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Mark Scheme (Results) January 2011

GCE

GCE Further Pure Mathematics FP1 (6667) Paper 1

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January 2011 Publications Code UA026332 All the material in this publication is copyright Š Edexcel Ltd 2011


General Instructions for Marking 1. The total number of marks for the paper is 75. 2. The Edexcel Mathematics mark schemes use the following types of marks: •

M marks: method marks are awarded for ‘knowing a method and attempting to apply it’, unless otherwise indicated.

A marks: Accuracy marks can only be awarded if the relevant method (M) marks have been earned.

B marks are unconditional accuracy marks (independent of M marks)

Marks should not be subdivided.

3. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes. •

bod – benefit of doubt

ft – follow through

the symbol

cao – correct answer only

cso - correct solution only. There must be no errors in this part of the question to obtain this mark

isw – ignore subsequent working

awrt – answers which round to

SC: special case

oe – or equivalent (and appropriate)

dep – dependent

indep – independent

dp decimal places

sf significant figures

¿ The answer is printed on the paper

will be used for correct ft

The second mark is dependent on gaining the first mark


January 2011 Further Pure Mathematics FP1 6667 Mark Scheme Question Number 1. (a)

Scheme

Marks

z = 5 − 3i, w = 2 + 2i z 2 = ( 5 − 3i )( 5 − 3i )

= 25 − 15i − 15i + 9i 2 = 25 − 15i − 15i − 9

An attempt to multiply out the brackets to give four terms (or four M1 terms implied). zw is M0

= 16 − 30i

16 − 30i

A1 (2)

Answer only 2/2 (b)

z ( 5 − 3i ) = w ( 2 + 2i )

=

( 5 − 3i ) ( 2 − 2i ) × ( 2 + 2i ) ( 2 − 2i )

=

10 − 10i − 6i − 6 4+4

=

4 − 16i 8

=

1 − 2i 2

Multiplies

z by w

( 2 − 2i ) ( 2 − 2i )

Simplifies realising that a real number is needed on the denominator and applies i 2 = − 1 on M1 their numerator expression and denominator expression.

1 1 − 2i or a = and b = −2 or 2 2 A1 equivalent Answer as a single fraction A0

GCE Further Pure Mathematics FP1 (6667) January 2011

M1

1

(3) [5]


Question Number 2. (a)

Scheme

Marks

⎛ 2 0⎞ ⎛ −3 −1⎞ A=⎜ ⎟,B=⎜ ⎟ ⎝ 5 3⎠ ⎝5 2⎠ ⎛ 2 0 ⎞⎛ −3 −1⎞ AB = ⎜ ⎟⎜ ⎟ ⎝ 5 3 ⎠⎝ 5 2 ⎠

A correct method to multiply out two matrices. Can be implied by M1 two out of four correct elements.

⎛ 2(−3) + 0(5) 2(−1) + 0(2) ⎞ = ⎜ ⎟ ⎝ 5(−3) + 3(5) 5(−1) + 3(2) ⎠ ⎛ −6 − 2⎞ =⎜ ⎟ 1 ⎠ ⎝ 0

Any three elements correct A1 Correct answer A1 Correct answer only 3/3

(3)

Reflection M1 y-axis (or x = 0 .) A1

(b) Reflection; about the y-axis.

(2) (c)

⎛1 0⎞ C100 = I = ⎜ ⎟ ⎝0 1⎠

⎛1 0⎞ ⎜ ⎟ or I B1 ⎝0 1⎠ (1) [6]

GCE Further Pure Mathematics FP1 (6667) January 2011

2


Question Number 3. (a)

Scheme 3

f ( x) = 5 x 2 − 4 x 2 − 6, x f (1.6) = − 1.29543081... f (1.8) = 0.5401863372...

α − 1.6 "1.29543081..."

=

Marks

0

awrt -1.30 B1 awrt 0.54 B1 Correct linear interpolation method with signs correct. Can be implied M1 by working below.

1.8 − α "0.5401863372..."

⎞ "1.29543081..." ⎟ 0.2 ⎝ "0.5401863372..." + "1.29543081..." ⎠

α = 1.6 + ⎜

= 1.741143899...

awrt 1.741 A1 (4) Correct answer seen 4/4 At least one of ± a x or ± b x M1 correct. Correct differentiation. A1 1 2

(b)

f ′( x) = 10 x − 6 x 2 1

(2) (c)

f (1.7) = − 0.4161152711... f ′(1.7) = 9.176957114...

f (1.7) = awrt − 0.42 f ′(1.7) = awrt 9.18

B1 B1

Correct application of NewtonRaphson formula using their M1 values.

⎛ "− 0.4161152711..." ⎞ ⎟ ⎝ "9.176957114..." ⎠

α 2 = 1.7 − ⎜

= 1.745343491... = 1.745 (3dp)

1.745 A1 cao (4) Correct answer seen 4/4

[10]

GCE Further Pure Mathematics FP1 (6667) January 2011

1


Question Number 4. (a)

Scheme

Marks

z 2 + p z + q = 0, z1 = 2 − 4i z2 = 2 + 4i

2 + 4i

B1 (1)

(b)

( z − 2 + 4i)(z − 2 − 4i) = 0 ⇒ z 2 − 2 z − 4iz − 2 z + 4 − 8i + 4iz − 8i + 16 = 0

⇒ z 2 − 4 z + 20 = 0

An attempt to multiply out brackets M1 of two complex factors and no i2. Any one of p = − 4, q = 20. A1 Both p = − 4, q = 20. A1 2 (3) ⇒ z − 4 z + 20 = 0 only 3/3 [4]

GCE Further Pure Mathematics FP1 (6667) January 2011

1


Question Number

Scheme

Marks

n

∑ r (r + 1)(r + 5)

5

r =1

(a)

=

n

∑r

3

+ 6 r 2 + 5r

r =1

(b)

Multiplying out brackets and an attempt to use at least one of the M1 standard formulae correctly.

=

1 2 1 1 n (n + 1) 2 + 6. n(n + 1)(2n + 1) + 5. n(n + 1) 4 6 2

=

1 2 5 n (n + 1) 2 + n(n + 1)(2n + 1) + n(n + 1) 4 2

=

1 n( n + 1) ( n(n + 1) + 4(2n + 1) + 10 ) 4

=

1 n(n + 1) ( n 2 + n + 8n + 4 + 10 ) 4

=

1 n(n + 1) ( n 2 + 9n + 14 ) 4

Correct 3 term quadratic factor A1

=

1 n(n + 1)(n + 2)(n + 7) * 4

Correct proof. No errors seen. A1

Sn =

Correct expression. A1

Factorising out at least n(n + 1)

dM1

(5)

50

∑ r (r + 1)(r + 5)

r = 20

= S50 − S19 =

1 4

(50)(51)(52)(57) − 14 (19)(20)(21)(26)

Use of S50 − S19

M1

1837 680

A1

= 1889550 − 51870 = 1837 680

Correct answer only 2/2

GCE Further Pure Mathematics FP1 (6667) January 2011

2

(2) [7]


Question Number 6. (a)

Scheme

C : y 2 = 36 x ⇒ a =

36 4

Marks

=9

S (9, 0)

(9, 0)

B1 (1)

(b)

(c)

x + 9 = 0 or x = − 9

x + 9 = 0 or x = − 9

or ft using their a from part (a).

B1 (1)

Either 25 by itself or PQ = 25 . Do not award if just PS = 25 is B1 seen.

PS = 25 ⇒ QP = 25

x = 16

(d) x-coordinate of P ⇒ x = 25 − 9 = 16

(1)

B1

Substitutes their x-coordinate into M1 equation of C.

y 2 = 36(16)

y = 24

y = 576 = 24

A1 (3)

Therefore P (16, 24) (e) Area OSPQ = 12 (9 + 25)24

1 2

( their a + 25)( their y )

M1

or rectangle and 2 distinct triangles, correct for their values. 408 A1

= 408 (units) 2

(2) [8]

GCE Further Pure Mathematics FP1 (6667) January 2011

3


Question Number 7.

Scheme

Marks

Im

(a)

-24

Re

Correct quadrant with (−24, − 7) B1 indicated.

-7

(1) (b)

arg z = − π + tan −1 ( 247 )

tan −1 ( 247 ) or tan −1 ( 247 )

= − 2.857798544... = − 2.86 (2 dp)

M1

awrt -2.86 or awrt 3.43 A1 (2)

(c)

w = 4 ,arg w =

⇒ r = 4 ,θ =

5π 6

5π 6

w = r cosθ + i r sin θ

Attempt to apply r cosθ + i r sin θ . M1 Correct expression for w. A1

w = 4cos ( 56π ) + 4i sin ( 56π ) =4

( ) + 4i ( − 3 2

1 2

)

= − 2 3 + 2i

either −2 3 + 2i or awrt − 3.5 + 2i

A1 (3)

a = − 2 3, b = 2 z = 25 or (d)

zw = (48 3 + 14) + (14 3 − 48) i or B1 awrt 97.1-23.8i

z = (−24) 2 + (−7) 2 = 25

zw = z × w = (25)(4)

Applies z × w or zw

= 100

M1

100 A1 (3) [9]

GCE Further Pure Mathematics FP1 (6667) January 2011

4


Question Number 8. (a)

Scheme

Marks

⎛ 2 −2 ⎞ A=⎜ ⎟ ⎝ −1 3 ⎠ det A = 2(3) − (−1)(−2) = 6 − 2 = 4

4 B1 (1)

(b)

(c)

A

−1

1 ⎛3 2⎞ ⎜ ⎟ det A ⎝ 1 2 ⎠ 1 ⎛3 2⎞ ⎜ ⎟ 4 ⎝1 2⎠

1 ⎛3 2⎞ = ⎜ ⎟ 4 ⎝1 2⎠

72 or 72 ( their det A ) their det A

72 = 18 (units)2 Area( R ) = 4

M1 A1 (2) M1

18 or ft answer. A1 (2)

(d)

AR = S ⇒ A −1 AR = A −1 S ⇒ R = A −1 S R=

=

1 ⎛ 3 2 ⎞ ⎛ 0 8 12 ⎞ ⎜ ⎟⎜ ⎟ 4 ⎝ 1 2 ⎠ ⎝ 4 16 4 ⎠

At least one attempt to apply A −1 by M1 any of the three vertices in S .

1 ⎛ 8 56 44 ⎞ ⎜ ⎟ 4 ⎝ 8 40 20 ⎠

⎛ 2 14 11⎞ =⎜ ⎟ ⎝ 2 10 5 ⎠

At least one correct column o.e. A1 At least two correct columns o.e.

A1

Vertices are (2, 2), (14, 10) and (11, 5). All three coordinates correct. A1

GCE Further Pure Mathematics FP1 (6667) January 2011

5

(4) [9]


Question Number 9.

Scheme

Marks

un +1 = 4un + 2 , u1 = 2 and un = 23 (4n − 1)

Check that un = 23 (4n − 1)

n = 1; u1 = 23 (41 − 1) = 23 (3) = 2 So un is true when n = 1.

yields 2 when n = 1.

B1

Assume that for n = k that, uk = 23 (4k − 1) is true for k ∈ Z + . Then uk +1 = 4uk + 2 Substituting uk = 23 (4k − 1) into

= 4 ( 23 (4k − 1) ) + 2 =

8 3

( 4)

=

2 3

( 4 )( 4 )

k

= 23 4k +1 −

8 3

k

un +1 = 4un + 2.

An attempt to multiply out the M1 brackets by 4 or 83

+2

M1

2 3

2 3

= 23 (4k +1 − 1)

2 3

Therefore, the general statement, un = 23 (4n − 1) is true when n = k + 1. (As un is true for n = 1, ) then un is true for all positive integers by mathematical induction

GCE Further Pure Mathematics FP1 (6667) January 2011

6

(4k +1 − 1)

A1

Require ‘True when n=1’, ‘Assume true when n=k’ and ‘True when A1 n = k + 1 ’ then true for all n o.e. (5) [5]


Question Number 10. (a)

Scheme

xy = 36 at ( 6t ,

6 t

Marks

). An attempt at

36 dy 36 = 36 x −1 ⇒ = − 36 x −2 = − 2 y= x x dx At ( 6t ,

6 t

),

So, mT =

T: T: T: T:

dy . dx

dy dx or and dt dt

dy 36 =− dx (6t ) 2

An attempt at

dy 1 =− 2 t dx

M1

dy . in terms of t M1 dx dy 1 =− 2 * A1 t dx

Must see working to award here

6 1 y − = − 2 ( x − 6t ) t t 6 1 6 y− = − 2x+ t t t 1 6 6 y = − 2x+ + t t t 1 12 y = − 2x+ * t t

Applies y −

6 = their mT ( x − 6t ) t

M1

Correct solution . A1 cso (5)

(b) Both T meet at ( −9, 12 ) gives

1 12 ( −9) + 2 t t 9 12 12 = 2 + (× t 2 ) t t 12t 2 = 9 + 12t 12 = −

Substituting (-9,12) into T. M1

An attempt to form a “3 term M1 quadratic”

12t 2 − 12t − 9 = 0

4t 2 − 4t − 3 = 0 (2t − 3)(2t + 1) = 0 t= t=

3 2

3 2

t=−

An attempt to factorise. M1

, − 12

t=

⇒ x = 6 ( 32 ) = 9 , y =

1 2

6

( ) 3 2

= 4 ⇒ ( 9, 4 )

⇒ x = 6 ( − 12 ) = − 3 , y=

6 = − 12 ⇒ ( −3, − 12 ) ( − 12 )

3 2

, − 12

A1

An attempt to substitute either their M1 t = 32 or their t = − 12 into x and y. At least one of A1 ( 9, 4 ) or ( −3, − 12 ) . Both ( 9, 4 ) and ( −3, − 12 ) .

A1 (7) [12]

GCE Further Pure Mathematics FP1 (6667) January 2011

7


Other Possible Solutions Question Number 4. (a) (i) Aliter (ii) Way 2

Scheme

Marks

z 2 + p z + q = 0, z1 = 2 − 4i

2 + 4i

z2 = 2 + 4i

B1

No i2. Attempt Sum and Product M1 of roots or Sum and discriminant

Product of roots = (2 − 4i)(2 + 4i) = 4 + 16 = 20

or b 2 − 4ac = (8 i)2 Sum of roots = (2 − 4i) + (2 + 4i) = 4

= z 2 − 4 z + 20 = 0

Any one of p = − 4, q = 20.

A1

Both p = − 4, q = 20.

A1 (4)

4.

z 2 + p z + q = 0, z1 = 2 − 4i

(a) (i) Aliter (ii) Way 3

z2 = 2 + 4i

2 + 4i

B1

An attempt to substitute either z1 or z2 into z 2 + p z + q = 0 M1 and no i2.

(2 − 4i) + p (2 − 4i) + q = 0 2

−12 − 16i + p (2 − 4i) + q = 0

Imaginary part: −16 − 4 p = 0 Real part: −12 + 2 p + q = 0 4 p = − 16 ⇒ p = − 4 q = 12 − 2 p ⇒ q = 12 − 2(−4) = 20

Any one of p = − 4, q = 20. A1 Both p = − 4, q = 20. A1 (4)

GCE Further Pure Mathematics FP1 (6667) January 2011

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Question Number Aliter 7. (c) Way 2

Scheme w = 4 ,arg w =

5π 6

Marks

and w = a + i b

Attempts to write down an equation in terms of a and b for either the M1 modulus or the argument of w.

w = 4 ⇒ a 2 + b 2 = 16

arg w =

5π 6

⇒ arctan ( ba ) =

5π 6

b a

=−

1 3

Either a 2 + b 2 = 16 or

b a

=−

1 3

A1

a = − 3 b ⇒ a 2 = 3b 2

So, 3b 2 + b 2 = 16 ⇒ b 2 = 4 ⇒ b = ± 2 and a = ∓ 2 3

As w is in the second quadrant w = − 2 3 + 2i

either −2 3 + 2i or awrt − 3.5 + 2i

A1 (3)

a = − 2 3, b = 2

GCE Further Pure Mathematics FP1 (6667) January 2011

9


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6667 FP1 MS Jan 2011