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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS
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Pre-U Certificate
MARK SCHEME for the May/June 2011 question paper
for the guidance of teachers
9795 FURTHER MATHEMATICS
9795/01
Paper 1 (Further Pure Mathematics), 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.
• Cambridge will not enter into discussions or correspondence in connection with these mark schemes.
Cambridge is publishing the mark schemes for the May/June 2011 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 2011
Syllabus
9795
Attempt at A2 and A3
1
k + 4 − k 
 , A3 =
−
1
k
+
9


A2 = 
1 0
 ⇔ k = – 7
0 1
A3 = 
Paper
01
M1
 k + 8 k + 7k 


 k + 7 − 4k − 27 


2
≥ 3 entries of A3 All entries of A3 must be if done this way
A1
A1
Otherwise, allow just one key element checked (since “given”)
2 − 7
 = – 6 – k = 1
 1 − 3
det 
ft numerical value consistent with their k
B1
[4]
ALTERNATIVE
det(A3) = (det A)3
A3 = I ⇒ det A = 1
Det A = – 6 – k
k=–7
M1
A1
B1
A1
[4]
2
(i)
Noting α + β + γ = –1 and αβ + βγ + γα = 7
α 2 + β 2 + γ 2 = (α + β + γ)2 – 2(αβ + βγ + γα) = –13
(αβγ = 1)
GIVEN ANSWER legit.
2
2
ALTERNATIVE Substitute x = y to find eqn. with roots α , β , γ
y3 + 13y2 + 51y – 1 = 0
α 2 + β 2 + γ 2 = (–13) / 1 = –13
(ii)
2
B1
M1
A1
M1
A1
A1
B1
B1
Eqn. has at least one non-real (complex) root
one real and two complex (conjugate) roots
[3]
[3]
[2]
3
(i)
f(r – 1) – f(r) =
1
1
8r
−
=
(denomr. may be factorised)
2
2
2
2
(2r − 1)
(2r + 1)
4r − 1
(
)
B1
[1]
n
(ii)
1 n
=
∑ { f (r − 1) − f (r )}
2
8 r =1
4r 2 − 1
∑(
r =1
r
)
Use of this result …
1 n
∑ {( f (0) − f (1)) + ( f (1) − f (2)) + ... + ( f (n − 1) − f (n))} & cancelling
8 r =1

1
1
1

= { f (0) − f (n)} = 1 −
2 
8
8  (2n + 1) 
=
1  (4n 2 + 4n + 1) − 1 

= 
8
(2n + 1) 2

=
n(n + 1)
2(2n + 1) 2
M1
M1
A1
Common denomr. with squaring attempted in numr.
GIVEN ANSWER from correct working
A1
[4]
© University of Cambridge International Examinations 2011
Page 3
4
Mark Scheme: Teachers’ version
Pre-U – May/June 2011
Syllabus
9795
Paper
01
(i)
B1
y = cosh x – 1
1
O
×
Grad. of tanhx
should be 1
α
y = tanh x
B1
–1
Curves cross twice (at x = 0 and x = α) so there are 2 roots to tanh x = cosh x – 1
i.e. 1 + tanh x – cosh x = 0
(ii)
B1
[3]
(a)
f(1) f(1.5) = 0.22… × (– 0.45…) < 0
⇒ 1 < α < 1.5 by the “Change-of-Sign” Rule
B1
[1]
(b)
f ′(x) = sech2x – sinh x
B1 B1
f ( xn )
at least once
Use of xn + 1 = xn –
f ′( xn )
x1 = 1.25, x2 = 1.219 625 3,
M1
x3 = 1.218 76 to 5 d.p.
A1
[4]
5
Aux.Eqn. m2 + 1 = 0 ⇒ m = ± i ⇒ Comp.Fn. is yc = A cos x + B sin x
For Part.Intgrl. trying y = ax2 + bx + c (with at least a non-zero)
dy
d2 y
= 2ax +b,
= 2a
dx 2
dx
Diffg. their yp to find y′ and y″ and substg. into the given d.e.
Equating terms to find a, b, c (with at least a and c non-zero)
a = 8, b = 0, c = – 16; i.e. yp = 8x2 – 16
Gen.Soln. is y = A cos x + B sin x + 8x2 – 16
ft their yc + yp provided yc has 2 arb. consts. and yp has none
M1
A1
M1
M1
M1
A1
B1
[7]
© University of Cambridge International Examinations 2011
Page 4
6
(i)
Mark Scheme: Teachers’ version
Pre-U – May/June 2011
Syllabus
9795
 p p  q q 


 p p  q q 
 2 pq 2 pq 
 ∈ S
= 
 2 pq 2 pq 
Paper
01
Good attempt to multiply 2 matrices of the appropriate form: 
M1
“Closure” noted or implied by correct product matrix
A1
Statement that ×M known to be associative
B1
Alt. [(p)(q)] (r) = (p) [(q)(r)] = (4pqr) shown
1
1
2
2
1
2
2
Identity is  1
−1
 41p
p
 =  1
p 
 4p
p

p

 (∈ S)

1
4p
1
4p
B1

 (∈ S as p ≠ 0)


B1
… and (S, ×M) is a group since all four group axioms are satisfied
(ii)
Attempt to look for a self-inverse element; i.e. solving p =
[5]
1
ft their (p) – 1 and
4p
M1
E
p= −
1
2
 12
and noting that H = {E, A} where E =  1
2

,A=
1
2
1
2
Looking for {E, B, B2} where B3 = E ; i.e. solving (4p3) =
Explaining carefully that p3 =
1
8
⇔ p=
1
2
 − 12
 1
− 2
− 12 

− 12 
1
2
and no such B (≠ E) exists
A1
M1
A1
[4]
7
y=
9
x 2 + 4 x 12 x(2 x − 1) + 94 (2 x − 1) + 94 1
=
= 2 x + 94 + 4
2x −1
2x −1
2x −1
For y = 12 x + c
For c = 94 (ignore remdr. term)
Vertical asymptote x =
1
2
A1 A1
B1
noted or clear from graph
dy (2 x − 1).(2 x + 4) − ( x 2 + 4 x).2
=
dx
(2 x − 1) 2
M1
Diffg. and setting num r. = 0
Solving a quadratic in x (x2 – x – 2 = 0)
TPs at (– 1, 1) and (2, 4) One each; or one for both x’s but y’s missing
Crossing-points on the axes at (0, 0) and (– 4, 0)
General shape
All correct
M1
M1
A1 A1
B1
B1
B1
[11]
© University of Cambridge International Examinations 2011
Page 5
8
(i)
(ii)
Mark Scheme: Teachers’ version
Pre-U – May/June 2011
Syllabus
9795
Paper
01
1 2 3
No unique soln. ⇔ 2 3 k = 0
1 k 6
M1
Gaining and solving a quadratic eqn. in k
0 = – (k2 – 8k + 15) = – (k – 3)(k – 5) ⇒ k = 3, 5
M1
A1
[3]
x + 2 y + 3z = 4
2 × (1) − (2) ⇒
k = 3 ⇒ 2 x + 3 y + 3z = 9 :
(3) − (1) ⇒
x + 3y + 6z = 1
y + 3 z = −1
y + 3 z = −3
Substg. back and eliminating one variable; inconsistency correctly
shown
x + 2 y + 3z = 4
k = 5 ⇒ 2 x + 3 y + 5z = 9 :
x + 5 y + 6z = 1
2 × (1) − (2) ⇒
y + z = −1
(3) − (1) ⇒ 3 y + 3 z = −3
Substg. back and eliminating one variable; consistency correctly shown
ALTERNATIVE (whole qn.)
1 2 3 4
1
2
3
4
→ 0
1
6−k
0 k −2
3
3
4
→ 0 1
6−k
−1
2
0 0 k − 8k + 15 k − 5
2
M1
A1
[4]
2 3 k 9 → 0 −1 k − 6 1
1 k 6 1
0 k −2
3
−3
1
2
3
4
1 2
M1
A1
−1
−3
convincing
R2 ' = R2 − 2 R1 attempt at
R3 ' = R3 − R1 Gaussian
elimination
or by Cramer’s Rule
R3 ' = R3 − (k − 2) R2
k – 8k + 15 = 0 for no unique solution
M1
Final R2
Final R3
k = 3, 5
Noting k = 3 ⇒ R3 = 0 0 0 | – 2 giving inconsistency
Noting k = 5 ⇒ R3 = 0 0 0 | 0 giving consistency
A1
A1
M1
A1
B1
B1
[7]
© University of Cambridge International Examinations 2011
Page 6
9
(a)
Mark Scheme: Teachers’ version
Pre-U – May/June 2011
Use of V =
1
6
and
 − 73 


a × b =  139 
 97 


M1
| a • b × c | formula
Answer 335 16 or
Paper
01
M1
Attempt at scalar triple product: a • b × c or a × b • c
 36 


NB b × c =  − 41
 145 


(b)
Syllabus
9795
A1
2011
6
[3]
 3  6  7 
 1 
Alt approach possible
    
 

(i) n =  1  ×  2  =  − 21 accept ±  − 3  etc. that eliminates and
 − 1  5   0 
 0 
from the original eqn.
    
 

 2   − 1
   
ft incorrect n
r • n = 1 • 3  =1 = d
 4  0 
   
(ii)
− x + 3y = 1
x + 4 y + 7 z = 13
M1
A1
M1
A1
[4]
Set y = λ ⇒
ft 1st eqn. from (i)
Parametrisation attempt
x = 3λ – 1, z = 2 – λ
M1
A1 A1
 − 1  3 
   
r =  0  + λ  1  or any other correct vector line eqn. form ft
 2   − 1
   
B1
MUST have r = at the start (allow r = )
[4]
ALTERNATIVE
 1   1   21 
     
d.v. =  − 3  ×  4  =  7 
 0  7 − 7
     
 3
 
accept ±  1  etc.
 − 1
 
 1 
 
ft  − 3 
 0 
 
Finding any point on the line; e.g. (– 1, 0, 2), (2, 1, 1), (5, 2, 0), etc.
Answer as above
ft point and d.v.
M1
A1
B1
A1
[4]
© University of Cambridge International Examinations 2011
Page 7
10
Mark Scheme: Teachers’ version
Pre-U – May/June 2011
Syllabus
9795
(i)
cos3θ = Re(cos3θ + i sin3θ) = Re(c + is)3
Use of de Moivre’s Thm.
= c3 – 3cs2 = c3 – 3c(1 – c2) = 4c3 – 3c ANSWER GIVEN
(ii)
(a) u2 = 2c2 – 1 and u3 = 4c3 – 3c
Paper
01
M1
A1
B1 B1
(b) Conjecture un = cos(nθ) 1
Base case, “true for n = 0 or 1 (or 2 or 3)”, may be taken as read
Assuming un = cos(nθ) for at least n = k and n = k – 1
Using given r.r. to generate uk + 1 = 2 cosθ cos(kθ) – cos([k – 1]θ)
= 2 cosθ cos(kθ) – {cos(kθ) cosθ + sin(kθ)sinθ}
= cosθ cos(kθ) – sinθ sin(kθ)
OR {cos(k + 1)θ – cos(k – 1)θ } – cos(k – 1)θ
= cos([k + 1]θ)
If statement is true for n = 0, 1, 2, …, k (but allow assumption for one term
only here) then it is also true for n = k + 1. Proof follows by (strong) induction
B1
[2]
[2]
M1
M1
M1
A1
B1
[6]
11
(i)
1π
6
In =
∫ sec
n
1π
6
∫ sec
t dt =
0
[
n−2
0
 2 
In = 

 3
2n − 2
=
t. tan t
n−2
( 3)
n −1
Correct splitting and use of parts
6
π
− ∫ tan t.(n − 2) secn − 3 t . sec t. tan t dt
0
M1
A1
A1
1π
6
1
.
– ( n − 2) ∫ sec n − 2 t.(sec 2 t − 1) dt
3
0
– (n − 2){I n − I n − 2 }
( 3)
n −1
(
1
]
2 n−2
(n – 1)In =
(ii)
t.sec 2 t dt
0
1
π
6
= sec
n−2
Substg. for tan …
… and reverting to I’s
+ (n – 2)In – 2 ANSWER GIVEN
) (
(
A1
)
2
M1
2
x& 2 + y& 2 = sec 2 t + sec 2 t. tan t
Attempted
4
2
6
= sec t 1 + tan t = sec t
)
∫ 2πy x& + y& dt with y =
= ∫ 2π . sec t.sec t dt = π I
2
Use of S =
1
2
2
2
3
1
2
sec2t and their x& and y&
M1
1
2
ln 3
Then, using the R.F., 2 I3 =
2
3
+
1
2
[4]
A1
ln 3 ⇒ I3 =
1
3
+
+ 3( 13 + 14 ln 3)
1
(68 + 27 ln 3)π
… leading to S = 144
Using the R.F. again, 4 I5 =
M1
A1
5
1 
 2
+
 − ln1 =
3
 3
[5]
A1
I1 = ln[sec t + tan t ]
= ln
M1
1
4
ln 3
M1
A1
M1
or exact equivalent
A1
8
9
[6]
© University of Cambridge International Examinations 2011
Page 8
12
(i)
Mark Scheme: Teachers’ version
Pre-U – May/June 2011
(a + ib)2 = 2 + 2i ⇔ a2 – b2 = 2 and ab = 1
Syllabus
9795
Squaring & equating Re/Im parts
1
a 2 − 2 − 2 = 0 ⇒ a4 – 2a2 – 1 = 0 ⇒ (a2 – 1)2 = 2 (or by the quadratic
a
formula)
Substg. for b (say) and solving a quadratic in a2
2 + 1 (AG) MUST note that a2 > 0 to explain choice of +ve sq.rt.
1
2 − 1 from b4 + 2b2 – 1 = 0 or b =
Similarly, b =
a
a=
(ii)
Paper
01
M1
M1
A1
M1
A1
[5]
(a)
z2 = −
2 +1 + i
2 −1
arg(z2) = π − tan −1
= π − tan −1
(b)
n
arg(z2 ) =
7
8
(
2 −1
2 −1
Attempt inclg. rationalising denomr.
×
2 +1
2 −1
)
2 − 1 = π − 18 π = 78 π
nπ
= (2k + 14 )π
n=
16k + 2
, giving least n = 14
7
M1
A1
B1
[2]
M1
A1
[Condone lack of convincing explanation that this IS the least such n.]
[3]
© University of Cambridge International Examinations 2011
Page 9
13
(i)
(a)
Mark Scheme: Teachers’ version
Pre-U – May/June 2011
Syllabus
9795
2t
2
in
, where t = tan x
2
2 − sin 2 x
1+ t
2
2
1+ t2
=
=
ANSWER GIVEN
2 − sin 2 x 2 − 1+2tt 2
1− t + t2
Use of sin2x =
( )
Paper
01
M1
A1
[2]
(b)

y = tan – 1 1 −

dy
1
2 tan x 
=
 ⇒
dx 1 + 1− 2t
3 
3
( )
2
−2
×
3
sec 2 x
M1
A1
or tan y = … and use of implicit diffn.
=
=
(ii)
(a)
− 3 1+ t2
.
2 1− t + t2
−2
(1 + t )×
2
3
4 − 4t + 4t 2
3
− 3
2
.
=
so that k = − 3
2 2 − sin 2 x
M1
A1
Use of x = r cosθ , y = r sinθ (and x2 + y2 = r2)
⇒ r2 = 72 + r2 sinθ cosθ i.e. x, y completely (and correctly) eliminated
M1
M1
⇒ r2 =
A1
144
2 − sin 2θ
or equivalent form
2 – sin 2θ ≥ 1 ⇒ r2 ≤ 144 so that rmax = 12
Calculus approach fine as an alternative
Then sin 2θ = 1 ⇒ θ = 14 π or 54 π
No M ft from incorrect differentiation
M1
A1
M1
A1
72
(b) A = 12 ∫ r 2 dθ = ∫
dθ
2 − sin 2θ
 −1
 1 − 2 tan x  π / 4
= 72 
tan −1 

3

 0
 3
=
72
3
(tan ( ) − tan (− ))
−1
1
3
−1
1
3
[4]
or
[7]
M1
Use of previous result
M1
72  π  π  
 −  −  
3  6  6 
A1
= 8π 3 CAO ft their k above
Area inside C in 1st quad. = 2A = 16π 3 since C is symmetric in y = x ft
A1
B1
[5]
© University of Cambridge International Examinations 2011