Mathematics 9709/13 October/November 2020: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 1 Pure Mathematics 1, October/November 2020. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) Express x2 + 6x + 5 in the form x + a2 + b, where a and b are constants. [2]
(b) The curve with equation y = x2 is transformed to the curve with equation y = x2 + 6x + 5.
Describe fully the transformation(s) involved. [2]
The function f is defined by f x = for x > −2.
2 x + 22
Mark scheme
1(a) ( x + 3) B1 B1
2
[ −4]
2
1(b) −3 B1 −their a
[Translation or shift] B1 FT Accept [translation/shift] their b
−4
OR translation ‒3 units in x-direction and (translation) ‒4 units in
y-direction.
2
∞
(a) Find Ó f x dx. [3]
1
= f x. It is given that the point −1, −1 lies on the dy
(b) The equation of a curve is such that dx curve.
Find the equation of the curve. [2]
Solve the equation 3 tan2 1 + 1 = for 0Å < 1 < 180Å.
2 tan2 1
Mark scheme
2(a) −2 B1 Integrate f(x). Accept −2 ( x + 2 )−1 . Can be unsimplified.
x+2
2 2 M1 A1 Apply limit(s) to an integrated expansion. CAO for A1
0−− =
3 3
3
2(b) −1 = − 2 + c M1 Substitute x = −1, y = −1 into their integrated expression (c
present)
−2 A1 Accept y = −2 ( x + 2 )−1 + 1 . ‒2 must be resolved.
y= +1
x+2
2
3 3tan θ + tan θ − 2 ( = 0 )
4 2 M1 SOI 3-term quartic, condone sign errors for this mark only
( 3tan θ − 2 )( tan θ + 1) ( = 0 )
2 2 M1 Attempt to factorise or solve 3-term quadratic in tan 2θ .
2 A1 SOI Implied by final answer = 39.2° after 1st M1 scored
tan θ = ( ± ) or ( ± ) 0.816 or ( ± ) 0.817
3
39.2°, 140.8° A1 FT for 2nd solution =180° ‒ 1st solution
A1 FT
5
A curve has equation y = 3x2 − 4x + 4 and a straight line has equation y = mx + m − 1, where m is a constant.
Find the set of values of m for which the curve and the line have two distinct points of intersection.
[5]
Mark scheme
4 3 x 2 − 4 x + 4 = mx + m − 1 → 3 x 2 − ( 4 + m ) x + ( 5 − m ) ( = 0 ) M1 3-term quadratic
b 2 − 4ac = ( 4 + m ) − 4 × 3 × ( 5 − m )
2 M1 Find b 2 − 4 ac for their quadratic
m 2 + 20m − 44 A1
(m + 22)(m ‒ 2) A1 Or use of formula or completing square. This step must be seen
m > 2 , m < −22 A1 Allow x > 2 , x < −22
5
In the expansion of a + bx7 , where a and b are non-zero constants, the coefficients of x, x2 and x4 are the first, second and third terms respectively of a geometric progression.
a
Find the value of . [5]
b
The function f is defined by f x = for x > 13 .
2x
3x − 1
Mark scheme
5 7C1a 6b ( x ) ,
( )
7C 2 a 5 b 2 x 2 ,
( )
[7C 4a 3b 4 x 4 ] B2, 1, 0 SOI, can be seen in an expansion.
( ) = 7C 4a b ( x ) → 21a b = 35a b
7C 2 a 5 b 2 x 2 3 4 4 5 2 3 4 M1 A1 M1 for a correct relationship OE (Ft from their 3 terms).
For A1 binomial coefficients must be correct & evaluated.
7C1a b ( x ) 7C 2 a b ( x )
6 5 2 2 6 5 2
7a b 21a b
a 5 A1 OE
=
b 9
5
(a) Find an expression for f −1 x. [3]
+
2 2 2x
3 3 3x − 1 3x − 1
(b) Show that can be expressed as . [2]
(c) State the range of f. [1]
tan2 1
−
1 cos2 1 cos2 1
Mark scheme
6(a) 2x *M1 For 1st two operations. Condone a sign error
y= → 3xy − y = 2 x → 3xy − 2 x = y (or − y = 2 x − 3xy )
3x − 1
y −y DM1 For 2nd two operations. Condone a sign error
x (3 y − 2) = y → x = (or x = )
3y − 2 2 − 3y
A1
( f ( x ) ) = 3xx− 2
−1
( ) 2 −−x3x
Allow f −1 ( x ) =
3
6(b) 2 ( 3 x − 1) + 2 6x 2x B1 B1 AG, WWW
= = First B1 is for a correct single unsimplified fraction.
3 ( 3 x − 1) 3 ( 3 x − 1) 3 x − 1
An intermediate step needs to be shown. Equivalent methods
accepted.
2
6(c) 2 B1 2 2
(f(x)) > Allow (y) > . Do not allow x >
3 3 3
1
The first and second terms of an arithmetic progression are and , respectively, where
0 < 1 < 12 π.
(a) Show that the common difference is −
1 cos4 1
. [4]
(b) Find the exact value of the 13th term when 1 = 16 π. [3]
The equation of a curve is y = 2x + 1 + for x > − 12 .
1
2x + 1
Mark scheme
7(a) 2
tan θ 1 B1 Allow sign error(s). Award only at form (d =)... stage
( d =) − −
cos 2θ cos 2θ
sin 2θ 1 −sec2θ M1 Allow sign error(s). Can imply B1
− − or
cos 4θ cos 2θ cos 2θ
1 M1
−
−sin 2θ − cos 2θ cos 2θ
or
cos 4θ cos 2θ
1 A1 AG, WWW
−
cos 4θ
4
7(b) 4 16 B1 1 1
a= , d =− SOI, both required. Allow a = ,d= −
3 9 3 9
4 16
1 12 4 −16 M1 Use of correct formula with their a and their d. The first 2 steps
u13 = − = + 12 could be reversed
cos 2θ cos 4θ 3 9
‒20 A1 WWW
3
dy d2 y
(a) Find and 2 . [3]
dx dx
(b) Find the coordinates of the stationary point and determine the nature of the stationary point. [5]
Mark scheme
8(a) dy B1 B1
= [ 2]
−2
[− 2 ( 2 x + 1) ]
dx
d2 y −3 B1
= 8 ( 2 x + 1)
dx 2
3
8(b) dy M1
Set their = 0 and attempt solution
dx
(2x + 1)2 = 1 → 2x + 1 = ( ± ) 1 or 4x2 + 4x = 0 → (4)x(x + 1) = 0 M1 Solving as far as x = ....
x=0 A1 WWW. Ignore other solution.
(0, 2) A1 One solution only. Accept x = 0, y = 2 only.
d2 y 1 B1 d2 y
> 0 from a solution x > − hence minimum Ignore other solution. Condone arithmetic slip in value of .
dx 2 2 dx 2
d2 y −3
Their must be of the form k ( 2 x + 1)
dx 2
5
B
12 cm
8 cm
A 8 cm O C
In the diagram, arc AB is part of a circle with centre O and radius 8 cm. Arc BC is part of a circle with centre A and radius 12 cm, where AOC is a straight line.
(a) Find angle BAO in radians. [2]
(b) Find the area of the shaded region. [4]
(c) Find the perimeter of the shaded region. [3]
−1
A curve has equation y = x + x 2 + 2 where x > 0 and k is a positive constant.
1 12 1
Mark scheme
9(a) 6 82 + 122 − 82 M1 Or other correct method
cos BAO = or
8 2 × 8 × 12
BAO = 0.723 A1
2
9(b) 2
Sector ABC = ½ × 12 × their 0.7227 *M1 Accept 52.1
Triangle AOB = ½ × 8 × 12sin ( their 0.7227 ) or ½×12×√28 *M1 or ½ × 8 × 8sin (π − 2 × their 0.7227 ) . Expect 31.7 or 31.8
Shaded area = their 52.0 − their 31.7 = 20.3 DM1 M1 dependent on both previous M marks
A1
4
9(c) Arc BC = 12 × their 0.7227 *M1 Expect 8.67
Perimeter = 8 + 4 + their 8.67 = 20.7 DM1
A1
3
k k
(a) It is given that when x = 14 , the gradient of the curve is 3.
Find the value of k. [4]
k2 @ A
−1
(b) It is given instead that Ô x + x 2 + 2 dx = 13
1 12 1
12
.
1 k2 k k
4
Find the value of k. [5]
Mark scheme
10(a) dy x −1/2 x −3/2 B2, 1, 0 ([0]) implies that more than 2 terms counts as an error
= − + ( [ 0 ])
dx 2k 2
dy 1 1 M1
Sub = 3 when x = Expect 3 = −4
dx 4 k
1 A1
k= (or 0.143)
7
4
10(b) 1 1 2 x 3/2 x B2, 1, 0 OE
x1/2 + x −1/2 + 2 = + 2 x + 2
1/2
k k 3 k k
2k 2 k2 1 M1 k2
+ 2k + 1 − + k + Apply limits → k 2 to an integrated expression.
3 12 4 4
7 2 3
Expect k +k+
12 4
7 2 3 13 M1 13
k +k + = Equate to and simplify to quadratic.
12 4 12 12
OE, expect 7 k 2 + 12k − 4 ( = 0 )
2 A1 Dependent on ( 7k − 2 )( k + 2 ) ( = 0 ) or formula or completing
k= only (or 0.286)
7 square.
5
A circle with centre C has equation x − 82 + y − 42 = 100.
(a) Show that the point T −6, 6 is outside the circle. [3]
Two tangents from T to the circle are drawn.
(b) Show that the angle between one of the tangents and CT is exactly 45Å. [2]
The two tangents touch the circle at A and B.
(c) Find the equation of the line AB, giving your answer in the form y = mx + c. [4]
(d) Find the x-coordinates of A and B. [3]
Additional Page
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
Mark scheme
11(a) ( −6 − 8) + ( 6 − 4 )
2 2 M1 OE
= 200 A1
200 > 10, hence outside circle A1 AG (‘Shown’ not sufficient). Accept equivalents of 200 > 10
Alternative method for question 11(a)
Radius = 10 and C = (8, 4) B1
Min(x) on circle = 8 ‒ 10 = ‒2 M1
Hence outside circle A1 AG
3
11(b) their 10 M1 Allow decimals for 10√2 at this stage. If cosine used, angle ACT
angle = sin −1 or BCT must be identified, or implied by use of 90°‒ 45°.
their 10 2
1 2 10 10 A1 AG Do not allow decimals
angle = sin −1 ( or or or ) = 45º
2 2 10 2 200
Alternative method for question 11(b)
2 M1
(10 2) = 102 + TA2
TA = 10 → 45º A1 AG
2
11(c) 1 B1 OE
Gradient, m, of CT = −
7
Attempt to find mid-point (M) of CT *M1 Expect (1, 5)
Equation of AB is y − 5 = 7 ( x − 1) DM1 1
Through their (1, 5) with gradient −
m
y = 7x − 2 A1
4
11(d) ( x − 8) + ( 7 x − 2 − 4) = 100 or equivalent in terms of y
2 2 M1 Substitute their equation of AB into equation of circle.
50 x 2 − 100 x ( = 0 ) A1
x = 0 and 2 A1 WWW
Alternative method for question 11(d)
7 M1
MC =
−1
1 −1 0 1 1 2 A1
+ = , + =
5 −7 −2 5 7 12
x = 0 and 2 A1
3
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On this page, free: the mark scheme for AS & A Level Mathematics 9709/13, October/November 2020 (9709_w20_ms_13), and the question paper (9709_w20_qp_13), to read or download, with each question's marking points written out underneath.
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Mathematics 9709/13 for October/November 2020 is out of 75 marks, with 1 hour 50 minutes to answer it.
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