UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS
Pre-U Certificate
MARK SCHEME for the May/June 2010 question paper
for the guidance of teachers
9794 MATHEMATICS
9794/02
Paper 2 (Pure Mathematics and Mechanics),
maximum raw mark 120
This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of
the examination. It shows the basis on which Examiners were instructed to award marks. It does not
indicate the details of the discussions that took place at an Examiners’ meeting before marking began,
which would have considered the acceptability of alternative answers.
Mark schemes must be read in conjunction with the question papers and the report on the
examination.
•
CIE will not enter into discussions or correspondence in connection with these mark schemes.
CIE is publishing the mark schemes for the May/June 2010 question papers for most IGCSE, Pre-U,
GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.
Page 2
Mark Scheme: Teachers’ version
Pre-U – May/June 2010
5
1
2
Syllabus
9794
3
Obtain the indefinite integral 4 x 2 − 2 x 2 .
Correctly substitute the limits in Ax a + Bx b
Obtain 110
Correctly combine the logs log 3
and simplify log 3
x
>1
x −1
(2 x
2
B1
M1
A1
)
−x
>1
2 x − 3x + 1
(
2
)
A1
x
>3
x −1
and solve: if x > 1 then x < 3/2; if x < 1, then x > 2/3.
The solution set is 1 < x < 3/2.
[Note: Equivalent marks for candidates who do not divide out the algebraic fraction and
attempt to solve a quadratic inequality.]
(i) Stating both of 60 = a + 12d and 141 = a + 30d
Obtain a = 6 and d = 4.5.
(ii) (a) The two sequences start: 6, 10.5, 15, 19.5 ... and 1.5, 4.5, 7.5, 10.5 ...
(b) Noting that 6 + (2n – 1)4.5 = 1.5 + (3n + 1 – 1)3,
the 2nth member of the first progression is identical to the (3n + 1)th member of the
second sequence. (Or equivalent convincing statement)
4
(i) Using difference of squares (or attempting to remove the sine or cosine)
cos 4 x − sin 4 x ≡ cos 2 x + sin 2 x cos 2 x − sin 2 x
2
(
[3]
M1
Rewrite as
3
Paper
02
)(
)
≡ 2 cos x − 1 using trig identities.
M1
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A1
[5]
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A1 A1
[3]
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[1]
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A1
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(ii) Required to solve 2 cos 2 x − 1 = cos x
which factorises as (2cos x + 1)(cos x – 1) (= 0), or use of the quadratic formula.
Solutions are: 120°, 240°,
0°, 360°
[A1 for any two correct, A1A1 for all correct.]
© UCLES 2010
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Page 3
5
Mark Scheme: Teachers’ version
Pre-U – May/June 2010
Syllabus
9794
2x
x−i+ x+i
.
= 2
2
2
x −i
x +1
Complete attempt to solve quadratic yx2 – 2x + y = 0 for x.
(i) Combine and simplify the two fractions y =
2
(ii) The larger solution for x satisfies xy = 1 +
M1
A1 AG
A1 AG
(1 − y )
2
A1
[2]
(iii) Sketch of region in the first quadrant,
bounded by the horizontal line y = 1 (region below)
and the rectangular hyperbola xy = 2 (curve acceptable)
B1
B1
B1
(i) Either using x – 1 = x + 1 – 2 and then dividing by (x + 1)2
1
2
or rationalising and equating coefficients, leading to
−
x + 1 (x + 1)2
M1
(ii) A correct process of division, leading to
3
2y − 2 +
, or equivalent statement.
y +1
M1
(iii) Attempt at separating the variables in the form
A1
A1
∫ f ( y ) d y = ∫ g ( x ) dx
or un-integrated equivalent form.
Correctly using the results of parts (i) and (ii) to obtain
 1

3 
2 
 dx
 2 y − 2 +
 dy = 
−
2 
y +1

 x + 1 (x + 1) 
∫
∫
Perform indefinite integrations y 2 − 2 y + 3 ln ( y + 1) = ln (x + 1) +
[The first A1 is for at least one ln term]
Correct substitution of x = 0 and y = 2
Obtain C = 3ln3 – 2.
© UCLES 2010
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Y 2 because 0 < y Y 1.
[Note: There are acceptable alternative methods]
6
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2
2 ± 4 − 4y
1± 1− y
=
.
2y
y
State that the reality of x ⇒ positivity of the discriminant
and that the positivity of y ⇒ y Y 1.
Obtain x =
Paper
02
[3]
[2]
[2]
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2
+ (C )
x +1
A1 A1
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[6]
Page 4
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Mark Scheme: Teachers’ version
Pre-U – May/June 2010
Syllabus
9794
Paper
02
2
(x − 1) = 2 − 22
2
x x
x
dy
2
4
1
= − 2 + 3 = 3 (4 − 2 x )
Attempt at differentiating:
dx
x
x
x
[Note: This may be seen later]
(i) When t = 0, y =
M1
(a) y is positive if x > 1
y is positive and increasing if 1 < x < 2
[Note: Accept alternative convincing arguments]
B1
A1
(b) y is stationary when x = 2
Consideration of the gradient either side, or the use of the second derivative, shows a
maximum.
B1
(ii) (a) Substitute t = 2 and y = –2 and show at least one line of manipulation
to obtain x3 + x – 1 = 0.
(b) f ′= 3x2 + 1.
This is positive definite ( ⇒ there are no roots)
xn3 + xn − 1  2 xn3 + 1 
=

3 xn2 + 1  3 xn2 + 1 
Using x0 = 1, obtain a succession of approximations
0.75, 0.68606, 0.68233, 0.68234.
State V = 0.6823 to 4 sfs.
(c) Derive iterative formula xn+1 = xn −
© UCLES 2010
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8
Mark Scheme: Teachers’ version
Pre-U – May/June 2010
Syllabus
9794
(i) Using distance-squared, the condition for a point (x, y) to be on the curve is
(x − 0)2 + ( y − a )2 = ( y − b )2
1
1
Expand, simplify and make y the subject, obtaining y =
x 2 + (a + b )
2(a − b )
2
1
(a + b )
2
This is the midpoint of the interval on the y-axis joining F to the point (0, b) on D.
The curve passes between these two points without crossing D.
(ii) As a > b the curve has a lowest point with y-coordinate
(iii) (a) The y-coordinate of P is a + b (or unsimplified expression).
x
.
The gradient of the tangent at P is
a −b
Attempt at forming the equation of the tangent
y − (a + b ) =
a2 − b2 
 x − a 2 − b 2 

a −b 
which simplifies to y =
a2 − b2
x , passing through the origin.
a−b
(b) The coordinates of P are ( 4 5 , 4)
The coordinates of Q are ( − 8 5 , –8)
The length of PQ is 144 × 5 + 144
This simplifies to − 12 6
Paper
02
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9
Mark Scheme: Teachers’ version
Pre-U – May/June 2010
Syllabus
9794
dv
= xn ,
dx
1 n+1
1 n+1 1
the indefinite integral is
dx
x ln x −
x
n +1
n +1
x
1 n+1
1
x n+1
((n + 1) ln x − 1)
x ln x −
x n+1 =
2
n +1
(n + 1)
(n + 1)2
Correctly substituting the limits in their integrated expressions,
a n+1
((n + 1) ln a − 1) + 1 2 .
obtain
2
(n + 1)
(n + 1)
(i) Using integration by parts, with u = lnx and
∫
(
(ii) (a) The curves intersect when x 2 ln x = x ln 2 x = x 2 ln 2
The solution is x = 2.
2
(b) The required area is given by
∫
)
Paper
02
M1 A1
A1
A1
M1
A1
[6]
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[2]
2
∫
x 2 ln 2 dx − x 2 ln x dx (condone reverse order)
1
M1
1
Attempt at using the result from part (i) and correctly substituting the limits for the
1
8
8 1
first integral, the value of this area is  −  ln 2 −  (3 ln 2 − 1) + 
M1 A1
9
9
 3 3
7 1
A1
This simplifies to − ln 2 .
9 3
(e )
(e )
(iii) The volume of revolution is (π ) y 2 (dx ) = (π ) x 3 ln x (dx )
∫
()
1
[4]
∫
()
B1
1
Using the result from part (i), the exact value of the volume is
 e4
1
π  (4 ln e − 1) +  = 32.36
16 
 16
10 (i) The net force opposing motion up the slope is mg sinφ, so the deceleration is g sinθ.
Using the formula v = u + at (v = 0, u = V, a = –g sinθ, t = 2V/g cosθ)
V = V 2sinθ cosθ ⇒ sin2θ = 1
θ = π/4.
(ii) Resolving forces perpendicular to the slope R – mgcosθ = 0
Applying Newton’s law to motion parallel to the slope ma = –mgsinθ – F
where F = 0.5 R.
Obtain a = –g(sinθ + 0.5cosθ ).
V2
V2 2
Using v2 = u2 + 2as, obtain s =
=
2 g (sin θ + 0.5 cos θ )
3g
© UCLES 2010
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Page 7
Mark Scheme: Teachers’ version
Pre-U – May/June 2010
Syllabus
9794
11 (i) | 2i – 2j + k | = 3
F3 = (±) 2(2i – 2j + k).
The possible resultants of the three forces are
15i + 6j and 7i + 14j – 4k
Both have magnitude 261 N
Paper
02
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A1
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[5]
(ii) The vector equation of motion is
| F1 + F2 | = | 11i + 10j – 2k | = 15
Equation of motion: 2 |a| = 15,
|a| = 7.5 ms–2
B1
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[3]
(iii) The particle moves in the straight line defined by the direction of the resultant force with
the constant acceleration 7.5 ms–2. Applying the formula s = ut + 0.5at2,
obtain t = 4 seconds.
M1
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[2]
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12 (i) (a) Attempt to obtain v(t) by integration:
For first stage v = 30t – 3t2 + (A) (constant of integration)
Show A = 0, using v(0) = 0.
Calculate v(10) = 0.
For second stage v = 3t2 – 90t + (B).
Show B = 600 using v(10) = 0
Calculate v(20) = 1200 – 1800 + 600 = 0
(b) Sketch of v(t): Inverted parabola above axis for 0 Y t Y 10
Upright parabola below axis for 10 Y t Y 20
(ii) Distance travelled obtained by integrating the modulus of each of the velocity formulae
10
and adding:
[15t
0
] [
10
20
∫ (30t − 3t ) dt − ∫ (3t
2
)
− 90t + 600 dt
M1 M1
10
− t 3 0 − t 3 − 45t 2 + 600t
= 500 – (–500) = 1000 m
2
2
]
20
10
13 (i) Applying Newton’s law to the particles, where a is the acceleration of A and T is the
tension in the string: 2a = T – 2g; 3a = 3g – T.
Adding and solving: a = 0.2g.
(ii) Using v2 = u2 + 2as (u = 0, s = 1), the common speed of A and B is
Using v2 = u2 + 2as for the motion of A, with a = –g, u =
2g
(= 2ms–1)
5
A1
A1
[4]
M1 A1
A1
[3]
M1 A1
2g
,
5
gives a total height of 2.2 m above the ground.
B1
If e is the coefficient of restitution, we use above formula with v = 0,
2g
a replaced by –g, s = 0.05, u = e
.
5
Obtain e = 0.5
M1
© UCLES 2010
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[3]
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