Mathematics 9709/12 October/November 2021: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 1 Pure Mathematics 1, October/November 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
Solve the equation 2 cos 1 = 7 − for −90Å < 1 < 90Å. [4]
cos 1
Mark scheme
1 2cos 2θ − 7 cos θ + 3[ = 0] M1 Forming a 3-term quadratic expression with all terms on the
same side or correctly set up prior to completing the square.
Allow ± sign errors.
( 2cosθ − 1)( cosθ − 3) = 0 DM1 Solving their 3-term quadratic using factorisation, formula or
completing the square.
1 A1
[cos θ = or cos θ = 3 leading to] θ = −60° or θ = 60°
2
θ = −60° and θ = 60° A1 FT π
FT for ± same answer between 0 ° and 90 ° or 0 and .
2
π
± or ± 1.05 AWRT scores maximum M1M1A0A1FT.
3
Special case: If M1 DM0 scored then SC B1 for
θ = −60° or θ = 60° , and SC B1 FT can be awarded for
± ( their 60°) .
4
The graph of y = f x is transformed to the graph of y = f 2x − 3.
(a) Describe fully the two single transformations that have been combined to give the resulting transformation. [3]
The point P 5, 6 lies on the transformed curve y = f 2x − 3.
(b) State the coordinates of the corresponding point on the original curve y = f x. [2]
Mark scheme
2(a) 1 B1
Stretch with [scale factor] either ± 2 or ±
2
1 B1
Scale factor in the x-direction
2
0 B1
Translation or translation of 3 units in negative y-direction
−3
3
2(b) ( 10 , 9) B1 B1 B1 for each correct co-ordinate.
2
The function f is defined as follows: x+3 f x = for x > 1. x−1
(a) Find the value of ff 5. [2]
(b) Find an expression for f −1 x. [3]
dy 8
Mark scheme
3(a) 2 + 3 M1 Clear evidence of applying f twice with x = 5.
f ( 5 ) = [ 2] and f ( their 2 ) = [5] OR ff ( 5 ) =
2 −1
x+3
+3
OR x − 1 and an attempt to substitute x =5.
x+3
−1
x −1
5 A1
2
3(b) x+3 y+3 *M1 Setting f(x) = y or swapping x and y, clearing of fractions and
= y x + 3 = xy − y OR = x y + 3 = xy − x expanding brackets. Allow ± sign errors.
x −1 y −1
y+3 x + 3 DM1 Finding x or y = . Allow ± sign errors.
xy − x = y + 3 x = OE OR y + 3 = xy − x y = OE
y −1 x −1
x+3 A1 4
[f −1 ( x ) or y ] = OE e.g. 1 + etc. Must be a function of x, cannot be x = .
x −1 x −1
3
The first, third and fifth terms of an arithmetic progression are 2 cos x, −6 3 sin x and 10 cos x respectively, where 21 π < x < π.
(a) Find the exact value of x. [3]
(b) Hence find the exact sum of the first 25 terms of the progression. [3]
Mark scheme
5(a) [(3rd term – 1st term) = (5th term – 3rd term) leading to…] *M1 OE. From the given terms, obtain 2 expressions relating to
−6 3 sin x − 2cos x = 10cos x + 6 3 sin x the common difference of the arithmetic progression, attempt
to solve them simultaneously and achieve an equation just
leading to − 12 3 sin x = 12cos x
involving sinx and cosx.
OR
[(1st term + 5th term) = 2 × 3rd term leading to…] 12cos x = −12 3 sin x
Elimination of sinx and cosx to give an expression in tanx DM1 sinx
For use of = tanx
1 cosx
tanx = −
3
5π A1 CAO. Must be exact.
[ x =] only
6
3
5(b) d = 2cosx or d = 2cos(their x) B1 FT Or an equivalent expression involving sinx and cosx e.g.
−3 3sin ( their x ) − cos ( their x ) = − 3
FT for their x from (a) only. If not ± 3 , must see
unevaluated form.
25 M1 25
S25 =
2
(
2 × ( 2cos ( their x ) ) + ( 25 − 1) × ( their d ) ) Using the correct sum formula with
2
, (25 — 1) and with
( ( )
= 12.5 2 × − 3 + 24 − 3
( ))
a replaced by either 2(cos(their x)) or ± 3 and
d replaced by either 2(cos(their x)) or ± 3 .
−325 3 A1 Must be exact.
3
The second term of a geometric progression is 54 and the sum to infinity of the progression is 243.
The common ratio is greater than 21 .
Find the tenth term, giving your answer in exact form. [5]
Mark scheme
6 a or their a B1 SOI
ar = 54 and = 243
1− r
54 *M1 Forming a 3-term quadratic expression in r or a using their
= 243 (1 − r ) leading to 243r 2 − 243r + 54 [ = 0] [ 9r 2 − 9r + 2 = 0] 2nd term and S∞. Allow ± sign errors.
r
OR a 2 − 243a + 13122 [ = 0]
k ( 3r − 2 )( 3r − 1) [ = 0] OR ( a − 81)( a − 162 ) [ = 0] DM1 Solving their 3-term quadratic using factorisation, formula or
completing the square. If factorising, factors must expand to
give ±their coefficient of r 2 .
2 DM1 May be implied by final answer.
54 ÷ their = a OR 54 ÷ ( their 81) = r
3
512 A1 OE. Must be exact.
2
9 8
2
Tenth term = OR 81 × OR 54 × Special case: If B1M1DM0DM1 scored then SC B1 can be
243 3 3 awarded for the correct final answer.
5
15 c m P B
Q
A
9 cm
15 cm
C
In the diagram the lengths of AB and AC are both 15 cm. The point P is the foot of the perpendicular from C to AB. The length CP = 9 cm. An arc of a circle with centre B passes through C and meets
AB at Q.
(a) Show that angle ABC = 1.25 radians, correct to 3 significant figures. [2]
(b) Calculate the area of the shaded region which is bounded by the arc CQ and the lines CP and
PQ. [4]
Mark scheme
7(a) EITHER M1
ˆ = 0.6435 … and ABC
By using trigonometry: BAC ˆ = π − 0.6435
2
OR
By Pythagoras: AP = 12 ⇒ BP = 3 so tan ABC ˆ = 9 3
= 0.9486…
10
= 0.3162…
3 10 10
OR
Using ∆PBC and either the sine or cosine rule
ˆ = 3 or cos ABC
sin ABC ˆ = 10
10 10
A1 AG. Final answer must be 1.25, more accurate value
ˆ = π − 0.6435 or tan-1 9 or sin −1 3 or cos −1 10 or
ABC 1.24904… with no rounding to 3sf seen as the final answer
2 3 10 10
gets M1A0.
1.249 ( 04…) or 71.56° = 1.25 radians (3 sf) If decimals are used all values must be given to at least 4sf
for A1.
2
7(b) 9 M1 Using correct method(s) to find BC.
BC = ( their 3)2 + 92 or [= 90 , 3 10 or 9.48697…]
sin1.25
1 M1 Using tan −1 3 or 1.25 and their BC, but not 9 or 15, in
× ( their BC ) × tan −1 3 [ = 56.207 or 56.25]
2
Area of sector =
2 correct area of sector formula.
1 B1
Area of triangle PBC = 13.4 to 13.6 or ×9×3
2
[Area = ( 56.207 or 56.25 ) – their 13.5 =] 42.7 or 42.8 A1 AWRT
4
(a) It is given that in the expansion of 4 + 2x 2 − ax5 , the coefficient of x2 is −15.
Find the possible values of a. [4]
(b) It is given instead that in the expansion of 4 + 2x 2 − ax5 , the coefficient of x2 is k. It is also given that there is only one value of a which leads to this value of k.
Find the values of k and a. [4]
Mark scheme
8(a) Terms required for x2: −5 × 24 × ax + 10 × 23 × a 2 x 2 = −80ax + 80a 2 x 2 B1 Can be seen as part of an expansion or in correct products.
2× (±their coefficient of x) + 4× (±their coefficient of x2) *M1
x2 coefficient is 320a2 – 160a = −15 DM1 Forming a 3-term quadratic in a, with all terms on the same
64a 2 − 32a + 3 ( 8a − 3)( 8a − 1) side or correctly setting up prior to completing the square and
solving using factorisation, formula or completing the square.
If factorising, factors must expand to give their coefficient of
a2 .
1 3 A1 OE.
a= or a = Special case: If DM0 for solving quadratic, SC B1 can be
8 8
awarded for correct final answers.
4
8(b) 320a2 – 160a = k 320a 2 − 160a − k [ = 0] M1 Forming a 3-term quadratic in a with all terms on the same
side. Allow ± sign errors.
Their b2 – 4ac [= 0], [1602 – 4×320×(–k) = 0] M1 Any use of discriminant on a 3-term quadratic.
k = –20 A1
1 B1 1
a= Condone a = from k = 20.
4 4
Alternative method for question 8(b)
a k M1 Allow ± sign errors.
320a2 – 160a = k and divide by 320 a 2 − =
2 320
1
2
1 k M1 1
2
Attempt to complete the square a − − = Must have a −
4 16 320 4
1 A1
a=
4
k = –20 B1
8(b) cont’d Alternative method for question 8(b)
320a2 – 160a = k and attempt to differentiate LHS [ 640a − 160] M1 Allow ± sign errors.
Setting their ( 640a − 160 ) = 0 and attempt to solve. M1
1 A1
a=
4
k = –20 B1
4
The volume V m3 of a large circular mound of iron ore of radius r m is modelled by the equation
3
V = 32 r − 12 − 1 for r ≥ 2. Iron ore is added to the mound at a constant rate of 1.5 m3 per second.
(a) Find the rate at which the radius of the mound is increasing at the instant when the radius is 5.5 m.
[3]
(b) Find the volume of the mound at the instant when the radius is increasing at 0.1 m per second.
[3]
k
Mark scheme
9(a) dV 9 1
2 B1 OE. Accept unsimplified.
dr = 2 r − 2
M1 Correct use of chain rule with 1.5, their differentiated
dV
dr dr dV 1.5 = 1.5 1.5 expression for and using r = 5.5 .
= × = = dr
dt dV dt their dV 9 1 112.5
2
dr 5.5 −
2 2
3 1 A1
0.0133 or or [ metres per second]
225 75
3
9(b) B1 FT dV dV
Correct statement involving or their , 1.5 and 0.1.
dr dr
dV
or their
dV 1.5
= or 15 OR 0.1 =
1.5 = 2 × 1.5 OE
dr dr 0.1 dV 1
2
their 9 r −
dr 2
9 1
2
1 10 B1 OE e.g. AWRT 2.3
r − = 15 r = + Can be implied by correct volume.
2 2 2 3
[Volume =] 8.13 AWRT B1 −3 + 5 30
OE e.g. . CAO.
3
3
The function f is defined by f x = x2 + + 2 for x > 0. x
(a) Given that the curve with equation y = f x has a stationary point when x = 2, find k. [3]
(b) Determine the nature of the stationary point. [2]
(c) Given that this is the only stationary point of the curve, find the range of f. [2]
Mark scheme
10(a) k B1
[f '(x) =] 2 x −
x2
k M1 Setting their 2-term f '(2) = 0, at least one term correct and
f '(2) = 0 2 × 2 − 2 = 0 k = … attempting to solve as far as k = .
2
k = 16 A1
3
10(b) 2k M1 Evaluate a two term f''(2) with at least one term correct.
f''(2) = e.g. 2+ Or other valid method.
23
2k A1 FT WWW. FT on positive k value.
2 + 23 > 0 ⇒ minimum or = 6 ⇒ minimum
2
10(c) When x = 2, f(x) = 14 B1 SOI
[Range is or y or f(x)] ⩾ their f(2) B1 FT Not x ⩾ their f(2)
2
y
1 y = 12 x + 10
7
− 1 x − 2 3
A 3, 65
x
O 5
2
7 − 1
The diagram shows the line x = 52 , part of the curve y = 12 x + 10 1 and the normal to the curve x − 2 3
at the point A 3, 65 .
(a) Find the x-coordinate of the point where the normal to the curve meets the x-axis. [5]
(b) Find the area of the shaded region, giving your answer correct to 2 decimal places. [6]
Mark scheme
11(a) dy 1 1 B1 OE. Allow unsimplified.
= + 4
dx 2
3( x − 2) 3
*M1 Substituting x = 3 into their differentiated expression –
dy 1 1 5 defined by one of 3 original terms with correct power of x.
Attempt at evaluating their at x = 3 + =
dx 2 4
6
3(3 − 2) 3
−1 6 *DM1 dy
Gradient of normal = =− Negative reciprocal of their evaluated .
dy 5 dx
their
dx
6 DM1 Using their normal gradient and A in the equation of a
Equation of normal y − = ( their normal gradient )( x − 3) straight line.
5
Dependent on *M1 and *DM1.
6
y = − 5 x + 4.8 5 y = −6 x + 24
[When y = 0,] x = 4 A1 or (4, 0)
5
11(b) M1 For intention to integrate the curve (no need for limits).
1 7 1
Area under curve = x + − dx
1 [ ] Condone inclusion of π for this mark.
2 10
( x − 2 ) 3
2 A1 For correct integral. Allow unsimplified.
1 2 7 3( x − 2) 3 Condone inclusion of π for this mark.
x + x−
4 10 2
2
M1 Clear substitution of 3 and 2.5 into their integrated
9 3 6.25 3 × 0.5 3 expression (with at least one correct term) and subtracting.
+ 2.1 − − + 1.75 −
4 2 4 2
0.48[24] A1 If M1A1M0 scored then SC B1 can be awarded for correct
answer.
[Area of triangle =] 0.6 B1 OE
[Total area =] 1.08 A1 Dependent on the first M1 and WWW.
6
y
B
P
A x
O
R
Q
The diagram shows the circle with equation x2 + y2 − 6x + 4y − 27 = 0 and the tangent to the circle at the point P 5, 4.
(a) The tangent to the circle at P meets the x-axis at A and the y-axis at B.
Find the area of triangle OAB, where O is the origin. [5]
(b) Points Q and R also lie on the circle, such that PQR is an equilateral triangle.
Find the exact area of triangle PQR. [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) Centre is (3, – 2) B1
( their −2 ) − 4 [= *M1 Finding gradient using their centre (not (0, 0)) and P (5,4).
Gradient of radius = 3]
( their 3) − 5
1 DM1 Using P and the negative reciprocal of their gradient to find
Equation of tangent y − 4 = − ( x − 5) the equation of AB.
3
17 A1
Sight of [x =]17 and [y =]
3
1 17 289 A1 1
Or 48 or AWRT 48.2.
Area = 2 × 3 × 17 = 6 6
Alternative method for question 12(a)
dy dy B1
2x + 2 y −6+4 =0
dx dx
dy dy dy 1 *M1 Find the gradient using P (5,4) in their implicit differential
At P: 10 + 8 − 6 + 4 = 0 = − (with at least one correctly differentiated y term).
dx dx dx 3
1 DM1 Using P and their value for the gradient to find the equation
Equation of tangent y − 4 = − ( x − 5) of AB.
3
17 A1
Sight of [x =]17 and [y =]
3
1 17 289 A1 1
Or 48 or AWRT 48.2.
Area = 2 × 3 × 17 = 6 6
12(a) Alternative method for question 12(a)
cont’d
1
dy 1 B1 OE. Correct differentiation of rearranged equation.
( )
y = −2 ± 40 − ( x − 3) 2 OE leading to
2
dx
(
−
= ( 3 − x ) 31 + 6 x − x 2 2 )
1 *M1 Find the gradient using x = 5 in their differential (with clear
dy
dx
( ) 2 − dy 1
= ( 3 − 5 ) 31 + 6 ( 5 ) − ( 5 ) 2 = −
dx 3
use of chain rule).
1 DM1 Using P and their value for the gradient to find the equation
Equation of tangent y − 4 = − ( x − 5) of AB.
3
17 A1
Sight of [x =]17 and [y =]
3
1 17 289 A1 1
Or 48 or AWRT 48.2.
Area = 2 × 3 × 17 = 6 6
5
12(b) Radius of circle = 40 , B1 Or 2 10 or 6.32 AWRT or r 2 = 40 .
1 1 3 M1 1 2
Area of ∆CRQ = × (their r ) 2 sin120 = × 40 × Using r sinθ with their r and 120 or 60 [ ×3 ]
2 2
2 2
OR 1
1 1 Using ×base ×height in a correct right-angled triangle
Area of ∆CQX = × 40cos30 × 40cos60 OE = × 30 × 10 2
2 2 [ ×6 ].
OR
π
Area of circle ‒ 3× Area of segment = 40π ‒ 3 × (40 ‒ 10 3)
3
OR
1
QR = 120 or 2 30 and area = QR 2 sin60 Use of cosine rule and area of large triangle
2
30 3 A1 AWRT 52[.0] implies B1M1A0.
3 See diagram for points stated in ‘Answer’ column.
C
X
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Mathematics 9709/12 for October/November 2021 is out of 75 marks, with 1 hour 50 minutes to answer it.
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