Partial

Question Answer Marks Guidance

Marks
9709/01

1(a) Gradient of CE = 2.5 1 B1

© UCLES 2017
Gradient of DE = 2.1 1 B1
2
1(b) f ′(2) = 2 1 B1 Accept reasonable conclusion following their
gradient

Partial
Question Answer Marks Guidance
Marks
2 x−2 1 B1
f –1(x) =
3
gf(x) = 4(3x + 2) – 12 1 B1
2 2 M1A1
Equate f –1(x) and gf(x) expressions, x = 7

Page 4 of 10
SPECIMEN

Partial
Question Answer Marks Guidance
Marks
3 7 + (n – 1)d = 84 and/or 7 + (3n – 1)d = 245 1 B1
(n – 1)d = 77, (3n – 1)d = 238 SOI OR 2nd = 161 explicitly stated 1 B1
n−1 1 M1 (must be from the correct un formula)
= 77
3n − 1 238 OR other attempt to eliminate d e.g.
Cambridge International AS & A Level – Mark Scheme

161
substitute d = .
2n
(If n is eliminated d must be found)
77 1 A1
n = 23 (d = 22
= 3.5)
4
from 2020
For examination
 Partial
Question Answer Marks Guidance
Marks
9709/01

4 Attempt integration 1 M1

© UCLES 2017
1 6 2 A1A1 Accept unsimplified terms, A1 for each term
f(x) = 2(x + 6) 2 – x (+ c)

6 1 M1 Substitute x = 3, y = 1. c must be present

2(3) – +c=1
3
[c = –3] 1 A1
1 6
f(x) = 2(x + 6) 2 – x –3

Partial
Question Answer Marks Guidance
Marks
5(a) y = (x – 2)2 + 3(x – 2) + 4 = x2 – x + 2 2 M1A1
5(b) Reflection [in] y axis 1 B1 In either order

Page 5 of 10
Stretch factor 3 in y direction 2 B1B1 B1 for stretch, B1 for factor 3 in y direction
SPECIMEN

Partial
Question Answer Marks Guidance
Marks
6(a) Coefficient of x2 is 240 1 B1
Coefficient of x3 is 20 × 8 × (–1) = –160 2 B2 B1 for +160
Cambridge International AS & A Level – Mark Scheme

3
6(b) Product needs exactly 2 terms 1 M1 3 × their 240 + their ‒160
720 – 160 = 560 1 A1FT FT for candidate's answers
2
from 2020
For examination
 Partial
Question Answer Marks Guidance
Marks
9709/01

7(a) sin x 1 M1 Correct formula

© UCLES 2017
Replace tan x by cos x
sin x 2
1 + cos x = 5 cos x

Replace sin x2 by 1 – cos x2 1 M1 Correct formula used in appropriate place

2
6 cos x – cos x – 1 (= 0) 1 A1 AG
3
7(b) 1 1 1 M1 Correct method seen
Solution of quadratic 8c = − 3 or 2B
x = 60° or 109.5° 2 A1A1
3

Partial
Question Answer Marks Guidance
Marks

Page 6 of 10
8(a) –12(3 – 2x)–2 × –2 2 B1B1 B1 for –12(3 – 2x)–2, B1 for –2
SPECIMEN

8(b) dy dy dx 1 M1 OE; chain rule used correctly

= ' = 0.4 ÷ 0.15
dx dt dt
24 8 1 M1 dy 8 3
2 = 3 Equates their with their or and method
(3 − 2x) dx 3 8
seen for solution of quadratic equation
x = 0 or 3 2 A1A1
Cambridge International AS & A Level – Mark Scheme

4
from 2020
For examination
e
Partial
Question Answer Marks Guidance
Marks
9709/01

9(a) 1 B1 AG

© UCLES 2017
BC2 = r2 + r2 = 2r2 → BC = r 2
9(b) 2 1 M1
Area sector BCFD = 14 r _r 2i seen or implied Expect 12 πr2. (F is intersection of large circle with
AE)
Area ∆BCD = 12 (2r)r 1 M1* Expect r2 (could be embedded)

Area segment CFDA = 12 πr2 – r2 1 A1 OE

Area semi circle CADE = 12 πr2 1 B1

Shaded area = 12 πr2 – ( 12 πr2 – r2) 1 DM1 Depends on the area ∆BCD
OR
1 2 2
2
πr2 – 91 rr 2 + `2 rr − r jC

= r2 1 A1
6

Page 7 of 10
SPECIMEN
Cambridge International AS & A Level – Mark Scheme
from 2020
For examination
 Partial
Question Answer Marks Guidance
Marks
9709/01

10(a) (‒2, 1) 1 B1

© UCLES 2017
10(b) Gradient of CD = 12 ' 3 12 = 17 1 B1

Gradient of AB = –7 1 M1 With gradient –1/their m

1 1 M1
Equation of AB is y – 1 12 = –7 `x - 1 2j
y = –7x + 12 1 A1
4
10(c) x2 + (12 – 7x)2 + 4x – 2(12 – 7x) – 20 (= 0) 1 M1 Substituting their AB equation into circle equation
(50)(x2 – 3x + 2) (= 0) 1 A1
x = 1, 2 1 A1 Dependent on method seen for solving quadratic
equation
3

Page 8 of 10
SPECIMEN
Cambridge International AS & A Level – Mark Scheme
from 2020
For examination
 Partial
Question Answer Marks Guidance
Marks
9709/01

11(a) x2 + 6x – 8 = (x + 3)2 – 17 2 B1B1 B1 for (x + 3)2, B1 for ‒17

© UCLES 2017
OR OR
2x + 6 = 0 → x = –3 → y = –17 B1 for x = –3, B1 for y = –17
Range f(x) ⩾ –17 1 B1FT FT; following through visible method
3
2
11(b) (x – k)(x + 2k) = 0 ≡ x + 5x + b = 0 1 M1 Realises the link between roots and the equation
k=5 1 A1 Comparing coefficients of x
b = –2k2 = –50 1 A1
3
11(c) (x + a)2 + a(x + a) + b = a 1 M1* Replaces ‘x’ by ‘x + a’ in 2 terms
Uses b2 – 4ac, 9a2 – 4(2a2 + b – a) 1 DM1 Any use of discriminant
2
a < 4(b – a) 1 A1 AG
3

Page 9 of 10
SPECIMEN
Cambridge International AS & A Level – Mark Scheme
from 2020
For examination
 Partial
Question Answer Marks Guidance
Marks
9709/01

12(a) a=2 1 B1

© UCLES 2017
3 2
12(b) y = x – 4x + 4x 1 B1
dy 2 B2FT FT B1 for 3x2, B1 for –8x + 4
= 3x2 – 8x + 4
dx
2 1 B1 Dependent on method seen for solving quadratic
(x – 2)(3x – 2) = 0 → b =
3 equation
4
12(c) R 4 3 V 2 B2
Sx 4 x 2W x4 4x3
+ B1 for , B1 for + 2x2
4 3 2x W 4 3
T X
Area = 8 y dx = S −

32 1 M1 Apply limits 0 → 2
4– +8
3
4 1 A1 Unsupported answer receives 0 marks
3
4

Page 10 of 10
SPECIMEN

12(d) d2y 4 2 M1*A1 Attempt 2nd derivative and set = 0

= 6x – 8 = 0, x =
dx 2 3

4 dy 4 2 DM1A1
When x = , (or m) = −
3 dx 3
4
Cambridge International AS & A Level – Mark Scheme
from 2020
For examination