Mathematics 9709/12 May/June 2021: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 1 Pure Mathematics 1, May/June 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) Express 16x2 − 24x + 10 in the form 4x + a2 + b. [2]
(b) It is given that the equation 16x2 − 24x + 10 = k, where k is a constant, has exactly one root.
Find the value of this root. [2]
Mark scheme
1(a) (4x – 3)2 or (4x +(– 3))2 or a = −3 B1 k(4x – 3)2 where k ≠ 1 scores B0 but mark final answer, allow
recovery.
+ 1 or b = 1 B1
2
1(b) [For one root] k = 1 or ‘their b’ B1 FT Either by inspection or solving or from
242 – 4 × 16 × (10 – k) = 0 WWW
3 B1 SC B2 for correct final answer WWW.
[ Root or x =] or 0.75
4
2
(a) The graph of y = f x is transformed to the graph of y = 2f x − 1.
Describe fully the two single transformations which have been combined to give the resulting transformation. [3]
(b) The curve y = sin 2x − 5x is reflected in the y-axis and then stretched by scale factor 13 in the x-direction.
Write down the equation of the transformed curve. [2]
Mark scheme
2(a) 1 B1 Allow shift and allow by 1 in x-direction or [parallel to/on/in/
Translation along/against] the x-axis or horizontally.
0 ‘Translation by 1 to the right’ only, scores B0
Stretch B1 Stretch. SC B2 for amplitude doubled.
Factor 2 in y-direction B1 With/by factor 2 in y-direction or [parallel to/on/in/along/against]
the y-axis or vertically or with x axis invariant
‘With/by factor 2 upwards’ only, scores B0.
Accept SF as an abbreviation for scale factor.
3 Note: Transformations can be in either order
2(b) [−sin 6x][ + 15x] or [sin(−6x)][ + 15x] OE B1 B1 Accept an unsimplified version. ISW.
B1 for each correct component – square brackets indicate each
required component.
If B0, SC B1 for either sin(−2x) + 5x or –sin(2x) + 5x or
2 5
sin 6x – 15x or sin − x + x
3 3
2
The equation of a curve is y = x − 3 x + 1 + 3. The following points lie on the curve. Non-exact values are rounded to 4 decimal places.
A 2, k B 2.9, 2.8025 C 2.99, 2.9800 D 2.999, 2.9980 E 3, 3
(a) Find k, giving your answer correct to 4 decimal places. [1]
(b) Find the gradient of AE, giving your answer correct to 4 decimal places. [1]
The gradients of BE, CE and DE , rounded to 4 decimal places, are 1.9748, 1.9975 and 1.9997 respectively.
(c) State, giving a reason for your answer, what the values of the four gradients suggest about the gradient of the curve at the point E. [2]
@ A
10 3 1
Mark scheme
3(a) 1.2679 B1 AWRT. ISW if correct answer seen. 3 – 3 scores B0
1
3(b) 1.7321 B1 AWRT. ISW if correct answer seen.
1
3(c) Sight of 2 or 2.0000 or two in reference to the gradient *B1
This is because the gradient at E is the limit of the gradients of the DB1 Allow it gets nearer/approaches/tends/almost/approximately 2
chords as the x-value tends to 3 or ꝺx tends to 0.
2
The coefficient of x in the expansion of 4x + is p. The coefficient of in the expansion of x x
@ A5 k
2x + 2 is q. x
Given that p = 6q, find the possible values of k. [5]
Mark scheme
4 [Coefficient of x or p =] 480 B1 SOI. Allow 480x even in an expansion.
1 3 k
2 M1 Appropriate term identified and selected.
Term in x or q = [10 ×] ( 2 x ) x 2
[10 × 23k2 =] 80k2 A1 80k 2
Allow
x
p = 6q used (480 = 6 × 80k2 or 80 = 80k2) M1 1
Correct link used for their coefficient of x and (p and q) with
x
no x’s.
[k2 = 1 ⇒] k = ±1 A1 A0 if a range of values given. Do not allow ± 1 .
5
The function f is defined by f x = 2x2 + 3 for x ≥ 0.
(a) Find and simplify an expression for ff x. [2]
(b) Solve the equation ff x = 34x2 + 19. [4]
Mark scheme
5(a) ff(x) = 2(2x2 + 3)2 + 3 M1 Condone = 0.
8x4 + 24x2 + 21 A1 ISW if correct answer seen. Condone = 0.
2
5(b) 8x4 + 24x2 + 21 = 34x2 + 19 ⇒ 8x4 + 24x2 – 34x2 + 21 – 19 [= 0] M1 Equating 34x3 + 19 to their 3-term ff(x) and collect all terms on
one side condone ± sign errors.
8x4 – 10x2 + 2[= 0] A1
[2](x2 – 1)(4x2 – 1) M1 Attempt to solve 3-term quartic or 3-term quadratic by
factorisation, formula or completing the square or factor theorem.
2 1 1 A1 If factorising, factors must expand to give 8x4 or 4x4 4 or their ax4
x = 1 or 4 leading to x = 1 or x = 2 otherwise M0A0 due to calculator use.
1 1
Condone ±1, ± but not or 1 .
2 4
4
Points A and B have coordinates 8, 3 and p, q respectively. The equation of the perpendicular bisector of AB is y = −2x + 4.
Find the values of p and q. [4]
Mark scheme
6 1 B1 SOI
Gradient AB =
2
1 *M1 Equating given perpendicular bisector with the line through
Lines meet when −2 x + 4 = ( x − 8) + 3 (8, 3) using their gradient of AB (but not -2) and solving.
2
Solving as far as x = Expect x = 2, y = 0.
Using mid-point to get as far as p = or q = DM1 8+ p 3+ q
Expect = 2 or =0
2 2
p = −4, q = −3 A1 Allow coordinates of B are (−4, −3).
Alternative method for Question 6
1 B1 SOI
Gradient AB =
2
q−3 1 *M1 Equating gradient of AB with their gradient of AB (but not -2) and
= [leading to 2q = p − 2] , using mid-point in equation of perpendicular bisector.
p −8 2
q+3 8+ p
= −2 +4 [leading to q = −11 − 2 p ]
2 2
Solving simultaneously their 2 linear equations DM1 Equating and solving 2 correct equations as far as p = or q = .
p = −4, q = −3 A1 Allow coordinates of B are (−4, −3).
6 Alternative method for Question 6
1 B1
Gradient AB =
2
q −3 1 *M1 Equating gradient of AB with their gradient of AB (but not -2) and
= [leading to p = 2q + 2] , using mid-point in equation of perpendicular bisector.
p −8 2
q+3 5q + 23
y− = −2 ( x − ( q + 5 ) ) leading to y = −2 x +
2 2
5q + 23 DM1 Equating and solving as far as q or p =
their =4q =
2
p = −4, q = −3 A1 Allow coordinates of B are (−4, −3).
4
The point A has coordinates 1, 5 and the line l has gradient − 23 and passes through A. A circle has
centre 5, 11 and radius 52.
(a) Show that l is the tangent to the circle at A. [2]
(b) Find the equation of the other circle of radius 52 for which l is also the tangent at A. [3]
Mark scheme
7(a) 11 − 5 M1 For substituting (1,5) into circle equation or showing
(5 – 1)2 + (11 – 5)2 = 52 or
5 −1 3
gradient = .
2
For both circle equation and gradient, and proving line is A1 Clear reasoning.
perpendicular and stating that A lies on the circle
Alternative method for Question 7(a)
2 M1 Both equations seen and attempt to solve.
( x − 5) + ( y − 11) = 52 and y − 5 = − ( x − 1)
2 2
3 2 17
May see y = − x +
3 3
Solving simultaneously to obtain (y – 5)2 = 0 or (x – 1)2 = 0 ⇒ 1 A1 Clear reasoning.
root or tangent or discriminant = 0 ⇒ 1 root or tangent
Alternative method for Question 7(a)
dy 10 − 2 x 10 − 2 M1 Attempting implicit differentiation of circle equation and
= = substitute x = 1 and y = 5.
dx 2 y − 22 10 − 22
2 A1 Clear reasoning.
Showing gradient of circle at A is −
3
2
7(b) Centre is (−3, −1) B1 B1 B1 for each correct co-ordinate.
Equation is (x + 3)2 + (y + 1)2 = 52 B1 FT FT their centre, but not if either (1, 5) or (5, 11). Do not accept
522 .
3
The first, second and third terms of an arithmetic progression are a, 32 a and b respectively, where a and b are positive constants. The first, second and third terms of a geometric progression are a, 18 and b + 3 respectively.
(a) Find the values of a and b. [5]
(b) Find the sum of the first 20 terms of the arithmetic progression. [3]
Mark scheme
8(a) 3 B1 SOI
a + b = 2 × a b = 2a
2
182 = a(b + 3) OE or 2 correct statements about r from the GP, B1 SOI
18 b+3
e.g. r = and b + 3 = 18r or r 2 =
a a
324 = a(2a + 3) ⇒ 2a2 + 3a – 324[= 0] M1 Using the correct connection between AP and GP to form a
or 3-term quadratic with all terms on one side.
b2 + 3b – 648[= 0]
or
6r2 – r – 12[= 0]
or
4d2 + 3d – 162[= 0]
(a – 12)(2a + 27)[= 0] M1 Solving their 3-term quadratic by factorisation, formula or
or completing the square to obtain answers for a, b, r or d.
( b − 24 )( b + 27 ) [ = 0]
or
( 2r − 3)( 3r + 4 ) [ = 0]
or
( d − 6 )( 4d + 27 ) [ = 0]
a = 12, b = 24 A1 WWW. Condone extra ‘solution’ a = −13.5, b = −27 only.
5
8(b) Common difference d = 6 B1 FT a
SOI. FT their
2
20 M1 Using correct sum formula with their a, their calculated d and 20.
S20 = ( 2 × 12 + 19 × 6 )
2
1380 A1
3
y
y2 = x − 2
1
x
0 5
The diagram shows part of the curve with equation y2 = x − 2 and the lines x = 5 and y = 1. The shaded region enclosed by the curve and the lines is rotated through 360Å about the x-axis.
Find the volume obtained. [6]
1 + sin x 1 − sin x 4 tan x
Mark scheme
9 Curve intersects y = 1 at (3, 1) B1 Throughout Question 9: 1 < their 3 < 5
Sight of x = 3
Volume = [ π ] ( x − 2 ) [ dx ] M1 M1 for showing the intention to integrate ( x − 2 ) .
Condone missing π or using 2 π.
A1 Correct integral. Condone missing π or using 2 π.
[ π ] x 2 − 2 x or [ π ] ( x − 2 )2
1 1
2 2
52 their 32 M1 Correct use of ‘their 3’ and 5 in an integrated expression.
= [ π ] − 2 × 5 − − 2 × their 3 Condone missing π or using 2 π. Condone +c.
2 2 Can be obtained by integrating and substituting between 5 and 2
5 3 and then 3 and 2 then subtracting.
= [ π ] + as a minimum requirement for their values
2 2
Volume of cylinder = π × 12 × ( 5 − their 3) [ = 2π ] B1 FT Or by integrating 1 to obtain x (condone y if 5 and their 3 used).
[Volume of solid = 4 π − 2π =] 2π or 6.28 A1 AWRT
9 Alternative method for Question 9
Curve intersects y = 1 at (3, 1) B1 Sight of x = 3
Volume of solid = π ( x − 2 ) − 1[ dx ] M1 B1 M1 for showing the intention to integrate ( x − 2 )
B1 for correct integration of –1.
Condone missing π or 2 π for M1 but not for B1.
A1 Correct integral, allow as two integrals.
[ π ] x 2 − 3x or [ π ] ( x − 3)2
1 1
Condone missing π or using 2 π.
2 2
52 their 32 M1 Correct use of ‘their 3’ and 5 in an integrated expression.
= [ π ] − 3 × 5 − − 3 × their 3 Condone missing π or using 2 π. Condone +c.
2 2 Can be obtained by integrating and substituting between 5 and 2
and then 3 and 2 then subtracting.
[Volume of solid = 4 π − 2π =] 2π or 6.28 A1 AWRT
6
(a) Prove the identity − . [4]
1 − sin x 1 + sin x cos x
1 + sin x 1 − sin x
(b) Hence solve the equation − = 8 tan x for 0 ≤ x ≤ 12 π. [3]
1 − sin x 1 + sin x
dy
Mark scheme
10(a) 1 + sinx 1 − sinx (1 + sinx ) − (1 − sinx )
2 2 *M1 For using a common denominator of (1 − sinx )(1 + sinx ) and
− ≡
1 − sinx 1 + sinx (1 − sinx )(1 + sinx ) reasonable attempt at the numerator(s).
≡
(
1 + 2sinx + sin 2 x − 1 − 2sinx + sin 2 x ) DM1 For multiplying out the numerators correctly. Condone sign errors
for this mark.
(1 − sinx )(1 + sinx )
4sinx 4sinx DM1 For simplifying denominator to cos 2 x .
≡ ≡
1 − sin x cos 2 x
2
4sinx 4tanx A1 AG.
≡ ≡ Do not award A1 if undefined notation such as s, c, t or missing
cosxcosx cosx
x’s used throughout or brackets are missing.
Alternative method for Question 10(a)
4tanx 4sinx 4sinx *M1 sinx
≡ ≡ Using tanx = and cos2 x = 1 − sin 2 x
cosx cos x 1 − sin 2 x
2
cosx
−2 2 DM1 Separating into partial fractions.
≡ +
1 + sinx 1 − sinx
−2 2 DM1 Use of 1-1 or similar
≡1 + + −1
1 + sinx 1 − sinx
1 − sinx 1 + sinx A1
≡− +
1 + sinx 1 − sinx
4
10(b) 1 *B1 OE. WWW.
cosx =
2
π DB1 Or AWRT 1.05
x=
3
x = 0 from tanx = 0 or sinx = 0 B1 π
WWW. Condone extra solutions outside the domain 0 to but
2
B0 if any inside.
3
The gradient of a curve is given by = 6 3x − 53 − kx2 , where k is a constant. The curve has a dx stationary point at 2, −3.5.
(a) Find the value of k. [2]
(b) Find the equation of the curve. [4]
d2 y
(c) Find . [2]
dx2
(d) Determine the nature of the stationary point at 2, −3.5. [2]
Mark scheme
11(a) dy M1 dy
= 0 so 6 ( 3 × 2 − 5) − k × 22 = 0
3
At stationary point Setting given = 0 and substituting x = 2 into it.
dx dx
3 A1 OE
[k = ]
2
2
11(b) 6 1 *M1 dy
[ y =] ( 3x − 5)4 − kx3 [+c] . A1FT Integrating (increase of power by 1 in at least one term) given dx
4×3 3
1 1
. Expect ( 3x − 5 ) 4 − x3 .
2 2
FT their non zero k.
7 1 1 3 DM1 Using (2,-3.5) in an integrated expression. + c needed.
− = ( 3 × 2 − 5) − × × 23 + c [leading to −3.5 + c = −3.5 ]
4
2 2 3 2 Substitution needs to be seen, simply stating c = 0 is DM0.
1 1 A1 y = or f(x)= must be seen somewhere in solution.
y= ( 3x − 5 )4 − x3
2 2
11(b) Alternative method for Question 11(b)
81 4 541 3 x3 *M1 dy
[ y =] x − x + 675 x 2 − 750 x ( +c ) or -270 x3 − k
3 2 2
A1 FT From dx = 162 x − 810 x − kx − 1350 x − 750 . FT their k
2 2 3
7 81 541 3 DM1 Using (2, -3.5) in an integrated expression. + c needed
− = × 24 − × 2 + 675 × 22 − 750 × 2 + c
2 2 2
81 4 541 3 625 A1 y = or f(x)= must be seen somewhere in solution.
y= x − x + 675 x 2 − 750 x +
2 2 2
4
11(c) B2,1,0 FT FT their k.
[3 ×] 18 ( 3x − 5)2 [ −2kx ] Square brackets indicate each required component.
B2 for fully correct, B1 for one error or one missing component,
B0 for 2 or more errors.
Alternative method for Question 11(c)
486 x 2 − 1623 x + 1350 or −1620 x − 2kx B2,1,0 FT FT their k.
B2 for fully correct, B1 for one error, B0 for 2 or more errors.
2
11(d) d2 y M1 OE. Substituting x = 2 into their second differential or other valid
[At x = 2] 2 = 54 ( 3 × 2 − 5 ) − 4k or 48
2
method.
dx
[> 0] Minimum A1 WWW
2
Q
P
A B
F C
E D
The diagram shows a cross-section of seven cylindrical pipes, each of radius 20 cm, held together by a thin rope which is wrapped tightly around the pipes. The centres of the six outer pipes are A, B, C, D,
E and F. Points P and Q are situated where straight sections of the rope meet the pipe with centre A.
(a) Show that angle PAQ = 13 π radians. [2]
(b) Find the length of the rope. [4]
(c) Find the area of the hexagon ABCDEF, giving your answer in terms of 3. [2]
(d) Find the area of the complete region enclosed by the rope. [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
12(a) ˆ = 2π], [ PAQ
[By symmetry] [6 × PAQ ˆ =] 2π÷6, M1
Explaining that there are six sectors around the diagram that make A1 AG
up a complete circle.
Alternative method for Question 12(a)
Using area or circumference of circle centre A ÷ 6 M1 400π 40π
or
6 6
Justification for dividing by 6 followed by comparison with the A1 AG
sector area or arc length.
Alternative method for Question 12(a)
Explain why ∆PAQ is an equilateral triangle M1 Assumption of this scores M0
ˆ = π A1 AG
Using ∆PAQ is an equilateral triangle ⸫ PAQ
3
Alternative method for Question 12(a)
2π M1
Using the internal angle of a regular hexagon =
3
ˆ + OAB
ˆ = 2π
Or FAO , equilateral triangles
3
A1 AG
ˆ = 2π — π + 2π + π = π
PAQ
2 3 2 3
12(a) Alternative method for Question 12(a)
20 M1
Sinθ = , with θ clearly identified
40
π π ˆ
ˆ and by similar triangles = PAQ A1 AG
θ = , 2θ = = FAO
6 3
2
12(b) Each straight section of rope has length 40 cm B1 SOI
π *M1 Use of rθ with r = 20 and θ in radians
Each curved section round each pipe has length rθ = 20 ×
3
Total length = 6 × ( ( their 40 ) + kπ ) DM1 6 × (their straight section + their curved section).
Their curved section must be from acceptable use of rθ – this
could now be numeric.
240 + 40π or 366 (AWRT) (cm) A1 Or directly: (6 × diameter) + circumference
4
12(c) 1 π 1 B1
[Triangle area =] × 40 × 40 × sin or × 40 × 20 3 or
2 3
2
400 3 or 693(AWRT)
[Total area of hexagon = 6 × 400 3 =] 2400 3 B1 3
Condone 4800
2
Alternative method for Question 12(c)
1 π B1
[Trapezium area =] × ( 40 + 80 ) × 40sin or 1200 3 or 2080
2 3
(AWRT)
[Total area of hexagon = 2 × 1200 3 =] 2400 3 B1 √3
Condone 4800
2
Alternative method for Question 12(c)
Area of triangle ABC = 400 3 or 693 (AWRT) or B1
4 × Area of half of triangle ABC = 4 ×200 3 or 1390 (AWRT)
or Area of rectangle ABDE = 1600 3 or 2770 (AWRT)
[Total area of hexagon = 2 × 400 3 +1600 3 =] 2400 3 B1 √3
Condone 4800
Or [= 4 ×200 3 +1600 =] 2400 3 2
If B0B0, SC B1 can be scored for sight of 4160 (AWRT) as final
answer.
2
12(d) Each rectangle area = 40 × 20 (= 800) B1 SOI, e.g. by sight of 4800
1 2 1 π 200π B1 SOI.
Each sector area = r θ = × 202 × =
2 2 3 3
Total area = 2400 3 + 4800 + 400π or 10 200 (cm2) (AWRT) B1 Or directly: part (c) + 6800 + area circle radius 20.
3
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Mathematics 9709/12 for May/June 2021 is out of 75 marks, with 1 hour 50 minutes to answer it.
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