to the plane

Dots v with a second vector in the plane

Shows v is perpendicular to two vectors in the plane

AO1.1a

AO2.1

M1

A1 [3]

27/03/18 v4final

(c) Method 2

i j k2 −1 68 − 4 10

= i−1 6−4 10

− j2 68 10

+ k2 −18 − 4

Considers cross product between two vectors in the plane and expands the

determinant correctly into 2x2 matrices

AO1.1a M1

i j k2 −1 68 − 4 10

= 14i − 28 j

Correctly does a cross product between two vectors in the plane

AO1.1a

M1

14i – 28j = 14(i – 2j), so v is perpendicular to the plane Convincing proof with conclusion AO2.1 A1 [3]

(d)

r ⋅120

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ = 0

LHS correct RHS correct

AO1.2 AO1.2

B1 B1

[2]

(e/i)

80−2

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ ⋅120

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ = 8 ≠ 0 (, so C does not lie in the plane)

Shows dot product isn’t 0. Condone no conclusion here

AO2.1 B1 [1]

(e/ii) Method 1

n̂ = 15

120

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

Correct unit normal, seen or implied AO1.1b B1

Shortest distance =

15

80−2

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ −

2−14

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥⋅120

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ = 1

56 + 2 = 8

5

Method to work out shortest distance Correct shortest distance

AO1.1a AO1.1b

M1 A1

[3]

27/03/18 v4final

(e/ii) Method 2

5 Correct normalisation factor, seen or implied

AO1.1a B1

Shortest distance=(8)(1)+ (0)(2)+ (−2)(0)

5= 8

5

Uses shortest distance formula with their normalisation factor and equation

of the plane Correct shortest distance

AO1.1a

AO1.1b

M1

A1 [3]

7

27/03/18 v4final

Question Scheme AO Marks

9

(a)

Let n = 1, then

LHS =

2 20 2

⎛⎝⎜

⎞⎠⎟

1

=2 20 2

⎛⎝⎜

⎞⎠⎟

RHS =

21 1× 21

0 21

⎛

⎝⎜⎞

⎠⎟=

2 20 2

⎛⎝⎜

⎞⎠⎟

So true for n = 1

Shows the statement is true for n = 1 AO2.1 B1

Assume true for n = k , i.e. 2 20 2

⎛⎝⎜

⎞⎠⎟

k

=2k k2k

0 2k

⎛

⎝⎜⎞

⎠⎟

Assumes true for n = k. Either can be shown explicitly or implied

AO2.1 M1

Then

2 20 2

⎛⎝⎜

⎞⎠⎟

k+1

=2 20 2

⎛⎝⎜

⎞⎠⎟

k 2 20 2

⎛⎝⎜

⎞⎠⎟

1

=2k k2k

0 2k

⎛

⎝⎜⎞

⎠⎟2 20 2

⎛⎝⎜

⎞⎠⎟

Begins inductive stage clearly using the assumption

AO2.1 dM1

=2k+1 2k+1 + k2k+1

0 2k+1

⎛

⎝⎜⎞

⎠⎟=

2k+1 (k +1)2k+1

0 2k+1

⎛

⎝⎜⎞

⎠⎟

Obtains the correct matrix for the case where n = k + 1 in the required form

AO2.1 A1

If true for n = k, then it has been shown that is is also true for n = k + 1. Since true for n = 1, it is true for all positive integers n.

Complete and convincing proof, with no errors seen and a complete

conclusion

AO2.4

A1 [5]

27/03/18 v4final

(b)

Let n = 2m {, then for m ≥ 0 ,}

2 20 2

⎛⎝⎜

⎞⎠⎟

2m

=22m 2m22m

0 2m

⎛

⎝⎜

⎞

⎠⎟

=22m 2m+2m

0 2m

⎛

⎝⎜

⎞

⎠⎟

Convincing proof. Need to use (a) and clearly show what they are doing

AO2.1

B1

[1]

6

27/03/18 v4final

Question Scheme AO Marks

10

(a/i) and (a/ii)

Circle in the correct quadrant Circle touches real axis

Straight vertical line Passes through 3

AO1.2 AO1.2 AO1.2 AO1.2

B1 B1 B1 B1

[4]

(b)

22 −12 = 3 Uses Pythagoras AO3.1a M1

z = 3+ i(2 + 3) , z = 3+ i(2 − 3) Correct complex solutions, one mark for each correct solution. Deduct 1 A

mark (max – 2) for any additional solutions given

AO2.2 AO2.2

A1 A1 [3]

(c)

k = 2, k = 6 Correct values of k, one mark for each correct value of k. Deduct 1 A mark

(max – 2) for any additional solutions given

AO2.2 AO2.2

B1 B1 [2]

9

27/03/18 v4final

Marks breakdown by AO

AO Number of marks %

AO1 48 61

AO2 23 24

AO3 9 15