OCR ADVANCED SUBSIDIARY GCE
IN MATHEMATICS (3840, 3841, 3842, 3843 and 3844)
OCR ADVANCED GCE
IN MATHEMATICS (7840, 7842 and 7844)
Specimen Question Papers and Mark Schemes
These specimen question papers and mark schemes are intended to accompany the OCR Advanced
Subsidiary GCE and Advanced GCE specifications in Mathematics for teaching from September
2000.
Centres are permitted to copy material from this booklet for their own internal use.
The GCE awarding bodies have prepared new specifications to incorporate the range of features
required by new GCE and subject criteria. The specimen assessment material accompanying the new
specifications is provided to give centres a reasonable idea of the general shape and character of the
planned question papers in advance of the first operational examination.
Specimen Materials - Mathematics
1
© OCR 2000
CONTENTS
Advanced Subsidiary GCE
Unit 2631: Pure Mathematics 1
P1
Question Paper
Mark Scheme
Unit 2637: Mechanics 1
Page
Page
M1
Question Paper
Mark Scheme
Unit 2641: Probability and Statistics 1
Page 53
Page 57
S1
Question Paper
Mark Scheme
Unit 2645: Discrete Mathematics 1
5
9
Page 85
Page 89
D1
Question Paper
Mark Scheme
Page 117
Page 121
A2
Unit 2632: Pure Mathematics 2
P2
Question Paper
Mark Scheme
Unit 2633: Pure Mathematics 3
Page 13
Page 17
P3
Question Paper
Mark Scheme
Unit 2634: Pure Mathematics 4
Page 21
Page 25
P4
Question Paper
Mark Scheme
Unit 2635: Pure Mathematics 5
Page 29
Page 33
P5
Question Paper
Mark Scheme
Unit 2636: Pure Mathematics 6
Page 37
Page 41
P6
Question Paper
Mark Scheme
Specimen Materials - Mathematics
Page 45
Page 49
2
© OCR 2000
A2 continued
Unit 2638: Mechanics 2
M2
Question Paper
Mark Scheme
Unit 2639: Mechanics 3
Page 61
Page 65
M3
Question Paper
Mark Scheme
Unit 2640: Mechanics 4
Page 69
Page 73
M4
Question Paper
Mark Scheme
Unit 2642: Probability and Statistics 2
Page 77
Page 81
S2
Question Paper
Mark Scheme
Unit 2643: Probability and Statistics 3
Page 93
Page 97
S3
Question Paper
Mark Scheme
Unit 2644: Probability and Statistics 4
Page 101
Page 105
S4
Question Paper
Mark Scheme
Unit 2646: Discrete Mathematics 2
Page 109
Page 113
D2
Question Paper
Insert
Mark Scheme
Page 125
Page 133
Page 137
Marking Instructions
Page 143
Specimen Materials - Mathematics
3
© OCR 2000
Specimen Materials
Mathematics
4
© OCR 2000
Oxford, Cambridge and RSA Examinations
General Certificate of Education
Advanced Subsidiary (AS) and Advanced Level
P1
MATHEMATICS
Pure Mathematics 1
Additional materials:
Answer paper
Graph paper
List of Formulae
TIME
1 hour 20 minutes
INSTRUCTIONS TO CANDIDATES
Write your name, Centre number and candidate number in the spaces provided on the answer paper.
Answer all the questions.
Give non-exact numerical answers correct to 3 significant figures, unless a different degree of
accuracy is specified in the question or is clearly appropriate.
You are permitted to use only a scientific calculator in this paper.
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 60.
Questions carrying smaller numbers of marks are printed earlier in the paper, and questions carrying
larger numbers of marks later in the paper.
You are reminded of the need for clear presentation in your answers.
This question paper consists of 3 printed pages and 1 blank page.
Specimen Materials - Mathematics
5
© OCR 2000
1
(i) Write down the exact value of 7 − 2 .
(ii) Simplify
2
[1]
(x x )3
.
2 x4
[2]
(i) Solve the simultaneous equations
y = x 2 − 3x + 2,
y = 3x − 7.
(ii) Interpret your solution to part (i) geometrically.
[4]
[1]
3
The point A has coordinates (7, 4) . The straight lines with equations x + 3 y + 1 = 0 and 2x + 5 y = 0
intersect at the point B. Show that one of these two lines is perpendicular to AB.
[6]
4
Show that the equation
15 cos 2 θ = 13 + sin θ
5
may be written as a quadratic equation in sin θ .
[2]
Hence solve the equation, giving all values of θ such that 0 ° ≤ θ ≤ 360 ° .
[5]
Sketch the graph of y = cos x ° , for values of x from 0 to 360.
[1]
Sketch, on the same diagram, the graph of y = cos(x − 60)° .
[2]
Use your diagram to solve the equation
cos x° = cos(x − 60)°
for values of x between 0 and 360. Indicate clearly on your diagram how the solutions relate to the graphs.
[3]
State how many values of x satisfying the equation
cos(10 x)° = cos(10 x − 60 )°
lie between 0 and 360. (You should explain your reasoning briefly, but no further detailed working or
sketching is necessary.)
[2]
Specimen Materials - Mathematics
6
© OCR 2000
6
(i) Evaluate
4
∫0 x(4 − x) dx .
[3]
(ii)
The diagram shows the curve y = x (4 − x ) , together with a straight line. This line cuts the curve at
the origin O and at the point P with x-coordinate k, where 0 < k < 4 .
(a) Show that the area of the shaded region, bounded by the line and the curve, is 16 k 3 .
[4]
(b) Find, correct to 3 decimal places, the value of k for which the area of the shaded region is half of
the total area under the curve between x = 0 and x = 4 .
[2]
7
A quadratic function is defined by
f( x) = x 2 + kx + 9 ,
where k is a constant. It is given that the equation f(x ) = 0 has two distinct real roots. Find the set of
values that k can take.
[3]
For the case where k = −4 3 ,
(i) express f(x) in the form ( x + a )2 + b , stating the values of a and b, and hence write down the least
value taken by f(x) ,
[4]
(ii) solve the equation f(x ) = 0 , expressing your answer in terms of surds, simplified as far as possible. [3]
8
The equation of a curve is y = 6 x2 − x3 . Find the coordinates of the two stationary points on the curve,
and determine the nature of each of these stationary points.
[6]
State the set of values of x for which 6 x 2 − x 3 is a decreasing function of x.
[2]
The gradient at the point M on the curve is 12. Find the equation of the tangent to the curve at M.
[4]
Specimen Materials - Mathematics
7
© OCR 2000
BLANK PAGE
Specimen Materials - Mathematics
8
© OCR 2000
General Certificate of Education
Advanced Subsidiary (AS) and Advanced Level
P1
MATHEMATICS
Pure Mathematics 1
MARK SCHEME
MAXIMUM MARK
60
For live examinations, each Mark Scheme
includes the General Instructions for Marking
set out on pages 143 to 145.
This mark scheme consists of 4 printed pages.
Specimen Materials - Mathematics
9
© OCR 2000
1
1
(i) 49
B1
1 Correct value stated as final answer
-------------------------------------------------------------------------------------------------------------------------------------------------------1
( x x )3 x 42
(ii)
= 4
M1
Power 3 × 1 12 or 3 + 1 12 in numerator
2x 4
2x
1
2
2
1
x 2 or
1
2
x
A1
(i) EITHER : 3x − 7 = x 2 − 3x + 2
x 2 − 6x + 9 = 0
x = 3 only
y = 2 only
M1
A1
A1
A1
2 Or any equally simple equivalent
Eliminate y to obtain an equation in x only
Correct 3-term equation in x
Obtained by any correct solution method
If two values of x are found both y-values
must follow correctly
y + 7
 y + 7 + 2
y = 
 − 3

 3 
 3 
M1
y2 − 4y + 4 = 0
y = 2 only
x = 3 only
A1
A1
A1
2
OR :
Eliminate x to obtain an equation in y only
Correct 3-term equation in y
Obtained by any correct solution method
4 If two values of y are found both x-values
must follow correctly
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii) The line y = 3x − 7 is the tangent to the curve
y = x 2 − 3 x + 2 at the point (3, 2)
3
Solve x + 3y + 1 = 0 and 2x + 5y = 0 simultaneously
x = 5, y = − 2 at B
B1
M1
A1
1 For identifying tangency
Attempt soln and obtain at least one answer
Identify correct coordinates with B, either
explicitly or implicitly
4 − ( −2)
=3
Gradient of AB is
7 −5
Gradients of the lines are − 13 and − 25
4
A1
For simplified follow-through value
B1
For either gradient correctly stated or used
Perpendicular lines require m1m2 = −1
M1
Any statement or use of the correct relation
AB is perpendicular to x + 3y + 1 = 0
A1
15 (1 − sin 2 θ ) = 13 + sin θ
M1
6 Correct use of 3 × − 13 = − 1 , or equivalent
Attempted relevant use of sin 2 θ + cos2 θ = 1
15sin 2 θ + sin θ − 2 = 0
A1
2 Any correct 3-term form
-------------------------------------------------------------------------------------------------------------------------------------------------------(5 sin θ + 2)( 3sin θ − 1) = 0
M1
Any recognisable solution method attempted
sin θ = − 25 or
1
3
θ = 19 .5 °, 160 . 5°, 203 .6 °, 336 .4 °
Specimen Materials - Mathematics
A1
A1
A1
A1
10
Both correct values
For any one correct value
For a second correct value
5 For both remaining values, and no others
© OCR 2000
5
B1
1 Correct y = cos x over (0, 360) ; ignore
anything outside this interval
--------------------------------------------------------------------------M1
Translation parallel to the x-axis recognised
A1
2 For correct y = cos(x − 60) throughout
(0, 360) ; ignore anything outside this
interval
-------------------------------------------------------------------------------------------------------------------------------------------------------Indicate use of points of intersection on diagram
B1
For identifying the points, not necessarily
the x-coordinates
x = 30 , 210
B1
For either correct value
B1
3 For second correct value and no others
-------------------------------------------------------------------------------------------------------------------------------------------------------Graphs are ‘squashed’ ×10 in the x-direction
M1
Or any equivalent method
There are 20 solutions
A1
2 10 × their number of solutions above
6
(i) Expand to 4 x − x 2 and integrate
[2x 2 − 13 x 3]04
M1
At least one integrated term with correct
power
A1
Correct indefinite integral
32 − 64
= 32
A1
3 Follow correct use of limits 4 and 0
3
3
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii) (a)
k
∫0 x(4 − x) dx = 2k 2 − 13 k 3
B1
Follow their earlier indefinite integral
Area of triangle = 12 × k × k (4 − k)
M1
To include attempt at y-coordinate of P in
terms of k
= 2k −
2
1 3
k
2
A1
Correct simplified expression
Shaded area = 2k −
− ( 2k −
=
A1
4 Given answer correctly obtained
-------------------------------------------------------------------------------------------------------------------------------------------------------M1
Using previous results correctly to form an
(b) 16 k 3 = 12 × 32
3
2
1 3
k
3
2
k = 3 .175
7
1 3
k )
2
1 3
k
6
A1
equation for k
2 Correct 3dp value
k 2 > 4 × 1× 9
B1
Correct condition stated in any form
( k − 6)(k + 6) > 0
M1
Factorise or carry out other solution method
k < −6, k > 6 or k > 6
A1
3 Do not allow 6 < k < −6
-------------------------------------------------------------------------------------------------------------------------------------------------------M1
May be implied by correct a and/or b
(i) EITHER : ( x − 2 3) 2 + 9 − ( 2 3) 2
a = −2 3
b = −3
Least value of f( x) is −3
A1
A1
B1
Their value of b
2a = −4 3 and a 2 + b = 9
M1
Expand and equate at least one pair of coeffs
a = −2 3
A1
b = −3
A1
B1
4 Their value of b
Least value of f( x) is −3
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii) Critical values where x − 2 3 = ± 3
M1
Or equivalent, via quadratic formula
x = 3 or 3 3
A1
For either critical value correctly obtained
A1
3 For completely correct answer; allow 27
OR :
Specimen Materials - Mathematics
11
© OCR 2000
8
dy
= 12 x − 3x2
dx
3x(4 − x) = 0
x = 0 and 4
Stationary points are (0, 0) and ( 4, 32)
B1
Correct derivative stated or used
M1
Equate to zero and factorise, or equivalent
A1
A1
when x = 0
d2 y
 + 12
= 12 − 6x = 
M1
Or other correct alternative method
dx 2
−
12
when x = 4

(0, 0) is a minimum and ( 4, 32) is a maximum
A1
6 Correct conclusion from correct working
-------------------------------------------------------------------------------------------------------------------------------------------------------x < 0, x > 4
B1
Either interval stated
B1
2 Both intervals, and no others, correct
-------------------------------------------------------------------------------------------------------------------------------------------------------dy
12x − 3x 2 = 12
M1
Equate
to 12 and solve for x
dx
x=2
A1
M1
Use of straight line equation with numerical
Tangent passes through ( 2, 16) with gradient 12
y = 12x − 8
Specimen Materials - Mathematics
A1
12
gradient and numerical coordinates of M
4 Any correct equivalent 3-term form
© OCR 2000
General Certificate of Education
Advanced Subsidiary (AS) and Advanced Level
P2
MATHEMATICS
Pure Mathematics 2
Additional materials:
Answer paper
Graph paper
List of Formulae
TIME
1 hour 20 minutes
INSTRUCTIONS TO CANDIDATES
Write your name, Centre number and candidate number in the spaces provided on the answer paper.
Answer all the questions.
Give non-exact numerical answers correct to 3 significant figures, unless a different degree of
accuracy is specified in the question or is clearly appropriate.
You are permitted to use a graphic calculator in this paper.
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 60.
Questions carrying smaller numbers of marks are printed earlier in the paper, and questions carrying
larger numbers of marks later in the paper.
You are reminded of the need for clear presentation in your answers.
This question paper consists of 4 printed pages.
Specimen Materials - Mathematics
13
© OCR 2000
1
2
The cubic polynomial 3x 3 − 7 x 2 − 18 x − 8 is denoted by f(x) . Use the factor theorem to show that ( x + 1)
is a factor of f(x) .
[2]
Hence factorise f(x) completely.
[3]
Solve the inequality x − 100 < 10 .
[2]
Hence find the set of integers n that satisfy the inequality
1.01n − 100 < 10 .
3
[3]
Expand (1 + x)5 in ascending powers of x, simplifying the coefficients.
[2]
Hence, by letting x = y + y 2 , find the coefficient of y4 in the expansion of (1 + y + y2 )5 in powers of y. [4]
4
The diagram shows the region R which is bounded by the curve y =
and x = 5 . Use integration
2
, the x-axis, and the lines x = 1
x +1
(i) to find the area of R, giving your answer as a single logarithm,
[4]
(ii) to show that the volume of the solid formed when R is rotated completely about the x-axis is 43 π . [4]
5
At time t minutes after an oven is switched on, its temperature θ °C is given by
θ = 200 − 180 e− 0.1t .
(i) State the value which the oven’s temperature approaches after a long time.
[1]
(ii) Find the time taken for the oven’s temperature to reach 150 °C .
[3]
(iii) Find the rate at which the temperature is increasing at the instant when the temperature reaches
150 °C .
[4]
Specimen Materials - Mathematics
14
© OCR 2000
6
The diagram shows a semicircle ABC on AC as diameter. The mid-point of AC is O, and angle
AOB = θ radians , where 0 < θ < 12 π . The area of the segment S1 bounded by the chord BC is twice the
area of the segment S2 bounded by the chord AB. Show that
3θ = π + sin θ .
[4]
Use an iterative method, based on the rearrangement
θ = 13 (π + sin θ ),
together with a suitable starting value, to find θ correct to 3 decimal places. You should show the value of
each approximation that you calculate.
[4]
7
The functions f and g are defined by
1
f : x a 1 + x2
g : x a x2
x ≥ 0,
x ∈ R.
(i) Find the domain of the inverse function f −1 .
[2]
(ii) Find an expression for f −1( x) .
[2]
(iii) Find and simplify an expression for fg(x) for the case where x ≥ 0 .
[2]
(iv) Explain clearly why the value of fg(−2) is 3.
[1]
(v) Sketch the graph of y = fg(x ) , for both positive and negative values of x, and give the equation of this
graph in a simplified form.
[3]
Specimen Materials - Mathematics
15
© OCR 2000
8
(i) A sequence of positive integers u1, u2 , u3, K is given by
u1 = 2
and
un +1 = 2un for n ≥ 1 .
(a) Write down the first four terms of this sequence.
[1]
(b) State what type of sequence this is, and express un in terms of n.
[2]
(ii) A sequence of positive integers v1, v2, v3, K is given by
v1 = 3
and
v n+1 = 2v n − 1 for n ≥ 1 .
(a) Show that the relation between v n+ 1 and vn may be written in the form
v n+1 − 1 = 2(v n − 1) .
(b) Hence, by using the results in part (i), show that vn = 2 n + 1 for n ≥ 1 .
[1]
[2]
(iii) The sum of the first N terms of the sequence v1, v2, v3, K is denoted by SN , i.e.
S N = v1 + v2 + v3 + K + vN .
Express SN in terms of N.
Specimen Materials - Mathematics
[4]
16
© OCR 2000
General Certificate of Education
Advanced Subsidiary (AS) and Advanced Level
P2
MATHEMATICS
Pure Mathematics 2
MARK SCHEME
MAXIMUM MARK
60
For live examinations, each Mark Scheme
includes the General Instructions for Marking
set out on pages 143 to 145.
This mark scheme consists of 3 printed pages and 1 blank page.
Specimen Materials - Mathematics
17
© OCR 2000
1
f( −1) = − 3 − 7 + 18 − 8 = 0
M1
Substitute x = −1 and evaluate
A1
2 Zero correctly obtained, and conclusion
Hence ( x + 1) is a factor
-------------------------------------------------------------------------------------------------------------------------------------------------------f( x) = ( x + 1)(3 x 2 − 10 x − 8)
M1
Carry out division (trinomial quotient) or
Answer ( x + 1)( 3x + 2)( x − 4)
2
3
4
attempt factorisation by inspection (3 terms
with at least 3x 2 and ±8 )
Correct quadratic factor
A1
A1
3
90 < x < 110
B1
Either end-point obtained
B1
2 Completely correct solution set
-------------------------------------------------------------------------------------------------------------------------------------------------------ln 90 < n ln 1 .01 < ln 110
M1
Correct use of logs in any equation or
inequality of the form 1.01n = c
452 . 2 < n < 472 .4
A1
Either value, in exact or decimal form
453 to 472 inclusive
A1
3 Allow any clear notation
5
5
5
5
5
M1
1 +   x +   x 2 +   x3 +   x4 +   x5
 1
 2
 3
 4
 5
1 + 5x + 10x2 + 10x3 + 5x 4 + x5
A1
2
-------------------------------------------------------------------------------------------------------------------------------------------------------1 + 5( y + y 2 ) + 10 ( y + y 2 ) 2 + 10 ( y + y 2 )3 + K
M1
May be implied
y 4 occurs in ( y + y2 ) 2 , ( y + y2 ) 3 and ( y + y2 ) 4
M1
Expand or pick out the relevant terms in at
10 y 4 + 30 y4 + 5 y 4
A1
least two of these cases
At least two of the three terms correct
Coefficient of y 4 is 45
A1
5
5
2
(i) ⌠
dx =  2 ln(x + 1) 


1
⌡1 x + 1
4 Allow answer 45 y 4
M1
For indefinite integral involving a log
A1
Correct indefinite integral 2 ln( x + 1)
2 ln 6 − 2 ln 2 = ln 9
M1
Use of both limits and at least one law of logs
A1
4 Correct simplified answer ln 9
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii) V = π ∫ y2 dx
B1
Correct formula stated or used
π [− 4( x + 1) −1] 1
5
π ( − 46 + 2) = 43 π
5
M1
Integration attempt with negative index result
A1
Correct indefinite integral − 4( x + 1) −1
A1
4 Given answer correctly shown
(i) 200 °C
B1
1 Allow answer 200 without units
-------------------------------------------------------------------------------------------------------------------------------------------------------5
(ii) 150 = 200 − 180 e− 0. 1t ⇒ e −0 .1t = 18
M1
Substitute θ = 150 and rearrange
5
− 0.1t = ln(18
)
M1
Take logs correctly
Answer 12.8 minutes
A1
3 Allow answer t = 12.8 without units
-------------------------------------------------------------------------------------------------------------------------------------------------------dθ
(iii) Rate of increase of temperature is
B1
Recognition may be implied
dt
dθ
dθ
= 18 e −0.1t
M1
Attempt at
involving a term k e− 0. 1t
dt
dt
A1
Correct differentiation
A1
4 Follow value of t or e −0.1t from (ii)
Answer 5 °C per minute
Specimen Materials - Mathematics
18
© OCR 2000
6
Area of S2 is
Area of S1 is
1 2
r (θ
2
1 2
r (π
2
− sin θ )
B1
− θ − sin(π − θ ))
B1
− θ − sin θ ) = 2 × 12 r 2 (θ − sin θ )
M1
Equate and attempt to simplify sin(π − θ )
i.e. 3θ = π + sin θ
A1
4 Given answer correctly shown
-------------------------------------------------------------------------------------------------------------------------------------------------------A suitable initial value is θ = 1 (e.g.)
B1
State or use any θ1 such that 0 ≤ θ1 ≤ 12 π
Hence
7
1 2
r (π
2
θ 2 = 13 (π + sin θ1) = 1. 327... (e.g.)
M1
θ ≈ 1. 374
A1
A1
For one iteration using correct formula
For correct θ 2 from their θ1
4 For correct answer and sufficient iterations
to justify 3dp accuracy
(i) Domain of f −1 is the range of f
M1
May be implied
A1
2 Allow any intelligible notation
i.e. x ≥ 1
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii) y = 1 + x ⇒ x = ( y − 1) 2
M1
Attempt to solve for x
f −1 ( x) = ( x − 1) 2
2 For answer ( x − 1) 2 or x 2 − 2 x + 1 ; do not
A1
allow final answer in terms of y
-------------------------------------------------------------------------------------------------------------------------------------------------------1
M1
Attempt composition in the correct order
(iii) fg(x) = 1 + ( x 2 ) 2
=1+ x
A1
2
-------------------------------------------------------------------------------------------------------------------------------------------------------(iv) g(− 2) = 4, f(4) = 3
B1
1 Intermediate value 4 must be seen
-------------------------------------------------------------------------------------------------------------------------------------------------------(v)
Sketch of y = 1 + x for x ≥ 0 only
Sketch reflection in the y-axis for x ≤ 0 only
B1
B1
y = 1+ x
8
B1
3
(i) (a) 2, 4, 8, 16
B1
1 All four terms correctly stated
-------------------------------------------------------------------------------------------------------------------------------------------------------(b) This is a geometric progression
B1
No need to state first term or ratio here
Hence u n = 2 × 2 n −1 = 2 n
B1
2 For simplified answer 2 n
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii) (a) vn +1 = 2vn − 1 ⇒ vn+1 − 1 = 2vn − 2 = 2( vn − 1) B1
1 Given result correctly shown
-------------------------------------------------------------------------------------------------------------------------------------------------------(b) vn − 1 satisfies same relation as un
M1
For making the connection to part (i)
Hence vn = un + 1 = 2 n + 1
A1
2 Given result correctly shown; no need for an
explicit check that 3 = 2 + 1
-------------------------------------------------------------------------------------------------------------------------------------------------------(iii) Σvn = Σ(2 n ) + Σ(1)
M1
For considering the two separate sums (use
of Σ notation is not expected in this module)
=
2(2 − 1)
+N
2 −1
=
2(2 N
N
− 1) + N or 2
Specimen Materials - Mathematics
N +1
+N−2
B1
For correct unsimplified GP sum
B1
For correct statement of Σ(1) = N
A1
4
19
© OCR 2000
BLANK PAGE
Specimen Materials – Mathematics
20
© OCR 2000
General Certificate of Education
Advanced Subsidiary (AS) and Advanced Level
P3
MATHEMATICS
Pure Mathematics 3
Additional materials:
Answer paper
Graph paper
List of Formulae
TIME
1 hour 20 minutes
INSTRUCTIONS TO CANDIDATES
Write your name, Centre number and candidate number in the spaces provided on the answer paper.
Answer all the questions.
Give non-exact numerical answers correct to 3 significant figures, unless a different degree of
accuracy is specified in the question or is clearly appropriate.
You are permitted to use only a scientific calculator in this paper.
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 60.
Questions carrying smaller numbers of marks are printed earlier in the paper, and questions carrying
larger numbers of marks later in the paper.
You are reminded of the need for clear presentation in your answers.
This question paper consists of 3 printed pages and 1 blank page.
Specimen Materials - Mathematics
21
© OCR 2000
1
2
Expand (1 − 2 x )− 2 in ascending powers of x, up to and including the term in x 2 .
[3]
State the set of values of x for which the expansion is valid.
[1]
1
Prove the identity
sin( x + 30 °) + 3 cos(x + 30 °) ≡ 2 cos x ,
3
where x is measured in degrees.
[3]
Hence express cos 15° in surd form.
[2]
By using the substitution u = sin x , or otherwise, find
⌠
 sin 3 x sin 2x dx ,
⌡
4
giving your answer in terms of x.
[6]
Find the centre and radius of the circle with equation x2 + y2 = 6 x .
[2]
The line x + y = k is a tangent to this circle. Find the two possible values of the constant k, giving your
answers in surd form.
[5]
5
The points A and B have coordinates (3, 2, 4) and (4, 4, − 3) respectively. The line l1 , which passes
through A, has equation
 3  5
r =  2  + t 1  .
 4  1
   
Show that AB is perpendicular to l1 .
[3]
The line l2 , which passes through B, has equation
 4  2
r =  4  + s 1 .
 − 3  − 2 
   
Show that the lines l1 and l2 intersect, and find the coordinates of their point of intersection.
Specimen Materials - Mathematics
22
[5]
© OCR 2000
6
The parametric equations of a curve are
x = a sin θ ,
y = aθ cos θ ,
where a is a positive constant and 0 < θ < 12 π . Find
of the curve is zero where tanθ =
dy
in terms of θ , and hence show that the gradient
dx
1
.
θ
[6]
By sketching a suitable pair of graphs, show that the equation tanθ =
in the relevant range.
1
is satisfied by just one value of θ
θ
[2]
Determine, with reasons, whether this value of θ is greater or less than 14 π .
7
Express
15 − 13x + 4 x2
in partial fractions.
(1 − x)2 (4 − x)
[2]
[5]
Hence show that
3
⌠ 15 − 13x + 4 x 2
 (1 − x )2(4 − x) dx = 1 + ln 4 .
⌡2
[5]
8
A cylindrical container has a height of 200 cm. The container was initially full of a chemical but there is a
leak from a hole in the base. When the leak is noticed, the container is half-full and the level of the
chemical is dropping at a rate of 1 cm per minute. It is required to find for how many minutes the
container has been leaking. To model the situation it is assumed that, when the depth of the chemical
remaining is x cm, the rate at which the level is dropping is proportional to
x . Set up and solve an
appropriate differential equation, and hence show that the container has been leaking for about 80 minutes.
[10]
Specimen Materials - Mathematics
23
© OCR 2000
BLANK PAGE
Specimen Materials - Mathematics
24
© OCR 2000
General Certificate of Education
Advanced Subsidiary (AS) and Advanced Level
P3
MATHEMATICS
Pure Mathematics 3
MARK SCHEME
MAXIMUM MARK
60
For live examinations, each Mark Scheme
includes the General Instructions for Marking
set out on pages 143 to 145.
This mark scheme consists of 4 printed pages.
Specimen Materials - Mathematics
25
© OCR 2000
1
1 + (− 12)( −2x) +
( − 12)( − 23)
(−2 x) 2 + K
2
1 + x + 32 x 2
M1
Either x or x 2 term correct (unsimplified)
A1
For 1 + x correct
A1
3 For + 32 x 2 correct
-------------------------------------------------------------------------------------------------------------------------------------------------------x < 12 or − 12 < x < 12
B1
1
2
}
sin(x + 30°) = sin x cos 30° + cos x sin 30°
cos(x + 30°) = cos x cos 30° − sin x sin 30°
1
2
3 sin x + 12 cos x + 3 ( 12 3 cos x − 12 sin x)
B1
Both expansions correct
B1
Both exact values substituted throughout
+
= 2 cos x
B1
3 Given result shown correctly
-------------------------------------------------------------------------------------------------------------------------------------------------------sin 45° + 3 cos 45 ° = 2 cos15°
M1
Substitute x = 15° and use exact values for
sin 45 ° and cos 45 °
1+ 3
A1
2 Allow any equivalent surd form
cos15 ° =
2 2
1
cos x
2
3
du
= cos x
dx
sin 2 x = 2 sin x cos x
∫
sin 3 x sin
2 5
u
5
4
3
cos x
2
+c =
EITHER :
OR :
2x dx = ∫
2
sin 5
5
u3
2u du = ∫
2u4
du
x+c
B1
Or equivalent; may be implied
B1
Stated or used
M1
Substituting for x and dx throughout
A1
Correct simplified integral in terms of u
M1
Integrate and substitute back
A1
6 Correct answer in terms of x (including +c )
( x − 3) 2 + y 2 = 3 2
M1
Complete the square to obtain standard form
Centre is (3, 0) and radius is 3
A1
Both correct
Centre ( − g, − f ) is (3, 0)
B1
Formula may be implied
Radius
g 2 + f 2 − c is 3
B1
2 Ditto; allow this mark for a correct radius
even if there appears to be a sign error in g
-------------------------------------------------------------------------------------------------------------------------------------------------------EITHER : x2 + ( k − x) 2 = 6 x or (k − y )2 + y 2 = 6 (k − y)
B1
Eliminate x or y completely
2 x2 − (2 k + 6) x + k 2 = 0 or
OR :
2 y 2 − (2 k − 6 ) y + (k 2 − 6 k ) = 0
M1
Simplify to quadratic form
( 2k + 6) 2 − 8k 2 = 0 or ( 2k − 6) 2 − 8(k 2 − 6 k) = 0
M1
Equate discriminant to zero
k 2 − 6k − 9 = 0
k = 3±3 2
A1
Or any equivalent 3-term quadratic in k
A1
Allow any equivalent exact form
Gradient of the tangent is −1
Perpendicular diameter is y − 0 = +1( x − 3)
B1
M1
Using their (3, 0) and perpendicular gradient
x2 + ( x −3) 2 = 6 x or ( y + 3)2 + y2 = 6( y + 3)
M1
Substitute, and solve for x or y
x = 3 ± 32 2 , y = ± 32 2
A1
At least one correct value of each
k = 3±3 2
A1
Specimen Materials - Mathematics
5 For both values, in any equivalent exact form
26
© OCR 2000
5
 1
AB =  2
− 7 
 
 5
Direction of l1 is  1
 1
 
B1
May be implied
B1
May be implied
1 × 5 + 2 × 1 − 7 × 1 = 0 , hence result
B1
3 Given result correctly shown
-------------------------------------------------------------------------------------------------------------------------------------------------------3 + 5t = 4 + 2 s, 2 + t = 4 + s, 4 + t = −3 − 2 s
B1
All three equations stated
Solve any pair of these equations simultaneously
M1
s = − 3, t = −1
A1
For both values correct
Check the solutions in the remaining equation
B1
Point of intersection is (−2, 1, 3)
A1
5 Allow answer as a position vector
6
x& = a cosθ , y& = a(cosθ − θ sin θ )
dy cosθ − θ sin θ
=
dx
cosθ
cosθ − θ sin θ
=0
cosθ
M1
Differentiate both, with product rule attempt
A1
Both correct, in any form
M1
Use of y& / x&
A1
Any correct form, involving θ only
dy
dy
Equate
or
to zero
dx
dθ
M1
1
A1
6 Given result correctly shown
θ
--------------------------------------------------------------------------------------------------------------------------------------------------------
1 − θ tanθ = 0 ⇒ tanθ =
B1
B1
Both curves correct
2 Correct conclusion, referring to single point
of intersection
-------------------------------------------------------------------------------------------------------------------------------------------------------1 4
M1
Or equivalent calculation(s)
EITHER : At θ = 14 π , tanθ = 1 and = > 1
θ π
OR :
7
Hence root is greater than 14 π
A1
Carry out any solution method
M1
θ ≈ 0.86 , so greater than 14 π
A1
A
B
C
+
+
1 − x (1 − x) 2 4 − x
B=2
C =3
15 − 13 x + 4 x 2 ≡ A(1 − x)(4 − x) + B (4 − x) + C (1 − x) 2
E.g. θ n +1 = tan− 1(1 /θ n )
2
B1
Correct form stated or implied
B1
B1
M1
May be obtained by the ‘cover-up’ rule
Ditto
Any correct use of this identity to give an
equation involving A
A =1
A1
5
--------------------------------------------------------------------------------------------------------------------------------------------------------
[− ln1 − x + 2(1 − x)
−1
− 3ln 4 − x
]
3
2
( − ln 2 − 1 − 3 ln1) − (− ln 1 − 2 − 3ln 2) = 1 + ln 4
B1
For both log terms, allowing omission of
B1
modulus signs at this stage
For + 2(1 − x) −1 , or equivalent
M1
Use both limits and combine terms
A1
A1
Specimen Materials - Mathematics
For correct reduction to given form of answer
5 Correct use of modulus throughout (not
dependent on correct answer being reached)
27
© OCR 2000
8
dx
dx
= −k x or
=k x
dt
dt
dx
x = 100 and
= −1 ⇒ k = ( ±)0.1
dt
dx
= −0. 1 x
dt
∫ x−
1
2
dx = −0.1∫ dt
1
2
2 x = −0.1t + A
A1
dx
for rate of change
dt
Correct form of equation (with either sign)
M1
Attempted evaluation of k from their DE
A1
Any equivalent correct form
M1
Separate and integrate both sides
A1
For both 2x 2 and (±)kt (the numerical
M1
Use of
1
evaluation of k may be delayed until after the
DE is solved)
For one arbitrary constant included, or
equivalent statement of both pairs of limits
B1
x = 200 , t = 0 ⇒ A = 2 200
M1
2 100 = −0.1t + 2 200
= 82 .84 K ≈ 80
M1
A1
Specimen Materials - Mathematics
If limits are used, this mark is combined with
the following one
Evaluating t
10 Given result correctly obtained
28
© OCR 2000
General Certificate of Education
Advanced Subsidiary (AS) and Advanced Level
P4
MATHEMATICS
Pure Mathematics 4
Additional materials:
Answer paper
Graph paper
List of Formulae
TIME
1 hour 20 minutes
INSTRUCTIONS TO CANDIDATES
Write your name, Centre number and candidate number in the spaces provided on the answer paper.
Answer all the questions.
Give non-exact numerical answers correct to 3 significant figures, unless a different degree of
accuracy is specified in the question or is clearly appropriate.
You are permitted to use a graphic calculator in this paper.
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 60.
Questions carrying smaller numbers of marks are printed earlier in the paper, and questions carrying
larger numbers of marks later in the paper.
You are reminded of the need for clear presentation in your answers.
This question paper consists of 4 printed pages.
Specimen Materials - Mathematics
29
© OCR 2000
1
A sequence of positive integers u1, u 2, u 3 K is defined by
u1 = 1 and
un +1 = 3u n + 2 for n ≥ 1 .
Prove by induction that
u n = 2(3n −1) − 1 .
2
[4]
Find the general solution of the differential equation
dy y
− = x,
dx x
giving y in terms of x in your answer.
3
[4]
For positive integers r, let
f(r) =
1
.
r(r + 1)
Verify that
f(r) − f(r + 1) =
2
.
r(r + 1)(r + 2)
[1]
Hence find the sum of the first n terms of the series
1
1
1
+
+K+
+ K.
1.2.3 2 .3 .4
r (r + 1)(r + 2)
4
[3]
Show that the series is convergent, and state the sum to infinity.
[2]
Find the first three terms of the Maclaurin series for ln( 2 + x) .
[4]
Write down the first three terms of the series for ln( 2 − x ) , and hence show that, if x is small, then
2 + x
ln 
≈x.
2 − x
Specimen Materials - Mathematics
[3]
30
© OCR 2000
5
The diagram shows a sketch of the graph of
y=
2 x2 + 3x + 3
.
x +1
Find
(i) the equations of the asymptotes of the curve,
[3]
(ii) the values of y between which there are no points on the curve.
[4]
6
The diagram shows the curve whose equation, in cartesian coordinates, is
( x2 + y 2 )2 = a 2( x2 − y 2 ) ,
where a is a positive constant. Show that the equation may be expressed, in polar coordinates, in the form
r 2 = a 2 cos 2θ .
[3]
Explain how you can deduce, from the polar form of the equation, that the line θ = 14 π is tangential to the
curve at the pole.
[2]
Find the area of the region enclosed by one loop of the curve.
[4]
Specimen Materials - Mathematics
31
© OCR 2000
7
(i) Find the value of a for which
∞
⌠
1
1
1⌠
 1 + x 2 dx = 2  1 + x 2 dx .
⌡0
⌡0
a
[3]
(ii) By means of the substitution x = tan θ , or otherwise, find the exact value of
∞
x2
⌠
 (1 + x 2)2 d x .
⌡0
8
[7]
The complex number z satisfies the equation z = z + 2 . Show that the real part of z is − 1 .
[2]
The complex number z also satisfies the equation z = 2 . By sketching two loci in an Argand diagram,
find the two possible values of the imaginary part of z, and state the two corresponding values of arg z . [5]
The two possible value of z are denoted by z1 and z2 , where Im z1 > Im z2 .
(i) Find a quadratic equation whose roots are z1 and z2 , giving your answer in the form az 2 + bz + c = 0
where the coefficients a, b and c are real.
[2]
(ii) Determine the square roots of z1 , giving your answers in the form x + iy .
Specimen Materials - Mathematics
32
[4]
© OCR 2000
General Certificate of Education
Advanced Subsidiary (AS) and Advanced Level
P4
MATHEMATICS
Pure Mathematics 4
MARK SCHEME
MAXIMUM MARK
60
For live examinations, each Mark Scheme
includes the General Instructions for Marking
set out on pages 143 to 145.
This mark scheme consists of 4 printed pages.
Specimen Materials - Mathematics
33
© OCR 2000
1
True for n = 1 since 1 = 2 × 30 − 1
u n +1 = 3{2(3 n −1 ) − 1} + 2
= 2(3 n ) − 1
=
2(3( n −1) +1 )
− 1 , hence result
B1
M1
Requires at least 1 = 2 − 1
Substitute given un into 3un + 2
A1
Correct simplification
A1
4 Correct conclusion, dependent on all previous
marks and a properly set out proof
2
Integrating factor is e ∫ −x
1
i.e. e −ln x =
x
d  y
y
= 1 ⇒ = ∫ 1 dx
dx  x 
x
−1
dx
M1
A1
M1
y = x2 + Ax
3
A1
4
r+ 2 − r
1
1
2
−
=
B1
1 Given result correctly shown
=
r (r + 1) (r + 1)( r + 2) r (r + 1)( r + 2) r (r + 1)( r + 2)
-------------------------------------------------------------------------------------------------------------------------------------------------------1  1
1   1
1 
1
 1

−
Using differences
 −  +  −  + K + 
 M1
2  1.2 2.3   2.3 3.4 
 n(n + 1) (n + 1)(n + 2) 
1
1
M1
Cancelling pairs of terms
−
4 2( n + 1)(n + 2)
A1
3 Correct answer; allow any equivalent form
-------------------------------------------------------------------------------------------------------------------------------------------------------1
→ 0 as n → ∞ ; convergent
B1
Or equivalent argument
2( n + 1)(n + 2)
Sum to infinity is
4
Requires integration attempt
1
Simplify to x −1 or
x
1
4
B1
EITHER : If f( x) = ln(2 + x) then f ′( x) =
M1
At least one differentiation attempt
A1
Correct first and second derivatives
f( 0) = ln 2, f ′( 0) = 12 , f ′′(0) = − 14
A1
All three correct
ln(2 + x) ≈ ln 2 + 12 x − 18 x 2
A1
Three correct terms
ln(2 + x) = ln{2(1 + 12 x)}
M1
and f ′′( x) = −
OR :
1
,
2+ x
2
1
( 2 + x) 2
= ln 2 + ln(1 + 12 x)
A1
( 12 x)2
M1
Use of standard series for ln(1 + kx)
2
= ln 2 + 12 x − 18 x2
A1
4 Three correct terms
-------------------------------------------------------------------------------------------------------------------------------------------------------ln(2 − x) ≈ ln 2 − 12 x − 18 x2
B1
Replacing x by − x
≈ ln 2 + 12 x −
(ln 2 + 12 x − 18 x2) − (ln 2 − 12 x − 18 x2)
M1
2 +x
ln 
≈ x
2− x
A1
Specimen Materials - Mathematics
Subtracting series
3 Given result correctly shown
34
© OCR 2000
5
2
M1
Algebraic division, or equivalent
x +1
y = 2x + 1 is an asymptote
A1
x = −1 is an asymptote
B1
3
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii) EITHER : Consider discriminant of
2 x2 + (3 − y) x + (3 − y)
M1
Form quadratic in x and refer to ∆
(i)
y = 2x + 1 +
(3 − y) 2 = 8(3 − y)
A1
Allow equation or inequality
(3 − y)(5 + y) = 0
y = 3 and − 5
M1
Or equivalent solution method
A1
dy
2
= 2−
or
dx
( x + 1) 2
OR :
( x + 1)( 4x + 2) − (2x 2 + 3x + 3)
( x + 1) 2
M1
Differentiate and equate to zero
( x + 1) 2 = 1 or 2 x 2 + 4 x = 0
A1
Correct simplification
x = − 2 and 0 ⇒ y = 3 and − 5
M1
Solve for x and substitute to find y
A1
6
x = r cosθ , y = r sin θ
( r 2 cos 2 θ
+
r 2 sin 2 θ ) 2
=
a 2 ( r 2 cos 2 θ
−
r 2 sin 2 θ )
4 Both values of y correct
B1
For both; may be implied
M1
Substitute, and use at least one trig identity
r = a cos2θ
A1
3 Given answer correctly shown
-------------------------------------------------------------------------------------------------------------------------------------------------------r = 0 ⇒ cos 2θ = 0
M1
Using r = 0 for form at pole
1
θ = 4 π is a solution
A1
2 Given result correctly shown
-------------------------------------------------------------------------------------------------------------------------------------------------------1 a 2 cos 2θ d θ = 1 a 2 sin 2θ
M1
Using correct formula 12 ∫ r 2 d θ
2∫
4
2
2
[
]
1
1 a 2 sin 2θ 4π
4
− 14π
or 2[
1 2
a sin
4
2θ ] 0
1π
4
Area = 12 a 2
7
B1
Indefinite integral of the form k sin 2θ
M1
Using correct limits
A1
(i) LHS = tan− 1 a
B1
1 1
× π
2 2
B1
RHS =
4
Recognising tan−1 ∞ = 12 π
tan−1 a = 14 π ⇒ a = 1
B1
3 Correct answer for a
-------------------------------------------------------------------------------------------------------------------------------------------------------dx
(ii)
= sec2 θ
B1
Or equivalent; may be implied
dθ
∞
1
π
x2
tan2 θ
⌠
⌠2
2
 (1 + x 2) 2 dx =  (1 + tan 2 θ ) 2 sec θ dθ
⌡0
⌡0
M1
Substitute for x and dx throughout
M1
Use trig identities to simplify integrand
A1
Correct simplification; ignore limits so far
M1
Use relevant double-angle formula
= [12 θ − 14 sin 2θ ] 20
A1
For correct indefinite integral
= 14 π
A1
1π
= ∫ 2 sin 2 θ dθ
0
=∫
1π
2
0
1 (1− cos 2θ ) dθ
2
1π
Specimen Materials - Mathematics
7 Correct answer and no previous error
35
© OCR 2000
8
EITHER : Locus z = z + 2 is a perp bisector
Hence Re z = −1
M1
A1
For recognis ing linear locus
Needs mention of points z = 0, z = −2 , or
equivalent
x 2 + y 2 = ( x + 2) 2 + y 2
M1
Hence x = −1 , i.e. Re z = −1
A1
2
--------------------------------------------------------------------------------------------------------------------------------------------------------
OR :
B1
Both loci correct
1 + y 2 = 22
M1
Using Pythagoras or equivalent
Im z = ± 3
arg z = ± (π − tan− 1(1 / 3) )
A1
M1
Or equivalent correct method for either case
2
π
3
=±
A1
5 Both correct
-------------------------------------------------------------------------------------------------------------------------------------------------------M1
Form equation and expand LHS; allow any
(i) ( z + 1 + i 3)( z + 1 − i 3) = 0
equivalent complete method
z 2 + 2z + 4 = 0
A1
2
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii) EITHER :
z1 = 2 ⇒ z1 = 2
B1
For 2
arg z1 = 23 π ⇒ arg( z1) = 13 π or − 23 π
B1
For either possibility
± 2 (cos 13 π + i sin 13 π )
M1
Convert either case to cartesian form
± 12 2 (1 + i 3)
A1
Both correct; allow any equivalent exact
x + i y expression
OR :
If
z1 = x + i y then
− 1 = x2 − y 2 and 3 = 2xy
B1
Both equations correct
4 x4 + 4 x 2 − 3 = 0 or
M1
Form and solve quadratic in x 2 or y 2
A1
Correct single value for x 2 or y 2
x2 =
1
2
(
z1 = ±
or y 2 =
1
2
+i
3
2
3
2
)
4 y 4 −4 y 2 −3 = 0
A1
4 Both correct; allow any equivalent exact
x + i y expression
Specimen Materials - Mathematics
36
© OCR 2000
General Certificate of Education
Advanced Subsidiary (AS) and Advanced Level
P5
MATHEMATICS
Pure Mathematics 5
Additional materials:
Answer paper
Graph paper
List of Formulae
TIME
1 hour 20 minutes
INSTRUCTIONS TO CANDIDATES
Write your name, Centre number and candidate number in the spaces provided on the answer paper.
Answer all the questions.
Give non-exact numerical answers correct to 3 significant figures, unless a different degree of
accuracy is specified in the question or is clearly appropriate.
You are permitted to use a graphic calculator in this paper.
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 60.
Questions carrying smaller numbers of marks are printed earlier in the paper, and questions carrying
larger numbers of marks later in the paper.
You are reminded of the need for clear presentation in your answers.
This question paper consists of 3 printed pages and 1 blank page.
Specimen Materials - Mathematics
37
© OCR 2000
1
The cubic equation x 3 + ax − b = 0 has roots α , β , γ . Given that γ = αβ , express each of a and b in terms
of γ only, and hence show that (a + b) 2 = b .
2
The part of the curve y = x2 between y = 0 and y = 2 is rotated completely about the y-axis. Show that
the area of the curved surface formed is
3
[5]
13
π
3
.
[5]
Starting from the definitions of sinh x and cosh x in terms of exponentials, show that
cosh 2 x ≡ cosh2 x + sinh 2 x .
Given that cosh 2 x = k , where k > 1 , express tanh x in terms of k.
4
[2]
[4]
The differential equation
dy
= 1 + xy ,
dx
with y = 1 when x = 0 , is to be solved numerically by a step-by-step method.
(i) Use two steps of Euler’s method, with step-length 0.1, to find an approximation for the value of y
when x = 0.2 .
[3]
(ii) Use one step of the modified Euler method, with step-length 0.2, to find an alternative approximation
for the value of y when x = 0.2 .
[3]
e
5
Given that I n = ⌠
 (ln x) n dx , show that, for n ≥ 1 ,
⌡1
I n = e − nI n−1 .
[4]
Hence find the exact value of I 4 .
6
[4]
The curve with equation
y=
x
cosh x
has one maximum point for x > 0 . Show that the x-coordinate of this maximum point satisfies the
equation x tanh x − 1 = 0 .
[2]
The positive root of the equation x tanh x − 1 = 0 is denoted by α . Use the Newton-Raphson method,
[4]
taking first approximation x1 = 1 , to find further approximations x2 and x3 for α .
By considering the approximate errors in x1 and x2 , estimate the error in x3 .
Specimen Materials - Mathematics
38
[3]
© OCR 2000
7
1
together with four rectangles of unit width and heights
x+1
respectively. Explain how the diagram shows that
The diagram shows the curve y =
1 1 1 1
, , ,
2 3 4 5
4
1
2
⌠ 1
dx .
+ 13 + 14 + 15 < 
⌡0 x + 1
[2]
1
passes through the top left-hand corner of each of the four rectangles shown. By
x+2
considering the rectangles in relation to this curve, write down a second inequality involving 12 + 13 + 14 + 15
The curve y =
and a definite integral.
[2]
By considering a suitable range of integration and corresponding rectangles, show that
∑
1000
ln( 500 .5) <
r =2
1
< ln(1000 ) .
r
[4]
∑
1000
Explain briefly how you can deduce that a reasonable estimate for the value of
r= 2
8
1
is 6.5.
r
[2]
The phenomenon of ‘resonance’ in a simple electrical circuit can be modelled by the differential equation
d 2V
+ 100V = 2 cos10t ,
dt 2
where V represents the voltage in the circuit and t represents time.
(i) Verify that kt sin 10t is a particular integral for this differential equation, where k is a constant whose
numerical value is to be found.
[4]
(ii) Find the general solution of the differential equation.
(iii) Find the particular solution for which both V and
[3]
dV
are zero when t = 0 .
dt
[3]
(iv) By considering the values of V when t becomes large, explain briefly why the mathematical model
cannot give an entirely satisfactory representation of the voltage in the circuit.
[1]
Specimen Materials - Mathematics
39
© OCR 2000
BLANK PAGE
Specimen Materials - Mathematics
40
© OCR 2000
General Certificate of Education
Advanced Subsidiary (AS) and Advanced Level
P5
MATHEMATICS
Pure Mathematics 5
MARK SCHEME
MAXIMUM MARK
60
For live examinations, each Mark Scheme
includes the General Instructions for Marking
set out on pages 143 to 145.
This mark scheme consists of 4 printed pages.
Specimen Materials - Mathematics
41
© OCR 2000
1
α + β + γ = 0, αβ + βγ + γα = a, αβγ = b
γ = αβ ⇒ b =
γ2
a = γ + γ (α + β ) = γ − γ
a = γ − b ⇒ (a +
b) 2
=
2
γ2
=b
B1
All three correct, at any stage
B1
Correct answer for b
B1
Correct answer for a
M1
Eliminating γ 2 , or equivalent
A1
2
EITHER :
dx
1
=
dy 2 y
5 Given result correctly shown
B1
Or equivalent; may be implied
M1
Stating 2π ∫ x ds all in terms of y
A1
Or equivalent simplification
A1
Correct indefinite integral
A1
Given result correctly shown
B1
Or equivalent; may be implied
M1
Stating 2π ∫ x ds all in terms of x
A1
Correct limits for x
A1
Correct indefinite integral
A1
5 Given result correctly shown
2
1
⌠
Area = 2π  y 1 +
dy
4y
⌡0
2
= 2π ⌠
 y + 14 dy
⌡0
[
= 43 π ( y + 14 )2
3
]
2
0
= 133 π
dy
= 2x
dx
OR :
2
Area = 2π ⌠
 x 1 + 4x 2 dx
⌡0
= 16 π (1 + 4x 2 ) 

 0
13
= 3π
3
2
3
2
RHS = 14 (e2 x + 2 + e−2x ) + 14 (e2x − 2 + e −2x )
1 2x
(e +
2
M1
Squaring and adding
−2 x
=
e ) = LHS
A1
2 Correct simplification and conclusion
-------------------------------------------------------------------------------------------------------------------------------------------------------cosh2 x − sinh2 x = 1
B1
May be implied
2
1
2
1
cosh x = 2 ( k + 1), sinh x = 2 (k − 1)
B1
Both correct
tanh x =
sinh x
k −1
=±
cosh x
k +1
sinh x
cosh x
4 Both values needed
Using tanh x =
M1
A1
4
y1 = 1 + 0.1 × 1
M1
Use of correct Euler formula for 1st step
= 1. 1
A1
y2 = 1. 1 + 0. 1 × (1 + 0. 1 × 1.1) = 1.211
A1
3 Allow 3sf value 1.21 correctly obtained
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii) ytemp = 1 + 0 .2 × 1 = 1 .2
B1
(i)
y1 = 1 + 12 × 0 .2 × {1 + (1 + 0 .2 × 1 .2)}
= 1.224
Specimen Materials - Mathematics
M1
Use of correct modified Euler method
A1
3 Allow 3sf value 1.22 correctly obtained
42
© OCR 2000
5
I n = [x(ln x)n ]1 − ∫ x n(ln x) n−1 x −1 dx
e
e
1
= e − nI n−1
M1
Using relevant integration by parts
A1
A1
Correct unsimplified result
First term shown to simplify to e
A1
4 Given result fully justified
-------------------------------------------------------------------------------------------------------------------------------------------------------I 4 = e − 4I 3 = e − 4(e − 3I 2 ) = − 3e + 12( e − 2I1)
M1
Reduction formula used at least twice
6
7
= 9e − 24(e − I 0) = −15e + 24 I 0
I 0 = e − 1 or I1 = 1
A1
For I 4 = −15e + 24 I 0 or 9e − 24 I 0
B1
For either correct
I 4 = 9e − 24
A1
4
dy cosh x − x sinh x
=
=0
M1
Differentiate and equate to zero
dx
cosh2 x
Maximum when cosh x = x sinh x , i.e. x tanh x = 1
A1
2 Given result correctly shown
-------------------------------------------------------------------------------------------------------------------------------------------------------xn tanh xn − 1
xn+1 = x n−
B1
tanh xn + xnsech2xn
x1 = 1 gives x2 = 1.20177K
M1
Newton-Raphson used at least once
A1
x2 correct to at least 3sf
x3 = 1.1996785 K
A1
4 x3 correct to at least 4sf
-------------------------------------------------------------------------------------------------------------------------------------------------------e1 ≈ 0.2, e2 ≈ −0.002
B1
For both; ignore signs of errors
e3 e2
≈ ⇒ e3 ≈ −2 × 10− 7
M1
Use of quadratic convergence property
e22 e12
A1
3 Ignore sign of answer
LHS is the total area of the four rectangles
B1
RHS is the corresponding area under the curve; hence
result
B1
2
-------------------------------------------------------------------------------------------------------------------------------------------------------4
1
1
dx
M1
Attempt at relevant new inequality
+ 13 + 14 + 15 > ⌠

2
⌡0 x + 2
A1
2 Correct statement
-------------------------------------------------------------------------------------------------------------------------------------------------------Σ is the area of 999 rectangles
M1
May be implied
999
999
1
1
Bounds are ⌠
dx and ⌠
dx
M1
Accuracy of 999 not essential here


⌡0 x + 2
⌡0 x + 1
Lower limit is ln(500 . 5)
A1
Given value correct shown
A1
4 Ditto
Upper limit is ln(1000 )
-------------------------------------------------------------------------------------------------------------------------------------------------------1
{ln(500 .5) + ln(1000 )} = 6 .56 K
M1
For considering the average
2
Round down to 6.5 as it’s an overestimate
Specimen Materials - Mathematics
A1
2 Some reason for given answer 6.5 rather than
6.6 needed
43
© OCR 2000
8
(i) V& = k sin 10t + 10kt cos10t and
V&& = 20k cos10t − 100 ktsin 10t
M1
Differentiate twice
Correct V& and V&&
A1
20 k cos10 t ≡ 2 cos10 t
M1
Substitute throughout the DE
k = 0. 1
A1
4
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii) m 2 + 100 = 0 ⇒ m = ±10 i
M1
State and solve auxiliary equation
A1
Or equivalent
CF is A sin 10 t + B cos10 t
A1
3 Follow their CF and k
GS is V = A sin 10 t + B cos10t + 0 .1t sin 10 t
-------------------------------------------------------------------------------------------------------------------------------------------------------(iii) V = 0, t = 0 ⇒ 0 = B
M1
V& = 0, t = 0 ⇒ 0 = 10 A
M1
Differentiate and substitute
Hence V = 0 .1t sin 10 t
A1
3
-------------------------------------------------------------------------------------------------------------------------------------------------------(iv) V is unbounded as t increases; hence unrealistic
B1
1 Follow errors if the result remains true
Specimen Materials - Mathematics
44
© OCR 2000
General Certificate of Education
Advanced Subsidiary (AS) and Advanced Level
P6
MATHEMATICS
Pure Mathematics 6
Additional materials:
Answer paper
Graph paper
List of Formulae
TIME
1 hour 20 minutes
INSTRUCTIONS TO CANDIDATES
Write your name, Centre number and candidate number in the spaces provided on the answer paper.
Answer all the questions.
Give non-exact numerical answers correct to 3 significant figures, unless a different degree of
accuracy is specified in the question or is clearly appropriate.
You are permitted to use a graphic calculator in this paper.
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 60.
Questions carrying smaller numbers of marks are printed earlier in the paper, and questions carrying
larger numbers of marks later in the paper.
You are reminded of the need for clear presentation in your answers.
This question paper consists of 3 printed pages and 1 blank page.
Specimen Materials - Mathematics
45
© OCR 2000
1
The matrices A and B are given by
1
A = 
0

cos α
B = 
 sin α
0
,
− 1
− sin α 
,
cos α 
respectively. Find the matrix BAB−1 , simplifying your answer.
2
[5]
The set G = {0, 1, 2, 3, 4, 5} is a cyclic group under addition modulo 6, and the set H = {e, p, q, r, s, t} is a
multiplicative group whose group table is shown below.
e
p
q
r
s
t
e
e
p
q
r
s
t
p
p
q
e
s
t
r
q
q
e
p
t
r
s
r
r
t
s
e
q
p
s
s
r
t
p
e
q
t
t
s
r
q
p
e
(i) Find an element of G of order 6.
[1]
(ii) Show that G and H are not isomorphic.
[2]
The set K = {1, 2, 3, 4, 5, 6} is a group under multiplication modulo 7. Determine the order of the element 3
in K, and hence or otherwise determine which of G and H is isomorphic to K.
[3]
3
G is a multiplicative group with identity element e. The group is not commutative, but the elements a and
b in G each commute with every element in G, i.e. for any x in G, ax = xa and bx = xb .
(i) Prove that the element ab commutes with every element in G.
[2]
(ii) Prove that the element a −1 commutes with every element in G.
[2]
(iii) Deduce that the set of all those elements in G which commute with every element in G forms a
subgroup of G.
[4]
4
Use de Moivre’s theorem to prove that
sin 5θ = sin θ (5 − 20 sin 2 θ + 16 sin 4 θ ) .
By letting θ = 15 π , show that the exact value of sin 2 (15 π ) is 18 (5 − 5) .
Specimen Materials - Mathematics
46
[4]
[4]
© OCR 2000
5
Write down the six 6th roots of unity.
[1]
By writing the equation ( z + 1)6 = z 6 in the form
6
 z + 1 = 1 ,


 z 
show that the values of z satisfying the equation are given by
z=
6
1
e
1 kπi
3
−1
,
where k is an integer, and state a set of values of k that gives all the roots of the equation.
[4]
Express each of these roots of ( z + 1)6 = z 6 in the form x + iy , where x and y are real.
[5]
Find the value of a for which the simultaneous equations
3x + 2 y − z = 10,
5x − y − 4 z = 17,
x + 5 y + az = b,
do not have a unique solution for x, y and z.
[4]
Show that, for this value of a, the equations are inconsistent unless b = 3 .
[2]
For the case where the equations represent three planes having a common line of intersection, L, find
x− p y −q z −r
equations for L, giving your answer in the form
.
[5]
=
=
l
m
n
7
The diagram shows a cuboid ABCDA'B'C'D' in which the lengths of AB, AD, and AA' are 3a, 2a and a
respectively. The point A is taken as origin, with unit vectors i, j, k in the directions of AB, AD, AA'
respectively.
(i) Find a normal vector for the plane through A', B and C', and find also the perpendicular distance from
D to this plane.
[6]
(ii) Find the shortest distance between the skew lines A'B and AD'.
Specimen Materials - Mathematics
47
[6]
© OCR 2000
BLANK PAGE
Specimen Materials - Mathematics
48
© OCR 2000
General Certificate of Education
Advanced Subsidiary (AS) and Advanced Level
P6
MATHEMATICS
Pure Mathematics 6
MARK SCHEME
MAXIMUM MARK
60
For live examinations, each Mark Scheme
includes the General Instructions for Marking
set out on pages 143 to 145.
This mark scheme consists of 4 printed pages.
Specimen Materials - Mathematics
49
© OCR 2000
1
cosα
B −1 = 
 − sin α
sin α 
cosα 
cosα
sin α 
= 
− cosα 
 sin α
B1
BA or AB −1
M1
Correct multiplication process
A1
Correct product of either pair
A1
Correct unsimplified product
cos 2 α
2 sin α cosα 
BAB −1 = 
sin 2 α − cos2 α 
 2sin α cosα
cos 2α
sin 2α 
= 
− cos 2α 
 sin 2α
2
− sin 2 α
A1
5
(i) 1 or 5
B1
1 For either answer
-------------------------------------------------------------------------------------------------------------------------------------------------------Element of H : e p q r s t
(ii) EITHER :
B1
Order :
1 3 3 2 2 2
No element of order 6, hence result
B1
OR :
H is not commutative,
e.g. pr = t, rp = s
B1
Counter-example is required
G is commutative, hence result
B1
2
-------------------------------------------------------------------------------------------------------------------------------------------------------32 = 2, 33 = 6, 34 = 4, 35 = 5, 36 = 1
M1
At least 3 calculations
The order of 3 in K is 6
A1
K is isomorphic to G
B1
3 If not deduced from orders of elements, a
reason is required (e.g. that K is
commutative)
3
(i) ( ab) x = a(bx) = a( xb) = (ax)b = ( xa)b = x(ab)
M1
At least one correct interchange
A1
2 Completely correct proof
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii) ax = xa ⇒ a −1 ( ax) a −1 = a −1 ( xa)a −1
M1
Pre- and/or post-multiply and simplify at
least once
A1
2 Completely correct proof
Hence xa = a x
-------------------------------------------------------------------------------------------------------------------------------------------------------(iii) Closure follows from (i)
B1
Associativity is true within G
B1
Identity is e, since ex = xe
B1
Inverses follow from (ii)
B1
4
−1
4
−1
sin 5θ = Im(cos θ + i sin θ )5
= 5 cos θ sin θ − 10 cos θ sin θ + sin θ
= 5(1 − s 2 ) 2 s − 10 (1 − s 2 )s 3 + s 5
4
2
3
5
M1
Expand and take imaginary part
A1
M1
All three terms correct
Use of cos2 θ = 1 − sin 2 θ throughout
= sin θ (5 − 20 sin 2 θ + 16 sin 4 θ )
A1
4 Given answer shown correctly
-------------------------------------------------------------------------------------------------------------------------------------------------------θ = 15 π ⇒ sin 5θ = 0
B1
Hence sin(15 π ) is a root of 16s 4 − 20s 2 + 5 = 0
B1
s2 = 18 (5 ± 5)
M1
For solving the quadratic in an exact form
A1
4 Given answer fully justified, e.g. negative
sin 2( 15 π ) = 18 (5
− 5)
sign chosen since sin 2( 15 π ) is certainly < 12
Specimen Materials - Mathematics
50
© OCR 2000
1
5
e3
kπ i
for k = 1, 2, 3, 4, 5, 6
1 Or equivalent values for k; allow exponential
trigonometrical or cartesian form for roots
-------------------------------------------------------------------------------------------------------------------------------------------------------z +1
= ω , where ω is a 6th root of unity
M1
ω ≠ 1 not required at this stage
z
1
z + 1 = zω ⇒ z =
M1
Solving for z
ω −1
1
A1
Given answer correctly shown
= 1
kπ i
e 3 −1
k = 1, 2, 3, 4, 5
B1
4 Or any correct set of 5 values
-------------------------------------------------------------------------------------------------------------------------------------------------------k = 3 gives real root − 12
B1
e
(e
1
kπ
3
i
− 1 kπ i
3
− 1)(e
B1
−1
− 1 kπ i
3
M1
− 1)
Use of conjugate for general, or for any one
specific, case
cos 13 kπ
− i sin 13 kπ
2 − 2 cos 13 kπ
−1
− 12 ± ( 12 3 ) i, − 12 ± ( 16 3) i
A1
Correct trigonometric or numerical form
A1
Any one complex root correct
A1
6
−1
− 4 = 0
a
3(− a + 20) − 2(5a + 4) − ( 25 + 1) = 0
3
det 5
1

2
−1
5
5 All four correct
B1
M1
Correct expansion method
A1
Correct unsimplified equation
a= 2
A1
4
-------------------------------------------------------------------------------------------------------------------------------------------------------EITHER : r3 = 2r1 − r2
M1
Complete method for relation between rows
b=3
A1
Given answer correctly shown
OR :
3
det5
1

2
−1
5
10 
17  = 0
b
M1
b=3
A1
2 Given answer correctly shown
------------------------------------------------------------------------------------------------------------------------------------------------------- 3  5
Direction of L is  2 ×  − 1
M1
For relevant vector product, or complete
 − 1  − 4
   
equivalent method
 − 9
i.e.  7 
A1
Or any multiple
 − 13 


3x + 2 y − z = 10, 5x − y − 4z = 17 ⇒ 13x − 9z = 44 (e.g.) M1
Solve two equations simultaneously
So ( 2, 1, − 2) lies on L
L is
x − 2 y −1 z + 2
=
=
−9
7
− 13
Specimen Materials - Mathematics
A1
Or any other correct point
A1
5 Or correct equivalent
51
© OCR 2000
7
(i) EITHER :
 3a 
 3a 
A′B =  0, A′ C′ =  2a 
− a 
 0
 
 
3
a
   3a
Normal n is  0 ×  2a
 − a  0
   
 2
i.e. n =  − 3
 6
 
 3a  2
DB.n =  − 2a .  − 3 = 12a
 0  6

  
Perpendicular distance is
OR :
12a
n
M1
Or equivalent pair of vectors
M1
Or equivalent complete method
A1
Or any multiple
M1
Or other relevant scalar product
M1
Calculate n and divide
= 12
a
7
A1
Equation px + qy + rz = s gives
ar = s, 3ap = s, 3ap + 2aq + ar = s
M1
Substitute all three points
ap = 13 s, aq = − 12 s, ar = s
M1
Solve for p, q, r or p : q : r
A1
Or any multiple
 2
Normal vector is  − 3
 6
 
Plane is 2 x − 3 y + 6z = 6a
M1
Perp distance from ( 0, 2a, 0) is
− 6a − 6a
M1
Relevant use of formula
22 + 32 + 62
= 12
a
A1
6
7
------------------------------------------------------------------------------------------------------------------------------------------------------- 3a 
 0
B1
Or equivalent
(ii) A′B =  0 , AD′ =  2a 
− a 
 a
 
 
Direction of common perpendicular is
 3a  0
 0 ×  2a
M1
 − a  a
   
 2
i.e. n =  − 3 is perpendicular to both
A1
or any multiple
 6
 
 0   2
AA′.n =  0  .  − 3 = 6a
M1
Or equivalent scalar product
 a   6
   
6a 6
= a
Shortest distance is
M1
Divide by n
n 7
A1
6 Correct answer
Specimen Materials - Mathematics
52
© OCR 2000
General Certificate of Education
Advanced Subsidiary (AS) and Advanced Level
M1
MATHEMATICS
Mechanics 1
Additional materials:
Answer paper
Graph paper
List of Formulae
TIME
1 hour 20 minutes
INSTRUCTIONS TO CANDIDATES
Write your name, Centre number and candidate number in the spaces provided on the answer paper.
Answer all the questions.
Give non-exact numerical answers correct to 3 significant figures, unless a different degree of
accuracy is specified in the question or is clearly appropriate.
–2
Where a numerical value for the acceleration due to gravity is needed, use 9.8 m s .
You are permitted to use a graphic calculator in this paper.
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 60.
Questions carrying smaller numbers of marks are printed earlier in the paper, and questions carrying
larger numbers of marks later in the paper.
You are reminded of the need for clear presentation in your answers.
This question paper consists of 4 printed pages.
Specimen Materials - Mathematics
53
© OCR 2000
1
Two forces, of magnitudes 1 N and 3 N, act on a particle in the directions shown in the diagram. Calculate
the magnitude of the resultant force on the particle and the angle between this resultant force and the force
of magnitude 3 N.
[5]
2
The diagram shows a railway engine of mass 50 tonnes pulling two trucks horizontally along a straight
track. The trucks are coupled together behind the engine and have masses 8 tonnes and 4 tonnes
respectively, starting with the truck nearer to the engine. The acceleration of the train is 0.5 m s−2 .
Assuming that there are no resistances to motion, find
(i) the driving force of the engine,
[2]
(ii) the tensions in the two couplings.
[4]
3
Two particles, of masses x kg and 0.1 kg, are moving towards each other in the same straight line and
collide directly. Immediately before the impact, the speeds of the particles are 2 m s−1 and 3 m s−1
respectively (see diagram).
(i) Given that both particles are brought to rest by the impact, find x.
[2]
(ii) Given instead that the particles move with equal speeds of 1 m s−1 after the impact, find the three
possible values of x.
[6]
4
A moving particle P travels in a straight line. At time t seconds after starting from the point O on the line,
the velocity of P is v m s−1 , where
v = t 2 (6 − t ) .
Show that the acceleration of P is zero when t = 4 .
[3]
After a certain time, P comes instantaneously to rest at the point A on the line. State the time taken for the
motion from O to A, and find the distance OA.
[5]
Specimen Materials - Mathematics
54
© OCR 2000
5
A heavy ring of mass 5 kg is threaded on a fixed rough horizontal rod. The coefficient of friction between
the ring and the rod is 12 . A light string is attached to the ring and is pulled downwards with a force of
magnitude T newtons acting at an angle of 30° to the horizontal (see diagram). Given that the ring is about
to slip along the rod, find the value of T.
[9]
6
The diagram shows an approximate (t, v ) graph for the motion of a parachutist falling vertically; v m s−1
is the parachutist’s downwards velocity at time t seconds after he jumps out of the plane. Use the
information in the diagram
(i) to give a brief description of the parachutist’s motion throughout the descent,
[4]
(ii) to calculate the height from which the jump was made.
[2]
The mass of the parachutist is 90 kg. Calculate the upwards force acting on the parachutist, due to the
[5]
parachute, when t = 7 .
Specimen Materials - Mathematics
55
© OCR 2000
7
(i)
Particles A, of mass 5m, and B, of mass 3m, are attached to the ends of a light inextensible string. The
string passes over a fixed peg, and the system is released from rest with both parts of the string taut
and vertical, and each particle a distance d above a fixed horizontal plane (see diagram). Neglecting
all resistances to motion,
(a) find the acceleration of A in terms of g and show that the tension in the string is
15 mg
4
,
(b) find an expression in terms of d and g for the time after release at which A hits the plane.
[6]
[2]
(ii) The results in part (i) are based on a mathematical model in which resistances to motion are
neglected. Describe briefly one resisting force, other than air resistance, which would be present in a
real system in which objects of unequal mass, hanging from a string passing over a fixed support, are
in motion.
[1]
When this force is taken into account, state with brief reasons whether each of the following would be
smaller or larger than the value calculated in part (i):
(a) the acceleration of A;
(b) the tension in the string acting on A;
(c) the tension in the string acting on B.
What can you conclude about the tension in the string in this case?
Specimen Materials - Mathematics
56
[4]
© OCR 2000
General Certificate of Education
Advanced Subsidiary (AS) and Advanced Level
M1
MATHEMATICS
Mechanics 1
MARK SCHEME
MAXIMUM MARK
60
For live examinations, each Mark Scheme
includes the General Instructions for Marking
set out on pages 143 to 145.
This mark scheme consists of 3 printed pages and 1 blank page.
Specimen Materials - Mathematics
57
© OCR 2000
1
EITHER : ‘Vertical’ component of resultant is 1sin 40o B1
B1
‘Horizontal’ component is 3 + 1cos 40 °
Magnitude is 3 .766 2 + 0 .6428 2
i.e. 3.82 N
0. 6428 
Angle is tan−1
 = 9.69 °
 3.766 
OR :
Vector triangle with sides 1, 3 and included
angle of 140º (not 40º)
R2 = 32 + 12 − 2 × 3 × 1 × cos140°
Hence magnitude is 3.82 N
sin θ sin 140 °
=
1
3. 82
Hence angle is 9.69°
M1
A1
Allow M mark for either Pythagoras or trig
For correct magnitude
A1
For correct angle
B1
M1
A1
May be implied
For use of cosine formula with 3, 1, 140 °
M1
For sin formula, or other complete method
A1
5
2
Force = (50000 + 8000 + 4000) × 0.5
M1
For use of NII applied to whole system
= 31000 N or 31 kN
A1
2
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii) For back t ruck: C1 = 4000 × 0.5
M1
Use of NII for the rear truck only
i.e. Force in rear coupling is 2000 N or 2 kN
A1
M1
Use of NII for the pair of trucks with one
For both trucks: C2 = 12000× 0. 5
force, or equivalent, i.e. C2 − C1 = 4000 or
31000 − C2 = 25000
i.e. Force in front coupling is 6000 N or 6 kN
A1
4 Follow through if earlier answer is used
3
x × 2 − 3 × 0. 1 = 0
M1
For relevant use of momentum conservation
A1
2
Hence x = 0. 15
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii) 2 x − 0.3 = ( x + 0.1) or − ( x + 0.1) or 0. 1 − x
M1
For any one relevant momentum equation
(i)
(i)
Hence x = 0. 4 or 0.0667 or 0.133
4
dv
= 12t − 3t 2 = 0
dt
A1
M1
A1
A1
A1
For any one correct (unsimplified) equation
For appreciating at least 2 correct cases
For any one correct value
For a second correct value
6 For all three correct answers
M1
For expanding v and differentiating
A1
For correct derivative equated to zero
3t(4 − t) = 0, so a = 0 when t = 4
A1
3 Given answer correctly found or verified
-------------------------------------------------------------------------------------------------------------------------------------------------------P reaches A when t = 6
B1
s = ∫ (6t2 − t 3) dt = [2t3 − 14 t4 ]0
M1
For integrating v
= 432 − 324
Distance OA = 108 m
A1
M1
A1
For correct indefinite integral
Use of limits or evaluation of arbitrary const
6
6
0
Specimen Materials - Mathematics
5
58
© OCR 2000
5
6
B1
Correct forces identified, by diagram or
otherwise
Resolving horizontally:
T cos 30° = F
Resolving vertically:
5g + T sin 30 ° = R
M1
A1
M1
A1
For attempting one resolution equation
For limiting equilibrium F = 12 R
B1
Available at any stage
T . 12 3 = 12 (5 g + 12 T )
M1
For eliminating F and R
T = 39. 8
A1
A1
For attempting a second resolution
Other correct equations are possible
Correct unsimplified equation in T only
9
(i) Initially the parachutist falls with constant acc
B1
Allow ‘free-fall’ etc here
Then decelerates at a constant rate
B1
Then falls with constant speed
B1
And finally hits the ground and comes to rest
B1
4
-------------------------------------------------------------------------------------------------------------------------------------------------------M1
For sensible attempt at total area under graph
(ii) Area is 12 × 4 × 40 + 12 (40 + 10) × 6 + 10 × 15
Height is 380 m
A1
2
-------------------------------------------------------------------------------------------------------------------------------------------------------10 − 40
= −5 (downwards) M1
Acceleration when t = 7 is
For use of gradient to find acceleration
10 − 4
A1
For value (±)5 even if sign/direction muddle
Hence 90g − T = 90 × ( −5)
M1
For use of NII with three relevant terms
B1
A1
Force from parachute is 1330 N
7
(i) (a)
Equations of motion for the particles are:
5mg − T = 5ma
T − 3mg = 3ma
Hence acceleration is
Tension is
15
mg
4
1
4
g and
For consistent signs in T and ma terms
5
M1
A1
A1
For use of NII for either particle separately
The ‘system’ equation 8mg = 2ma is an
alternative for one of these A marks
M1
For finding T or a from sufficient equation(s)
A1
For correct acceleration
A1
6 For obtaining given tension correctly
-------------------------------------------------------------------------------------------------------------------------------------------------------(b) d = 12 × 14 g × t 2
M1
Use appropriate uvast equation and solve for t
8d
A1
2
g
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii) Friction between the string and the support
B1
1
-------------------------------------------------------------------------------------------------------------------------------------------------------(a) Acceleration is smaller, as the resistance
opposes the motion
B1
(b) Tension at A is larger, because
TA = 5mg − 5ma , and a is less than before
B1
(c) Tension at B is smaller, because
TB = 3ma + 3mg and a is less than before
B1
The tensions in the two parts of the string are now
unequal
B1
4
Time is
Specimen Materials - Mathematics
59
© OCR 2000
BLANK PAGE
Specimen Materials – Mathematics
60
© OCR 2000
General Certificate of Education
Advanced Subsidiary (AS) and Advanced Level
M2
MATHEMATICS
Mechanics 2
Additional materials:
Answer paper
Graph paper
List of Formulae
TIME
1 hour 20 minutes
INSTRUCTIONS TO CANDIDATES
Write your name, Centre number and candidate number in the spaces provided on the answer paper.
Answer all the questions.
Give non-exact numerical answers correct to 3 significant figures, unless a different degree of
accuracy is specified in the question or is clearly appropriate.
–2
Where a numerical value for the acceleration due to gravity is needed, use 9.8 m s .
You are permitted to use a graphic calculator in this paper.
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 60.
Questions carrying smaller numbers of marks are printed earlier in the paper, and questions carrying
larger numbers of marks later in the paper.
You are reminded of the need for clear presentation in your answers.
This question paper consists of 4 printed pages.
Specimen Materials - Mathematics
61
© OCR 2000
1
The diagram shows the cross-section of a uniform solid rectangular block. This cross-section has
dimensions 20 cm by 10 cm and lies in a vertical plane. The block rests in equilibrium on a rough plane
whose inclination α to the horizontal can be varied. The coefficient of friction between the block and the
plane is 0.7. Given that α is slowly increased from zero, determine whether equilibrium is broken by
toppling or sliding.
[5]
2
A small ball of mass 0.2 kg is dropped from rest at a height of 1.2 m above a horizontal floor. The ball
rebounds vertically from the floor, reaching a height of 0.8 m. Assuming that air resistance can be
neglected, calculate
(i) the coefficient of restitution between the ball and the floor,
[4]
(ii) the impulse exerted by the floor on the ball when the ball bounces.
[2]
If air resistance were taken into account, would the value calculated for the coefficient of restitution be
larger or smaller than the value calculated in part (i)? Justify your conclusion.
[1]
3
A uniform lamina ABCD has the shape of a square of side a adjoining a right-angled isosceles triangle
whose equal sides are also of length a. The weight of the lamina is W. The lamina rests, in a vertical
plane, on smooth supports at A and D, with AD horizontal (see diagram).
(i) Show that the centre of mass of the lamina is at a horizontal distance of
11 a
9
(ii) Find, in terms of W, the magnitudes of the forces on the supports at A and D.
Specimen Materials - Mathematics
62
from A.
[4]
[4]
© OCR 2000
4
Fig. 1 shows the cross-section of a hollow container. The base of the container is circular, and is
horizontal. The sloping part of the side makes an angle of 15° with the horizontal, and the vertical part of
the side forms a circular cylinder of radius 0.4 m. A small steel ball of mass 0.1 kg moves in a horizontal
circle inside the container, in contact with the vertical and sloping parts of the side at A and B respectively,
as shown in Fig. 2.
It is assumed that all contacts are smooth and that the radius of the ball is negligible compared to 0.4 m.
(i) Given that the ball is moving with constant speed 3 m s−1 , find the magnitudes of the contact forces
acting on the ball at A and at B.
[5]
(ii) Calculate the least speed that the ball can have while remaining in contact with the vertical part of the
side of the container.
[3]
5
A car of mass 650 kg is travelling on a straight road which is inclined to the horizontal at 5°. At a certain
point P on the road the car’s speed is 15 m s −1 . The point Q is 400 m down the hill from P, and at Q the
car’s speed is 35 m s −1 .
(i) Assume that the car’s engine produces a constant driving force on the car as it moves down the hill
from P to Q, and that any resistances to the car’s motion may be neglected. By considering the
change in energy of the car, or otherwise, calculate the magnitude of the driving force of the car’s
engine.
[4]
(ii) Assume instead that resistance to the car’s motion between P and Q may be represented by a constant
force of magnitude 900 N. Given that the acceleration of the car at Q is zero, show that the power of
the car’s engine at this instant is approximately 12.1 kW.
[4]
Given that the power of the car’s engine is the same when the car is at P as it is when the car is at Q,
calculate the car’s acceleration at P.
[2]
Specimen Materials - Mathematics
63
© OCR 2000
6
A uniform rectangular box of weight W stands on a horizontal floor and leans against a vertical wall. The
diagram shows the vertical cross-section ABCD containing the centre of mass G of the box. AD makes an
angle θ with the horizontal, and the lengths of AB and AD are 2a and 8a respectively.
(i) By splitting the weight into components parallel and perpendicular to AD, or otherwise, show that the
anticlockwise moment of the weight about the point D is Wa (4 cos θ − sin θ ) .
[3]
(ii) The contact at A between the box and the wall is smooth. Find, in terms of W and θ , the magnitude
of the force acting on the box at A.
[3]
(iii) The contact at D between the box and the ground is rough, with coefficient of friction µ . Given that
4
.
[4]
the box is about to slip, show that tanθ =
8µ + 1
7
A shell is fired from a stationary ship O which is at a distance of 1000 m from the foot of a vertical cliff AB
of height 100 m. The shell passes vertically above B and lands at a point C on horizontal ground, level
with the top of the cliff (see Fig. 1). The shell is fired with speed 300 m s− 1 at angle of elevation θ , and
air resistance to the motion of the shell may be neglected.
(i) Given that θ = 30 ° , find the time of flight of the shell and the distance BC.
[6]
(ii)
Given instead that the shell just passes over B, as shown in Fig. 2, find the value of θ , correct to the
nearest degree.
[6]
Specimen Materials - Mathematics
64
© OCR 2000
General Certificate of Education
Advanced Subsidiary (AS) and Advanced Level
M2
MATHEMATICS
Mechanics 2
MARK SCHEME
MAXIMUM MARK
60
For live examinations, each Mark Scheme
includes the General Instructions for Marking
set out on pages 143 to 145.
This mark scheme consists of 4 printed pages.
Specimen Materials - Mathematics
65
© OCR 2000
1
Topples when CG is above lowest corner
i.e. when tanα = 12
B1
B1
May be implied
Slides when R = mg cosα and 0.7R = mg sin α
i.e. when tanα = 0. 7
M1
A1
Both equations attempted
Allow B2 in place of M1 A1 if µ = tanα is
1
2
Hence it topples, since
2
< 0. 7
2g × 1. 2 and
(i) Speed before impact is
Hence e =
1. 6g
=
2.4g
2
3
quoted
5 Conclusion and reason needed
B1
2g × 0.8
≈ 0.816
M1
For one relevant use energy or const acc
A1
For both correct (unsimplified) values
M1
For use of v′ = ev to calculate e
A1
4 For correct exact or decimal answer
-------------------------------------------------------------------------------------------------------------------------------------------------------M1
Allow M mark even if there is a sign error
(ii) Impulse= 0.2( 2. 4g + 1. 6g ) ≈ 1.76 N s
A1
2
-------------------------------------------------------------------------------------------------------------------------------------------------------With air resistance, speed before impact is smaller, and
speed after impact is larger; hence e is larger
B1
1 For correct conclusion with correct reasons
3
(i) CG of triangle is
Moments: 13 W ×
2
a horizontally from A
3
2
a + 23 W × 32 a = W × x
3
B1
M1
For equating moments about A, or equivalent
A1
For a correct unsimplified equation
Hence x
A1
4 Given answer correctly shown
-------------------------------------------------------------------------------------------------------------------------------------------------------7
(ii) RA × 2a = W × 79 a ⇒ RA = 18
W
M1
For one moments equation
= 119 a
RA + RD = W ⇒
RD = 11
W
18
A1
M1
For one correct answer
For resolving, or a second moments equation
A1
4
(i)
RB cos15 ° = 0. 1g
B1
Hence RB = 1.01 N
B1
32
RA + RB sin 15 ° = 0. 1×
0.4
M1
4 For a second correct answer
For using NII horizontally (3 terms needed)
A1
Correct equation
A1
5
Hence RA = 1. 99 N
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii) RA = 0, and RB = 1. 01 as before
B1
May be implied
RB sin 15° = 0.1×
v2
0. 4
v = 1. 02 m s −1
Specimen Materials - Mathematics
M1
A1
3
66
© OCR 2000
5
(i) EITHER : ∆ KE = 12 × 650 × (35 2 − 15 2 )
∆ PE = 650 g × 400 sin 5°
400 T = 325 000 − 222 073
T = 257 N
OR :
352 − 152
= 1. 25
2 × 400
T + 650 g sin 5° = 650 a
a=
B1
For correct unsimplified expression
B1
M1
A1
Sign of change not required
For equating work and energy change
B1
For correct unsimplified expression for a
M1
For 3-term NII equation || slope
A1
Correct equation
T = 257 N
A1
4
-------------------------------------------------------------------------------------------------------------------------------------------------------H
(ii) Driving force at Q is
B1
For any correct statement of ‘ P = Fv ’
35
H
900 =
+ 650 g sin 5°
M1
For 3-term force equation || slope at Q
35
A1
Correct equation
Hence H ≈ 12069 W i.e. 12 .1 kW
A1
4 Given answer correctly shown
-------------------------------------------------------------------------------------------------------------------------------------------------------H
+ 650 g sin 5° − 900 = 650 a
M1
For 4-term NII equation
15
a = 0.707 m s −2
A1
2
6
(i) EITHER : Moments of weight components are:
W cosθ × 4a (anticlockwise)
W sin θ × a (clockwise)
Hence total anticlockwise is
Wa ( 4 cosθ − sin θ )
OR :
OR :
Horizontal distance G – D is
−a sin θ + 4a cos θ
Hence anticlockwise moment is
Wa ( 4 cosθ − sin θ )
B1
B1
B1
Given answer correctly shown
M1
A1
For using horizontal projections
Correct expression
A1
Given answer correctly shown
Horizontal distance G – D is
(4 a ) 2 + a 2 cos(θ + α ) where tanα = 14 B1
Anticlockwise moment is
W × 17a 2
( 417 cosθ −
1
sin θ
17
)
M1
A1
3 Given answer correctly shown
i.e. Wa ( 4 cosθ − sin θ )
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii) Wa ( 4 cosθ − sin θ ) = R A × 8a sin θ
M1
For moments equation about D, using (i)
A1
Correct equation
− 1)
A1
3 For correct answer, in any form
-------------------------------------------------------------------------------------------------------------------------------------------------------(iii) RD = W
B1
FD = RA
B1
4 cotθ − 1
µ=
M1
For use of F = µR in equation involving θ
8
4
Hence tanθ =
A1
4 For showing given result correctly
8µ + 1
RA = 18 W (4 cotθ
Specimen Materials - Mathematics
67
© OCR 2000
7
(i) EITHER : 100 = 150t − 4. 9t2
150 ± 20540
t=
9.8
Time to C is 29.9 s
x = 300 cos 30° × 29 .9
BC = x − 1000 ≈ 6780 m
OR :
100 =
9. 8x2
x
−
3 2 × 300 2 × 34
M1
For use of s = ut + 12 at 2 vertically
A1
For correct equation
M1
For any solution method
A1
M1
A1
For trajectory equation with y = 100 ,
M1
V = 300 , θ = 30°
For correct unsimplified equation
For any solution method
A1
x ≈ 7776
M1
A1
Hence BC ≈ 6780 m
x
t=
M1
300 cos 30 °
= 29. 9 s
A1
6
-------------------------------------------------------------------------------------------------------------------------------------------------------9.8 × 1000 2
(ii) 100 = 1000 tanθ −
(1 + tan2 θ )
M1
For trajectory equation with y = 100 ,
2 × 300 2
x = 1000 , V = 300
M1
For use of sec2 θ = 1 + tan 2 θ
2
A1
Correct simplified quadratic
49 tan θ − 900 tan θ + 139 = 0
tanθ =
900 ± 900 2 − 27244
98
θ ≈ 9°
Specimen Materials - Mathematics
M1
For any solution method
M1
A1
For taking arctan of the smaller root
6
68
© OCR 2000
General Certificate of Education
Advanced Subsidiary (AS) and Advanced Level
M3
MATHEMATICS
Mechanics 3
Additional materials:
Answer paper
Graph paper
List of Formulae
TIME
1 hour 20 minutes
INSTRUCTIONS TO CANDIDATES
Write your name, Centre number and candidate number in the spaces provided on the answer paper.
Answer all the questions.
Give non-exact numerical answers correct to 3 significant figures, unless a different degree of
accuracy is specified in the question or is clearly appropriate.
–2
Where a numerical value for the acceleration due to gravity is needed, use 9.8 m s .
You are permitted to use a graphic calculator in this paper.
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 60.
Questions carrying smaller numbers of marks are printed earlier in the paper, and questions carrying
larger numbers of marks later in the paper.
You are reminded of the need for clear presentation in your answers.
This question paper consists of 4 printed pages.
Specimen Materials - Mathematics
69
© OCR 2000
1
A ball of mass 0.2 kg falls vertically onto a sloping grass bank, and rebounds horizontally (see diagram).
Immediately before the bounce the speed of the ball is 8 m s−1 , and immediately after the bounce the speed
is 3 m s−1 . Calculate the magnitude and direction of the impulse on the ball due to the impact.
[4]
2
A light elastic string of modulus 28 N and natural length 0.8 m has one end attached to a fixed point O. A
particle of mass 0.5 kg is attached to the other end.
(i) The particle hangs in equilibrium at the point E. Calculate the distance OE.
[2]
(ii) The particle is held at O and is released from rest. Calculate the speed of the particle as it passes the
point E.
[4]
3
Two uniform smooth spheres A and B, of equal radius, are free to move on a smooth horizontal table. The
mass of B is twice the mass of A. Initially B is at rest and A is moving with speed 5 m s−1 . The spheres
collide, and immediately before impact the direction of motion of A makes an angle of 30° with the line of
centres. After the collision A moves at right angles to its original direction (see diagram). Show that
(i) the speed of A immediately after the collision is
5
3
3 m s− 1 ,
(ii) the speed of B immediately after the collision is also
5
3
3 m s− 1 ,
(iii) the collision is perfectly elastic.
Specimen Materials - Mathematics
[2]
[3]
[3]
70
© OCR 2000
4
A particle of mass 0.2 kg is connected by two equal light elastic springs, each of natural length 0.5 m and
modulus of elasticity 5 N, to two points A and B on a smooth horizontal table. The mid-point of AB is O
and the length of AB is 1 m. The particle is displaced from O, towards B, through a distance of 0.3 m to
the point C and released from rest. In the subsequent motion air resistance may be neglected. After
t seconds the displacement of the particle from O is x metres. Show that
d2 x
= −100 x .
dt 2
[3]
The particle moves a distance 0.1 m from C to D. Find
5
(i) the speed of the particle at D,
[3]
(ii) the time taken to reach D.
[3]
A particle of mass m is attached to one end of a light inextensible string of length 10a. The other end of
the string is attached to a fixed point O. The particle is released from rest with the string taut and
horizontal. Assuming there is no air resistance, find
(i) the speed of the particle when the string has turned through 30°,
[2]
(ii) the tension in the string at this instant.
[3]
When the string reaches the vertical position, it comes into contact with a small fixed peg A which is a
distance 7a below O. The particle begins to move in a vertical circle of radius 3a with centre A (see
diagram). Determine, showing your working, whether the particle describes a complete circle about A. [5]
Specimen Materials - Mathematics
71
© OCR 2000
6
Two uniform beams AB and BC, each of length 5a, have masses 3m and 2m respectively. The beams are
freely jointed to fixed points at A and C, and to each other at B. The points A and C are on the same
horizontal level at a distance 8a apart, and the beams are in equilibrium with B vertically below the midpoint of AC, as shown in the diagram.
(i) Find the vertical component of the force acting on BC at C, and show that the horizontal component
of this force is 53 mg .
[6]
(ii) Find the magnitude and direction of the force acting on AB at B.
7
[5]
A body falls vertically, the forces acting being gravity and air resistance. The air resistance is proportional
to v, where v is the body’s speed at time t. The value of v for which the acceleration is zero is known as the
‘terminal velocity’ for the motion, and is denoted by U. Show that the equation of motion of the body may
be expressed as
dv g
= (U − v ) .
dt U
[3]
A parachutist jumps from a helicopter which is hovering at a height of several hundred metres, and falls
vertically. Assume that, before the parachute is opened, the terminal velocity for the motion is 50 m s −1 .
The parachutist opens the parachute 10 s after jumping. Find the speed at which the parachutist is falling
just before the parachute opens.
[5]
The diagram shows a (t, v ) graph for the parachutist’s motion, as modelled using the above differential
equation.
(i) Explain the significance of the speed of 10 m s −1 in relation to the differential equation.
[1]
(ii) What has been assumed about the opening of the parachute?
[1]
(iii) Find the decele ration of the parachutist just after the parachute opens.
[2]
Specimen Materials - Mathematics
72
© OCR 2000
General Certificate of Education
Advanced Subsidiary (AS) and Advanced Level
M3
MATHEMATICS
Mechanics 3
MARK SCHEME
MAXIMUM MARK
60
For live examinations, each Mark Scheme
includes the General Instructions for Marking
set out on pages 143 to 145.
This mark scheme consists of 3 printed pages and 1 blank page.
Specimen Materials - Mathematics
73
© OCR 2000
1


1
.
6



Direction above horizontal = tan −1  
0
.
6
  
Impulse is 1.71 N s at 69. 4° to horizontal
B1
For identifying impulse vector with the
change of momentum, by means of a triangle
or otherwise
M1
For either Pythagoras or trig calculation
Magnitude = 0.62 + 1.62
2
3
A1
A1
For magnitude
4 For angle to horizontal, or equivalent
28 x
M1
For equilibrium equation and Hooke
0 .8
x = 0. 14 ⇒ OE = 0.94 m
A1
2 Correct answer for OE
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii) Conservation of energy:
M1
For equation involving KE, PE and EE
2
28 × 0.14
1
× 0. 5v2 +
= 0.5g × 0.94
B1
For correct EE term
2
2 × 0.8
B1
For PE term, correct apart possibly from sign
0.25v2 + 0.343 = 4.606 , hence v = 4.13 m s−1
A1
4
(i) 0 .5 g =
(i) v A cos30° = 5 cos60°
Equating components ⊥ line of centres
M1
v A = ÷ ( 3) =
3
A1
2 Given answer correctly shown
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii) m × 5 cos30° = 2mvB − mvA cos60°
M1
Using momentum || line of centres
A1
Correct equation ( v A need not be numerical)
5
2
1
2
5
3
2 vB = 52 3 + 56 3 ⇒ vB = 53 3
A1
3 Given answer correctly shown
-------------------------------------------------------------------------------------------------------------------------------------------------------M1
Using restitution || line of centres
(iii) v A cos60° + vB = e × 5 cos30°
5
3
3 cos 60 ° + 53 3 = 5e cos 30 °
Hence e = 1 , as required
4
A1
Correct equation
A1
3 Given result correctly shown
5x
B1
Correct expression for a general position
0.5
M1
For relevant use of NII
Equation of motion is 10 x + 10x = −0.2x&&
&
&
A1
3 Given answer correctly shown
i.e. x = − 100 x
-------------------------------------------------------------------------------------------------------------------------------------------------------(i) Motion is SHM with amplitude 0.3 m
B1
Allow at any stage in the question
2
2
2
vD = 100 (0 .3 − 0. 2 )
M1
Force in each spring is
Speed at D is 2. 24 m s −1
A1
3
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii) 0.2 = 0.3cos(10tD )
B1
For correct SHM equation involving tD
tD = 0. 1 cos−1(23 )
M1
Time to reach D is 0.0841 s
A1
Specimen Materials - Mathematics
Or equivalent complete solution method
3
74
© OCR 2000
5
(i)
1
2
mv2 = mg × 10a sin 30°
M1
For relevant use of conservation of energy
Hence v = 10ga
A1
2
-------------------------------------------------------------------------------------------------------------------------------------------------------10 ga
(ii) T − mg cos 60 ° = m ×
M1
3-term NII equation || string
10 a
A1
Correct unsimplified equation
T = 32 mg
A1
3
-------------------------------------------------------------------------------------------------------------------------------------------------------Critical case is T = 0 at highest point
B1
May be implied
1
2
mvH = mg × 4a
M1
Use of energy to find v H at highest point
2
vH2 = 8 ga
mg + T =
A1
2
mv
3a
Resolving to find T ( = 53 mg ) when v = vH or
M1
to find critical v 2 ( = 3 ga) for T = 0
Hence it does make a complete circle
6
(i) Moments about A for the system:
3mg × 2a + 2mg × 6a = YC × 8a
YC = 94 mg
Moments about B for BC:
2 mg × 2a + X C × 3a =
9
mg
4
× 4a
A1
5 Correct result and reason
M1
A1
Equation with 3 terms needed
For correct unsimplified equation
A1
Correct answer for the vertical component
M1
Equation with 3 terms needed
A1
For correct unsimplified equation
XC =
A1
6 Given answer correctly shown
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii) X B = 53 mg
B1
5
mg
3
YB = 14 mg
B1
Follow the answer for YC in (i)


Y



B
−1
Dir above horizontal = tan 
 
X
 B 
M1
For numerical Pythagoras or trig calculation
1
Magnitude = 12
409 mg ≈ 1 .69 mg
A1
Correct exact or approximate value
A1
5 Correct exact or approximate angle
Magnitude =
X B2 + YB2
Dir above horizontal =
7
( ) ≈ 8.5°
3
tan−1 20
dv
= mg − kv
dt
At terminal velocity mg = kU
Equation of motion is m
B1
B1
dv g
= (U − v)
B1
3 Given answer correctly shown
Hence
dt U
-------------------------------------------------------------------------------------------------------------------------------------------------------⌠ 1 dv = ⌠ g dt
M1
For separation and attempt at integration


⌡ 50 − v
⌡ 50
− ln(50 − v) = 0.196t + c
A1
For both indefinite integrals correct
v = 0, t = 0 ⇒ − ln 50 = c
− ln(50 − v10) = 1.96 − ln 50
M1
A1
Evaluation of constant or equiv use of limits
Correct equation for v10
v10 = 50 (1 − e−1 .96 ) ≈ 43 .0 m s-1
A1
5 For correct exact or approximate answer
-------------------------------------------------------------------------------------------------------------------------------------------------------(i) 10 m s −1 is the terminal velocity (value of U) after
the parachute opens
B1
1
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii) The parachute is assumed to open instantaneously B1
1
-------------------------------------------------------------------------------------------------------------------------------------------------------d v 9 .8
(iii)
=
(10 − 43 .0 )
M1
d t 10
Hence deceleration is 32.3 m s −2
A1
2
Specimen Materials - Mathematics
75
© OCR 2000
BLANK PAGE
Specimen Materials – Mathematics
76
© OCR 2000
General Certificate of Education
Advanced Subsidiary (AS) and Advanced Level
M4
MATHEMATICS
Mechanics 4
Additional materials:
Answer paper
Graph paper
List of Formulae
TIME
1 hour 20 minutes
INSTRUCTIONS TO CANDIDATES
Write your name, Centre number and candidate number in the spaces provided on the answer paper.
Answer all the questions.
Give non-exact numerical answers correct to 3 significant figures, unless a different degree of
accuracy is specified in the question or is clearly appropriate.
–2
Where a numerical value for the acceleration due to gravity is needed, use 9.8 m s .
You are permitted to use a graphic calculator in this paper.
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 60.
Questions carrying smaller numbers of marks are printed earlier in the paper, and questions carrying
larger numbers of marks later in the paper.
You are reminded of the need for clear presentation in your answers.
This question paper consists of 4 printed pages.
Specimen Materials - Mathematics
77
© OCR 2000
74
1
The region bounded by the part of the curve y = x from x = 0 to x = a , the x-axis and the line x = a is
rotated completely about the x-axis to form a uniform solid of revolution. Find by integration the
x-coordinate of the centre of mass of the solid.
[6]
2
The diagram shows a uniform circular disc, of mass m and radius a, which is free to rotate in a vertical
plane about a smooth fixed horizontal axis through its centre O. A light inextensible string is wrapped
round the circumference of the disc, and has one end attached to the circumference. A particle of mass m
hangs freely at the other end of the string. The system is released from rest. Show that the angular
2g
acceleration of the disc is
, and find the tension in the string.
[6]
3a
3
A rigid square frame consists of four uniform rods, each of mass m and length 2a, joined at their ends to
form a square. Show that the moment of inertia of the frame, about an axis through one of its corners and
perpendicular to its plane, is 403 ma 2 .
[3]
The frame is suspended from one corner, and can rotate in a vertical plane about a smooth horizontal axis
through that corner. Show that the motion in which the frame makes small oscillations about its
equilibrium position is approximately simple harmonic, and find the period of this simple harmonic
motion.
[5]
4
A uniform circular disc, of mass m and radius a, can rotate in a vertical plane about a fixed horizontal axis
passing through its centre O. When the disc is at rest, a particle of mass 2m is released, from rest, at a
height 2a vertically above one end A of the horizontal diameter of the disc. The particle falls freely, strikes
the disc at A, and adheres to the disc. Find the angular speed with which the disc starts to rotate.
[5]
While the disc (with the attached particle) rotates, a constant frictional couple C acts on the disc. The disc
comes to rest after one quarter of a revolution, when the particle is at the lowest point of the disc. Find C.
[3]
Specimen Materials - Mathematics
78
© OCR 2000
75
5
A uniform rod AB, of mass m and length 2a, is free to rotate in a vertical plane about a smooth horizontal
axis through A. When the rod is hanging at rest in equilibrium with B vertically below A, it is given an
angular speed Ω, where
Ω2 =
3g
.
4a
Show that the rod comes instantaneously to rest when it has turned through an angle 13 π .
[2]
At this position of instantaneous rest, the force acting on the rod at A has horizontal and vertical
components X and Y respectively. Find X and Y in terms of m and g.
[8]
6
A light aircraft flies flies from A due east to B and then flies directly back to A. The distance AB is d, and
the speed of the aircraft relative to the air is 4V. During the flight, there is a steady wind, of speed V in a
direction making an angle θ with AB (see diagram). Show that the total flying time for the journey from A
to B and back is
2d 15 + cos 2 θ
.
15V
Specimen Materials - Mathematics
79
[10]
© OCR 2000
76
7
A small smooth bead B, of mass m, is threaded on a circular wire with centre O and radius a. The wire is
fixed in a vertical plane. A light elastic string, of natural length a and modulus of elasticity λ , has one end
fixed at O. The string passes through a small smooth ring A fixed at the highest point of the wire, and the
other end of the string is attached to B. The diagram shows the system at an instant when OB makes an
angle θ with the downward vertical at O.
(i) Taking the horizontal level of O as the reference level for gravitational potential energy, show that the
total potential energy of the system in the position shown is
λa + a(λ − mg ) cos θ .
[4]
(ii) Hence show that there is a position of stable equilibrium with θ = 0 so long as λ < mg .
[3]
(iii) Given that λ = 12 mg , show that the approximate period of small oscillations about the equilibrium
position is
2π
Specimen Materials - Mathematics
2a
.
g
[5]
80
© OCR 2000
General Certificate of Education
Advanced Subsidiary (AS) and Advanced Level
M4
MATHEMATICS
Mechanics 4
MARK SCHEME
MAXIMUM MARK
60
For live examinations, each Mark Scheme
includes the General Instructions for Marking
set out on pages 143 to 145.
This mark scheme consists of 4 printed pages.
Specimen Materials - Mathematics
81
© OCR 2000
1
a
Total mass = πρ ∫ x dx
0
= 12 πρ a 2
a
Total moment = πρ ∫ x 2 dx
0
= 13 πρ a 3
Hence 12 πρ a 2 x = 13 πρ a 3 ⇒ x = 23 a
For particle: mg − T = mf
For disc: Ta =
dx
A1
Introduction of ρ is required
M1
For relevant use of
A1
No further penalty if ρ omitted
M1
For equating and solving for x
∫ xy
2
dx
6 For correct final answer
B1
1
ma 2α
2
B1
String moves with disc, so f = aα
B1
Stated or used at any stage
Hence mg = mf + 12 ma α = 32 ma α
M1
For obtaining an equation in α (or T)
A1
Given answer shown correctly
2g
3a
1
T = 3 mg
α =
3
2
For relevant use of
A1
2
∫y
M1
A1
M.I. for one of the rods opposite the axis is:
1
ma 2 + m(a 2 + 4a 2) = 16
ma 2
3
3
6
M1
A1
Relevant use of parallel axes theorem
For correct unsimplified expression
M.I. of frame = 2( 43 ma 2 + 16
ma 2 ) = 40
ma 2
A1
3 Given answer shown correctly
3
3
-------------------------------------------------------------------------------------------------------------------------------------------------------B1
Weight 4mg acts at a 2 from the axis
40
2 &&
M1
For relevant use of C = I θ&& at general posn
Equation of motion is: 4 mga 2 sin θ = − ma θ
3
3g 2
i.e. θ&& ≈ −
θ
10a
Hence SHM with period 2π
4
10a
3g 2
Speed of particle before impact is
2 m 4 ga × a = (12 ma 2 + 2ma 2 )ω
A1
Correct equation
A1
Reduction to standard SHM form
A1
4ga
5
B1
M1
Equating ang mom before and after
B1
For total M.I.
A1
Correct equation
(12 ma 2 + 2ma 2 ) , or equivalent
8 g
A1
5
5 a
-------------------------------------------------------------------------------------------------------------------------------------------------------B1
Work done against friction is C × 12 π
Hence ω =
Hence 12 πC = 54 ma 2ω 2 + 2mga
C=
52mga
5π
Specimen Materials - Mathematics
M1
Equating energy (KE and PE) to work
A1
3
82
© OCR 2000
5
3g
1 4
. ma 2.
2 3
4a
= mga(1 − cosθ )
M1
For relevant use of conservation of energy
LHS = 12 mga, so θ = 13 π
A1
2 Given answer obtained or verified correctly
--------------------------------------------------------------------------------------------------------------------------------------------------------
They may have X and/or Y reversed; any
consistent work is acceptable
mga sin (13 π ) = 43 ma 2α
M1
3 3g
8a
Acceleration component of CG || rod is zero
3 3g
and perpendicular to rod is
8
i.e. α =
EITHER : Res hor: X = m.
i.e. X =
3 3g
cos(13π )
8
3 3
mg
16
Res vert: mg − Y = m.
Res || rod:
1
2
3 3g
sin (13π )
8
B1
May be implied
A1
M1
To include attempt at resolving transverse acc
M1
A1
3 X + 12 Y = 12 mg
Res ⊥ rod:
1
X − 12 3Y + 12 3mg =
2
X =
A1
A1
7
mg
Hence Y = 16
OR :
Taking moments about A to find ang acc
3
8
3mg
3 3
7
mg , Y = 16
mg
16
B1
M1
A1
4 terms required
Correct equation
A1
8 For both answers
6
B1
For correct triangle; may be implied
sin α = 14 sin θ or 16V 2 = R 2 + V 2 − 2RV cosθ
B1
For appropriate use of sine or cosine rule
R = V cosθ + 4V 1 − 14 sin 2 θ or
M1
Any method for R in terms of V and θ
R = V cos θ + V cos2 θ + 15 or equivalent
A1
For any correct (unsimplified) expression
B1
For correct return triangle; may be implied
M1
A1
Any method for S in terms of V and θ
For any correct (unsimplified) expression
M1
For expressing this in terms of d, V, θ
d (R + S )
2dV cos2 θ + 15
= 2
RS
V (cos2 θ + 15) − V 2 cos2 θ
M1
For combining the terms algebraically
2d 15 + cos2 θ
15V
A1
S = −V cosθ + V cos2 θ + 15 or equivalent
Total time is
i.e.
d d
+
R S
=
Specimen Materials - Mathematics
10 Given answer correctly shown
83
© OCR 2000
7
(i)
AB = 2a cos 12 θ or AB 2 = 2a 2 + 2a 2 cosθ
2a 2λ (1 + cosθ )
2a
PE of B is − mgacosθ
Elastic energy =
M1
A1
Or equivalent
B1
A1
4 Given answer correctly shown
Total energy V = λ a + a(λ − mg ) cosθ
-------------------------------------------------------------------------------------------------------------------------------------------------------dV
(ii)
= −a(λ − mg ) sin θ = 0
M1
Differentiate and equate to zero, or equivalent
dθ
argument from properties of cos graph
A1
Given answer correctly shown
Hence θ = 0
2
dV
= a( mg − λ ) > 0 if λ < mg
A1
3 Stability explained via identification of min V
dθ 2 θ =0
-------------------------------------------------------------------------------------------------------------------------------------------------------(iii) 12 m (a θ&) 2 + V = constant
B1
ma 2θ&& + 12 mga sin θ = 0
g
i.e. θ&& ≈ − θ , so SHM
2a
2a
Period is 2π
g
Specimen Materials - Mathematics
M1
Differentiate with respect to t
A1
Correct simplified equation
M1
Reduction to standard form
A1
5 Given answer correctly shown
84
© OCR 2000
General Certificate of Education
Advanced Subsidiary (AS) and Advanced Level
S1
MATHEMATICS
Probability & Statistics 1
Additional materials:
Answer paper
Graph paper
List of Formulae
TIME
1 hour 20 minutes
INSTRUCTIONS TO CANDIDATES
Write your name, Centre number and candidate number in the spaces provided on the answer paper.
Answer all the questions.
Give non-exact numerical answers correct to 3 significant figures, unless a different degree of
accuracy is specified in the question or is clearly appropriate.
You are permitted to use a graphic calculator in this paper.
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 60.
Questions carrying smaller numbers of marks are printed earlier in the paper, and questions carrying
larger numbers of marks later in the paper.
You are reminded of the need for clear presentation in your answers.
This question paper consists of 4 printed pages.
Specimen Materials - Mathematics
85
© OCR 2000
1
Janet and John wanted to compare their daily journey times to work, so they each kept a record of their
journey times for a few weeks. Janet’s daily journey times, x minutes, for a period of 25 days, were
summarised by Σx = 2120 and Σx 2 = 180 044 . Calculate the mean and standard deviation of Janet’s
journey times.
[3]
John’s journey times had a mean of 79.7 minutes and a standard deviation of 6.22 minutes. Describe
briefly, in everyday terms, how Janet and John’s journey times compare.
[2]
2
Two independent assessors awarded marks to each of 5 projects. The results were as shown in the table.
Project
A
B
C
D
E
First assessor
Second assessor
38
56
91
84
62
41
83
85
61
62
Calculate Spearman’s rank correlation coefficient for the data.
[4]
Show, by sketching a suitable scatter diagram, how two assessors might have assessed 5 projects in such a
way that Spearman’s rank correlation coefficient for their marks was + 1 while the product moment
correlation coefficient for their marks was not + 1 . (Your scatter diagram need not be drawn accurately to
scale.)
[2]
3
Each packet of the breakfast cereal Fizz contains one plastic toy animal. There are five different animals in
the set, and the cereal manufacturers use equal numbers of each. Without opening a packet it is impossible
to tell which animal it contains. A family has already collected four different animals at the start of a year
and they now need to collect an elephant to complete their set. The family is interested in how many
packets they will need to buy before they complete their set.
(i) Stating any necessary assumption, name an appropriate distribution with which to mode l this
situation. What is the expected number of packets that they family will need to buy?
[3]
(ii) Find the probability that the family will complete their set with the third packet they buy after the start
of the year.
[2]
(iii) Find the probability that, in order to complete their collection, the family will need to buy more than 4
packets after the start of the year.
[3]
4
A sixth-form class consists of 7 girls and 5 boys. Three students from the class are chosen at random. The
number of boys chosen is denoted by the random variable X. Show that
(i) P( X = 0) =
7 ,
44
[2]
(ii) P( X = 2 ) =
7 .
22
[3]
The complete probability distribution of X is shown in the following table.
x
P(X = x )
0
1
2
3
7
44
21
44
7
22
1
22
Calculate E(X ) and Var( X ) .
Specimen Materials - Mathematics
[5]
86
© OCR 2000
5
The diagram shows the cumulative frequency graphs for the marks scored by the candidates in an
examination. The 2000 candidates each took two papers; the upper curve shows the distribution of marks
on paper 1 and the lower curve shows the distribution on paper 2. The maximum mark on each paper was
100.
(i) Use the diagram to estimate the median mark for each of paper 1 and paper 2, and the interquartile
range for paper 1.
[6]
(ii) State with a reason which of the two papers you think was the easier one.
[2]
(iii) The candidates’ marks for the two papers could also be illustrated by means of a pair of box-and
whisker plots. Give two brief comments on any advantages or disadvantages in using cumulative
frequency graphs and box-and-whisker plots to represent the data.
[2]
6
Items from a production line are examined for any defects. The probability that any item will be found to
be defective is 0.15, independently of all other items.
(i) A batch of 16 items is inspected. Using tables of cumulative binomial probabilities, or otherwise, find
the probability that
(a) at least 4 items in the batch are defective,
[2]
(b) exactly 4 items in the batch are defective.
[2]
(ii) Five batches, each containing 16 items, are taken.
(a) Find the probability that at most 2 of these 5 batches contain at least 4 defective items.
[4]
(b) Find the expected number of batches that contain at least 4 defective items.
[2]
Specimen Materials - Mathematics
87
© OCR 2000
7
An experiment was conducted to see whether there was any relationship between the maximum tidal
current, y cm s−1 , and the tidal range, x metres, at a particular marine location. [The tidal range is the
difference between the height of high tide and the height of low tide.] Readings were taken over a period
of 12 days, and the results are shown in the following table.
x
2.0
2.4
3.0
3.1
3.4
3.7
3.8
3.9
4.0
4.5
4.6
4.9
y
15.2
22.0
25.2
33.0
33.1
34.2
51.0
42.3
45.0
50.7
61.0
59.2
[Σx = 43 .3, Σy = 471 .9, Σx2 = 164 .69 , Σy2 = 20 915 .75 , Σxy = 1837 .78.]
The scatter diagram below illustrates the data.
(i) Calculate the product moment correlation coefficient for the data, and comment briefly on your
answer with reference to the appearance of the scatter diagram.
[4]
(ii) Calculate the equation of the regression line of maximum tidal current on tidal range.
[3]
(iii) Estimate the maximum tidal current on a day when the tidal range is 4.2 m, and indicate briefly how
reliable an estimate you consider your answer to be.
[3]
(iv) It is suggested that the equation found in part (ii) could be used to predict the maximum tidal current
on a day when the tidal range is 15 m. Comment briefly on the validity of this suggestion.
[1]
Specimen Materials - Mathematics
88
© OCR 2000
General Certificate of Education
Advanced Subsidiary (AS) and Advanced Level
S1
MATHEMATICS
Probability & Statistics 1
MARK SCHEME
MAXIMUM MARK
60
For live examinations, each Mark Scheme
includes the General Instructions for Marking
set out on pages 143 to 145.
This mark scheme consists of 3 printed pages and 1 blank page.
Specimen Materials - Mathematics
89
© OCR 2000
1
Mean is 84.8 minutes
B1
180044
Standard deviation =
− 84.82
M1
May be implied
25
= 3.27 minutes
A1
3
-------------------------------------------------------------------------------------------------------------------------------------------------------John’s average time is about 5 mins less than Janet’s
B1
John’s times are more variable than Janet’s
B1
2
2
Ranks are:
2
3
B1
Or with ranks reversed
Values of d are − 1, 1, 2, − 1, − 1
M1
Or reversed, or values of d 2
M1
Correct formula for Spearman used
rs = 1 −
1
2
5
4
3
1
4
5
6 ×8
= 0. 6
5 × 24
A1
4 Correct answer (fraction or decimal)
--------------------------------------------------------------------------------------------------------------------------------------------------------
3
(e.g.)
B2
(i) Each packet is equally likely to contain any of the
5 animals, independently of other packets
Geometric distribution
B1
B1
2 For 5 points, showing any non-linear
‘increasing’ relationship
Allow either ‘equally likely’ or ‘independent’
No need to state p = 15 here
Expected number is 5
B1
3
-------------------------------------------------------------------------------------------------------------------------------------------------------16
(45 )2 × (15) = 125
(ii)
or 0.128
Any numerical ‘ q n p ’ calculation
M1
A1
2 Correct answer
--------------------------------------------------------------------------------------------------------------------------------------------------------
(45 )4
(iii)
256
625
4
or 1 −
{ + ( )( )+ ( ) ( )+ ( ) ( )}
1
5
4 1
5 5
4 2 1
5 5
43 1
5 5
Allow M mark even if there is an error of 1 in
the number of terms
For correct expression for the answer
3 Correct fraction or decimal
A1
A1
or 0 .4096 or 0 .410
7
12
35 =
P( X = 0) =   ÷   = 220
3
  3
M1
n
For ratio of relevant   terms, or equivalent
r
A1
2 Given answer correctly shown
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii) P( X = 2) = P(2 boys and 1 girl)
B1
May be implied
(i)
7
44
7
5
12
7 × 10 7
=   ×   ÷   =
= 22
 1  2  3  220
M1
n
For use of relevant   terms, or complete
r
alternative multiplication/addition of probs
A1
3 Given answer correctly shown
-------------------------------------------------------------------------------------------------------------------------------------------------------7
7
1
E( X ) = 0 × 44
+ 1 × 21
+ 2 × 22
+ 3 × 22
= 54
M1
For correct calculation process
44
E( X ) = 0 ×
2
7
44
21
7
1
+ 1 × 44
+ 4 × 22
+ 9 × 22
= 95
44
5 2
105
= 176 or 0. 597 (to 3dp)
4
Var ( X ) = 95
−(
44
)
M1
A1
B1
Correct answer (fraction or decimal form)
For correct numerical expression for Σx 2 p
M1
For correct overall method for variance
A1
Specimen Materials - Mathematics
5 For correct answer
90
© OCR 2000
5
6
(i) Medians correspond to 1000 candidates
m1 = 38, m2 = 63
M1
Reading off at 1000; may be implied
A1
Correct value for either median
A1
For both correct
Quartiles correspond to 1500 and 500 candidates M1
Reading off at either; may be implied
q3 = 56, q1 = 26
A1
Both correct
IQR = 30
A1
6
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii) Paper 2 was easier
B1
Marks were higher on paper 2
B1
2 Or similar statement, e.g. ‘higher median’
-------------------------------------------------------------------------------------------------------------------------------------------------------(iii) Possible valid comments include:
Box plots give quick direct comparisons of
medians and IQRs
Box plots don’t include all the information that
CF graphs do
CF graphs can be used to read off values both
ways round
B1
For any one valid comment
etc
B1
2 Any other valid comment
(i) (a)
1 − 0.7899 = 0.210 (1)
M1
Complement of relevant tabular value
A1
2 Correct answer
-------------------------------------------------------------------------------------------------------------------------------------------------------M1
Subtracting relevant tabular values
(b) 0.9209 − 0. 7899 − 0.131
A1
2 Correct answer
-------------------------------------------------------------------------------------------------------------------------------------------------------Any use of B(5, 0.210)
(ii) (a) 0.7905 + 5 × 0.7904 × 0.210 + 10 × 0.7903 × 0.2102 M1
M1
Correct 3 cases identified
A1
Correct numerical expression for required
probability, with their value from (i)(a)
= 0.934
A1
4
-------------------------------------------------------------------------------------------------------------------------------------------------------(b) 5 × 0.210 = 1.05
M1
For relevant use of np
A1
2 For correct answer
7
(i) r =
43 .3 × 471 .9
12
43.32 
471 . 92 

164 . 69 −
 20915 .75 −

12 
12 

1837 . 78 −
M1
Or equivalent; may be implied
= 0. 956
A1
B1
For relating the value to 1
The value is close to +1 ,
and the points in the diagram lie (fairly) close to
a straight line with positive gradient
B1
4 For a reasonable comment about linearity
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii) Gradient of regression line is
43 .3 × 471 .9
1837 . 78 −
12
= 15 .9789
M1
May be implied if calculator routine used
43 .32
164 .69 −
12
471 .9
43 .3

y−
= 15. 9789  x −
M1
May similarly be implied

12
12 

y = 16. 0x − 18.3
A1
3
-------------------------------------------------------------------------------------------------------------------------------------------------------(iii) y = 16. 0 × 4. 2 − 18.3
M1
Current is 48.8 cms −1
A1
Units required in answer
Diagram indicates some uncertainty, e.g. ± 5 cm s−1 B1
3 Allow any reasonable comment
-------------------------------------------------------------------------------------------------------------------------------------------------------(v) The prediction would be (very) unreliable because
of the extrapolation involved
B1
1 For conclusion and idea of extrapolation
Specimen Materials - Mathematics
91
© OCR 2000
BLANK PAGE
Specimen Materials – Mathematics
92
© OCR 2000
General Certificate of Education
Advanced Subsidiary (AS) and Advanced Level
S2
MATHEMATICS
Probability & Statistics 2
Additional materials:
Answer paper
Graph paper
List of Formulae
TIME
1 hour 20 minutes
INSTRUCTIONS TO CANDIDATES
Write your name, Centre number and candidate number in the spaces provided on the answer paper.
Answer all the questions.
Give non-exact numerical answers correct to 3 significant figures, unless a different degree of
accuracy is specified in the question or is clearly appropriate.
You are permitted to use a graphic calculator in this paper.
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 60.
Questions carrying smaller numbers of marks are printed earlier in the paper, and questions carrying
larger numbers of marks later in the paper.
You are reminded of the need for clear presentation in your answers.
This question paper consists of 3 printed pages and 1 blank page.
Specimen Materials - Mathematics
93
© OCR 2000
90
1
Before being packed in boxes, apples in a fruit-packing plant have to be checked for bruising. The apples
pass along a conveyor belt, and an inspector removes any of the apples that are badly bruised. Badly
bruised apples arrive at random times, but at a constant average rate of 1.8 per minute.
(i) Find the probability that at least one badly bruised apple arrives in a one-minute period.
[3]
(ii) In a period of a minutes, the probability of at least one badly bruised apple arriving is 0.995. Find the
value of a.
[3]
2
A student answers a test consisting of 16 multiple-choice questions, in each of which the correct response
has to be selected from the four possible answers given. The student only gets 2 of the questions correct,
and the teacher remarks that this ‘shows that the student did worse than anyone would do just by guessing
the answers’. The probability of the student answering a question correctly is denoted by p, assumed to be
the same for all the questions in the test.
(i) State suitable null and alternative hypotheses, in terms of p, for a test to examine whether the
teacher’s remark is justified.
[2]
(ii) Carry out the test, using a 10% significance level, and state your conclusion clearly.
3
[4]
Lessons in a school are supposed to last for 40 minutes. However, a Mathematics teacher finds that pupils
are usually late in arriving for his lessons, and that the actual length of teaching time available can be
modelled by a normal distribution with mean 34.8 minutes and standard deviation 1.6 minutes.
(i) Find the probability that the length of teaching time available will be less than 37.0 minutes.
[2]
(ii) The probability that the length of teaching time available exceeds m minutes is 0.75. Find m.
[3]
The teacher has a weekly allocation of 5 lessons with a particular class. Assuming that these 5 le ssons can
be regarded as a random sample, find the probability that the mean length of teaching time available in
these 5 lessons will lie between 34.0 and 36.0 minutes.
[4]
4
It is given that 93% of children in the UK have been immunised against whooping cough. The number of
children in a random sample of 60 UK children who have been immunised is X, and the number who have
not been immunised is Y. State, with reasons, which of X or Y has a distribution which can be
approximated by a Poisson distribution.
[3]
Using a Poisson approximation, find the probability that at least 58 children in the sample of 60 have been
immunised against whooping cough.
[3]
Three random samples, each of 60 UK children, are taken. Find the probability that in one of these
samples exactly 59 children have been immunised while in each of the other two samples exactly 58
children have been immunised. Give your answer correct to 1 significant figure.
[3]
Specimen Materials - Mathematics
94
© OCR 2000
5
The continuous random variable X has probability density function f given by
k x2 (3 − x)
f( x ) = 
0
0 ≤ x ≤ 3,
otherwise,
where k is a constant.
4 , and find E(X ) .
(i) Show that k = 27
[5]
(ii) Find P(X < 2) .
[2]
(iii) Use your answer to part (ii) to state, with a reason, whether the median of X is less than 2, equal to 2
or greater than 2.
[2]
6
The ‘reading age’ of children about to start secondary school is a measure of how good they are at reading
and understanding printed text. A child’s reading age, measured in years, is denoted by the random
variable X. The distribution of X is assumed to be N(µ, σ 2 ) . The reading ages of a random sample of 80
children were measured, and the data obtained is summarised by Σ x = 892 .7 , Σx2 = 10 266 .82 .
(i) Calculate unbiased estimates of µ and σ 2 , giving your answers correct to 2 decimal places.
[3]
(ii) Previous research has suggested that the value of µ was 10.75. Determine whether the evidence of
this sample indicates that the value of µ is now different from 10.75. Use a 10% significance level
for your test.
[5]
(iii) State, giving a brief reason, whether your conclusion in part (ii) would remain valid if
7
(a) the distribution of X could not be assumed to be normal,
[1]
(b) the 80 children were all chosen from those starting at one particular secondary school.
[1]
The breaking strength of a certain type of fishing line has a normal distribution with standard deviation
0.24 kN. A random sample of 10 lines is tested. The mean breaking strengths of the sample and of the
population are x kN and µ kN respectively. The null hypothesis µ = 8 .75 is tested against the
alternative hypothesis µ < 8 .75 at the 2 12 % significance level.
(i) Show that the range of values of x for which the null hypothesis is rejected is given by x < 8.60 ,
correct to 2 decimal places.
[4]
(ii) Explain briefly what is meant, in the context of this question, by a Type I error, and state the
probability of making a Type I error.
[2]
(iii) Explain briefly what is meant, in the context of this question, by a Type II error, and find the
[5]
probability of making a Type II error when µ = 8 .50 .
Specimen Materials - Mathematics
95
© OCR 2000
BLANK PAGE
Specimen Materials - Mathematics
96
© OCR 2000
General Certificate of Education
Advanced Subsidiary (AS) and Advanced Level
S2
MATHEMATICS
Probability & Statistics 2
MARK SCHEME
MAXIMUM MARK
60
For live examinations, each Mark Scheme
includes the General Instructions for Marking
set out on pages 143 to 145.
This mark scheme consists of 3 printed pages and 1 blank page.
Specimen Materials - Mathematics
97
© OCR 2000
1
(i)
P( at least one) = 1 − e−1 .8
B1
For any use of Po(1.8)
M1
For use of complementary probability
= 0. 835
A1
3
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii) 1 − e− 1. 8a = 0.995
B1
−1.8a = ln 0.005
M1
For correct use of logs in solving for a
a = 2. 94
A1
3
2
(i) Null hypothesis: p = 14
B1
Alternative hypothesis: p < 14
B1
2
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii) Under the NH, numbercorrect~ B(16, 14)
M1
May be implied
3
(i)
Hence P(2 or fewer correct) = 0. 1971
A1
Using tables or direct calculation
This is greater than 0.1 (not significant)
There is insufficient evidence to justify the
teacher’s suggestion that the score was worse
than would be produced by pure guesswork
M1
For comparing with the significance level
37.0 − 34.8 
P(T < 37. 0) = Φ
 = Φ(1.375) = 0.915
1.6


A1
4
M1
Standardising and using tables
A1
2 Correct answer 0.915
-------------------------------------------------------------------------------------------------------------------------------------------------------m − 34 . 8
= − 0. 674
(ii)
M1
Equating standardised m to a z value
1 .6
B1
For use of (± )0. 674 in an equation
A1
3
Hence m = 33 .7
-------------------------------------------------------------------------------------------------------------------------------------------------------36.0 − 34. 8 
 34. 0 − 34 .8  M1
P( 34.0 < T < 36. 0) = Φ
For using N( 34 . 8, 1 .6 2 / 5)
 − Φ

 1.6 / 5 
 1.6 / 5 
A1
For both end-points standardised correctly
= Φ(1.677) − {1 − Φ(1. 118)}
M1
Correct process for prob between end-points
= 0. 821
A1
4
4
X ~ B( 60, 0.93), Y ~ B( 60, 0. 07)
M1
For either binomial distribution identified
Hence Y is suitable for a Poisson approximation,
A1
For correct conclusion
since n is large and p is small
A1
3 For correct justification
-------------------------------------------------------------------------------------------------------------------------------------------------------Y ~ Po(4.2)
B1
May be implied
Required probability is P(Y ≤ 2)
M1
For attempted evaluation of relevant Po prob
i.e. 0.210
A1
3 Using tables or direct calculation
-------------------------------------------------------------------------------------------------------------------------------------------------------2
 4.22 −4.2 
Required probability is 4.2 e− 4. 2× 
e
B1
For p1 × p2 × p2 , using Poisson or binomial
 ×3
 2

B1
For correct factor of 3
i.e. 0.003 to 1 sf
B1
3 Follow through wrong value of 4.2 only
Specimen Materials - Mathematics
98
© OCR 2000
5
3
(i) k ∫ (3x2 − x 3) dx = 1
M1
Equating to 1 and attempting to integrate
k [x3 − 14 x4 ]0 = 1
A1
For correct integration
81
4
k  27 −  = 1 ⇒ k = 27
4

A1
Given answer correctly obtained
M1
For correct application of
0
3
3
E( X ) = 274 ∫ (3x3 − x 4) dx
0
∫ x f(x ) dx
= 95
A1
5
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii)
4
P( X < 2) = 27
[x3 − 14 x 4 ]0
2
M1
= 16
A1
2 Or equivalent decimal answer
27
-------------------------------------------------------------------------------------------------------------------------------------------------------(iii) 16
is greater than 12
M1
For comparison of answer (ii) with 12
27
Hence the median is less than 2
6
(i)
892 .7
= 11 .16 to 2dp
80
1 
892 .72 
s2 = 10 266. 82 −

79 
80 
x=
A1
2
B1
For correct value 11.16
M1
For this expression, or equivalent
= 3.87 to 2dp
A1
3
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii) H0 : µ = 10.75 , H1 : µ ≠ 10.75
B1
Both hypotheses
Test statistic is z =
11 .16 − 10 .75
= 1. 86
3. 87 / 80
Standardising attempt using s 2 / 80
M1
A1
This is greater than critical (2 tail) value z = 1. 645 M1
There is evidence to suggest that the value of µ is
Correct value; follow their s
Or comparing Φ (1.86) with 5%
now different
A1
5
-------------------------------------------------------------------------------------------------------------------------------------------------------(iii) (a) Still valid, since the sample size (80) is large
Allow any reasoned conclusion mentioning
enough to appeal to the CLT
B1
1 the CLT
-------------------------------------------------------------------------------------------------------------------------------------------------------(b) Not valid, since the children starting at one
school may not be representative of all
children of this age.
B1
1 For conclusion and reason
7
(i) Critical value is 8.75 − 1. 96 ×
0.24
10
M1
Calculation of correct form 8.75 − z × S.E.
B1
Relevant use of − 1. 96
A1
Relevant use of 0.24 / 10
i.e. reject null hypothesis when x < 8.60
A1
4
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii) NH µ = 8. 75 would be rejected when the mean
breaking strength is in fact 8.75 kN
B1
P ( Type I error) = 0.025
B1
2
-------------------------------------------------------------------------------------------------------------------------------------------------------(iii) NH µ = 8. 75 would be accepted when the mean
breaking strength is in fact less than 8.75 kN
Type II error occurs when x > 8.60
8.60 − 8.50 
Probability is 1 − Φ 

 0. 24 / 10 
= 0. 0938
Specimen Materials - Mathematics
B1
B1
May be implied
M1
Using normal distribution with mean 8.50
A1
A1
Correct standardising, and use of tables
5
99
© OCR 2000
BLANK PAGE
Specimen Materials - Mathematics
100
© OCR 2000
General Certificate of Education
Advanced Subsidiary (AS) and Advanced Level
S3
MATHEMATICS
Probability & Statistics 3
Additional materials:
Answer paper
Graph paper
List of Formulae
TIME
1 hour 20 minutes
INSTRUCTIONS TO CANDIDATES
Write your name, Centre number and candidate number in the spaces provided on the answer paper.
Answer all the questions.
Give non-exact numerical answers correct to 3 significant figures, unless a different degree of
accuracy is specified in the question or is clearly appropriate.
You are permitted to use a graphic calculator in this paper.
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 60.
Questions carrying smaller numbers of marks are printed earlier in the paper, and questions carrying
larger numbers of marks later in the paper.
You are reminded of the need for clear presentation in your answers.
This question paper consists of 4 printed pages.
Specimen Materials - Mathematics
101
© OCR 2000
1
A test of the null hypothesis of independence of two characteristics is required for the following
contingency table.
Characteristic 1
Characteristic 2
Totals
Totals
12
9
19
40
28
61
71
160
40
70
90
200
(i) Find the expected frequency corresponding to the cell with observed frequency 12.
[1]
(ii) Given that the value of the χ 2 test statistic is 4.80, correct to 3 significant figures, carry out the test at
the 10% significance level.
[3]
2
The ‘customer services’ section in a department store deals both with enquiries by telephone and also with
enquiries in person by shoppers in the store. Telephone enquiries occur at random times at an average rate
of 4 per half hour. Shoppers in the store arrive to make enquiries at random times at an average rate of 5
per half hour. Assuming that the two types of enquiries occur independently of each other, find the
probability that a total of between 10 and 20 enquiries (inclusive) have to be dealt with in a randomly
chosen half hour period.
[5]
3
A supermarket sells 2 kg bags of new potatoes, in which the potatoes have been selected to be all roughly
the same size. The potatoes used to fill the bags may be assumed to be randomly chosen items from a
population in which the mass, in grams, of an individual potato is normally distributed with mean 90 and
standard deviation 4.
(i) Show that the probability that the total mass of 21 of these potatoes exceeds 2 kg is very small.
[4]
(ii) Find the probability that the total mass of 22 of these potatoes exceeds 2 kg.
[2]
(iii) The machine filling the bags delivers potatoes one by one until a total mass of at least 2 kg is reached.
Show that the bags are almost certain to contain either 22 or 23 potatoes.
[2]
4
A random sample of six observations of the random variable X gave the following values:
3 .2,
1 .8,
4 .0,
[Σx = 14 .7,
− 2.1,
6.1,
2
Σx = 73 .99 .]
1.7.
The population mean of X is µ . Calculate a 90% confidence interval for µ ,
(i) assuming that X has a normal distribution with variance 8.5,
(ii) assuming instead that X has a normal distribution with unknown variance.
[9]
Specimen Materials - Mathematics
102
© OCR 2000
5
State what distributional assumptions are necessary for it to be valid to use
(i) the two-sample t-test,
(ii) the paired-sample t-test,
to test for a difference in population means.
[3]
Two different types of nylon fibre were tested for the amount of stretching under tension. Ten random
samples of each fibre, of the same length and diameter, were stretched by applying a standard load. For
Fibre 1 the increases in length, x mm, were as follows.
12.84
14.26
13.23
14.75 15.13 14.15 13.37 12.96
[Σx = 140 .09, Σx2 = 1969 .0513 .]
15.02
14.38
15.14
14.81
For Fibre 2 the increases in length, y mm, were as follows.
14.27
13.25
14.17
13.11
14.92
[Σy = 139 .68 ,
12.12
Σy 2
14.21
13.68
= 1958 .9794 .]
Assuming that any necessary conditions for the validity of your test hold, test whether the mean increase in
length of the two types of fibre is different. Use a 10% significance level.
[7]
6
Six hens are observed over a period of 20 days and the number of eggs laid each day is summarised in the
following table.
Number of eggs
3
4
5
6
Number of days
2
2
10
6
Show that the mean number of eggs per day is 5.
[2]
It may be assumed that a hen never lays more than one egg in any day. State one other assumption that
needs to be made in order to consider a binomial model, with n = 6 , for the total number of eggs laid in a
day.
[1]
Calculate the expected frequencies using a binomial model for the above data and carry out a χ 2 goodness
of fit test, using a 10% significance level.
[9]
Specimen Materials - Mathematics
103
© OCR 2000
7
The continuous random variable X has a triangular distribution with probability density function given by
1 + x

f( x ) = 1 − x
0
− 1 ≤ x ≤ 0,
0 ≤ x ≤ 1,
otherwise.
Show that, for 0 ≤ a ≤ 1 ,
P( X ≤ a ) = 2a − a 2 .
[3]
The random variable Y is given by Y = X 2 . Express P(Y ≤ y ) in terms of y, for 0 ≤ y ≤ 1 , and hence
show that the probability density function of Y is given by
g( y ) =
1
−1 ,
y
for 0 < y ≤ 1 .
[3]
Use the density function of Y to find E(Y ) , and show how the value of E(Y ) may also be obtained directly
using the probability density function of X.
[4]
Determine E( Y ) .
Specimen Materials - Mathematics
[2]
104
© OCR 2000
General Certificate of Education
Advanced Subsidiary (AS) and Advanced Level
S3
MATHEMATICS
Probability & Statistics 3
MARK SCHEME
MAXIMUM MARK
60
For live examinations, each Mark Scheme
includes the General Instructions for Marking
set out on pages 143 to 145.
This mark scheme consists of 4 printed pages.
Specimen Materials - Mathematics
105
© OCR 2000
1
2
40 × 40
=8
B1
1 For correct answer; working not needed
200
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii) 4.80 is greater than critical value 4.605
M1
Comparing with a tabular value
A1
Using correct figure 4.605
There is evidence to suggest that the characteristics
are not independent
A1
3
(i)
Model for all enquiries is Po(9)
B1
M1
A1
M1
A1
Probabilit y = 0. 9996 − 0. 5874
= 0. 412
3
(i) T21 ~ N(1890 , 336)
For any mention of a Poisson distribution
Summing two Poisson distributions
For statement of correct parameter 9
Subtracting relevant tabular values
5
M1
For normal distribution with correct mean
A1
For variance 21 × 4 2 (both these first two
marks may be implied by later working)
2000 − 1890 
P(T21 > 2000 ) = 1 − Φ 
M1
For correct processes for relevant tail
 = 1 − Φ( 6)

336

Hence prob is very small since Φ( 6) ≈ 1
A1
4 Correct conclusion, based on suff large z
-------------------------------------------------------------------------------------------------------------------------------------------------------2000 − 22 × 90 
(ii) P(T22 > 2000 ) = 1 − Φ 
M1
For relevant new normal calculation


22 × 16 
= 1 − Φ (1.066) = 0.143
A1
2 For correct probability
-------------------------------------------------------------------------------------------------------------------------------------------------------2000 − 2070 
(iii) P(T23 > 2000 ) = 1 − Φ
For relevant calculation for 23 potatoes
 = Φ( 3. 649 ) ≈ 1 M1

368

Hence 23 potatoes is almost always enough
A1
2 For correct conclusion based on correct figs
4
(i)
x=
14 .7
= 2 .45
6
Interval is 2.45 ± 1.645 ×
8.5
= 2. 45 ± 1.9579
6
B1
At any stage; may be implied
M1
For calculation of the form x ± z σ 2 / n
A1
For relevant use of z = 1. 645
i.e. 0.49 < µ < 4.41
A1
4 For correct interval
-------------------------------------------------------------------------------------------------------------------------------------------------------1
14 .72 
(ii) s2 =  73 .99 −
M1
For correct unsimplified expression for s2
 = 7.595
5
6 
A1
For correct value (may be implied)
7.595
Interval is 2.45 ± 2.015 ×
= 2.45 ± 2. 267
6
M1
For calculation of the form x ± t s2 / n
i.e. 0.18 < µ < 4.72
A1
A1
Specimen Materials - Mathematics
For relevant use of t = 2.015
5 For correct interval
106
© OCR 2000
5
(i) The distributions must be normal
The distributions must have equal variances
B1
B1
(ii) The differences must be normally distributed
B1
3
-------------------------------------------------------------------------------------------------------------------------------------------------------H 0 : µ x − µ y = 0; H 1 : µ x − µ y ≠ 0
B1
For both
140 .09 2  
139 .68 2 

1969 . 0513 −
 + 1958 .9794 −

10  
10 

2
Sp =
18
= 0. 8033
14.009 − 13.968
Test statistic t =
= 0.1023
1
1
0.8033 (10
+ 10
)
M1
Use of correct formula
A1
May be implied
M1
Use of correct formula
A1
This is less than the critical value 1.734
M1
There is not significant evidence of a difference in mean
increase
A1
6
Mean =
3 × 2 + 4 × 2 + 5 × 10 + 6 × 6
=5
20
Correct value of t
Comparing to tabular value of t
7
M1
A1
2 Obtain given answer correctly
-------------------------------------------------------------------------------------------------------------------------------------------------------Assume that the probability of any hen laying an egg on
any day is constant and that hens lay eggs
independently of each other
B1
1 For either constant prob or independence
-------------------------------------------------------------------------------------------------------------------------------------------------------Distribution to be fitted is B(6, 56 )
B1
May be implied
i
6− i
6
Expected frequencies are fi = 20 ×   × (56) × (16)
i
 
f6 = 6.70, f5 = 8.04, f4 = 4.02
f3 = 1. 07, f 2 = 0.16, f1 = 0. 02, f0 = 0.00
Combining cells: fo
fe
χ2 =
≤4
4
5.27
5
10
8.04
1.27 2 1.96 2 0.70 2
+
+
= 0.857
5.27
8. 04
6.70
This is less than the critical value 2.706
Hence there is a satisfactory fit
Specimen Materials - Mathematics
6
6
6.70
M1
For any one calculation using correct method
A1
All three correct to at least 2dp
A1
For all four correct, or f≤ 3 = 1.25 stated
M1
Use of criterion fe < 5 for combining
M1
A1
M1
A1
Allow anything between 0.85 and 0.86 (inc)
Compare with tabular value
9
107
© OCR 2000
7
P( X < a) = P( − a < X < a)
0
a
= ∫ (1 + x ) dx + ∫ (1 − x) dx
−a
0
M1
For consideration of two areas, or equiv
A1
Or equivalent trapezium areas
= [x + 12 x 2]− a + [x − 12 x2 ]0 = 2a − a2
A1
3 Given answer correctly shown
-------------------------------------------------------------------------------------------------------------------------------------------------------P(Y ≤ y) = P( X 2 ≤ y) = P( X ≤ y ) = 2 y − y
B1
For correct expression
a
0
Hence the pgf of Y is
d
1
(
2 y − y) =
−1
dy
y
M1
For differentiation of previous expression
A1
3 Given answer correctly shown
-------------------------------------------------------------------------------------------------------------------------------------------------------1
E( Y ) = ∫ y 2 − y d y =
1
0
[y
2
3
3
2
1
]
1
− 12 y 2 = 16
0
0
1
−1
0
E( X 2 ) = ∫ ( x 2 + x3) dx + ∫ ( x 2 − x3) dx
M1
For correct integral
A1
For answer
1
6
M1
1
1
= [13 x3 + 14 x 4]−1 + [13 x3 − 14 x4 ]0 = 12
+ 12
= 16
A1
4
-------------------------------------------------------------------------------------------------------------------------------------------------------0
1
1
1
[
]
1
E( Y ) = ∫ y 2 g( y ) d y = ∫ (1 − y 2 ) dy = y − 23 y 2 = 13
0
1
0
1
3
0
M1
For forming the correct integral
A1
Specimen Materials - Mathematics
2 Correct answer
108
© OCR 2000
General Certificate of Education
Advanced Subsidiary (AS) and Advanced Level
S4
MATHEMATICS
Probability & Statistics 4
Additional materials:
Answer paper
Graph paper
List of Formulae
TIME
1 hour 20 minutes
INSTRUCTIONS TO CANDIDATES
Write your name, Centre number and candidate number in the spaces provided on the answer paper.
Answer all the questions.
Give non-exact numerical answers correct to 3 significant figures, unless a different degree of
accuracy is specified in the question or is clearly appropriate.
You are permitted to use a graphic calculator in this paper.
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 60.
Questions carrying smaller numbers of marks are printed earlier in the paper, and questions carrying
larger numbers of marks later in the paper.
You are reminded of the need for clear presentation in your answers.
This question paper consists of 4 printed pages.
Specimen Materials - Mathematics
109
© OCR 2000
1
To compare the effect of a new drug on men and women, a small-scale trial was conducted, involving 6
randomly chosen men and 4 randomly chosen women. For each person, the time, in minutes, before the
drug began to take effect was recorded, with the results shown in the table below.
Men
19
32
24
33
Women
14
13
18
27
31
28
Use a suitable non-parametric test to determine if there is evidence, at the 5% significance level, that the
drug acts more quickly in women than it does in men.
[5]
2
The continuous random variable X has probability density function given by
λe −λ x
f( x ) = 
0
x ≥ 0,
x < 0,
where λ is a positive constant. Show that the moment generating function of X is given by
M X (t ) =
λ
.
λ −t
[3]
Use this moment generating function to find the mean and variance of X.
3
Events A and B are such that
P( A) = 12 ,
P( B ) = 13 ,
[4]
P( A ∪ B) = 34 .
(i) Determine, giving your reasons clearly, whether A and B are
(a) mutually exclusive,
[1]
(b) independent.
[3]
(ii) Find the values of
(a)
P( A | B) ,
[2]
(b) P( A | B′) , where B′ denotes the complement of B.
Specimen Materials - Mathematics
110
[3]
© OCR 2000
4
A bin contains a large number of seeds of which 20% will produce a plant with a red flower and 30% will
produce a plant with a yellow flower. The remaining 50% will fail to germinate. Two seeds are chosen at
random from the bin and planted in a pot. The random variables R and Y denote the number of red and
yellow flowers, respectively, that will be produced from the seeds. The table below shows the joint
probability distribution of R and Y.
Y
0
R
1
2
0
1
0.25
0.30
0.20
0.12
0.04
0
2
0.09
0
0
(i) Find the marginal distributions of R and Y.
[2]
(ii) Show that E (R ) = 0.4 and find E (Y ) .
[2]
(iii) Find Cov ( R, Y ) .
[3]
(iv) Find the distribution of Y conditional on R = 1 , and hence state the expected value of the number of
yellow flowers in a pot in which there is one red flower.
[2]
5
The discrete random variable X denotes the score obtained in a single throw of an ordinary fair die. Show
that the probability generating function of X may be expressed as
t (1 − t 6 )
.
6(1 − t )
[2]
Write down the probability generating function for the total score obtained when three fair dice are thrown.
[1]
Hence show that the probability of obtaining a total score of 10 when three fair dice are thrown is 18 .
6
[6]
Explain briefly the circumstances under which a non-parametric test of significance should be used in
preference to a parametric test.
[1]
The acidity of soil can be measured by its pH value. As a part of a Geography project a student measured
the pH values of 14 randomly chosen samples of soil in a certain area, with the following results.
5.67
5.73
6.64
6.76
6 .10
5.41
5 .80
6 .52
5.16
5 .10
6.71
5.89
5.68
5.37
Use a suitable non-parametric test to test whether the average pH value for soil in this area is 6.24. Use a
10% level of significance.
[4]
Some time later, the pH values of soil samples taken at exactly the same locations as before were again
measured. It was found that, for 3 of the 14 locations, the new pH value was higher than the previous
value, while for the other 11 locations the new value was lower. Test, at the 5% significance level,
whether there is evidence that the average pH value of soil in this area is lower than previously.
[4]
Specimen Materials - Mathematics
111
© OCR 2000
7
The continuous random variable X has a uniform distribution on the interval 0 ≤ x ≤ a , where the value of
the parameter a is unknown. Three independent observations, X1 , X 2 , X 3 , of X are taken.
(i) An estimator θ is defined by θ = 23 ( X 1 + X 2 + X 3 ) . Show that θ is an unbiased estimator of a, and
find Var(θ ) in terms of a.
[6]
(ii) Another estimator φ is based on the greatest of the three values X1 , X 2 , X 3 . Denoting the greatest of
the three values by the variable G, use the fact that, for any value x between 0 and a,
G < x ⇔ ( X1 < x and X 2 < x and X 3 < x )
to write down the cumulative distribution function of G, and hence to obtain the probability density
function of G.
[2]
Hence show that, if φ = 43 G , then φ is an unbiased estimator of a, and determine which of θ and φ
is the more efficient estimator.
Specimen Materials - Mathematics
[4]
112
© OCR 2000
General Certificate of Education
Advanced Subsidiary (AS) and Advanced Level
S4
MATHEMATICS
Probability & Statistics 4
MARK SCHEME
MAXIMUM MARK
60
For live examinations, each Mark Scheme
includes the General Instructions for Marking
set out on pages 143 to 145.
This mark scheme consists of 4 printed pages.
Specimen Materials - Mathematics
113
© OCR 2000
1
H0 : same distributions of times for men and women,
H1 : lower average for women
M
4 9 5 10 8 7
Ranks are:
W
2 1 3 6
Test statistic is 1 + 2 + 3 + 6 = 12
Critical Wilcoxon rank-sum value is 13
Hence reject H0 and conclude that there is evidence to
B1
For both NH and AH
M1
For ranking all 10 values
A1
M1
For sum of women’s ranks
For comparing with correct tabular value
suggest that drug acts more quickly in women than it
does in men
A1
2
∞
M X (t) = ∫ etx λ e− λx dx
0
5
B1
Correct integral stated
M1
For correct integration method
∞
λ (t − λ) x 
= 
e

t − λ
0
λ
A1
3 Given answer correctly shown
λ −t
-------------------------------------------------------------------------------------------------------------------------------------------------------2
t
t
M X (t) = 1 + +   + K
B1
Three correct terms
λ λ 
1
E( X ) =
B1
λ
=
Var( X ) =
2
2!  1 
1
−  = 2
λ2  λ 
λ
M1
For using correct variance formula
A1
3
4 Correct answer for variance
(i) (a) Not exclusive, since 12 + 13 ≠ 34
B1
1 Conclusion and reason both required
-------------------------------------------------------------------------------------------------------------------------------------------------------1
M1
Use of correct formula
(b) P( A ∩ B) = 34 − 12 − 13 = 12
A1
Correct value
1
12
1
Not independent, as 12 × 13 ≠ 12
A1
3 Conclusion and reason both required
-------------------------------------------------------------------------------------------------------------------------------------------------------P( A ∩ B) 1
=4
(ii) (a) P( A | B) =
M1
Use of correct formula
P( B)
A1
2 Follow answer to (i)(b)
-------------------------------------------------------------------------------------------------------------------------------------------------------(b) P( A ∩ B′) = P( A) − P( A ∩ B )
M1
Or equivalent, e.g. from Venn diagram
1
5
= 12 − 12
= 12
5
P( A | B′) = 12
÷ 23 =
Specimen Materials - Mathematics
5
8
A1
A1
3
114
© OCR 2000
4
0
1
2
M1
Adding rows (or columns)
0 .64 0 .32
0. 04
0
1
2
A1
2 Both distributions correct
Y:
0 .49 0 .42
0. 09
-------------------------------------------------------------------------------------------------------------------------------------------------------M1
Correct process for either mean
(ii) E( R) = 0 × 0. 64 + 1 × 0.32 + 2 × 0.04 = 0.4
(i) R :
E( Y ) = 0 × 0. 49 + 1 × 0.42 + 2 × 0. 09 = 0.6
A1
2 Both means correct
-------------------------------------------------------------------------------------------------------------------------------------------------------(iii) E( RY ) = 0. 12
B1
Cov(R, Y ) = 0.12 − 0.4 × 0.6
M1
= −0.12
A1
3
-------------------------------------------------------------------------------------------------------------------------------------------------------0 1 2
(iv) Conditional distribution is: 5
B1
3
0
8
8
Expected value is
5
PGF is
1
(t
6
3
8
B1
+ t 2 + t 3 + t 4 + t5 + t 6)
2
B1
t(1 − t 6 )
B1
2 Use of GP sum to deduce given answer
6(1 − t )
-------------------------------------------------------------------------------------------------------------------------------------------------------t3 (1 − t 6 ) 3
B1
1 For cube of answer (i), in any form
216 (1 − t)3
-------------------------------------------------------------------------------------------------------------------------------------------------------1 3
t (1 − 3t6 + K)(1 + 3t + 6t 2 + K)
M1
For using both binomial expansions
216
i.e.
A1
M1
A1
A1
A1
7
Terms in t from product of brackets are:
− 3t 6 × 3t
1 × 36t 7
1
Probability = 216
(36 − 9) = 18
6
Both correct at least as far as shown
Picking out relevant term(s)
6 Given answer correctly shown
A non-parametric test is needed when there is no
information (or reasonable assumption) available
about an underlying distribution
B1
1
-------------------------------------------------------------------------------------------------------------------------------------------------------Deviations from NH value 6.24 are:
− 0.57
0.28
− 0.51
− 1.08
Signed ranks are:
0.40
− 1.14
−10
2
0.52
0.47
−7
− 13
− 0.14
− 0.35
4
− 14
8
6
− 0.83
− 0.56
−1
−3
− 0.44
− 0.87
−11
−9
M1
For calculating signed differences from 6.24
−5
A1
− 12
For calculating correct signed ranks
Test statistic is 2 + 4 + 6 + 8 = 20 < 25
M1
For calculating T and comparing
Conclude that there is evidence to suggest that the
average pH value is not 6.24
A1
4 For correct conclusion based on correct work
-------------------------------------------------------------------------------------------------------------------------------------------------------H0 : same average pH as before; H1 : lower value
B1
For both hypotheses
Under H0 , tail probability for 3 or fewer out of 14 is
M1
For use of relevant binomial distribution
× (1 + 14 + 91 + 364 ) = 0.0287 < 0.05
A1
For correct value tail probability 0.0287
Hence reject H0 and accept that average pH is lower
A1
()
1 14
2
Specimen Materials - Mathematics
4 For correct conclusion based on correct work
115
© OCR 2000
7
(i)
E( X ) = 12 a
B1
E( θ ) = 23 × 3 E( X )
M1
= a , as required
May be implied
A1
For conclusion correctly shown and stated
a 2
x
1 2
Var( X ) = ⌠
 a dx − (2 a)
⌡0
=
1 2
a
12
M1
A1
3 2
Var( θ ) = 49 × 12
a = 19 a 2
A1
6
-------------------------------------------------------------------------------------------------------------------------------------------------------3
x
B1
(ii) P( G < x) =  
 a
3x 2
B1
2
a3
-------------------------------------------------------------------------------------------------------------------------------------------------------4 a 3 x3
E(φ ) = ⌠
dx
M1
3
⌡0 a 3
= a , as required
A1
For conclusion correctly shown and stated
Hence pdf is
Var( φ ) =
16 ⌠ a 3x 4
dx − a2
9
⌡0 a3
=
1 2
a
15
M1
< 19 a 2
Hence φ is more efficient
Specimen Materials - Mathematics
4 Correct Var(φ ) and conclusion
A1
116
© OCR 2000
General Certificate of Education
Advanced Subsidiary (AS) and Advanced Level
D1
MATHEMATICS
Discrete Mathematics 1
Additional materials:
Answer paper
Graph paper
List of Formulae
TIME
1 hour 20 minutes
INSTRUCTIONS TO CANDIDATES
Write your name, Centre number and candidate number in the spaces provided on the answer paper.
Answer all the questions.
Give non-exact numerical answers correct to 3 significant figures, unless a different degree of
accuracy is specified in the question or is clearly appropriate.
You are permitted to use a graphic calculator in this paper.
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 60.
Questions carrying smaller numbers of marks are printed earlier in the paper, and questions carrying
larger numbers of marks later in the paper.
You are reminded of the need for clear presentation in your answers.
This question paper consists of 4 printed pages.
Specimen Materials - Mathematics
117
© OCR 2000
1
Andy wants to record the following twelve TV programmes onto video tape. Each video tape has space for
up to three hours of programmes.
Programme
A
B
C
Length (hours)
1
2
1
2
3
4
D
1
E
1
F
1
G
H
I
J
K
L
1
1 12
1 12
1 34
2
2
(i) Suppose that Andy records the programmes in the order A to L using the first fit algorithm. Find the
number of tapes needed, and show which programmes are recorded onto which tape.
[3]
(ii) Suppose instead that Andy is transferring the programmes from previously recorded tapes, so that
they can be copied in any order, and that Andy uses the first fit decreasing algorithm. Find the
number of tapes needed, and show which programmes are recorded onto which tape.
[3]
2
(a)
(b)
(c)
(i) By considering the order of each node, classify each of the graphs (a), (b) and (c) shown in the
diagram as Eulerian, semi-Eulerian or neither.
[4]
(ii) Explain briefly how your classification in part (i) relates to the problem of finding a route through the
graph that includes each arc exactly once. Are there any restrictions on where such a route can start
and finish?
[3]
3
A company has offices is six towns, A, B, C, D, E and F. The costs, in £, of travelling between these towns
are shown in the table.
Town
A
B
C
D
E
F
A
–
15
26
13
14
25
B
15
–
16
16
25
13
C
26
16
–
38
16
15
D
13
16
38
–
15
19
E
14
25
16
15
–
14
F
25
13
15
19
14
–
Use Prim’s algorithm, starting by deleting row A, to find the cheapest way of connecting the six towns.
You should show all your working and indicate the order in which the towns were included.
[7]
Specimen Materials - Mathematics
118
© OCR 2000
4
An express delivery pizza company promises to deliver pizzas within 30 minutes of an order being
telephoned in. Four customers telephone in orders at the same time. While the pizzas are cooking, which
takes 10 minutes, a delivery route must be planned for the person who will deliver all four pizzas. The
diagram below shows the pizza company P and the four customers A, B, C and D, and the accompanying
table shows the travel times for each possible leg of a journey between them.
P
A
B
C
D
P
–
5
3
4
2
A
5
–
1
4
4
B
3
1
–
3
5
C
4
4
3
–
7
D
2
4
5
7
–
The travel times shown exclude the time taken to stop at a customer’s house and deliver the pizza; this
stopping and delivery time is 2 12 minutes.
(i) Explain why a travelling salesperson solution taking less than 13 minutes guarantees that all four
customers will get their pizzas (including stopping and delivery times) within 30 minutes of their
telephone call.
[4]
(ii) The pizzas are delivered using the nearest neighbour algorithm, as follows.
The first delivery is to the customer who is nearest to P (in the sense of having the shortest travel
time).
The second delivery is to the customer who is nearest to the first one.
The third delivery is to the customer nearest to the second who is still waiting for a pizza.
The fourth delivery is to the remaining customer.
Write down the order in which the pizzas are delivered using this algorithm, and calculate how long
the fourth customer has to wait for their pizza (including the stopping and delivery time).
[4]
Specimen Materials - Mathematics
119
© OCR 2000
5
Lou Zitt has a budget of £2000 to spend on storage units for his office. The storage units must not cover
more than 50 m 2 of floor space, and Lou wants to maximise the storage capacity. The three types of
storage unit that he can choose from are shown in the following table.
Type
Storage capacity (m3 )
Floor space covered (m2 )
Cost (£)
3
8
6
1
4
3
100
500
200
Antique pine units
Beech wood units
Cedar wood units
Suppose that Lou buys a antique pine units, b beech wood units and c cedar wood units.
6
(i) Write down two constraints that must be satisfied by a, b and c, other than a ≥ 0 , b ≥ 0 , c ≥ 0 .
[2]
(ii) Write down the objective function for this problem.
[1]
(iii) Set up the problem as an LP formulation. (You are not expected to solve the problem.)
[4]
(iv) Identify which aspect of the original problem has been overlooked in the LP formulation.
[1]
A graph has five vertices, P, Q, R, S, T, and each vertex is directly connected to every other vertex.
Describe how to apply Dijkstra’s algorithm to find the shortest path from P to T, and explain why this
requires 6 addition calculations in the worst case.
[6]
Show that when Dijkstra’s algorithm is used on a graph with six vertices it requires 10 addition
calculations in the worst case.
[2]
The number of additions affects the amount of time that Dijkstra’s algorithm takes to run on a computer.
(i) Assuming that the problem has already been put into a suitable format, what is the other main factor
that would affect the time that Dijkstra’s algorithm takes to run on a computer?
[1]
(ii) Dijkstra’s algorithm is of quadratic order (order n 2 ). Explain what this means.
7
[2]
A linear programming problem gives the following LP formulation.
Maximise
P = 3x + 4 y + 5 z
subject to
2x + 8y + 5z ≤ 3
9 x + 3 y + 6z ≤ 2
and
x ≥ 0, y ≥ 0, z ≥ 0.
(i) Set up an initial simplex tableau for this problem. Perform two iterations, choosing first to pivot on
an element chosen from the z column.
[9]
(ii) State the values of x, y, z and P that result from each of the two iterations carried out in part (i).
[2]
(iii) Explain how you know whether or not the optimal solution has been achieved.
[2]
Specimen Materials - Mathematics
120
© OCR 2000
General Certificate of Education
Advanced Subsidiary (AS) and Advanced Level
D1
MATHEMATICS
Discrete Mathematics 1
MARK SCHEME
MAXIMUM MARK
60
For live examinations, each Mark Scheme
includes the General Instructions for Marking
set out on pages 143 to 145.
This mark scheme consists of 4 printed pages.
Specimen Materials - Mathematics
121
© OCR 2000
1
2
(i) A, B, C, D all fit onto one tape but E doesn’t
M1
Applying the algorithm with first tape
Tapes hold ABCD, EFG, HI, J, K, L
A1
So 6 tapes are needed
A1
3
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii) First tape contains K and D (or other ‘2 + 1’ pair) M1
Applying the algorithm with first tape
Tapes hold KD, LE, JF, HI, GCAB (or equiv)
A1
So 5 tapes are needed
A1
3
A B C
D
E
5
5
4
5
5
A B C
D
E
(b)
M1
For recognisable attempt for any one case
5
4
4
4
3
A B C
D
E
(c)
4
4
4
4
4
Hence (a) is neither, since there are 4 odd nodes
A1
(b) is semi-Eulerian, since there are 2 odd nodes
A1
(c) is Eulerian, since all the nodes are even
A1
4
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii) No such route exists for graph (a)
B1
There are routes for graph (b), but they have to
start and finish at odd nodes, i.e. A and E
B1
There are routes for graph (c), where all routes
start and finish at the same node, which can be
any one of the five
B1
3
(i) Orders are: (a)
3
A
B
C
D
E
F
A
B
C
D
E
F
A
B
C
D
E
F
A
B
C
D
E
F
A
B
C
D
E
F
A1
—
15
26
13
14
25
A1
—
15
26
13
14
25
A1
—
15
26
13
14
25
A1
—
15
26
13
14
25
A1
—
15
26
13
14
25
B
15
—
16
16
25
13
B
15
—
16
16
25
13
B
15
—
16
16
25
13
B
15
—
16
16
25
13
B5
15
—
16
16
25
13
C
26
16
—
38
16
15
C
26
16
—
38
16
15
C
26
16
—
38
16
15
C
26
16
—
38
16
15
C
26
16
—
38
16
15
D
13
16
38
—
15
19
D2
13
16
38
—
15
19
D2
13
16
38
—
15
19
D2
13
16
38
—
15
19
D2
13
16
38
—
15
19
E
14
25
16
15
—
14
E
14
25
16
15
—
14
E3
14
25
16
15
—
14
E3
14
25
16
15
—
14
E3
14
25
16
15
—
14
F
25
13
15
19
14
—
F
25
13
15
19
14
—
F
25
13
15
19
14
—
F4
25
13
15
19
14
—
F4
25
13
15
19
14
—
Arcs: AD = 13, AE = 14 , EF = 14 , FB = 13, FC = 15
Order of adding is A, D, E, F, B, C
Total cost is 13 + 14 + 14 + 13 + 15 = £69
Specimen Materials - Mathematics
B1
Deleting row A and choosing AD
M1
Deleting row D and choose from cols A, D
A1
Correct choice at this stage
M1
For carrying out the complete algorithm
A1
B1
B1
All arcs correct
7
122
© OCR 2000
4
(i) Travel time available is 30 − 10 − 4 × 2 12 = 10 mins B1
Return time from last delivery is not relevant
M1
This can be at least 3 mins (since any route can be
followed in either direction)
A1
4 Given answer correctly explained
Hence TSP solution of 10 + 3 = 13 or less suffices A1
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii) The order is DABC
B2
Allow B1 for DACB
Total time is 10 + 2 + 4 + 1 + 3 + 4 × 2 12
M1
i.e. 30 minutes
A1
Cooking and stopping times required
4
5
(i) a + 4b + 3c ≤ 50
B1
a + 5b + 2c ≤ 20
B1
2 Allow any correct unsimplified version
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii) 3a + 8b + 6c
B1
1 Allow in the form of an equation
-------------------------------------------------------------------------------------------------------------------------------------------------------(iii) Maximise P = 3a + 8b + 6 c
B1
B1
Subject to a + 4b + 3c ≤ 50
B1
and a + 5b + 2c ≤ 20
B1
4
Together with a ≥ 0, b ≥ 0, c ≥ 0
-------------------------------------------------------------------------------------------------------------------------------------------------------(iv) The variables a, b, c must be integers
B1
1
6
Label Q, R, S, T with direct distances from P, and select
smallest (say Q)
B1
Correct description of labelling and selecting
Update distances to R, S, T if route from P via Q is
shorter
B1
Correct description of first stage of Dijkstra
The initial 3 possible updates require 3 additions so far B1
Select the smallest of R, S, T and repeat the updating
process; this stage requires another 2 additions
B1
Explanation of 2 additions at this stage
The final stage similarly requires 1 addition, so the total
B1
Given total of 6 correctly explained
is 3 + 2 + 1 = 6
Trace back to find the shortest path, including an arc XY
whenever the distance to X plus the length of XY is
the same as the distance to Y
B1
6 For explanation of tracing back
-------------------------------------------------------------------------------------------------------------------------------------------------------M1
Worst case number is 4 + 3 + 2 + 1
i.e. 10
A1
2
-------------------------------------------------------------------------------------------------------------------------------------------------------(i) The number of comparisons
B1
1
-------------------------------------------------------------------------------------------------------------------------------------------------------(ii) The number of operations needed to complete the
algorithm is (approximately) proportional to the
square of the number of nodes
B1
Allow e.g. ‘time taken prop to square of size’
So, for example, doubling the number of nodes
will (roughly) quadruple the amount to be done B1
2
Specimen Materials - Mathematics
123
© OCR 2000
7
(i) Simplex tableau is:
P
x
y
z
1 −3 − 4 −5 0 0 0
0
2
8
5
1 0
3
0
9
3
6 0 1 2
Select 6 in the z column as pivot
Result of first iteration is:
P
x
y z
5
1
4 12
− 1 12 0 0
123
6
0 − 5 12
5 12 0
1 − 56 113
1
1
1
1
0
12
1 0
2
6
3
Select 5 12 in y column for second pivot
B1
Result of second iteration is:
P
x y
z
3
20
1
3 0 0
11
33
2
5
−
0 −1
1 0
11
33
M1
67
33
8
33
7
33
M1
Table with correct numbers of rows and cols
A1
Correct initial values throughout
B1
M1
For correct row operations
A1
Correct values throughout
For correct row operations
A2
9 All values correct; allow A1 if one error
8
0
2
0
1 − 111
33
-------------------------------------------------------------------------------------------------------------------------------------------------------B1
(ii) After 1 iteration: x = 0, y = 0, z = 13 , P = 1 23
8
7
After 2 iterations: x = 0, y = 33
, z = 33
, P = 67
B1
2
33
-------------------------------------------------------------------------------------------------------------------------------------------------------(iii) There are no negative numbers in the top row
of the tableau
B1
Hence the solution is optimal
B1
2
Specimen Materials - Mathematics
124
© OCR 2000
General Certificate of Education
Advanced Subsidiary (AS) and Advanced Level
D2
MATHEMATICS
Discrete Mathematics 2
Additional materials:
Answer paper
Graph paper
List of Formulae
TIME
1 hour 20 minutes
INSTRUCTIONS TO CANDIDATES
Write your name, Centre number and candidate number in the spaces provided on the answer paper.
Answer all the questions.
Give non-exact numerical answers correct to 3 significant figures, unless a different degree of
accuracy is specified in the question or is clearly appropriate.
You are permitted to use a graphic calculator in this paper.
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 60.
Questions carrying smaller numbers of marks are printed earlier in the paper, and questions carrying
larger numbers of marks later in the paper.
You are reminded of the need for clear presentation in your answers.
This question paper consists of 5 printed pages, 3 blank pages and an insert.
Specimen Materials – Mathematics
125
© OCR 2000
1
The diagram represents a system of pipes. The weights show the (directed) maximum capacity for each
pipe in litres per minute.
(i) Calculate the capacity of the cut C marked in the diagram.
[2]
(ii) Draw a diagram showing a flow from S to T of 230 litres per minute in which each of the pipes JL,
KM and MT is carrying its full capacity.
[1]
(iii) Explain carefully what can be deduced from parts (i) and (ii) about flows through this network.
2
[2]
Granny has bought Christmas presents for her five grandchildren.
The teddy bear is suitable for Cathy, Daniel or Elvis;
the book is suitable for Annie or Ben;
the football is suitable for Daniel or Elvis;
the money box is suitable for Annie or Daniel;
the drum is suitable for Cathy or Elvis.
Draw a bipartite graph, G, to show which present is suitable for which grandchild.
[2]
Granny decides to give Annie the book, Cathy the teddy bear, Daniel the money box and Elvis the drum.
This leaves Ben without a present, since the football is not suitable for him.
(i) Show the incomplete matching, M, that describes which present Granny had decided to give to each
child.
[1]
(ii) Use a matching algorithm to construct an alternating path for M in G, and hence find a maximal
matching between the presents and the grandchildren.
[2]
Specimen Materials – Mathematics
126
© OCR 2000
3
Richard and Carol play a two-person zero-sum simultaneous play game. The table shows Richard’s payoff matrix for the game.
Richard
Strategy A
Strategy B
Strategy C
Carol
Strategy Y
3
–1
0
Strategy X
2
–4
–5
Strategy Z
–2
–1
1
(i) Find the play-safe strategy for each player, and hence show that this game does not have a stable
solution.
[4]
(ii) Richard’s optimal mixed strategy can be found by using linear programming. Set up a suitable LP
formulation, defining the symbols you use. (You are not required to solve the LP problem.)
[4]
4
A relay team consists of four runners, A, B, C and D, each of whom runs one leg of the race. The best
training times, in seconds, for each of the runners over each of the legs are given in the table.
A
B
C
D
1st leg
47
48
46
50
2nd leg
45
44
45
47
3rd leg
45
45
44
46
4th leg
43
44
43
44
Use the Hungarian algorithm to decide which runner should be allocated to which leg of the race.
5
[8]
The Rolling Pebbles have been playing as a band for many years. When they tour they sometimes play old
songs, they sometimes play new songs and they sometimes play a mixture of old and new songs. Their
choice depends on the age of the audience. The table shows the audience reaction (as a score out of 10) for
each of the possible combinations. High scores are good.
Songs played
Old
Mixture
New
Young
1
3
8
Audience
Mixed
3
1
5
Older
8
2
3
Explain why, according to this data, the band should never choose to play a mixture of old and new songs.
[1]
The band do not know whether their audience will be young, older or of mixed ages. Suppose that they
choose to play old songs with probability p and new songs with probability 1 − p .
(i) Calculate, in terms of p, the expected reaction from each of the three types of audience.
[4]
(ii) Use a graphical method to decide what value p should take to maximise the minimum expected
reaction from part (i). Mark clearly on your graph the vertex where the optimal value occurs.
[5]
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6
[Answer both parts of this question on the Insert Sheet provided.]
Jim is playing a space adventure game. He has found the alien headquarters and now has four turns to
escape from the planet, or else perish. On each turn, Jim must play one of three tactics. He can attack, run
away or dodge the aliens. The number of energy pods used up during the turn is shown in Table 1.
Table 1
Tactic
Energy pods used
Attack
Run away
Dodge
2
1
0
The number of squares that Jim travels with each of these tactics is shown in Table 2.
Table 2
Energy pods remaining at start of turn
5
4
3
2
1
Attack
6
7
6
4
–
Run away
5
4
4
3
1
Dodge
1
2
1
1
0
Jim currently has five energy pods, and needs to maximise the number of squares that he travels in the four
turns that he has left. He should finish the game with no energy pods remaining, but he needs at least one
energy pod at the start of each of his four turns.
(i) The diagram on the Insert Sheet shows the action (A, R or D) for each possible transition. Mark on
the diagram the cost associated with each action, i.e. the number of squares travelled.
[3]
(ii) The dynamic programming tabulation on the Insert Sheet shows stages (turns), states (numbers of
energy pods remaining), actions (A, R or D) and a column for costs (numbers of squares travelled).
Complete this tabulation to find Jim’s best strategy, and the number of squares that Jim travels using
this strategy.
[8]
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7
The table shows the six activities in a project, together with their durations, precedences and the number of
people required for each activity.
Activity
A
B
C
D
E
F
Duration (days)
3
2
1
5
4
5
Preceded by
—
—
A, B
B
B, C
D, E
People required
2
1
3
2
1
2
Draw an activity network to represent these activities and their precedences.
[2]
In addition to the precedences shown in the table, activity E must not start until at least 2 days after activity
B has finished.
(i) Determine the earliest and latest starting times for each activity, for completion of the project in the
minimum time.
[4]
(ii) Identify the critical activities, and state the minimum time for completion of the project, assuming that
there are sufficient people available.
[2]
(iii) Find the least number of people required for the project to be completed in the minimum time,
explaining your reasoning.
[3]
(iv) By how many days must the project over-run if the least possible number of people are used? Justify
your answer.
[2]
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Centre Number
Candidate
Number
Candidate Name ___________________________________________
D2
MATHEMATICS
Discrete Mathematics 2
INSERT
INSTRUCTIONS TO CANDIDATES
This insert contains a diagram for use in Question 6 (i) and a table for use in Question 6 (ii).
Write your name, Centre number and candidate number in the spaces at the top of this page and
attach it to your answers.
This insert consists of 3 printed pages and 1 blank page.
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6
(i) Write your answer on this sheet and hand it in with your answer booklet.
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6
(ii) Write your answer on this sheet and hand it in with your answer booklet.
Stage
4
State
Action
1
2
1
R
A
D
R
D
A
R
A
A
R
D
A
R
D
A
R
A
R
D
2
3
3
4
3
2
4
5
1
5
Cost
Current maximum
Best strategy is: ..............................................................................................................................
......................................................................................................................................................
......................................................................................................................................................
Number of squares travelled: ...........................................................................................................
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General Certificate of Education
Advanced Subsidiary (AS) and Advanced Level
D2
MATHEMATICS
Discrete Mathematics 2
MARK SCHEME
MAXIMUM MARK
60
For live examinations, each Mark Scheme
includes the General Instructions for Marking
set out on pages 143 to 145.
This mark scheme consists of 5 printed pages and 1 blank page.
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1
(i) Capacity of C is 150 + 80
M1
For adding appropriate weights
i.e. 230 litres per minute
A1
2 For correct answer
--------------------------------------------------------------------------------------------------------------------------------------------------------(ii)
B1
1 For all the flows correct
--------------------------------------------------------------------------------------------------------------------------------------------------------(iii) The maximum flow is 230 litres per minute
B1
For stating the correct maximum flow
Max flow is at least 230 from (ii) and min cut is
at most 230 from (i), and result follows from
max flow = min cut
B1
2 For explanation using max flow = min cut
2
M1
A1
For labels and appropriate connections
2 For completely correct graph
--------------------------------------------------------------------------------------------------------------------------------------------------------(i)
B1
1 For correct incomplete matching
--------------------------------------------------------------------------------------------------------------------------------------------------------(ii)
M1
Alternating path is B–bk–A–mb–D–fb, and so a
maximal matching is (Annie, Money Box),
(Ben, Book), (Cathy, Teddy Bear),
(Daniel, Football), (Elvis, Drum)
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A1
138
Correct labelling procedure
2 Alternating path and maximal matching
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3
(i) The row minima for A, B, C are − 2, − 4, − 5
and the negatives of the column maxima for X,
Y, Z are −2, − 3, − 1
M1
For row minima and/or (neg) column maxima
Hence Richard’s play-safe strategy is A
A1
Carol’s play-safe strategy is Z
A1
Not stable, since ( −2) + (− 1) < 0
A1
4 Or equivalent demonstration of given result
--------------------------------------------------------------------------------------------------------------------------------------------------------8 9 4
(ii) Convert to positive values, e.g. 2 5 5
M1
For adding 6 to each entry
1 6 7
Maximise P subject to a + b + c ≤ 1 and
P − 8a − 2b − c ≤ 0 ,
P − 9a − 5b − 6c ≤ 0 ,
A1
For at least two correct constraints
P − 4a − 5b − 7c ≤ 0 .
A1
For all four correct
a, b, c are the probabilities with which Richard
plays strategy A, B, C, and P represents the
value of the game.
B1
4
4
Row reduction (or column reduction) gives:
A
B
C
D
1st
4
4
3
6
2nd
2
0
2
3
3rd
2
1
1
2
4th
0
A
0 or B
0
C
0
D
1st
1
2
0
4
2nd
1
0
1
3
3rd
1
1
0
2
4th
0
1
0
1
Column reduction (or row reduction) now gives:
A
B
C
D
1st
1
1
0
3
2nd
2
0
2
3
3rd
1
0
0
1
4th
0
A
0 or B
0
C
0
D
1st
1
2
0
3
2nd
1
0
1
2
3rd
1
1
0
1
4th
0
1
0
0
Matching is incomplete (3 lines will cover zeros)
Augmenting the matrix (there are 2 possible results)
A
B
C
D
1st
0
1
0
2
2nd
1
0
2
2
3rd
0
0
0
0
4th
0
A
1 or B
1
C
0
D
1st
0
2
0
2
2nd
0
0
1
1
3rd
0
1
0
0
4th
0
2
1
0
Possible complete matchings are:
1A, 2B, 3C, 4D or 1C, 2B, 3A, 4D or 1C, 2B, 3D, 4A
5
M1
For correct row (column) process
A1
For correct reduced matrix
M1
For correct subsequent column (row) process
A1
For correct reduced matrix
B1
M1
May be implied
For correct augmentation process
A1
Their reduced matrix correctly augmented
B1
8 For any one correct solution
Because ‘mixture’ is dominated by ‘new’
B1
1
--------------------------------------------------------------------------------------------------------------------------------------------------------(i) For ‘young’ expectation is : 1 × p + 8 × (1 − p)
M1
Correct calculation method for any one case
i.e. 8 − 7 p
A1
For ‘mixed’: 3 × p + 5 × (1 − p) = 5 − 2 p
A1
For ‘older’: 8 × p + 3 × (1 − p) = 3 + 5 p
A1
4
--------------------------------------------------------------------------------------------------------------------------------------------------------(ii)
B2
B1
For correct lines plotted/sketched
For identification of ‘optimal’ vertex
Optimum occurs where 5 − 2 p = 3 + 5 p
M1
For equating relevant expectations
i.e. where p =
A1
2
7
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6
(i)
M1
A1
A1
For showing costs appropriately
For all Turn 1 and Turn 2 costs correct
3 For all remaining costs correct
--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) Stage State Action
Cost
1
R
1
B1
Correct details for stage 4
←
4
2
1
2
3
3
4
3
2
4
5
1
5
A
D
R
D
A
R
A
A
R
D
A
R
D
A
R
A
R
D
4
1+0=1
1+3=4
4+1=5
1+6=7
4+4=8
4 + 7 = 11
1+6=7
5+4=9
8+1=9
5 + 7 = 12
8 + 4 = 12
11 + 2 = 13
8 + 6 = 14
11 + 5 = 16
9 + 6 = 15
13 + 5 = 18
16 + 1 = 17
←
←
*
←
←
←
A1
Correct details for stage 2
A1
Correct details for stage 1
*
←
←
*
The best strategy is 1:Run, 2:Dodge, 3:Attack,
4:Attack
The number of squares is 18
Specimen Materials – Mathematics
For method of updating costs
For sub-optimization
Correct details for stage 3
*
←
←
←
M1
M1
A1
A1
A1
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7
M1
For correct structure and labels (diagram may
alternatively be activity on arc)
A1
2 For all arrows correctly shown (start and
finish nodes may be omitted)
--------------------------------------------------------------------------------------------------------------------------------------------------------(i) Earliest starting times are:
M1
Carrying out forward pass
A B C
D
E
F
A1
0
0
3
2
4
8
Latest starting times are:
M1
Carrying out reverse pass
A B C
D
E
F
A1
4
0
0
3
3
4
8
--------------------------------------------------------------------------------------------------------------------------------------------------------(ii) Critical events are A, B, C, E and F
B1
Minimum completion time is 8 + 5 = 13 days
B1
2
--------------------------------------------------------------------------------------------------------------------------------------------------------(iii)
M1
A1
For scheduling attempt, either diagrammatic
or otherwise
For a correct schedule, or equivalent
Since C and D overlap, 5 people are needed
A1
3
--------------------------------------------------------------------------------------------------------------------------------------------------------(iv)
Least number of people is 3 (for activity C)
One extra day is required
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M1
A1
141
2 A correct schedule or argument is needed
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MARKING INSTRUCTIONS
1.
Mark in red. Correct answers should be ticked, errors which determine marks should be indicated by
ringing or by a cross or by underlining, and omissions by . Do not cross out or obliterate any work. In
cases of particular difficulty, brief comments written on the script may be helpful should the script be
reviewed at a later stage. Each page of the script (including unused pages in an answer booklet) must have
some indication that it has been seen, e.g. a tick in the margin.
2.
Marks are of the following three types.
M
Method mark, awarded for a valid method applied to the problem. Method marks are not lost for
numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate
just to indicate an intention of using some method or just to quote a formula; the formula or idea must
be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula.
A
Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks
cannot be given unless the associated Method mark is earned (or implied).
B
Mark for a correct result or statement independent of Method marks.
The marks indicated in the scheme may not be subdivided. If there is genuine doubt whether a candidate
has earned a mark, allow the candidate the benefit of the doubt. Unless otherwise indicated, marks once
gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored.
3.
When a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless
the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The
notation ‘dep *’ is used to indicate that a particular M or B mark is dependent on an earlier, asterisked,
mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a
part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the
other hand, when two or more steps are run together by the candidate, the earlier marks are implied and full
credit must be given.
4.
The symbol implies that the A or B mark indicated is allowed for work correctly following on from
previously incorrect results. Otherwise, A and B marks are given for correct work only — differences in
notation are of course permitted. A and B marks are not given for ‘correct’ answers or results obtained
from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution,
there may be various alternatives that are equally acceptable. In such cases, exactly what is acceptable will
be agreed at the standardisation meeting.
5.
Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specifically
indicates otherwise. Candidates are expected to give numerical answers to an appropriate degree of
accuracy, with 3 significant figures being the norm. Small variations in the degree of accuracy to which an
answer is given (e.g. 2 or 4 significant figures where 3 is expected) should not be penalised, while answers
which are grossly over- or under-specified should result in the loss of a mark. The situation regarding any
particular cases where the accuracy of the answer may be a marking issue will be decided at the
standardisation meeting.
6.
If work is deleted and replaced, mark the replacement. If work is deleted without replacement, mark the
deleted work provided that it is legible. When two solutions are offered (neither crossed out), count what
appears the more serious attempt or the more complete attempt at the question. If attempts are
indistinguishable in these respects, count the better.
7.
For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the
question remain unaltered, mark according to the scheme but following through from the candidate’s data.
(Note that a miscopy of the candidate’s own working is not a misread but an accuracy error.) All M marks
are available for a ‘misread’ solution, and A or B marks are initially given, as per the scheme, but for
Specimen Materials – Mathematics
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© OCR 2000
results as modified by the misread. At the end of each part of the question affected, deduct 0, 1 or 2
according as the number of ‘misread’ A and B marks earned in that part is 0, 1–4 or >4. If the misread
makes the question easier, a further deduction of 1 or more marks may be made at your discretion; this
deduction can include M marks.
8.
For a partially correct part of a question, exhibit the detailed marks, e.g. M1 A0, in the margin at the point
where the marks have been first earned. Please give sufficient detail to allow your marking to be
understood. For a completely correct part of a question, only the total mark for that part need be given, in
the margin . Do NOT use subtotals (underlined or otherwise). The question total should be ringed and
placed in the margin at the end of the question. This total MUST equal the sum of all the marks in the
margin for that question and should be entered against the question number in the question grid on the
front of the script. (N.B. Addendum to the booklet ‘Instructions for Examiners’: please use the left hand
margin of left hand pages.)
If a candidate’s answer is in two instalments, indicate the carried forward total at the end of the first part
by, for example,
and the brought forward total at the start of the second instalment by, for example,
.
The total mark for the paper should be obtained (a) by adding all the unringed marks through the script
(checking at the same time that all pages have been marked) and (b) by adding the question marks in the
grid in reverse order. The two totals must, of course, tally, and the resulting figure should be written,
ringed, on the front of the script.
9.
The following abbreviations may be used in a mark scheme, or may be found useful for notating a script.
AEF
Any Equivalent Form (of answer or result is equally acceptable).
AG
Answer Given on the question paper (so extra care is needed in checking that the detailed working
leading to the result is valid).
BOD Benefit Of Doubt (allowed for work whose validity may not be absolutely plain).
CAO Correct Answer Only (emphasising that no ‘follow through’ from a previous error is allowed).
ISW
Ignore Subsequent Working.
MR
Misread.
PA
Premature Approximation (resulting in basically correct work that is numerically insufficiently
accurate).
SOS
See Other Solution (the candidate makes a better attempt at the same question).
SR
Special Ruling (detailing the mark to be given for a specific wrong solution, or a case where some
standard marking practice is to be varied in the light of a particular circumstance).
10. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions get
full marks. Be alert for correct but unfamiliar or unexpected methods — often signalled by a correct result
following an apparently incorrect method. Such work must be carefully assessed. On the other hand, work
must not be judged on the answer alone, and answers that are given in the question, especially, must be
validly obtained. Key steps in the working must always be looked at and anything unfamiliar must be
investigated thoroughly. If a method is not catered for in the scheme, mark at discretion, imitating the
scheme as closely as possible. If a number of candidates are involved, or you are not sure what to do,
telephone your Team Leader.
11. For papers in which graphic (and programmable) calculators are allowed, some answers may be obtained
with little or no working visible. Allow full marks for correct answers (provided, of course, that there is
nothing in the wording of the question specifying that analytical methods are required). Where an answer is
wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no
supporting method score zero. If in doubt, consult your Team Leader.
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12. If in any case the scheme operates with considerable unfairness, mark at discretion but please give a brief
reason and initial the mark. This discretion should only be used very rarely.
13. If there is any suspicion of cheating or copying, mark according to the scheme and enter the marks on the
marksheet as usual. Send the script(s) to your Team Leader, as per OCR instructions. Notes concerning
illness etc should be sent to OCR with the marksheets. Scripts should be marked as per the scheme.
14. Examiners are reminded of the VITAL importance of checking the accuracy of the addition of marks and of
the transcriptions onto the marksheets; in particular that the marks are entered against the right candidates.
Do not assume that the scripts are in the same order as the names on the marksheet. As detailed in §8
above, each Examiner must check the paper total, obtaining the same figure twice by different methods.
The transcription to the marksheet should also be checked; ideally, the Checker should read out the
candidate’s name and mark from the marksheet, while the Examiner checks with the front of the script.
The Examiner has final responsibility for the accuracy of the mark recorded on the marksheet.
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