INSTRUCTIONS TO CANDIDATES

• Write your name, centre number and candidate number in the spaces provided on the answer booklet.

• Answer all the questions.

• You are permitted to use a graphical calculator in this paper.

• Final answers should be given to a degree of accuracy appropriate to the context.

INFORMATION FOR CANDIDATES

• The number of marks is given in brackets [ ] at the end of each question or part question.

• The total number of marks for this paper is 72.

ADVICE TO CANDIDATES

• Read each question carefully and make sure you know what you have to do before starting youranswer.

• You are advised that an answer may receive no marks unless you show sufficient detail of theworking to indicate that a correct method is being used.

NOTE

• This paper will be followed by Paper B: Comprehension.

This document consists of 6 printed pages and 2 blank pages.

HN/5 © OCR 2007 [T/102/2653] OCR is an exempt Charity [Turn over

ADVANCED GCE UNIT 4754(A)/01MATHEMATICS (MEI)Applications of Advanced Mathematics (C4)

Paper A

THURSDAY 14 JUNE 2007 AfternoonTime: 1 hour 30 minutes

Additional materials:Answer booklet (8 pages)Graph paperMEI Examination Formulae and Tables (MF2)

Section A (36 marks)

1 Express in the form where R and a are constants to be determined,and

Hence solve the equation for [7]

2 Write down normal vectors to the planes and

Hence show that these planes are perpendicular to each other. [4]

3 Fig. 3 shows the curve and part of the line

Fig. 3

The shaded region is rotated through 360° about the y-axis.

(i) Show that the volume of the solid of revolution formed is given by [3]

(ii) Evaluate this, leaving your answer in an exact form. [3]

4 A curve is defined by parametric equations

Show that the cartesian equation of the curve is [4]

5 Verify that the point lies on both the lines

Find the acute angle between the lines. [7]

r r=-

Ê

ËÁÁ

ˆ

¯˜˜

+-Ê

ËÁÁ

ˆ

¯˜˜

=Ê

ËÁÁ

ˆ

¯˜˜

+-

Ê

ËÁÁ

ˆ

¯˜˜

121

123

063

102

l mand .

(�1, 6, 5)

y �3 � 2x

2 � x.

x �1t � 1, y �

2 � t

1 � t.

p e d2

0

2y yÛ

ıÙ.

O

2

y

x

y � 2.y � ln x

x � 2y � z � 5.2x � 3y � 4z � 10

0° � q � 360°.sin q � 3 cos q � 1

0° � a � 90°.R sin (q � a),sin q � 3 cos q

2

© OCR 2007 4754A/01 June 07

6 Two students are trying to evaluate the integral

Sarah uses the trapezium rule with 2 strips, and starts by constructing the following table.

(i) Complete the calculation, giving your answer to 3 significant figures. [2]

Anish uses a binomial approximation for and then integrates this.

(ii) Show that, provided is suitably small, [3]

(iii) Use this result to evaluate approximately, giving your answer to 3 significant

figures. [3]

11

2

+ÛıÙ

-e dx x

1 112 1

218

2+( ) ª + -- - -e e ex x x .e�x

1 + -e x

x 1 1.5 2

1.1696 1.1060 1.06551 + e- x

11

2

+ÛıÙ

-e dx x.

3

© OCR 2007 4754A/01 June 07 [Turn over

Section B (36 marks)

7 Data suggest that the number of cases of infection from a particular disease tends to oscillatebetween two values over a period of approximately 6 months.

(a) Suppose that the number of cases, P thousand, after time t months is modelled by the equation

Thus, when

(i) By considering the greatest and least values of , write down the greatest and leastvalues of P predicted by this model. [2]

(ii) Verify that P satisfies the differential equation [5]

(b) An alternative model is proposed, with differential equation

As before,

(i) Express in partial fractions. [4]

(ii) Solve the differential equation (*) to show that

[5]

This equation can be rearranged to give

(iii) Find the greatest and least values of P predicted by this model. [4]

Pt

=-

1

212e sin

.

ln sin .2 1 1

2PP

t-Ê

ˈ¯ =

1P(2P�1)

P � 1 when t � 0.

dP

dt� 1

2 (2P 2 � P) cos t. (*)

dP

dt� 1

2 P 2 cos t.

sin t

t � 0, P � 1.P �2

2 � sin t.

4

© OCR 2007 4754A/01 June 07

8

Fig. 8

In a theme park ride, a capsule C moves in a vertical plane (see Fig. 8). With respect to the axesshown, the path of C is modelled by the parametric equations

where x and y are in metres.

(i) Show that

Verify that when Hence find the exact coordinates of the highest point A on

the path of C. [6]

(ii) Express in terms of Hence show that

[4]

(iii) Using this result, or otherwise, find the greatest and least distances of C from O. [2]

You are given that, at the point B on the path vertically above O,

(iv) Using this result, and the result in part (ii), find the distance OB. Give your answer to 3 significant figures. [4]

2 cos 2q � 2 cos q � 1 � 0.

x 2 � y 2 � 125 � 100 cos q.

q.x 2 � y 2

q � 13 p.

dy

dx� 0

dy

dx� �

cos q � cos 2qsin q � sin 2q

.

x � 10 cos q � 5 cos 2q, y � 10 sin q � 5 sin 2q , (0 � q � 2p),

x

y

C(x, y)

O

B A

5

© OCR 2007 4754A/01 June 07

4754 Mark Scheme June 2007

Section A

1 sin θ − 3 cos θ = R sin(θ − α) = R(sin θ cos α − cos θ sin α) ⇒ R cos α = 1 , R sin α = 3 ⇒ R2 = 12 + 32 = 10 ⇒ R = √10 tan α = 3 ⇒ α = 71.57° √10 sin(θ − 71.57°) = 1 ⇒ θ − 71.57° = sin (1/√10) 1−

θ − 71.57°= 18.43°, 161.57° ⇒ θ = 90° , 233.1°

M1 B1 M1 A1 M1

B1 A1 [7]

equating correct pairs oe ft www cao (71.6°or better) oe ft R, α www and no others in range (MR-1 for radians)

2 Normal vectors are and 234

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

12

1

⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟⎝ ⎠

⇒ 2 13 . 2 2 6 4 04 1

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟− = − + =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⇒ planes are perpendicular.

B1 B1 M1 E1 [4]

3 (i) y = ln x ⇒ x = e y

⇒ 2 2

0V xπ= ∫ dy

= 2 22 2

0 0( ) *y ye dy e dyπ π=∫ ∫

B1 M1 E1 [3]

(ii) 2

2 2 2

00

12

y ye dy eπ π ⎡ ⎤= ⎢ ⎥⎣ ⎦∫

= ½ π(e4 – 1)

B1 M1 A1 [3]

½ e2y

substituting limits in kπe 2 y

or equivalent, but must be exact and evaluate e0 as 1.

4 1 11 1x xt t

= − ⇒ = +

⇒ 11

tx

=+

⇒ 12 2 2 1 2 311 1 1 21

1

x xxyx x

x

+ + + ++= = =+ + ++

+

M1 A1 M1 E1

Solving for t in terms of x or y Subst their t which must include a fraction, clearing subsidiary fractions/ changing the subject oe www

or 2 233 212 2

tx t

txt

−++

=−+ +

= 3 2 22 1t tt t+ −+ −

= 21

tt++

= y

M1 A1 M1 E1 [4]

substituting for x or y in terms of t

clearing subsidiary fractions/changing the subject

14

4754 Mark Scheme June 2007

5 ⇒ 1 12 2

1 3λ

−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠

r12 2

1 3

xyz

λλλ

−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− +⎝ ⎠ ⎝ ⎠

When x = −1, 1 − λ = −1, ⇒ λ = 2 ⇒ y = 2 + 2λ = 6, z = −1 + 3λ = 5 ⇒ point lies on first line

⇒ 0 16 03 2

μ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠

r 63 2

xyz

μ

μ

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟=⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠

When x = −1, μ = −1, ⇒ y = 6, z = 3 − 2μ = 5 ⇒ point lies on second line

Angle between and is θ, where 1

23

−⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

10

2

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟−⎝ ⎠

1 1 2 0 3 2cos14. 5

θ − × + × + × −=

= 770

−

⇒ θ = 146.8° ⇒ acute angle is 33.2°

M1 E1

E1 M1

M1 A1 A1cao [7]

Finding λ or μ checking other two coordinates checking other two co-ordinates Finding angle between correct vectors use of formula

770

±

Final answer must be acute angle

6(i) (1.1696 1.06550.5[ 1.1060]2

A +≈ +

= 1.11 (3 s.f.)

M1

A1 cao [2]

Correct expression for trapezium rule

(ii) 1/ 2 2

1 1.1 2 2(1 ) 1 ( ) ...2 2!

x x xe e e− − −−

+ = + + +

21 112 8

x xe e− −≈ + − *

M1 A1 E1 [3]

Binomial expansion with p = ½ Correct coeffs

(iii) I = 2 2

1

1 1(1 )2 8

x xe e− −+ −∫

dx

= 2

2

1

1 12 16

x xx e e− −⎡ ⎤− +⎢ ⎥⎣ ⎦

= 2 4 11 1 1 1(2 ) (1 )2 16 2 16

e e e e− − −− + − − + 2−

= 1.9335 − 0.8245 = 1.11 (3 s.f.)

M1 A1 A1 [3]

integration substituting limits into correct expression

15

4754 Mark Scheme June 2007

Section B

7 (a) (i) Pmax = 2

2 1− = 2

Pmin = 22 1+

= 2/3.

B1 B1 [2]

(ii) 12 2(2 sin )2 sin

P tt

−= = −−

⇒ 22(2 sin ) . cosdP t tdt

−= − − −

= 2

2cos(2 sin )

tt−

22

1 1 4cos cos2 2 (2 sin )

P tt

=−

t

= 2

2cos(2 sin )

tt−

= dPdt

M1 B1 A1 DM1 E1 [5]

chain rule −1(…)−2 soi (or quotient rule M1,numerator A1,denominator A1) attempt to verify or by integration as in (b)(ii)

(b)(i) 1

(2 1) 2 1A B

P P P P= +

− −

= (2 1)(2 1)

A P BP P

− +−

P

⇒ 1 = A(2P − 1) + BP P = 0 ⇒ 1 = −A ⇒ A = −1 P = ½ ⇒ 1 = A.0 + ½ B ⇒ B = 2 So 1 1

(2 1) 2 1P P P P= − +

− −2

M1 M1

A1 A1 [4]

correct partial fractions substituting values, equating coeffs or cover up rule A = −1 B = 2

(ii) 21 (2 )cos2

dP P P tdt

= −

⇒ 2

1 1 cos2 2

dP tdtP P

=−∫ ∫

⇒ 2 1 1( ) co2 1 2

dP tdtsP P

− =−∫ ∫

⇒ ln(2P − 1) − ln P = ½ sin t + c When t = 0, P = 1 ⇒ ln 1 − ln 1 = ½ sin 0 + c ⇒ c = 0 ⇒ 2 1 1ln( ) sin

2P tP−

= *

M1 A1 A1 B1 E1 [5]

separating variables ln(2P − 1) − ln P ft their A,B from (i) ½ sin t finding constant = 0

(iii) Pmax =

1/ 2

12 e−

= 2.847

Pmin = 1/ 2

12 e−−

= 0.718

M1A1 M1A1 [4]

www www

16

4754 Mark Scheme June 2007

8 (i) 10cos 10cos 210sin 10sin 2

dydx

θ θθ θ+

=− −

= cos cos 2sin sin 2

θ θθ θ

+−

+ *

When θ = π/3, dydx

= cos / 3 cos 2 / 3sin / 3 sin 2 / 3

π ππ π

+−

+

= 0 as cos π/3 = ½ , cos 2π/3 = −½ At A x = 10 cos π/3 + 5 cos 2π/3 = 2½ y = 10 sin π/3 + 5 sin 2π/3 = 15√3/2

M1 E1 B1 M1 A1 A1 [6]

dy/dθ ÷dx/dθ or solving cosθ+cos2θ=0 substituting π/3 into x or y 2½ 15√3/2 (condone 13 or better)

(ii) 2 2 2(10cos 5cos 2 ) (10sin 5sin 2 )x y 2θ θ θ+ = + + + θ

θθ

= 2 2100 cos 100 cos cos 2 25cos 2θ θ θ+ + 2 2100sin 100sin sin 2 25sin 2θ θ θ+ + + = 100 + 100cos(2θ − θ) + 25 = 125 + 100cos θ *

B1 M1 DM1 E1 [4]

expanding cos 2θ cos θ + sin 2θ sin θ = cos(2θ − θ) or substituting for sin 2θ and cos 2θ

(iii) Max 125 100+ = 15 min 125 100− = 5

B1 B1 [2]

(iv) 2cos2 θ +2cos θ − 1 = 0

cos θ = 2 12 2 2 34 4

− ± − ±=

At B, cos θ = 1 32

− +

OB2 = 125 + 50(−1 + √3) = 75 + 50√3 = 161.6… ⇒ OB = √161.6… = 12.7 (m)

M1 A1 M1 A1 [4]

quadratic formula or θ=68.53° or 1.20radians, correct root selected or OB=10sinθ+5sin2θ ft their θ/cosθ oe cao

17