January 2016 Mark Scheme - WordPress.com · January 2016 WFM01 Further Pure Mathematics F1 Mark Scheme Question Number Scheme Notes Marks 1. (a) ... Correct equation in any form A1

January 2016 WFM01 Further Pure Mathematics F1

Mark Scheme

Question Number

Scheme Notes Marks

1. (a) 3 2i 1 i 3 3i 2i 2 At least 3 correct terms M1

5 i cao

(Correct answer only scores both marks)A1

(2)(b) w 1 i Understanding that w

1 i B1 z

w

3 2i

1 i

1 i

1 i

Multiplies top and bottom by the conjugate of the denominator

M1

3 3i 2i 2

11

5

2

1

2i

5

2

1

2i or 2.5 0.5i A1

(3)

(c) 3 2i k 53 3 k 2 4; 53

Substitutes for z and uses Pythagoras correctly. M1; Correct equation in any form A1

3 k 2 4 53 3 k 2

49 k

or

3 k 2 4 53 k 2 6k 40 0

k 4 k 10 0 k

dependent on

the previous M markAttempt to solve for k

dM1

{k = } 4, 10 Both {k } 4 , 10 A1 (4)

9

Question 1 Notes

1. (b) Note Alternative acceptable method: z

w

w

w

zw

w2

5 i

2

5

2

1

2i

(b) Note Give A0 for writing down 5 i

2 without reference to

5

2

1

2i or 2.5 0.5i

Note Give B0M0A0 for writing down

5

2

1

2i from no working in part (b).

Note Give B0M1A0 for

3 2i

1 i

1 i

1 i

Note Simplifying a correct

5

2

1

2i in part (b) to a final answer of 5 i is A0

(c) Note Give final A0 if a candidate rejects one of k 4 or k 10

(b) ALT

3 2i

1 i a bi B1;

3 2i (a bi)(1 i) 3 a b, 2 a b a ..., b ... for M1 and

5

2

1

2i for A1

MARK SCHEMES FOR RECENT EDEXCEL "IAL" PAPERS

Question Number

Scheme Notes Marks

2. f (x) x2 3

x

4

3x2

(a) f (1.6) 0.3325....

f (1.7) 0.1277....

Attempts to evaluate both f (1.6) and f (1.7) and either

f(1.6) = awrt 0.3 or f(1.7) = awrt 0.1M1

Sign change (positive, negative) (and f ( )x is

continuous) therefore (a root) is between

x 1.6 and x 1.7

Both f(1.6) = awrt 0.3 and f(1.7) = awrt 0.1, sign change and

conclusion.A1 cso

(2)

(b) f (x) 2x 3

2x

32

8

3x3

At least one of either

x2 Ax or 3

x Bx

3

2 or 4

3x2 Cx3

where A, B and C are non-zero constants.

M1

At least 2 differentiated terms are correct A1 Correct differentiation A1

dependent on the previous M mark Valid attempt at Newton-Rapshon using

their values of f (1.6) and f (1.6) dM1

1.672414... 1.672 dependent on all 4 previous marks

1.672 on their first iteration (Ignore any subsequent applications)

A1 cso cao

Correct derivative followed by correct answer scores full marks in (b) Correct answer with no working scores no marks in (b)

(5) 7

Question 2 Notes 2. (a) A1 correct solution only.

Candidate needs to state both f(1.6) = awrt 0.3 and f(1.7) = awrt 0.1 along with a reason and conclusion. Reference to change of sign or f (1.6) f (1.7) 0 or a diagram or < 0 and > 0 or one positive, one negative are sufficient reasons. There must be a (minimal, not incorrect) conclusion, eg. root is in between 1.6 and 1.7, hence root is in interval, QED and a square are all acceptable. Ignore the presence or absence of any reference to continuity. A minimal acceptable reason and conclusion is “change of sign, hence root”.

(b) Note Incorrect differentiation followed by their estimate of with no evidence of applying the NR formula is final dM0A0. Note If the answer is incorrect it must be clear that we must see evidence of both f (1.6) and f (1.6)

being used in the Newton-Raphson process. So that just 1.6 f 1.6 f 1.6 with an incorrect answer

and no other evidence scores M0.

Question Number

Scheme Notes Marks

3. x2 2x 3 0

(a) (i) 2, 3 Both 2, 3 B1

(ii) 2 2 2

2 ...... Use of a correct identity for 2 2

(May be implied by their work) M1

22 6 2 *

2 from a correct solution only A1 *

(iii) 3 3 3 3 ......

or 2 2 ...

Use of a correct identity for 3 3

(May be implied by their work) M1

8 3(3)(2) 10 or 2(2 3) 10

10 from a correct solution only A1

(5)

(b)(i) 2 2 2 2 2

4 2 2 4 2 2

4 4 Correct algebraic proof B1 *

(ii) Sum 3 3 10 2 12

Correct working without using explicit roots leading to a correct sum.

B1

Product 3 3 3 4 4 Attempts to expand

giving at least one termM1

3 ( 2 2 )2 2( )2

27 418 3 44 Correct product A1

x2 sum x product 0 x2 12x 44 0 Applying x2 – (sum)x + product M1

x2 12x 44 0 A1

(6) 11

Question 3 Notes

(a) (i) 1st A1 2, 3 2 2 4 6 2 is M1A0 cso

(b) (ii) 1st A1 2, 3 3 4 4 44 is first M1A1

(a) Note Applying 1 2 i, 1 2 i explicitly in part (a) will score B0M0A0M0A0 (b) Note Applying 1 2 i, 1 2 i explicitly in part (b) will score a maximum of B1B0M0A0M1A0

(a) Note Finding 2, 3 by writing down or applying 1 2 i, 1 2 i but then writing

2 2 2

2 22 6 2 and 3 3 3 3 8 3(3)(2) 10

scores B0M1A0M1A0 in part (a). Such candidates will be able to score all marks in part (b) if they use the method as detailed on the scheme in part (b).

(b)(ii) Note A correct method leading to a candidate stating p 1, q 12, r 44 without writing a final

answer of x2 12x 44 0 is final M1A0

Question Number

Scheme Notes Marks

4. (a) Rotation Rotation B1

225 degrees (anticlockwise) 225 degrees or 54 (anticlockwise)

or 135 degrees clockwiseB1 o.e.

about (0, 0)

This mark is dependent on at least one of the previous B marks being awarded.

About (0, 0) or about O or about the origindB1

Note: Give 2nd B0 for 225 degrees clockwise (3)

(b) n 8 8 B1 cao

(1)(c)

Way 1 A-1 1

2 1

2

12

12

or 2

2 22

22 2

2

Correct matrix B1

B CA-1 2 4

3 5

12

12

12

12

...

Attempts CA-1 and finds at least one element

of the matrix BM1

2 3 2

2 4 2

dependent on the previous B1M1 marks

At least 2 correct elementsA1

All elements are correct A1

(4)

(c) Way 2

BA a b

c d

12

12

12

12

2 4

3 5

Correct statement using 2 2 matrices.

All 3 matrices must contain four elements. (Can be implied).

(Allow one slip in copying down C)

B1

a

2

b

2 2,

a

2

b

2 4 or

c

2

d

2 3,

c

2

d

2 5

and finds at least one of either a or b or c or d

Applies BA C and attempts simultaneous equations in a and b

or c and d and finds at least one of either a or b or c or d

M1

2 3 2

2 4 2

or a 2, b 3 2, c 2, d 4 2

dependent on the previous B1M1 marksAt least 2 correct elements

A1

All elements are correct A1

(4) 8 Question 4 Notes

4. (a) Note Condone “Turn” for the 1st B1 mark. (c) Note You can ignore previous working prior to a candidate finding CA-1

(i.e. you can ignore the statements C BA or C AB ).

A1 A1 You can allow equivalent matrices/values, e.g.

2

2

6

2

2

2

8

2

Question Number

Scheme Notes Marks

5. (a) 8r3 3rr1

n

8

1

4n2 n1 2

3

1

2n n1

Attempt to substitute at least one of the standard formulae correctly into the

given expressionM1

Correct expression A1

1

2n n 1 4n n1 3

dependent on the previous M mark Attempt to factorise at least n(n + 1) having

used both standard formulae correctlydM1

1

2n n1 4n2 4n 3 {this step does not have to be written}

1

2n n1 2n 3 2n1 Correct completion with no errors A1 cso

(4)

(b) Let f (n) 1

2n n 1 2n 3 2n1 , g(n)

8

4n2(n 1)2 & h(n)

3

2n(n1)

8r3 3rr5

10

1

2(10)(11)(23)(19)

1

2(4)(5)(11)(7)

24035 770 23265

Attempts to find either f (10) and f (4) or f (5)

g(10) and g(4) or g(5)

and h(10) and h(4) or h(5)

M1

kr 2

r5

10

k1

6(10)(11)(21)

1

6(4)(5)(9)

k(385 30) 355k

or k 52 62 72 82 92 102 355k

Correct attempt

at kr 2

r5

10

M1

23265 355k 22768 k 497

355or

7

5

dependent on both previous M marks.Uses both previous method mark results to

form a linear equation in k using 22768 and solves to give k ...

ddM1

k 497

355or

7

5or 1.4 or equivalent A1 o.e.

(4) 8 Question 5 Notes

5. (a) Note Applying eg. n 1, n 2 to the printed equation without applying the standard formula to give a = 2, b = 1 is M0A0M0A0 Alt Alternative Method: Using 2n4 4n3 1

2 n2 32 n an4 (b 5

2 a)n3 ( 52 b 3

2 a)n2 32 bn o.e.

dM1 Equating coefficients to give both a = 2, b = 1 A1 cso Demonstrates that the identity works for all of its terms

(b) Note f (10) f (5) 1

2(10)(11)(23)(19)

1

2(5)(6)(13)(9) 24035 1755 22280

Note Applying 8r 3 3r k

r 5

10

r5

10

r 2

r5

10

gives either

(24200 165 385k) (800 30 30k) 22768 23400135 355k 22768 Note 985 25k 1710 36k 2723 49k 4072 64k 580581k 7970 100k 23265 355k

is fine for the first two M1M1 marks with the final ddM1A1 leading to k 1.4

Question Number

Scheme Notes Marks

6. (a) y c2

x c2x1

dy

dx c2x2

c2

x2

2d

d

yk x

x

M1

2 d0

d

yxy c x y

x Correct use of product rule. The sum of

two terms, one of which is correct.

2

d d d 1. .

d d d

y y p c

x p x p c

their d 1dd theird

yxpp

2 2

2

d d d 1or 0 or .

d d d

y y y cc x x y

x x x p c Correct differentiation A1

2S o , Nm p

Perpendicular gradient rule where

mN

( mT

) is found from using calculus.M1

2c

y p x cpp

or y p2x c

p cp3

Correct line method

where mN is found from using calculus.

M1

3 41 *py p x c p A1*

(5)(b)

y

c2

x p

c2

x p3x c 1 p4 or x

c2

y py p3 c2

y c 1 p4

Substitutes y c2

p or x

c2

y into the printed equation

to obtain an equation in either x, c and p only or in y, c and p only.

M1

p3x2 c 1 p4 x c2 p 0 or py2 c 1 p4 y c2 p3 0

x cp p3x c 0 x ... or y cp

yp cp4 0 y ...

Correct attempt of solving a 3TQ to find the x or y coordinate of Q

M1

Q c

p3, cp3

Can be simplified or

un-simplified.

At least one correct coordinate. A1

Both correct coordinates A1

Note: If Q is stated as coordinates then they must be correct for the final A1 mark. (4)

(b) ALT

Let Q be cq,c

q

soc

qp p3cq c 1 p4

Substitutes x cq or y cq into the printed equation to obtain an equation in only p, c and q.

M1

cp p3cq2 cq cqp4 p q p3q2 qp4 0

p q 1 p3q 0 q ...

Correct attempt to find q in terms of p M1

Q

c

p3, cp3

Can be simplified or

un-simplified.

At least one correct coordinate A1

Both correct coordinates A1

(4)

9

Question Number Scheme Notes

Marks

7. f (x) x4 3x3 15x2 99x 130(a) 3 – 2i is also a root 3 – 2i B1

x2 6x + 13

Attempt to expand (x (3 2i))(x (3 2i)) or any valid method to establish the quadratic factor

e.g. x 3 2i x 3 2i x2 6x 9 4 or sum of roots 6, product of roots 13

M1

x2 6x 13 A1

f (x) x2 6x 13 x2 3x 10 Attempt other quadratic factor.

Note: Using long division to get as far as x2 kx is fine for this mark.

M1

2 3 10x x A1 x2 3x 10 x 5 x 2 x ... Correct method for solving a 3TQ

on their 2nd quadratic factorM1

x 5, x 2 Both values correct A1

(7) Note: Writing down 2, 5, 3 2i, 3 2i with no working is B1M0A0M0A0M0A0

(a) Alternative using Factor Theorem 3 – 2i

3 – 2i B1

f (2) 24 3 23 15 22 99 2130 0

Attempts to find f (2) M1 Shows that f (2) 0 A1

f (5) 5 4 3 5 3

15 5 2 99 5 130 0 Attempts to find f (5) M1

Shows that f (5) 0 A1

x 2, x 5

Either shows that f (2) 0 and states x 2

or shows that f (5) 0 and states x 5M1

Shows both f (2) 0 & f (5) 0

and states both x 5, x 2A1

(7)

(b)

3 2i plotted correctly in quadrants 1 and 4 with some evidence of symmetry

dependent on the final M mark being awarded in part (a). Their other two roots plotted correctly.

Satisfies at least one of the criteria.

B1ft

Satisfies both criteria with some indication of scale or coordinates stated. All points (arrows) must

be in the correct positions relative to each other.

B1ft

(2) 9

Question Number

Scheme Notes Marks

8.

S(a ,0), B(q,r), C a, 2ar

q a

or C(a , 3ar)

(a) m r 0

q a

Correct gradient using (a, 0) and (q, r) (Can be implied)

B1

y r

q ax a or

Correct straight line method

M1 y r

r

q ax q

0 ra

q a "c" "c"

ra

q aand y

r

q ax

ra

q a

leading to (q a)y r(x a)* cso A1* (3)

(b) C a , 2ar

q a

or height OCS 2ar

q a

2ar

q a or 2ar

q aB1

2arq a

3r or12

(a)2arq a

3

12

(a)(r) ...

Applies height OCS 3r or applies Area(OSC) 3Area(OSB)

and rearranges to give a q

where , are numerical values.

M1

5a 3q 5a 3q or a

3

5q A1

Area(OBC) 4

1

2

3q

5

r

or 1

2

3q

5

r 3

2

3q

5

r

dependent on the previous M mark

Uses their a 3

5q and applies a correct

method to find Area(OBC) in terms of only q and r

dM1

6

5qr (*)

6

5qr A1* cso

(5) 8 Alternative Method (Similar Triangles)

(b) 3r

2a

r

q a 3r

2a

r

q a or equivalent B1

3r

2a

r

q a ...

3r

2a

r

q a or equivalent and rearranges

to give a q where , are numerical values.M1

… then apply the original mark scheme. Question 8 Notes

8. (a) Note The first two marks B1M1 can be gained together by applying the formula y y

1

y2 y

1

x x

1

x2 x

1

to give y 0

r 0

x a

q a

(b) Note If a candidate uses either 2ar

q a or 3r they can get 1st M1 but not 2nd M1 in (b).

Question Number

Scheme Notes Marks

9. f (n) 4n1 52n1

f (1) 42 5 21 f(1) = 21 is the minimum B1

f (k 1) f (k) 4k2 52(k1)1 (4k1 52k1) Attempts f (k 1) f (k) M1

f (k 1) f (k) 3(4k1) 24(52k1)

3(4k1 52k1) 21(52k1)

or 24(4k1 52k1) 21(4k1)Either

3(4k1 52k1)or 3f (k); 21(52k1)A1; A1

24(4k1 52k1)or 24f (k) ;21(4k1)

f (k 1) 3f (k) 21(52k1) f (k)

or f (k 1) 24f (k) 21(4k1) f (k)

dependent on at least one of the previous accuracy marks being awarded.

Makes f (k 1) the subjectdM1

If the result is true for n k,then it is true for n k 1, As the result has been shown to be

true for n 1,then the result is is true for all n» . A1 cso

(6) 6

WAY 2 General Method: Using f (k 1) mf (k) f (1) 42 5 21 f(1) = 21 is the minimum B1

f (k 1) mf (k) 4k2 52(k1)1 m(4k1 52k1) Attempts f (k 1) f (k) M1

f (k 1) mf (k) (4 m)(4k1) (25 m)(52k1)

(4 m)(4k1 52k1) 21(52k1)

or (25 m)(4k1 52k1) 21(4k1)

(4 m)(4k1 52k1) or (4 m)f (k); 21(52k1)A1; A1

(25 m)(4k1 52k1) or (25 m)f (k) ;21(4k1)

f (k 1) (4 m)f (k) 21(52k1) mf (k)

or f (k 1) (25 m)f (k) 21(4k1) mf (k)




true for n 1,then the result is is true for all n» .A1 cso

WAY 3 f (1) 42 5 21 f(1) = 21 is the minimum B1

f (k 1) 4k2 52(k1)1

Attempts f (k 1) M1

f (k 1) 4(4k1) 25(52k1)

4(4k1 52k1) 21(52k1)

or 25(4k1 52k1) 21(4k1)Either

4(4k1 52k1)or 4f (k) ; 21(52k1)A1; A1

25(4k1 52k1)or 25f (k);21(4k1)

f (k 1) 4f (k) 21(52k1)

or f (k 1) 25f (k) 21(4k1)




true for n 1,then the result is is true for all n» .A1 cso

Note Some candidates may set f (k) 21M and so may prove the following general results

June 2015 Further Pure Mathematics F1 WFM01

Mark Scheme

Question Number

Scheme

Marks

1.(a) 3 2 22 5 7 6 (2 3)( )z z z z z az b

a = 1 and b = 2 B1: One of a = 1 or b = 2

B1 B1 B1: Both a = 1 and b = 2

Values may be implied by a correct quadratic e.g. sight of 2 2z z

(2)

(b) z = 121 z = 1.5 or equivalent B1

1 12 2 7 iz

M1: Solves their 3 term quadratic (usual rules) as far as z = ...

M1A1 A1: Allow 1 i 7

2z

or equivalent

e.g. 12

7i

4z

Answers must be exact and accept correct answers only for both marks. Answers that are not exact with no working score M0A0

(3) [5 marks]

Question Number

Scheme

Marks

2. 2 2(3 2) 9 12 4r r r Correct expansion B1

2 2

1 1 1 1

(3 2) 9 12 4n n n n

r r r r

r r r

9 ( 1)(2 1) 12 ( 1) 46 2

n nn n n n

B1ft: “4” = “4”n

B1ft M1 M1: Uses valid formulae for sum of squares and sum of integers (their 9 or 12 may be followed through from their coefficients)

3( 1)(2 1) 12( 1) 82

or

9( 1)(2 1) 36( 1) 246

nn n n

nn n n

Takes out factor 2

n or

6

n. Dependent

on the B1ft having been scored. dM1

26 3 12

nn n

Correct result or states a = 6, b = 3,

c = 1 A1

(5) You should always award marks as in the scheme but generally there are no

marks for proof by induction [5 marks]

Question Number

Scheme

Marks

3.(a) 7

2 and 2 Allow

4

2for 2 B1

2 2 2( ) 2

2

7 332 2

2 4

M1: Uses 2 2 2( ) 2 M1 A1

A1: 33 1

or 8 or 8.254 4

(3)(b) Sum of roots is

2 2 33 3342 8

and product of roots is 1

M1: Attempts sum or product of new roots correctly (may be implied)

M1 A1

A1: Sum = 33

8 and product = 1

2 2331 0 8 33 8 0

8x x x x

28 33 8 0x x or any integer multiple including the “= 0”

A1

(3) [6 marks]

Alternative – finds roots explicitly: (a) 1

, 7 174

Correct exact roots including 17 B1

2 2 49 17 66 332 2 2

16 16 16 4

M1: Squares and adds their roots M1 A1 A1: cao

33 1or 8 or 8.25

4 4

(3)(b)

7 17 7 17...

7 17 7 17x x

Uses x x

with numerical

and

and attempts to expand. There are

no marks until numerical values are used.

M1

2 331

8x x A1

28 33 8 0x x This answer with no errors or any integer multiple including the “= 0”

A1cso

(3)Note:

Roots of the form 17 17 , 4k

k will give a correct answer – in this case lose

the final mark as not cso.

Question Number

Scheme

Marks

4.(a) (PQ =) 13 Sight of 13 (Must be seen in (a)) B1

(1)(b) 9 13 ...a a

or 2(9 ) 36 169 ...a a a

M1: Uses 9 13a or 2(9 ) 36 169a a

to obtain a value for a M1 A1 A1: 4a only

(2)(c)

y = 12 Correct y coordinate of P. M1

Uses Area of triangle = 12 13 " "y or

12

4 9 4 4

0y y y

A correct triangle area method M1

= 78 cao A1 (3)

Alternative method for area of triangle using midpoint of QS (M) Area = 1 1

2 2 208 117QS MP

The method for QS and MP must be correct for their values There are other methods for the area and the method should be correct for their

values to score the M1 e.g. Box – Triangles = 156 – 48 – 30 = 78 [6 marks]

Question Number

Scheme Marks

5.(a)

f (2) ... and f (3) ...

Attempts to evaluate both f (2) and f (3) (ignore use of degrees for this mark) NB degrees usually scores M1A0M0A0 NB f (2) 2 and f (3) 3 for degrees

M1

f (2) 2.3... , f (3) 1.4... Needs accuracy to 1 figure truncated or rounded

A1

f(2.5) = 0.5….and f(2.75)= -0.4…. Evaluates both f(2.5) and f(2.75) (and not f(2.25))

M1

(2.5, 2.75)

2.5 2.75x or 2.5 2.75x or 2.5 2.75 or

2.5 2.75 or 2.5, 2.75or 2.5, 2.75 or equivalent in words. Allow a mixture of ‘ends’ but not incorrect statements such as 2.75 2.5x . Needs accuracy to 1 figure truncated or rounded for f(2.5) and f(2.75) and conclusion

A1

(4) Note that some candidates only indicate the sign of f not its value. In this case the

M’s can still score as defined but not the A’s. However if f(2) and f(3) are correctly evaluated in (b) then the first A1 can be given retrospectively.

Common Approach in the form of a table:

a f(a) b f(b) 2a b 2f a b

2 2.31576... 3 -1.428... 2.5 0.5151... 2.5 0.5151... 3 -1.428... 2.75 -0.4472...

2.5 2.75 Would score full marks in (a)

(b) 2 3

2.3158 1.4280

or

2 3 2

2.3158 3.7438

Correct equation involving or x and their values even in degrees. Use of negative lengths scores M0

M1

(1.4280 2.3158) 3 2.3158 2 1.4280 so ...

Makes or x the subject. Dependent on the previous M but condone poor algebra.

dM1

2.62 cao and cso (Allow x = ) A1

A correct statement followed by 2.62 scores 3/3 (3)

Using y = mx + c: f (2) f (3) 3.74.....m

f (2) 2 9.80...c m Correct method to find equation of straight line

M1

0 ...y x Substitutes y = 0 and makes x or the subject. Dependent on the previous M

dM1

2.62 cao and cso (Allow x = ) A1

Also allow candidates to find the value of e.g 3 or 2 and then add to 2 or subtract from 3: M1 for a correct method for 3 or 2 , dM1 for adding to 2

or subtracting from 3 and A1 for 2.62 cao and cso.

[7 marks]

Question Number

Scheme Marks

6(a) Gradient of PQ is

6 6 1

6 6p q

p q pq

Correct gradient in any form B1

Equation of PQ is 6 66

66 6

p qy x qq p q

M1: Uses straight line equation in any form correctly for their gradient or uses y = mx + c and attempts to find c in terms of p and q

M1 A1

A1: Correct line in any form

6 16y x q

q pq

6( ) ( 6 ) 6( )*qpq y x q pqy x p q Cso. Reaches the given answer with at least one intermediate step.

A1*

(4)

Alternative simultaneous equations 6 6

(6 ) , (6 )m p c m q cp q Correct equations B1

6 6

,q p

m cpq p q p q

M1: Solves simultaneously to obtain either “m” or “c” in terms of p and q

M1A1 A1:

6 6q p

m cpq p q p q

and

6( )*pqy x p q Cso. Reaches the given answer with at least one intermediate step.

A1*

(b)

236 d36

d

yy x

x x and uses x = 6r

d d0

d d

y y yx y

x x x and use x = 6r, y = 6/r

2

d d 66

d d

x y

r r r and uses

d d d

d d d

y y x

x r r

2d

d

ykx

x and uses x = 6r

M1

d

d

y yk

x x and uses x = 6r, y = 6/r

2

d d

d d

x y kk

r r r and uses

d d d

d d d

y y x

x r r

So at R gradient of curve =

2

1

r

Allow any un-simplified correct form

e.g. 6

2

2

636 6 , , 6

6rrr r

A1

So gradient of normal = 2r Correct use of perpendicular gradient rule

M1

1 11

pr qr M1: Uses gradient PR perpendicular to gradient QR

M1 A1 A1: Correct equation connecting p, q and r

So 2 1r

pq

which is the gradient of PQ so the

normal at R is parallel to PQ

Conclusion with all previous marks scored. Must see the word ‘parallel’ used.

A1cso

(6) [10 marks]

Question Number

Scheme Marks

7.(a) 13z k

Accept 213k but not 2 29 4k k

B1

argz = 2

arctan 0.5883

= 3.73 or 2.55

M1: Uses

2arctan 0.588 ... / 33.6 ...

3

3or arctan 0.98 ... / 56.3 ...

2

c o

c o

M1 A1

A1: 3.73 or 2.55 only

(3)(b)(i)

4 4 2i

3 2 iz k k k

M1: Substitutes z and multiplies numerator and denominator by conjugate of denominator or equivalent

M1 A1 A1: 2

ik

oe (Allow un-simplified e.g. 2

8i

4

k

k). Allow

20 i

k

(ii) 2 2 2 2 2( 3 2 i)( 3 2 i) 9 12ik 4iz k k k k k k Multiplies out obtaining 3 term quadratic in i M1

2 25 12 ik k M1: Uses 2i = 1 (may be implied) M1A1 A1: cao

(5)(c)

Plots z in 3rd quadrant and z* as mirror image in 2nd quadrant and both correctly labelled

B1

Plots a complex number on positive imaginary axis and correctly labelled B1

Plots and labels D in the first quadrant, positioned correctly relative to the other points and further from the origin than all the other points.

B1

Notes: 1. Penalise the omission of labels once and penalise it the first time it occurs. 2. For labels allow letters, in terms of z, coordinates or labels on axes. 3. If there are separate Argand Diagrams, imagine them superimposed. 4. Accept points, lines or arrows.

(3)

(3) [11 marks]

I

R

D or z2

B or z*

A or z

C

Question Number

Scheme Marks

8.(a)

1P = 2

3 41

4 325

a a

a aa

or

3 41

4 325a

M1: Switches signs on minor diagonal

M1 B1 A1

B1: Correct determinant. Allow simplified or un-simplified e.g. 3a(3a)-(-4a)(4a), score when first seen. A1: Completely correct inverse with determinant simplified.

(3)(b) 3 4 3 6 201

4 3 4 8 1525

a a a

a a aa

Sets up correct multiplication

including 1

25aor equivalent M1

1 2 0

0 0 5

Correct matrix A1

( 1,0), (2,0) and (0,5) Follow through their matrix but must be written as coordinates

A1ft

(3)(c)

Area of triangle 1T = 12 3 5 o.e.

15

2

Correct area for triangle 1T . M1

Area scale factor is 225a

so Area of triangle 2T = 2 21525 187.5

2a a oe

M1: Multiplies their area of 1T by their detP to find required area M1A1

A1: cao (3)

Alternative 1: Shoelace method

area 2T = 1

2

3 6 20 3

4 8 15 4

a a a a

a a a a

Correct statement. M1

12

3 8 6 15 20 4

4 6 8 20 15 3

a a a a a a

a a a a a a

Correct calculation M1

2187.5a oe cao A1

Alternative 2: Encloses T2 by a rectangle and subtracts triangles: Rectangle area = 494a2 and one triangle area of

161.5a2, 91a2 or 54a2 Correct values M1

2 2 2 2494 161.5 91 54a a a a Complete method for area M1 2187.5a oe cao A1

(d)

Q = 3 45 5

345 5

M1: 35

35

3

5 in both entries of main diagonal and

0and 0 M1A1

A1: Correct matrix (2)

(e)

3 45 5

345 5

R

3 4

4 3

a a

a a

=

5 0

0 5

a

a

oe

M1: Sets up correct multiplication in correct order. “Their Q” P M1 A1 A1: cao

(2) [13 marks]

Question Number

Scheme Marks

9.(i) If n =1, 2 2

1

1(2 1) 1 and ( 1)(3 1) 1

6

n

r

r r n n n n

, LHS=RHS so true for n = 1. B1

12 2 2

1

1(2 1) ( 1)(3 1) ( 1) (2( 1) 1)

6

k

r

r r k k k k k k

(Adds the (k + 1)th term to the sum of the first k terms)

M1

3 21

( 1)(3 13 17 6)6

k k k k dM1: Attempt factor of 1

6 ( 1)k dM1A1

A1: 3 216 ( 1)(3 13 17 6)k k k k

216 ( 1)( 2)(3 7 3)k k k k 21

6 ( 1)( 2)(3( 1) ( 1) 1)k k k k A1

Achieves this result with no errors and 23 7 3k k seen Allow work that shows equivalence between

e.g. 3 216 ( 1)(3 13 17 6)k k k k and 21

6 ( 1)( 2)(3( 1) ( 1) 1)k k k k

True for n = k + 1 if true for n = k, and as true for n = 1 true by induction for all n. A1cso Full conclusion and all previous marks scored (6)

(ii) n = 1:

17 12 6 1 12

3 5 3 1 6

so true for n = 1 Shows true for n = 1 B1

17 12 6 1 12 7 12 7 12 6 1 12

or3 5 3 1 6 3 5 3 5 3 1 6

kk k k k

k k k k

M1

Either statement scores M1 6 7 12 12

3 3 6 5

k k

k k

or e.g.

7 6 1 3 12 12 6 1 12 5

3 7 3 1 6 12 3 1 6 5

k k k k

k k k k

M1: Correct attempt at multiplication (if unclear, at least 2 terms must be correct)

M1A1 A1: Correct matrix possibly un-simplified

If the previous A1 was awarded for 6 7 12 12

3 3 6 5

k k

k k

then allow the next A

mark for the matrix as shown. If the previous A1 was awarded for e.g.

7 6 1 3 12 12 6 1 12 5

3 7 3 1 6 12 3 1 6 5

k k k k

k k k k

then this must be simplified to

6 7 12 12

3 3 6 5

k k

k k

before the next A mark can be awarded.

6( 1) 1 12( 1)

3( 1) 1 6( 1)

k k

k k

States or shows by equivalence that the result is true for n = k + 1

A1

True for n = k + 1 if true for n = k, and as true for n = 1 true by induction for all n. A1

Full conclusion and all previous marks scored (6) [12 marks]

Question Number

Scheme Notes

Marks 1. 4 3 2f ( ) 9 29 60x x x x x

1 – 2i is also a root Seen anywhere B1

x2 2x + 5

M1: Attempt to expand ( (1 2i))( (1 2i))x x or any valid

method to establish the quadratic factor M1A1

A1: 2 2 5x x

2 2f ( ) 2 5 12x x x x x M1: Attempt other quadratic factor

M1A1 A1: 2 12x x

2 12 4 3 ...x x x x x Attempt to solve their other quadratic factor.

M1

x = 4 and x = 3 Both values correct A1 (7)

Total 7 Alternative using Factor Theorem f (3) ...... or f ( 4) ...... M1: Attempts f(3) or f(-4) M1

f (3) 0 or f ( 4) = 0 A1: Shows or states f(3) = 0 or f(-4) = 0 A1

f (3) ...... and f ( 4) ......

M1: Attempts f(3) and f(-4) or f(3) and g(-4) where g(x) = f(x)/(x - 3) or f(-4) and h(3) where h(x) = f(x)/(x + 4)

M1

f (3) 0 and f ( 4) 0

A1: Shows or states f(3) = 0 and f(-4) = 0 or shows or states f(3) = 0 and g(-4) = 0 where g(x) = f(x)/(x - 3) or shows or states f(-4) = 0 and h(3) = 0 where h(x) = f(x)/(x + 4)

A1

NB 3 2 3 2g 2 3 20, h 5 11 15x x x x x x x x

x = 3 or x = 4 One of x = 3 or x = 4 clearly stated as a root M1

x = 3 and x = 4 Both x = 3 and x = 4 clearly stated as roots A1

1 2ix B1

MARK SCHEME JANUARY 2015

Question Number

Scheme Notes Marks

2 3 2

5

1f ( ) 3 2

2x x x

x

(a) f (2) ... and f (3) ...

Attempts both f(2) and f(3) M1

f (2) 1.9116.. , f (3) 2.032... Sign change (and f ( )x is continuous) therefore

a root exists between 2x and 3x

Both values correct : f (2) 1.9116.. (awrt 1.9 ), and f (3) 2.032... (awrt

2.0 or e.g. 3

254

) , sign change (or

equivalent) and conclusion

A1

(2)(b)

2 3.55f ( ) 3 6

4x x x x

M1: 1n nx x

M1A1A1 A1: 23 6x x

A1: 3.55

4x or equivalent un-simplified

and no other terms (+ c loses this mark) 2.032075015

38.973270821

Correct attempt at Newton-Raphson using their values of f (3) andf (3). M1

2.774 Cao (Ignore any subsequent applications) A1


(5) NB if the answer is incorrect it must be clear that both f (3) andf (3) are being used in

the Newton-Raphson process. So that just 3

3' 3

f

f with an incorrect answer and no

other evidence scores M0. Total 7

Question Number Scheme Notes Marks

3 2i * 2i 21 12iz z

* iz x y B1

i 2i i 2i ....x y x y Substitutes for z and their z* and attempts to expand

M1

2 22 i 2 i 4x x y x y y

2 2 4 4 ix y x

2 2 4 21

and 4 12

x y

x

Compares real and imaginary parts (allow sign errors only)

M1

4 12 ....x x Solves real and imaginary parts to obtain at least one value of x or y

M1

3, 4x y 3x cso A1, A1 4y cso

(6) Total 6

Way 2 2i * 2i * 2i * 4z z zz z z Attempt to expand M1

i i 2i i i 4x y x y x y x y * iz x y (may be implied) B1

2 2 4 i 4x y x 2 2 4 21

and 4 12

x y

x

Compares real and imaginary parts (allow sign errors only)

M1

4 12 ....x x Solves real and imaginary parts to obtain at least one value of x or y

M1

3, 4x y 3x cso

A1, A1 4y cso

Total 6

Way 3 2i * 2i * 2i * 4z z zz z z Attempt to expand M1

* 2i * 4 21 12izz z z

* 4 21, 2 * 12zz z z Compares real and imaginary parts (allow sign errors only)

M1

22 * *6 25 0 or 6 25 0z z z z Correct quadratic B1

22 * *

*

6 25 0 or 6 25 0

..... or .....

z z z z

z z

Solves to obtain at least one value of z

or z* M1

3, 4iz

3x cso A1, A1 4y cso

Total 6

Question Number Scheme

Notes

Marks

4(a) 1 12 22 d 1

12 12 12d 2

yy x y x x

x

12

d

d

yk x

x

M1

2 d12 2 12

d

yy x y

x

d

d

yy

x

d d d 1. 6.

d d d 6

y y p

x p x p

their d 1

dd theird

yxpp

12

d 1 d d 112 or 2 12or 6.

d 2 d d 6

y y yx y

x x x p

or equivalent expressions

Correct differentiation A1

1T Nm m p

p Correct perpendicular gradient rule M1

26 3y p p x p

2

2

6 their 3 or

with their and (3 ,6 )in

an attempt tofind ' '.

N

N

y p m x p

y mx c m p p

c

Their mN must have come from calculus and should be a function of p which is not their tangent gradient.

M1

36 3 *y px p p Achieves printed answer with no errors

A1*

(5)(b) 2 2 12 24p y x Substitutes the given value of p into

the normal M1

2

366

yy

Substitutes to obtain an equation in one variable (x, y or “q”)

M1

2 6 216 0y y

18 12 0y y y Solves their 3TQ M1

18 27y x A1: One correct coordinate A1, A1

A1: Both coordinates correct (5)

(c)

Focus is (3, 0) or a = 3 or OS = 3 Must be seen or used in (c) B1 0 18y x

1 118 3 12 18 3 18

2 2A

M1: Correct attempt at area M1A1

A1: Correct expression A = 225 Correct area A1

(4) Total 14


5(a) 3 14 4, B1, B1 (2)

(b) 22 2 9 1 1

216 2 16

M1:Use of 22 2 2

M1 A1 A1: 1

16 cso (allow 0.0625)

(2)(c) 9

4Sum 4 4 3 Attempt numerical sum M1

2 2

17 14 4

Product 4 4 17 4

4

Attempt numerical product M1

2 94 4 0x x

Uses x2 – (sum)x + (prod) with sum, prod numerical (= 0 not reqd.)

M1

24 9 16 0x x Any multiple (including = 0) A1 (4)

Total 8 Alternative: Finds roots explicitly

(a) 3 7i

8 8x

3 7 3 7 3i i=

8 8 8 8 4

B1

3 7 3 7 1i i =

8 8 8 8 4

B1

(2)(b)

2 2

2 2 3 7 3 7 1i i

8 8 8 8 16

M1: Substitutes their α and β and attempt to square and add both brackets M1 A1

A1: 116 cso (allow 0.0625)

(2)(c) 9 5 7 9 5 7

4 i , 4 i8 8 8 8

9 5 7 9 5 7f ( ) i i

8 8 8 8x x x

Uses

4 4x x

With numerical values (May expand first)

M1

2 9 5 7 9 5 7 9 5 7 9 5 7

f ( ) i i i i8 8 8 8 8 8 8 8

x x x x

Attempt to expand (may occur in terms of α and β but must be numerical for both M’s)

M1

2 94 4 0x x Collects terms (= 0 not reqd.) M1

24 9 16 0x x Any multiple (including = 0) A1 (4) Total 8

Question Number

Scheme Notes Marks

6(i)(a) A: Stretch scale factor 3 parallel to the x-axis

B1: Stretch

B1B1 B1: SF 3 parallel to (or along) x-axis Allow e.g. horizontal stretch SF 3 (Ignore any reference to the origin)

(2)(b)

B: Rotation 210 degrees (anticlockwise) about (0, 0) or about O

B1: Rotation about (0, 0)

B1B1 B1: 210 degrees (anticlockwise) (or equivalent e.g. -150o or 150o clockwise). Allow equivalents in radians.

(2)(c) 3 1

2 2

312 2

3 0

0 1

C = BA Attempts BA (This statement is sufficient) M1

3 3 12 2

332 2

Correct matrix A1

(2)(ii) detM =

22 5 . 1 4 2 5 4k k k k

M1: Correct attempt at determinant M1A1 A1: Correct determinant (allow un-

simplified) 2 4 25 32b ac

Attempts discriminant or uses quadratic formula

M1

2 4 0b ac So no real roots so detM ≠ 0

Convincing explanation and conclusion with no previous errors

A1

(4) Total 10

(ii) Way 2 22 5 . 1 4 2 5 4k k k k


simplified) 25 7

4 82 k Attempts to complete the square: M1

detM> 0 k Therefore detM ≠ 0


A1

(ii) Way 3 22 5 . 1 4 2 5 4k k k k


simplified) d 5

d 44 5 0k

det k k M

4

78

5 d tk e M Attempts coordinates of turning point M1

Minimum detM is 7

8 therefore

detM ≠ 0


A1

Question Number

Scheme Notes

Marks

7 1

12 11 1

6

n

r

r a r b n n n

(a) 2r a r b r ra rb ab B1

1

1 11 2 1 1

6 2

n

r

r a r b n n n a b n n abn

M1A1B1

M1: Attempt to use one of the standard formulae correctly

A1: 1 11 2 1 1

6 2n n n a b n n

B1: abn

1 11 2 1 3 1 6 2 11 1

6 6n n n a b n ab n n n

21 2 1 3 1 6 2 9 11n n a b n ab n n

2 22 3 1 3 1 6 2 9 11n n a b n ab n n

3 3 3 9, 3 3 1 6 11

2, 3

a b a b ab

a b ab

M1: Compares coefficients to obtain at least one equation in a and b

M1M1M1 M1: One correct equation M1: Both equations correct

1, 3b a Both values correct. This can be withheld if 3, 1b a is not rejected. A1

(8)(b)

20

9r

r a r b

20

9

f (20) f (8or 9)r

r a r b

Use of f (20) f (8or 9) M1

1 120 51 19 8 27 7

6 6

Correct (possibly un-simplified) numerical expression

A1

=3230 – 252 = 2978 cao A1 (3)

Total 11


8(i) When n = 1 1 11 2 3 5u

When n = 2 2 22 2 3 13u

Both B1

True for n = 1 and n = 2

Assume 2 3k kku and 1 1

1 2 3k kku

1 12 15 6 5 2 3 6 2 3k k k k

k k ku u u

M1: Attempts uk+2 in terms of uk+1

and uk M1A1 A1: Correct expression

1 15.2 5.3 6.2 6.3k k k k

1 1 1 15.2 3.2 5.3 2.3k k k k Attempt uk+2 in terms of 2f(k) and 3f(k) only M1

So 1 12 2.2 3.3k k

ku

1 1 1 1 2 22 3 or 2 3k k k k Correct expression with no errors

A1

If true for k and k + 1 then shown true for k + 2 and

as true for n = 1 and n = 2, true for n ℤ+ Full conclusion with all previous marks scored

A1

(6)(ii) 4f (2) 7 48 2 1 2304

So true for n = 2 Shows true for n = 2 B1

Assume 2f ( ) 7 48 1 2304kk k p

for some integer p

2 2 2f ( 1) f ( ) 7 48 1 1 7 48 1k kk k k k Attempt f(k+1) – f(k) M1 2 2 27 7 48k k

27 49 1 48k

248f ( ) 48k k

M1: Attempt rhs in terms of f(k) or 27 48 1k k M1A1 A1: Correct expression which is a multiple of 2304

48 2304 2304p k

f ( 1) 49 2304 2304k p k Obtains f(k+1) as a correct multiple of 2304 with no errors

A1

If true for k then shown true for k + 1 and as true for

n = 2, true for 2n (n ℤ) Full conclusion with all previous marks scored

A1

(6) Total 12

Question Number

Scheme Marks

1. 2( 1)( 1) 1r r r

Correct expansion.

Allow 2 1r r r

B1

2002

1

1200(201)(401)

6r

r

Use of 1

( 1)(2 1)6

n n n with n = 200 M1

200

1

1 200r

Cao (May be implied by their work) B1

2002

1

( 1)r

r

= 2686700 – 200 = 2686500 2686500 A1

Note use of 200

1

1 1r

gives a sum of 2686699 and usually scores B1M1B0A0

Total 4

MARK SCHEME JUNE 2014

Question Number

Scheme Marks

Mark (a) and (b) together

2.(a) 2 3i cao B1 (1)

Way 1 (b) p = sum of roots = 2 3i 2 3i

or

q = product of roots = (– 2 + 3i)( – 2 – 3i)

A correct approach for either p or q M1

p = 4, q = 13

1st A1: One value correct 2nd A1: Both values correct Can be implied by a correct equation or expression e.g 2 4 13z z

A1A1

(3) Total 4

(b) Way 2

2 3i 2 3iz z 2 3iz and 2 3iz and

attempt to expand (condone invisible brackets)

M1

Equation is 2 4 13 0z z

or p = 4, q = 13

1st A1: One value correct 2nd A1: Both values correct Condone use of x instead of z

A1 A1

(b) Way 3 2

2 3i 2 3i 0 3 12 i 2 5 0p q p q p

3 12 i 2 5 0

3 12 0, 2 5

... or ...

p q p

p q p

p q

Substitutes – 2 + 3i or – 2 – 3i into the given equation, compares real and imaginary parts and obtains a real value for p or a real value for q

M1

p = 4, q = 13

1st A1: One value correct 2nd A1: Both values correct

A1A1

(b) Way 4 2 42

20 p p qz pz q z

2

2

42 3i 2 ...

2pp p q

p

M1

Correct method to find a value for p p = 4 A1 2 4 36 13p q q Correct value for q A1

(b) Way 5 2 2

2 3i 2 3i 0 and 2 3i 2 3i 0p q p q

24 6 0 ...i pi p

M1: Substitutes both roots into the given equation and attempts to solve simultaneously to obtain a real value for p or a real value q

M1

p = 4, q = 13

1st A1: One value correct 2nd A1: Both values correct

A1A1

Question Number

Scheme Marks

3.(a) detA = 4 3 2 2 12a a Any correct form (possibly un-

simplified) of the determinant B1

3 2adj

4a

A

Correct attempt at swapping elements in the major diagonal and changing signs in the minor diagonal. Three or four of the numbers in the matrix should be correct e.g. allow one slip

M1

1 3 21

42 12 aa

A Correct inverse A1

(3)(b) 4

3

2

a

+

3 22

42 12 aa

=1 0

0 1

Correct statement for A, “their” inverse and use of the correct identity matrix

M1

64

1 02 12 2 122 8 0 1

32 12 2 12

42

a aa

aa a

So e.g. 6

4 1 ....2 12

aa

Adds their A and 2 x theirA-1 and compares corresponding elements to form an equation in a and attempts to solve as far as a = … or adds an element of their A and the corresponding element of 2 x theirA-1 to form an equation in a and attempts to solve as far as a = …

M1

a = 7 only

Cao (from a correct equation i.e. their A-1 might be incorrect) If they solve a second equation and get a different value for a, this mark can be withheld.

A1

(3)

Total 6

(b) Way 2 (does not use the inverse)

1 22 2 A A I A I A 16 2

9 2

2a

a a

+

1 02

0 1

=4 2

3a

Correct statement for 2 2 A I A using A, “their” A2 and use of the correct identity matrix

M1

16 2 2 4 2

9 2 2 3

2a

a a a

So 16 – 2a + 2 = 4 or 11 – 2a = –3

Adds their 2A and 2I , compares elements, forms an equation in a and attempts to solve as far as a = …

M1

a = 7 cao A1

Question Number

Scheme Marks

4.(a) f (4) ... and f (5) ...

Attempt to evaluate both f(4) and f(5)

NB f(5) = 2 5 3 but this must be evaluated to score the A1

M1

f (4) 1 , f (5) 1.472... Sign change (and f ( )x is continuous) therefore

a root exists between 4x and 5x

Both values correct f (4) 1 , andf (5) 1.472... (awrt 1.5) , sign change (or equivalent) and conclusion E.g. f (4) 1 < 0 and f(5) = 1.472 > 0 so

4 5 scores M1A1

A1

(2)(b)

1 12 23 3

2 2f ( )x x x

M1: 1n nx x

M1A1A1 A1: Either 123

2 x or 123

2 x

A1: Correct derivative

1

f (4.5) 0.1819805153..4.5 4.5

f (4.5) 2.474873734...x

Correct attempt at Newton-Raphson Can be implied by a correct answer or their working provided a correct derivative is seen or implied.

M1

4.426 Cao (Ignore any subsequent applications)

A1


(5)(c)

5 4

1.472 1

or

4 5 4

1 1.472 1

A correct statement for or 5 - or

- 4 M1

(1.472 1) 5 4 1.472 so … Attempt to make “α” the subject (allow poor manipulation). Dependent on the previous M1.

dM1

4.405 cao A1 There are no marks for interval bisection (3) Total 10

Question Scheme Marks 5. (a)

M1: One point in third quadrant and one in the fourth quadrant. Can be vectors, points or even lines.

M1A1 A1: The points representing the complex numbers plotted correctly. The points must be indicated by a scale (could be ticks on axes) or labelled with coordinates or as complex numbers.

(2)(b) M1 requires a correct strategy e.g.

1. Gradient OP = 43 , Gradient OQ = 3

4 34

3 4 ............

M1

2. Angles with Im axis are 1 34tan and 1 4

3tan . 1 13 44 3tan tan .....

3. Angles with Re axis are 1 43tan and 1 3

4tan . 1 13 44 3180 tan tan .....

4. 2 2 2 2 2 2 2 2 23 4 , 3 4 , 1 7OP OQ PQ 2 2 .....OP OQ

5. 3 4

, . .....4 3

OP OQ OP OQ

343 4 1 so right angle or 53.1 + 26.9 = 90 (accept 53 + 27) or radians or

1 13 44 3tan tan

2

2 2 2 50OP OQ PQ so right angle or . 0OP OQ

so right angle

Correct work with no slips and conclusion

A1

(2)

(c) (c) 1z + 2z = 1 – 7 i

B1

(1)(d) Writes down another fact about OPQR other than OP being perpendicular to OQ:

e.g. OP = OQ, OP is parallel to QR, QR = PR, QR is perpendicular to PR B1

Sufficient justification that OPQR is a square and conclusion If their explanation could relate to something other than a square score B0 B1

(2) Total 7

P(-3, -4) or z1

Q(4, -3) or z2

New point as shown. It must be the point 1 – 7 i and it must be correctly plotted. The point must be indicated by a scale (could be ticks on axes) or labelled with coordinates or as a complex number. May be on its own axes.R

1 – 7 i

Question Number

Scheme Marks

6.(a) 5

3 and

1

3 Both correct statements B1

3 3 3( ) 3 ( ) ...... Use of a correct identity for 3 3 (may be implied by their work)

M1

33 3 5 5 170

3 3 27

Correct value (allow exact equivalent – even the correct recurring decimal -6.296296......)

A1

Special Case – but must be a complete method – generally there are no marks for finding the roots explicitly

3 3

3 35 37 5 37 5 37 5 37 170,

6 6 6 6 27

Could score 3/3 in (a) B1: Both correct roots M1: Cube and add A1: Correct value

(3)(b)

2 2 3 3 17027

13

170+

9

M1: Uses the identity 2 2 3 3

+

M1 A1 A1: Correct sum (or equivalent)

2 2

1

3 Correct product B1

2 170 1

9 3x x

Uses 2x their sum x their product

(= 0 not needed here) M1

29 170 3 0x x This equation or any integer multiple including = 0. Follow through their sum and product.

A1ft

(5) Total 8

(b) Alternative using explicit roots

5 37 5 37,

6 6

2 285 14 37 85 14 37,

9 9

2 2 170+

9

M1: Adds their

2 2

and

M1A1

A1: Correct sum 2 2

1

3 Correct product B1

2 170 1

9 3x x

Uses 2x their sum x their product

(= 0 not needed here) M1

29 170 3 0x x This equation or any integer multiple including = 0. Follow through their sum and product.

A1ft

Question Number

Scheme Marks

7. (a) Rotation, 30 degrees (anticlockwise), about O

Allow 6

(radians) for 30 degrees.

Anticlockwise may be omitted but do not allow -30 degrees or 30 degrees clockwise B1: Rotation B1: 30 degrees B1: About O

B1, B1, B1

(3)(b) 1 0

0 1

Correct matrix B1

(1)(c)

R = PQ=

3 1

2 2

1 3

2 2

-

1 0

0 1

Multiplies P by their Q This statement is sufficient in correct order M1

=

3 1

2 2

1 3

2 2

Correct matrix

A1

(2)(d)

3 1

2 2

1 3

2 2

1 1k k

R 1 1

k k

A correct statement but allow poor notation provided there is an indication that the candidate understands that the point (1, k) is mapped onto itself. This Method mark could be implied by a correct equation or correct follow through equation below.

M1

3

21

2

k +

or

1

2

3

2

kk

One correct equation (not a matrix equation)

A1

3

2

1

2

31 or ...

2 2

k kk k +

Attempts to solve their equation for k. Dependent on the first M.

dM1

2 3k cao A1

Solves both 3

2

1

2

31 and

2 2

k kk +

or checks other component

Solves both equations explicitly to obtain the same correct value for k or clearly

verifies that 2 3k is valid for the other equation

B1

(5) Total 11

Question Scheme Marks

8.(a) 28 16 16t t

or 216 32x x

or 2 1632 yy

Attempts to obtains an equation in one variable x, y or t

M1

12t or x = 2 or y = 8 A correct value for t, x or y A1

(2, 8) Correct coordinates following correct work with no other points

B1

(3)(b) 216 d

16d

yy x

x x

or d

0d

yy x

x

d

d

y y

x x

or 2 2

4 d 14,

d

yx y

t x t

Correct derivative in terms of x, y and x, or t B1

at (8, 2) 2

d 16 1

d (8) 4

y

x Uses x = 8, x = 8 and y = 2, or t = 2 M1

gradient of normal is 4 Correct normal gradient A1

2 4( 8)y x or y = 4x + c and uses x = 8 and y = 2 to find c

Correct straight line method using the point (8, 2) and a numerical gradient from their d

d

y

x which is not the tangent gradient.

M1

4 30y x Correct equation A1 (5)

(c) 216 32 30t t or2

8 30yy

or 16

84 30x x

Uses their straight line from part (b) and the parabola to obtain an equation in one variable ( x, y or t )

M1

(4 3)(4 5) 0 ...t t t ( 20)( 12) 0 ...y y y (2 25)(2 9) 0 ...x x x

Attempts to solve three term quadratic (see general guidance) to obtain t = ... or y = ... or x = ... Dependent on the previous M

dM1

Note if they solve the tangent with the parabola this gives 2 544 256 0x x which has roots x = 543.53... and 0.47...... (Seeing these values would imply a correct attempt to solve their 3TQ)

3 54 4andt or 20 and 12y

or 25 92 2andx

Correct values for t or y or x A1

3 5, or ,

4 4t x y t x y

20 or 12y x y x 25 9

or2 2

x y x y

Uses their values of t to find at least one point or uses their values of y to find at least one x or uses their values of x to find at least one y. Not dependent on previous method marks.

M1

252( , 20) , 9

2( , 12) A1: One correct pair of coordinates A1 A1

A1: Both pairs correct (6) Total 14

Question Scheme Marks

9.(i)

1

( 1)( 2)( 3) 1 2 3 4( 1)( 2) 6 6

4 4

n

r

n n n nr r r

and

Minimum: lhs = rhs = 6

B1

Assume result true for n = “k” so 1

( 1)( 2)( 3)( 1)( 2)

4

k

r

k k k kr r r

Minimum Assume result true

M1

1

1

( 1)( 2)( 3)( 1)( 2) ( 1)( 2)( 3)

4

k

r

k k k kr r r k k k

Adds the (k + 1)th term to the given result

M1

1( 1)( 2)( 3)( 4)

4k k k k

Achieves this result with no errors Note this may be written down directly from the line above.

A1

If the result is true for ,n k then it is true for 1.n k As the result has been shown

to be true for 1,n then the result is true for all n. A1cso

(5) (ii) 1f (1) 4 6 1 8 18 f(1) = 18 is the minimum B1

1f ( 1) f ( ) 4 6 1 8 4 6 8k kk k k k M1: Attempts f(k + 1) – f(k) M1

3 4 6 3 4 6 8 18 18k k k k A1: 3 4 6 8 or 3f ( )k k k

A1A1 A1: 18 18 or 18( 1)k k f ( 1) 3 4 6 8 18 1 f ( )kk k k k Makes f(k + 1) the subject dM1



(6) Total 11

(ii) ALT 1

1f (1) 4 6 1 8 18 B1

1f ( 1) 4f ( ) 4 6 1 8 4 4 6 8k kk k k k M1: Attempts f(k + 1) – 4f(k) M1

18 18k A1: – 18k

A1A1 A1: – 18

f ( 1) 4f ( ) 18 1k k k Makes f(k + 1) the subject dM1



(ii) ALT 2

1f (1) 4 6 1 8 18 B1

1f ( 1) 4 6 1 8kk k M1: Attempts f(k + 1) M1

4 4 6 8 18 18k k k A1: 4 4 6 8 or 4f ( )k k k

A1A1 A1: 18 18 or 18( 1)k k

f ( 1) 4f ( ) 18 1k k k Makes f(k + 1) the subject (implicit with first M)

dM1



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January 2016 Mark Scheme - WordPress.com · January 2016 WFM01 Further Pure Mathematics F1 Mark Scheme Question Number Scheme Notes Marks 1. (a) ... Correct equation in any form A1