–7– M15/5/MATHL/HP1/ENG/TZ1/XX/M
Section A
1. (a) METHOD 1
1 2
area 22 2 3 M1A1
2
Note: Award M1 for using area formula.
2 A1
2
Note: Degrees loses final A1
METHOD 2
let x 2
1 2
area 2 x( 3 M1
2
3
x A1
2
A1
2
METHOD 3
Area of circle is 4π A1
Shaded area is of the circle (R1)
4
A1
2
[3 marks]
3
(b) arc length 2 A1
2
3
perimeter 2 2 2
2
3 4 A1
[2 marks]
Total [5 marks]
–8– M15/5/MATHL/HP1/ENG/TZ1/XX/M
1 0 19 10
2. (a) x 9 .5 (M1)A1
20
[2 marks]
(b) median is 10 A1
[1 mark]
(c) (i) 19 A1
(ii) 1 A1
[2 marks]
Total [5 marks]
3. (a) 1 tan 2 x dx sec 2 x dx tan x ( c) M1A1
[2 marks]
2 1 cos2 x
(b) sin x dx dx M1A1
2
x sin2 x
( c) A1
2 4
Note: Allow integration by parts followed by trig identity.
Award M1 for parts, A1 for trig identity, A1 final answer.
[3 marks]
Total [5 marks]
4. (a) ( x h)3 x3 3x 2 h 3xh 2 h3 (M1)A1
[2 marks]
( x h) 3 x 3
(b) f ( x) lim (M1)
h 0 h
x 3 x h 3 xh 2 h3 x 3
3 2
lim
h 0 h
lim 3 x 3 xh h 2
2
A1
h 0
3x 2 A1
Note: Do not award final A1 on FT if 3x 2 is not obtained
Note: Final A1 can only be obtained if previous A1 is given
[3 marks]
Total [5 marks]
– 10 – M15/5/MATHL/HP1/ENG/TZ1/XX/M
3x 2 1
6. (a) f :x y f :y x
2x 1
3x 2
y 3 x 2 2 xy y M1
2x 1
3x 2 xy y 2 M1
x (3 2 y ) 2 y
2 y
x A1
3 2y
2 y
( f 1 ( y) )
3 2y
2 x 3
f 1 ( x) x A1
3 2x 2
Note: x and y might be interchanged earlier.
Note: First M1 is for interchange of variables second M1 for
manipulation
Note: Final answer must be a function of x
[4 marks]
3x 2 B
(b) A 3x 2 A (2 x 1) B
2x 1 2x 1
equating coefficients 3 2A and 2 A B (M1)
3 1
A and B A1
2 2
Note: Could also be done by division or substitution of values.
[2 marks]
3 1
(c) f ( x) dx x ln 2 x 1 c A1
2 4
Note: accept equivalent e.g. ln 4 x 2
[1 mark]
Total [7 marks]
– 13 – M15/5/MATHL/HP1/ENG/TZ1/XX/M
9. (a) g f ( x) g f ( x) M1
g 2x
5
3sin 2 x 4 AG
5
[1 mark]
(b) since 1 sin 1 , range is 1, 7 (R1)A1
[2 marks]
3
(c) 3sin 2 x 4 7 2x 2n x n (M1)
5 5 2 20
23
so next biggest value is A1
20
Note: Allow use of period.
[2 marks]
(d) Note: Transformations can be in any order but see notes below.
stretch scale factor 3 parallel to y axis (vertically) A1
vertical translation of 4 up A1
4
Note: Vertical translation is up if it occurs before stretch
3
parallel to y axis.
1
stretch scale factor parallel to x axis (horizontally) A1
2
horizontal translation of to the left A1
10
Note: Horizontal translation is to the left if it occurs before
5
stretch parallel to x axis.
Note: Award A1 for magnitude and direction in each case.
Accept any correct terminology provided that the
meaning is clear eg shift for translation.
[4 marks]
Total [9 marks]
– 13 – M15/5/MATME/SP1/ENG/TZ1/XX/M
Section B
8. (a) (i) correct approach A1
eg B − A , AO + OB
1
→
AB = −1 AG N0
−2
(ii) correct substitution (A1)
eg (1) + (−1) + (−2) , 1 + 1 + 4
2 2 2
→
AB = 6 A1 N2
[3 marks]
(b) any correct equation in the form r = a + tb (any parameter for t )
−2 −1 1
where a is 4 or 3 and b is a scalar multiple of −1 A2 N2
3 1 −2
−2 1 −1 + t
eg r = 4 + t −1 , ( x , y , z ) = (−1, 3, 1) + t (1, − 1, − 2) , r = 3 − t
3 −2 1 − 2t
Note: Award A1 for the form a + tb , A1 for the form L = a + tb , A0 for the form r = b + ta .
[2 marks]
continued…
– 14 – M15/5/MATME/SP1/ENG/TZ1/XX/M
Question 8 continued
(c) METHOD 1
valid approach (M1)
−1 1 0 0 −2 1
eg 3 + t −1 = y , y = 4 + s −1
1 −2 −1 −1 3 −2
one correct equation from their approach A1
eg −1 + t = 0, 1 − 2t = −1, − 2 + s = 0, 3 − 2s = −1
one correct value for their parameter and equation A1
eg t = 1, s = 2
correct substitution A1
eg 3 + 1(−1) , 4 + 2 (−1)
y=2 AG N0
METHOD 2
valid approach (M1)
→ →
eg AC = k AB
correct working A1
2 2 1
→
eg AC = y − 4 , y − 4 = k −1
−4 −4 −2
k =2 A1
correct substitution A1
eg y − 4 = −2
y=2 AG N0
[4 marks]
(d) (i) correct substitution A1
eg 0 (1) + 2 (−1) − 1(−2) , 0 − 2 + 2
→ →
OC AB = 0 A1 N1
π
(ii) 90 or A1 N1
2
[3 marks]
continued…
– 15 – M15/5/MATME/SP1/ENG/TZ1/XX/M
Question 8 continued
(e) METHOD 1 (area = 0.5 × height × base)
(= 5 )
→
OC = 0 + 22 + (−1) 2 (seen anywhere) A1
valid approach (M1)
1 → → →
eg × AB × OC , OC is height of triangle
2
correct substitution A1
1 1
× 6 × 0 + ( 2 ) + ( −1) , × 6 × 5
2 2
eg
2 2
30
area is A1 N2
2
METHOD 2 (difference of two areas)
one correct magnitude (seen anywhere) A1
(= 5), (= )
→ → →
eg OC = 22 + ( −1) 2 AC = 4 + 4 + 16 24 , BC = 6
valid approach (M1)
eg ∆OAC − ∆OBC
correct substitution A1
1 1
eg × 24 × 5 − × 5 × 6
2 2
30
area is A1 N2
2
1
METHOD 3 (area = ab sin C for ∆OAB )
2
→ →
one correct magnitude of OA or OB (seen anywhere) A1
(= ) (= )
→ →
( −2 )
2
eg OA = + 42 + 32 29 , OB = 1 + 9 + 1 11
valid attempt to find cos θ or sin θ (M1)
−1 − 3 − 2 −6 sin θ sin 90
eg cos C = = , 29 = 6 + 11 − 2 6 11 cos θ , =
6 × 11 66 5 29
1
ab sin C
correct substitution into A1
2
1 36 5
eg × 6 × 11 × 1 − , 0.5 × 6 × 29 ×
2 66 29
30
area is A1 N2
2
[4 marks]
Total [16 marks]
– 19 – M15/5/MATHL/HP1/ENG/TZ1/XX/M
12. (a) (i) METHOD 1
vn 1 2u n 1
M1
vn 2u n
2un 1 un
2d A1
METHOD 2
vn 1 2a nd
M1
vn 2a ( n 1) d
2d A1
(ii) 2a A1
u
Note: Accept 2 1 .
(iii) EITHER
vn is a GP with first term 2a and common ratio 2d
vn 2 a (2 d ) ( n 1)
OR
un a (n 1)d as it is an AP
THEN
vn 2a ( n 1 )d
A1
[4 marks]
2 a (2 d ) n 1 2a 2d n 1
(b) (i) Sn = M1A1
2d 1 2d 1
Note: Accept either expression.
(ii) for sum to infinity to exist need 1 2d 1 R1
log 2d 0 d log 2 0 d 0 (M1)A1
d
Note: Also allow graph of 2 .
2a
(iii) S A1
1 2d
continued…
– 20 – M15/5/MATHL/HP1/ENG/TZ1/XX/M
Question 12 continued
2a 1 1
(iv) 2a 2 M1
1 2d 1 2d
1 2 2d 1
2d 1
1
d 1 A1
[8 marks]
(c) METHOD 1
wn pq n 1 , zn ln pq n 1
(A1)
zn ln p (n 1) ln q M1A1
zn 1 zn (ln p n ln q ) (ln p ( n 1) ln q ) ln q
which is a constant so this is an AP
(with first term ln p and common difference ln q )
n
n
zi 2ln p (n 1) ln q M1
i 1 2
n 1 n 1
2 2
n ln p ln q n ln pq (M1)
n n 1
ln p n q 2
A1
METHOD 2
n
zi ln p ln pq ln pq 2 ln pq n 1
(M1)A1
i 1
ln p n q 1 2 3 ( n 1)
(M1)A1
n n 1
ln p n q 2
(M1)A1
[6 marks]
Total [18 marks]
