Mathematics 9709/11 October/November 2023: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 1 Pure Mathematics 1, October/November 2023. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) Expand 1 + 3x6 in ascending powers of x up to, and including, the term in x2 . [2]
(b) Hence find the coefficient of x2 in the expansion of 1 − 7x + x2 1 + 3x6 . [2]
2 A line has equation y = 2cx + 3 and a curve has equation y = cx2 + 3x − c, where c is a constant.
Showing all necessary working, determine which of the following statements is correct.
A The line and curve intersect only for a particular set of values of c.
B The line and curve intersect for all values of c.
C The line and curve do not intersect for any values of c. [4]
Mark scheme
1(a) 1 + 18 x + 135 x 2 B2, 1, 0 Accept 1, 18 x, 135 x 2 listed horizontally or vertically
or 1x0 + 18x + 135x 2 .
2
1(b) Coefficient of x 2 is 135 − 7 ×18 + 1 = 10 M1 A1 3 products, allow 10x 2 .
If full expansion given, like terms must be collected
for M1.
2
2
( ) (
cx 2 + 3x − c = 2cx + 3 leading to cx + 3 − 2c x − c + 3
2
) [= 0] M1 Forming a 3-term quadratic, all terms on one side.
b2 − 4ac = (3 − 2c ) + 4c ( c + 3) M1 2nd M1 for b2 − 4ac correct for their a, b, c
2
i.e. no sign errors.
= 8c2 + 9 A1
> 0 [for all values of c] leading to B [Intersects for all values of c] A1 WWW
4
x
x x
The diagram shows a cubical closed container made of a thin elastic material which is filled with water and frozen. During the freezing process the length, x cm, of each edge of the container increases at the constant rate of 0.01 cm per minute. The volume of the container at time t minutes is V cm3.
Find the rate of increase of V when x = 20. [3]
Mark scheme
3 dV B1 SOI
= 3x 2
dx
dV dV dx M1 Correct use of chain rule with x = 20 substituted into
= × = 3 × 202 × 0.01
dt dx dt dV
.
dx
12 A1
3
The@ transformation
A R denotes a reflection in the x-axis and the transformation T denotes a translation
3 of .
−1
(a) Find the equation, y = g x, of the curve with equation y = x2 after it has been transformed by the sequence of transformations R followed by T. [2]
(b) Find the equation, y = h x, of the curve with equation y = x2 after it has been transformed by the sequence of transformations T followed by R. [2]
(c) State fully the transformation that maps the curve y = g x onto the curve y = h x. [2]
Mark scheme
4(a) − ( x − 3) −1
2 B1 B1 OE. Must be a quadratic e.g. 3x – 1 B0 B0.
SC B1 for correct use of generalised function notation.
2
4(b) − ( x − 3) +1
2 B1 B1 OE. Must be a quadratic.
SC B1 for correct use of generalised function notation.
2
4(c) 0 B2, 1, 0 FT from (a) and (b) if a translation parallel to the y
Translation axis.
2 B2 for fully correct, B1 with two elements correct. {}
indicates different elements.
2
(a) Show that the equation
5 2
4 sin x + + =0 tan x sin x may be expressed in the form a cos2 x + b cos x + c = 0, where a, b and c are integers to be found.
[3]
5 2
(b) Hence solve the equation 4 sin x + + = 0 for 0Å ≤ x ≤ 360Å. [3]
tan x sin x
Mark scheme
5(a) 4sin2 x + 5cos x + 2 [= 0] *M1 Multiply by sin x (or writing as a single fraction) and
sin x
using tan x = .
cos x
( )
4 1 − cos2 x + 5cos x + 2 [ = 0] DM1 Correctly obtaining a quadratic in cos x (allow sign
errors).
4cos 2 x − 5cos x − 6 = 0 A1 Condone missing x.
Must be = 0 unless 0 appears on RHS earlier.
3
5(b) (4cos x + 3)(cos x − 2) [ = 0] M1 Or use of formula or completing square.
138.6º, 221.4º A1 B1 FT FT on 360º ‒ 1st solution from quadratic in cos x.
Use of radians (2.42) A0 but allow B1 FT for 2π: 1st
solution if use of radians is clear.
SC If M0 scored SC B1 B1 for correct final answer(s).
If extra incorrect solutions in the range 0 → 360º are
given award A1 B0.
3
O
r
A r B
C
The diagram shows a motif formed by the major arc AB of a circle with radius r and centre O, and the minor arc AOB of a circle, also with radius r but with centre C. The point C lies on the circle with centre O.
(a) Given that angle ACB = kπ radians, state the value of the fraction k. [1]
(b) State the perimeter of the shaded motif in terms of π and r. [1]
(c) Find the area of the shaded motif, giving your answer in terms of π, r and 3. [5]
Mark scheme
6(a) 2 B1 2π
k= Allow ACB = .
3 3
1
6(b) Perimeter of shaded area = 2πr B1
1
6(c) 1 2 4π *M1 Expect 2 πr 2 . Finds area of any relevant sector or
Major sector OAB = r × 3
2 3 triangle. Can be embedded in segment formula.
One or both segments = [ 2] × 1 r 2 × π − 1 r 2 sin π
*M1
2 3 2 3
π 3 A1
= [ 2] r 2 − r 2
6 4
2 1 r2 3 DM1
Shaded area = πr 2 − 2 πr 2 −
3 6 4
πr 2 r 2 3 A1
= +
3 2
6(c) Alternative method for Question 6(c)
1 1 *M1 1 2
Sector CAOB = [ 2] × r 2their π Expect [2] × πr .
2 3 6
Can be embedded in segment formula.
One or both segments = [ 2] × 1 r 2 × π − 1 r 2 sin π
*M1
2 3 2 3
π 3 A1
= [ 2] r 2 − r 2
6 4
1 2 2 π 2 3 DM1
Shaded area = πr − πr + 2 r
2 −r
3 6 4
πr 2 r 2 3 A1
= +
3 2
6(c) Alternative method for Question 6(c)
1 π M1
Area of rhombus AOBC = [ 2] × r 2 sin Expect [2] ×
3
.
2 3 4
Can be embedded in segment formula.
One or both segments = [ 2] × 1 r 2 × π − 1 r 2 sin π
M1
2 3 2 3
π 3 A1
= [ 2] r 2 − r 2
6 4
3 2 π 3 DM1
Shaded area = πr 2 − r − 4 r 2 − r 2
2 6 4
πr 2 r 2 3 A1
= +
3 2
5
The sum of the first two terms of a geometric progression is 15 and the sum to infinity is 125
7
. The common ratio of the progression is negative.
Find the third term of the progression. [7]
Mark scheme
7 a (1 + r ) = 15 B1
Accept
(
a 1− r2 ) = 15 for first B1.
1− r
a 125 B1
=
1− r 7
125 M1 Eliminate a.
(1 − r )(1 + r ) = 15
7
105 M1
1− r2 =
125
4 2 A1 2 2
r2 = leading to r = − Condone or ± .
25 5 5 5
125 7 A1 Ignore 2nd answer.
a= × = 25
7 5
4 A1 CAO
3rd term = 25 × =4
25
7 Alternative method for Question 7
a (1 + r ) = 15 B1
a 125 B1
=
1− r 7
7(15-15r) = (125 – 125r)(1 – r2) M1
125r3 – 125r2 – 20r +20 = 0 M1
−2 2 A1 Condone extra ‘answer’ of r = 1.
r= [1, ]
5 5
a = 25 A1 Ignore 2nd answer.
3rd term = 4 A1 CAO
7 Alternative method for Question 7
a (1 + r ) = 15 B1
a 125 B1
=
1− r 7
a 125 M1 Eliminate r.
=
15 7
1− − 1
a
7a2 – 250a + 1875 [= 0] M1
a = 25, 75 Condone extra ‘answer’ of r = 75 .
A1
7 7
−2 A1 Ignore 2nd answer.
r=
5
4 A1 CAO
3rd term = 25 × =4
25
7
y
y = 2x − 32 + 1
A B
y = 2 2x − 34
x
O
The diagram shows the curves with equations y = 2 2x − 34 and y = 2x − 32 + 1 meeting at points
A and B.
(a) By using the substitution u = 2x − 3 find, by calculation, the coordinates of A and B. [4]
(b) Find the exact area of the shaded region. [5]
Mark scheme
8(a) u = 2x − 3 leading to 2u 4 = u 2 + 1 leading to 2u4 − u2 −1[= 0] B1
( 2u + 1)(u − 1)[= 0]
2 2 M1 Factors or formula or completing square must be
shown.
u = ±1 leading to 2x − 3 = ±1 leading to x = 1 or 2 A1
(1, 2), (2, 2) A1 Special case: If B1 M0 scored then SC B2 can be
awarded for correct coordinates or SC B1 for correct x
values only.
Special case
2(2x – 3)4 = (2x – 3)2 + 1
32x4 – 192x3 + 428x2 – 420x + 152 = 0
x = 1, 2 finding both from a correct quartic SC B1
(1, 2), (2, 2) SC DB1
Special case: Trial and improvement without quartic.
Both x values correct B1, both coordinates correct B2.
4
8(b) ( 2 x − 3)3 2 ( 2 x − 3)5 B1 B1 Integrate the 2 functions.
+ x [ −]
3 × 2 5 × 2
1 1 1 1 M1 Apply their limits 1 → 2 (must be shown) to an
+ 2 − − + 1 − − − integral.
6 6 5 5 Some evidence of substitution. Minimum
13 5 1 1
( − ) – ( + ) or equivalent.
6 6 5 5
Allow 1 sign error for 1st M1.
4 2 M1 Subtract (at some point) the 2 areas.
− Must subtract areas and not just integrals.
3 5
14 A1 Special case: If M0 for substitution of limits can
15 award SC B1 for correct answer.
14
Condone − if corrected.
15
If subtraction is the wrong way round award B1 B1
M1 M1 A0.
y 2 dx or x dy scores 0 /5.
π y dx used. Award B1 B1 M1 M1 A0.
8(b) Alternative method for Question 8(b)
u = 2x – 3 B2,1,0
( )
u 2 + 1 − 2u 4 du
1 1 3 2 5
u + u − u
2 3 5
1 1 2 −1 2 M1 Applies limits –1 → 1.
+ 1 − − −1 +
2 3 5 3 5
M1 Subtract (at some point) the 2 areas.
1 14 14 A1
+
2 15 15
14
15
5
(a) Express 4x2 − 12x + 13 in the form 2x + a2 + b, where a and b are constants. [2]
The function f is defined by f x = 4x2 − 12x + 13 for p < x < q, where p and q are constants. The function g is defined by g x = 3x + 1 for x < 8.
(b) Given that it is possible to form the composite function gf, find the least possible value of p and the greatest possible value of q. [3]
(c) Find an expression for gf x. [1]
The function h is defined by h x = 4x2 − 12x + 13 for x < 0.
(d) Find an expression for h−1 x. [3]
d2 y
Mark scheme
9(a)
( 2x − 3)2 + 4 B1 B1 Or a = ‒3, b = 4.
2
their ( 2x − 3) + 4[<]8 OR 4 x − 12 x + 13[<]8
9(b) 2 2 *M1 Linking quadratic with 8.
( 2x − 3)2 < 4 leading to −2 < 2x − 3 < 2 DM1 Simplify to 3-term quadratic and solve. Condone no
method shown.
OR 4 x 2 − 12 x + 5 < 0 leading to ( 2 x − 1)( 2 x − 5 ) < 0
1 1 1 1 A1
< x < 2 leading to [LEAST] p = , [GREATEST] q = 2
2 2 2 2
3
9(c) gf(x) = 12 x 2 − 36 x + 40 ( )
B1 OE gf(x) = 3 2x − 3 2 + 13 .
1
9(d) y = ( 2x − 3) + 4 leading to ( 2x − 3) = y − 4 leading to *M1
2 2
2 x − 3 = [ ±] y − 4
2x = 3[±] y − 4 leading to x = 3 [ ± ] y−4 DM1
2 2
3 x−4 A1
h −1 ( x ) = −
2 2
3
A curve has a stationary point at 2, −10 and is such that = 6x. dx2
dy
(a) Find . [3]
dx
(b) Find the equation of the curve. [3]
(c) Find the coordinates of the other stationary point and determine its nature. [3]
(d) Find the equation of the tangent to the curve at the point where the curve crosses the y-axis. [2]
Mark scheme
10(a) dy B1
= 3x 2 [ +c ]
dx
3 × 22 + c = 0 M1 dy
Substitute x = 2 and = 0 into an integral (c must be
dx
present).
dy A1
= 3x 2 −12
dx
3
10(b) y = x3 −12x [+k ] B1 FT FT on their non-zero c (dependent on c being found at
some stage).
−10 = 23 − 12 × 2 + k M1 Substitute x = 2, y = ‒10 (k present).
y = x3 − 12 x + 6 A1
Must be y = (unless y = x − 12 x + k stated earlier).
3
3
10(c) 3x 2 − 12 = 0 [leading to x = −2 ] M1 dy
Set their two term = 0 . Expect x = −2 .
dx
Ignore x = 2 given in addition.
y = ( −2) − 12 × ( −2) + 6 = 22 leading to (‒2, 22) A1
3
2 A1 dy
When x = −2, d y2 < 0 (or ‒12) hence Maximum Can be from correct conclusion from sign diagram
dx dx
dy
if calculated correctly.
dx
Do not allow concave downward for final A1.
Can be awarded if the only error is incorrect or missing
y-coordinate.
3
10(d) dy M1 dy
At x = 0, = −12, y = 6 Both required. FT on their and y.
dx dx
y − 6 = −12 x A1 OE
2
y x − 42 + y + 12 = 40
A
x
O
B
The diagram shows the circle with equation x − 42 + y + 12 = 40. Parallel tangents, each with gradient 1, touch the circle at points A and B.
(a) Find the equation of the line AB, giving the answer in the form y = mx + c. [3]
(b) Find the coordinates of A, giving each coordinate in surd form. [4]
(c) Find the equation of the tangent at A, giving the answer in the form y = mx + c, where c is in surd form. [2]
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) Gradient of AB = ‒1 B1 SOI
Centre of circle = (4, ‒1) B1 SOI
Equation of AB is y + 1 = −1( x − 4 ) leading to y = − x + 3 B1 FT FT their centre with gradient ‒1.
3
11(b)
( x − 4)2 + ( −x + 3 + 1)2 = 40 *M1 Substitute their AB into circle equation.
2 ( x − 4) = 40 OR [2] ( x 2 − 8 x − 4 ) leading to DM1 Forming and solving 3-term quadratic.
2
8 ± 64 + 16 16 ± 256 + 64
or
2 4
x = 4[±] 20 A1 OE. No fractions.
( 4 − 20 , −1 + 20 ) A1 OE
Special case: If M1 M0 scored then SCB2 can be
awarded for correct coordinates or SCB1 for correct x
values only. Ignore other coordinate
4
11(c) (
y − their −1 + ) (
20 = 1 x − their 4 − 20 ) M1 OE
y = x − 5 + 2 20 or y = x − 5 + 80 or y = x − 5 + 4 5 A1
2
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Mathematics 9709/11 for October/November 2023 is out of 75 marks, with 1 hour 50 minutes to answer it.
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