Friday 20 May 2016 – MorningAS GCE MATHEMATICS (MEI)

4755/01 Further Concepts for Advanced Mathematics (FP1)

QUESTION PAPER

*6386695432*

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of the working to indicate that a correct method is being used.• The total number of marks for this paper is 72.• The Printed Answer Book consists of 16 pages. The Question Paper consists of 4 pages.

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Candidates answer on the Printed Answer Book.

OCR supplied materials:• Printed Answer Book 4755/01• MEI Examination Formulae and Tables (MF2)

Other materials required:• Scientific or graphical calculator

Duration: 1 hour 30 minutes

Oxford Cambridge and RSA

2

4755/01 Jun16© OCR 2016

Section A (36 marks)

1 The matrix M is given by M p8 2

1=

-c m, where p 4!- .

(i) Find the inverse of M in terms of p. [2]

(ii)

x

y

1

10 2 3 4 5 6

–1

–1

–2

–3

2

3

4

5

6

Fig. 1

The triangle shown in Fig. 1 undergoes the transformation represented by the matrix 8

3

2

1

-c m. Find the area of the image of the triangle following this transformation. [2]

2 The complex number z1 is j2 5- and the complex number z2 is ( ) ( ) ja b1 2- + - , where a and b are real.

(i) Express zz *

1

1 in the form jx y+ , giving x and y in exact form. You must show clearly how you obtain your answer. [4]

(ii) Given that zz

z*

1

1

2= , find the exact values of a and b. [2]

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4755/01 Jun16 Turn over© OCR 2016

3 You are given that A 2

1

6

5

4

4

1

3

m

=

-

-

-f p, B19

5

13

34

5

18

14

5

3

=

-

-

-

-

-f p and AB In= , where I is the 3 3# identity

matrix.

(i) Find the values of m and n. [4]

(ii) Hence find B 1- . [2]

4 (i) Use standard series to show that

( ) ( ) ( ( ) )r r p n n n p n p26

11 3 3 2

r

n2

1

2- = + + - -=

/ ,

where p is a constant. [4]

(ii) Given that the coefficients of n3 and n4 in the expression for ( )r r p2

r

n2

1

-=

/ are equal, find the value of p. [2]

5 The loci C1 and C2 are given by jz 3 4 5+ - = and arg ( )jz 3 62

1r+ - = respectively.

(i) Sketch, on a single Argand diagram, the loci C1 and C2 . [5]

(ii) Write down the complex number represented by the point of intersection of C1 and C2 . [1]

(iii) Indicate, by shading on your sketch, the region satisfying

jz 3 4 5H+ - and ( )arg jz2

13 6

4

3G Gr r+ - . [2]

6 A sequence is defined by u 81= and u u n3 2 5n n1

= + ++

. Prove by induction that ( )u n4 3 3nn= - - . [6]

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4755/01 Jun16© OCR 2016

Oxford Cambridge and RSA

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Section B (36 marks)

7 The function ( )f z z z z BzA2 9 264 3 2= - + + - has real coefficients. The equation ( )f z 0= has two real roots,

a and b, where 2a b , and two complex roots, c and d, where j3 2c = + .

(i) Show that 2

3a b+ =- and find the value of ab. [5]

(ii) Hence find the two real roots a and b. [3]

(iii) Find the values of A and B. [3]

(iv) Write down the roots of the equation fj

w0=c m . [2]

8 A curve has equation yx xx3 4

3 92

2

=+ -

- .

(i) Find the equations of the two vertical asymptotes and the one horizontal asymptote of this curve. [3]

(ii) State, with justification, how the curve approaches the horizontal asymptote for large positive and large negative values of x. [3]

(iii) Sketch the curve. [3]

(iv) Solve the inequality x x

x3 4

3 90

2

2

H+ -

- . [3]

9 You are given that ( ) ( ) ( ) ( ) ( )r r r r r r

r4 2 1

3

2 1

1

4 2 3

1

2 1 2 1 2 3

2 5

--++

+=

- + ++ .

(i) Use the method of differences to show that

( ) ( ) ( ) ( ) ( )r r r

rn n2 1 2 1 2 3

2 5

3

2

4 2 1

3

4 2 3

1

r

n

1- + +

+= -

++

+=

/ . [6]

(ii) Write down the limit to which( ) ( ) ( )r r r

r2 1 2 1 2 3

2 5

r

n

1- + +

+

=

/ converges as n tends to infinity. [1]

(iii) Find the sum of the finite series

39 41 43

45

4 4

47

4

49

99 1

105

41 3 5 43 45 7 10 103# # # # # # # #f+ + + + ,

giving your answer to 3 significant figures. [4]

END OF QUESTION PAPER

4755 Mark Scheme June 2016

Question Answer Marks Guidance B1 Correct determinant correctly used 1 (i)

1 1 2188 2 pp

− = −+

M

B1 [2]

Correct re-arrangements of elements

1 (ii) Area of image = triangle area × (8 + 2p) = ( )12 8 2 p+ M1

Multiplying an area by their determinant with 3p = (accept 8 + 2 × 3 only)

= 168 (square units) Or new coords and valid method to area A1 cao [2] 2 (i) *

1 2 5jz = + B1

( )( )

*1

1

(2 5j) 2 5j 21 20 j2 5j (2 5j) 29 29

zz

+ += = − +

− +

M1 A1 A1

Correct use of conjugate 29 in denominator All correct

[4] 2 (ii) 211

29a − = − and

20229

b− =

M1 Equating real and imaginary parts

829

a = , 3829

b =

A1 [2]

Both, ft their (i)

3 (i) Either (2)(34) (5)( 5) ( 1)(18) 25µ = + − + − = Or ( 1)( 14) (4)(5) (3)( 3)µ = − − + + −

19 (6)(5) ( 4)( 13) 25λ− + + − − = Or 34 λ – 102 = 0 Or -14 λ + 42 = 0

M1

M1

A1 A1 [4]

Multiplying a row of A by a column of B to find λ or µ Multiplying another row of A by a column of B to find the other unknown µ = 25 cao λ = 3 cao

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4755 Mark Scheme June 2016

Question Answer Marks Guidance 3 (ii)

M1 1 1µ

− =B A with their λ and theirµ

1

3 6 41 2 5 125

1 4 3

−

− = − −

B

B1 cao NB µ = 1/25 in (i) then 25A earns M1 B0 whereas 1/25 A earns M0 B1

[2] 4 (i)

( )2 3 2

1 1 12 2

n n n

r r rr r p r p r

= = =

− = −∑ ∑ ∑

M1*

Splitting into sums a∑r3 ± b∑r2

( ) ( )( )221 11 1 2 1

2 6n n pn n n= + − + + o.e.

A1

Use of standard results in terms of n

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

6n n n n p n= + + − +

M1dep

Attempt to factorise with ( 1)n n + If from quartic in n all steps justified, otherwise M0

( )( )21 1 3 (3 2 )

6n n n p n p= + + − −

A1 [4]

AG

4 (ii) 3 6 2 p= −

32

p =

M1

A1 [2]

Equating their coefficients of 3n and 4n

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4755 Mark Scheme June 2016

Question Answer Marks Guidance 5 (i)

B1 B1

B1

B1

B1

[5]

Accept un-numbered evenly spaced marks on axes to show scale Circle Centre 3 4j− + Radius = 5 (check that the circle passes through O or any valid point or explicitly shown) allow for centre

Half line 12π from 3 6 j±

Fully correct

5 (ii) 3 9j− + B1 [1]

Not (-3, 9j) nor (-3, 9)

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4755 Mark Scheme June 2016

Question Answer Marks Guidance 5 iii

B1

B1

[2]

Region outside circle, must have a border with the circumference

Correct region shown, boundary at 34π

reasonably well drawn

6

When 1n = , ( )14 3 1 3 8nu = − − = , (so true for n = 1) Assume ( )4 3 3k

ku k= − −

B1

E1

Showing use of ( )4 3 3n

nu n= − − Assuming true for n k= Allow “let n = k and (result)” or “If n = k and (result)” Do not allow “n=k” or “let n=k” without the result quoted

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4755 Mark Scheme June 2016

Question Answer Marks Guidance

( )( )1 3 2 5 3 4 3 3 2 5kk ku u k k k+⇒ = + + = − − + +

( ) 14 3 ( 1) 3k k+= − + − But this is the given result with 1k + replacing k . Therefore if it is true for n k= then it is also true for 1n k= + . Since it is true for 1n = , it is true for all positive integers.

M1

A1

E1 E1 [6]

1ku + , using ku and attempting to simplify Correct simplification or identification with a ‘target’ expression using 1n k= + The “target” shows this Dependent on A1 and previous E1 Dependent on B1 and previous E1

7 (i) 3 2jδ = − and used B1 9sum of roots =

2∑ (correct sign in RHS)

( ) ( )9 33 2j 3 2j2 2

α β⇒ + = − + − − = −

(3 2 j)(3 2j) 13αβ + − = − (correct sign in RHS)

1αβ = −

M1

A1

M1

A1 [5]

Or through obtaining two quadratic factors (AG) Or through obtaining two quadratic factors

7 (ii) 23 1 2 3 2 02

α α α α − − = − ⇒ + − =

( )( ) 12 1 2 0 , 22

α α α− + = ⇒ = −

1 , 22

α β= = −

M1

A1

A1

[3]

Attempt at solution for α or β (from their αβ) Or from quadratic found earlier Or z2 - (sum of roots)z + product of roots = 0 One root correct Both roots correct (condone vice-versa)

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4755 Mark Scheme June 2016

Question Answer Marks Guidance 7 (iii) f ( ) (2 1)( 2)( (3 2j))( (3 2j))z z z z z= − + − + − −

( )( )2 22 3 2 6 13z z z z= + − − + 4 3 22 9 6 51 26z z z z= − + + −

M1

A1

A1 [3]

A valid method seen to find A or B (by factors or root relations or substitution) May be seen earlier

6A =

51B = both cao

7 (iv) 1f 0 j, 2 j, 2 3j, 2+3jj 2

w w = ⇒ = − − +

M1

A1 [2]

Their roots j× FT

8 (i) 1x = 4x = −

3y =

B1 B1 B1 [3]

SC “ x = 1, -4 ” is B1 B0

(ii) Evidence of method needed e.g. evaluating using ‘large’ values Large positive x , 3y −→

Large negative x , 3y +→

M1

A1 A1 [3]

SC B1 for correct approaches without valid method seen

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4755 Mark Scheme June 2016

Question Answer Marks Guidance (iii)

B1

B1 B1

[3]

3 branches correct relative to their asymptotes and carefully drawn Asymptotes correct and labelled Intercepts correct and labelled (allow 1.7 and -1.7)

(iv) 4, 3 1, 3x x x< − − ≤ < ≥ B3 [3]

One mark for each. Correct inequality signs. Allow 1.73 for 3 (B3 then 1− if more than 3 inequalities)

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4755 Mark Scheme June 2016

Question Answer Marks Guidance 9 (i)

( )( )( ) ( ) ( )1 1

2 5 3 1 12 1 2 1 2 3 4 2 1 2 1 4 2 3

n n

r r

rr r r r r r= =

+= − + − + + − + +

∑ ∑

3 1 1 3 1 1 3 1 14 3 20 12 5 28 20 7 36

= − + + − + + − + +

3

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

4 2 3 2 1 4 2 1 4 2 1 2 1 4 2 3n n n n n n

+ − + + − + − − + − + + 3

( ) ( )3 1 1 1 1 14 3 4 4 2 1 2 1 4 2 3n n n

= − + + − ++ + +

( ) ( )2 3 13 4 2 1 4 2 3n n

= − ++ +

as required

M1

M1 A1

A1

M1

A1

[6]

Use of the given result (may be implied) Terms in full (first and at least one other) First term correct and one other correct Any 3 consecutive terms fully correct Or 2 consecutive algebraic terms fully correct (need not be simplified) Valid attempt to cancel, including algebraic terms. Need to see the three algebraic fractions that remain after cancelling Convincingly shown (AG)

9 (ii)

1

23r

∞

=

=∑

B1

[1] 9 (iii) 2 5 45 20r r+ = ⇒ = B1 May be implied (eg by using sum to 19

terms)

2 5 105 50r r+ = ⇒ = B1 May be implied 50 19

1 1

2 3 1 2 1 13 404 412 3 52 164r r= =

− = − + − − +

∑ ∑ M1 Difference of their sum from 1r = to 50

and their sum from 1r = to 19

= 0.00813 A1 Cao Invalid method A0 Unseen method SC B1

[4]

15