Mathematics 9709/11 May/June 2024: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 1 Pure Mathematics 1, May/June 2024. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) Express 3y 2 - 12y - 15 in the form 3 (y + a) 2 + b , where a and b are constants. [2]
(b) Hence find the exact solutions of the equation 3x 4 - 12x 2 - 15 = 0 . [3]
Mark scheme
1(a) 3( y − 2) − 27 or a = −2, b = −27 B1 B1
2
2
1(b)
( x − 2 ) = 9 leading to x − 2 = ±3 M1 Must be x 2 unless substitution is clear.
2 2 2
x 2 = − 1 or x 2 = 5 M1 Allow omission of -1 if ±3 seen.
x = ± 5 A1 B1 SC if M1M1 not awarded. Ignore ± i, i, –i, √–1.
Use of calculator with no working scores 0/3.
Alternative method for Question 1(b)
( )( )
3 x 4 – 12 x 2 – 15 = 0 leading to 3 x 2 − 5 x 2 + 1 [ = 0] (M1)
x 2 = − 1 or x 2 = 5 (M1) Allow omission of –1 if factors seen. Factorising or other
valid method.
x = ± 5 (A1) B1 SC if M1M1 not scored. Ignore ± i, i, –i, √–1.
Use of calculator with no working scores 0/3.
3
y
1
O 1 r 3 2r 5 3r 7 x
2r 2r 2r 2r
–1
–2
–3
–4
–5
The diagram shows two curves. One curve has equation y = sin x and the other curve has equation y = f (x) .
(a) In order to transform the curve y = sin x to the curve y = f (x) , the curve y = sin x is first reflected in the x-axis.
Describe fully a sequence of two further transformations which are required. [4]
(b) Find f (x) in terms of sin x. [2]
Mark scheme
2(a) {Stretch}{factor 3}{ in y-direction} B2,1,0 2 out of 3 scores B1.
0 B2,1,0 Accept shift.
{Translation}
−2
Alternative Method for Question 2(a)
0 (B2,1,0) 2 out of 3 scores B1.
{Translation} 2 Accept shift.
− 3
{Stretch}{factor 3}{ in y-direction} (B2,1,0)
4
2(b) f ( x ) = {−3sin x}{−2 } B1 B1 No marks awarded if extra terms seen.
2
The coefficient of x 3 in the expansion of (3 + ax) 6 is 160.
(a) Find the value of the constant a. [2]
(b) Hence find the coefficient of x 3 in the expansion of (3 + ax) 6 (1 - 2x) . [3]
Mark scheme
3(a) 20 × 27 × a3 [ = 160] [ ]
M1 Allow 6C3 33 ×a3 =160 .
×
3
Accept 540a with no other working for M1.
2 A1 Allow 0.667 AWRT.
[a] = 2
3 SC B1 is a = with no other working.
3
2
3(b) 2 B1 FT May be in a list. 6C2 and 34 must be evaluated but may be
2
Coefficient of x is 15 × 81× their [ = 540]
2
implied by later work. Condone 540 with no working.
3
160 ×1 − 2 ×their 540 M1
= −920 A1 Condone − 920 x 3 .
3
The equation of a curve is y = f (x) , where f (x) = (2x - 1) 3x - 2 - 2 . The following points lie on the curve. Non-exact values have been given correct to 5 decimal places.
A(2, 4), B(2.0001, k), C(2.001, 4.00625), D(2.01, 4.06261), E(2.1, 4.63566), F(3, 11.22876)
(a) Find the value of k. Give your answer correct to 5 decimal places. [1]
The table shows the gradients of the chords AB, AC, AD and AF.
Chord AB AC AD AE AF
Gradient of
6.2501 6.2511 6.2608 7.2288 chord
(b) Find the gradient of the chord AE. Give your answer correct to 4 decimal places. [1]
(c) Deduce the value of f l (2) using the values in the table. [1]
sin 2 x - cos x - 1
Mark scheme
4(a) [k] = 4.00063 B1 CAO
1
4(b) [Gradient AE] = 6.3566 B1 CAO
1
4(c) Suggests that f' ( 2) = 6.25 B1 CAO
1
(a) Prove the identity / - cos x . [3]
1 + cos x
sin 2 x - cos x - 1 1
(b) Hence solve the equation = for 0° G x G 360° . [3]
2 + 2 cos x 4
Mark scheme
5(a) sin 2 x − cos x − 1 1 − cos 2 x − cos x − 1 − cos 2 x − cos x M1 For use of sin 2 x + cos 2 x = 1 .
= or
1 + cos x 1 + cos x 1 + cos x Allow use of s, c, t or omission of x throughout.
− cos x (1 + cos x ) M1 For factorising.
=
1 + cos x
= −cos x A1
3
5(b) 1 1 M1
− cos x = ⇒ x = cos − 1 − 1
2 4 2
x = 120° or x = 240° A1
A1 FT FT for 360 – their answer. A1 A0 if extra solution(s) in
2π 4π
range. SC B1 if answer in radians for both , .
3 3
3
y
O x
2
The function f is defined by f (x) = + 4 for x 1 0 . The diagram shows the graph of y = f (x) . x2
(a) On this diagram, sketch the graph of y = f -1 (x) . Show any relevant mirror line. [2]
(b) Find an expression for f -1 (x) . [3]
(c) Solve the equation f (x) = 4.5 . [1]
(d) Explain why the equation f -1 (x) = f (x) has no solution. [1]
Mark scheme
6(a) B1 For curve in correct quadrant.
B1 Fully correct including line y = x .
Horizontal asymptote closer to x axis than vertical
asymptote is to y axis.
2
6(b) 2 M1 Allow x and y swapped around.
x=
2
+ 4 leading to y
2
( x − 4) = 2 or y2 = x − 4
y 2
( )
2 2 M1
y2 = leading to y = [ ± ] 2
or x = [ ±]
( x − 4) x−4 y−4
2 A1
f −1 ( x ) = −
x−4
3
6(c) [ x] =−2 B1
1
6(d) Because f −1 is always negative and f is always positive or curves do not B1 Accept other correct answers e.g. ‘f is only defined for
intersect positive values of x and f–1 is only defined for negative
values of x’ or ‘domains do not overlap’ or ‘the y values
cannot be the same’ or ‘the x values cannot be the same’.
1
B C
i rad
A 15 cm O 10 cm D
In the diagram, AOD and BC are two parallel straight lines. Arc AB is part of a circle with centre O and radius 15 cm . Angle BOA = i radians. Arc CD is part of a circle with centre O and radius 10 cm .
Angle COD = 12 r radians.
(a) Show that i = 0.7297 , correct to 4 decimal places. [1]
(b) Find the perimeter and the area of the shape ABCD. Give your answers correct to 3 significant figures. [7]
Mark scheme
7(a) π 10 10 B1 Condone working in degrees if converted to radians at the
Angle θ = − cos −1 or sin −1 = 0.7297 end.
2 15 15
AG
1
7(b) BC = 15 2 − 10 2 = 11.18... or 5 √ 5 B1
[
Arc AB = 15×0.7297 =10.9455 ] B1
Perimeter = their BC + their arc AB + 25 + 5π M1
Perimeter = 62.8 A1 AWRT
1 B1
Area sector AOB = × 152 × 0.7297 [ = 82.09]
2
1 π M1
Area = × 10 × their BC + their sector AOB + × 10 2
2 4
Area = 217 A1 AWRT
7
(a) The first three terms of an arithmetic progression are 25, 4p - 1 and 13 - p , where p is a constant.
Find the value of the tenth term of the progression. [4]
(b) The first three terms of a geometric progression are 25, 4q - 1 and 13 - q , where q is a positive constant.
Find the sum to infinity of the progression. [4]
Mark scheme
8(a) 2( 4p −1) = 25 +13− p *M1
40 A1
p=
9
40 74 DM1 Using their p to find d.
d = 4× − 1 − 25 = −
9 9
10th term = 25 + 9d = −49 A1
Alternative Method for first 3 marks of Question 8(a)
d = 4 p − 26 , d = 14 − 5 p , p + 2d = -12 Any two (*M1) Allow unsimplified or equivalent.
Solving simultaneously to find p or d (DM1)
40 , 74 (A1)
p = 9
d =−
9
4
8(b)
( 4q − 1)2 = 25(13 − q ) ⇒ 16q2 +17q − 324[ = 0] M1
( q − 4)(16q +81)[ = 0] leading to [q = 4] M1 Solve 3 term quadratic with real solutions.
3 A1 −17
[ r =] Ignore .
5 20
25 125 A1 Ignore extra solution.
Sum to infinity = = SC B1 if no method shown for solving quadratic.
3 2
1−
5
Alternative Method for Question 8(b)
25r = 4q – 1, 25r2 = 13 – q leading to 100r 2 + 25r – 51 = 0 (M1)
( 5r − 3)( 20r +17) = 0 (M1) Solve 3 term quadratic with real solutions.
3 (A1) −17
r= Ignore .
5 20
25 125 (A1) Ignore extra solution.
Sum to infinity = = SC B1 if no method shown for solving quadratic.
3 2
1−
5
4
y
1
(1, 1)
O 2.4 x
1
The diagram shows part of the curve with equation y = 1 and the lines x = 2.4 and y = 1. The
( 5x - 4 ) 3 curve intersects the line y = 1 at the point (1, 1).
Find the exact volume of the solid generated when the shaded region is rotated through 360° about the x-axis. [6]
Mark scheme
9 7 7 B1 2.4
Volume of cylinder = π × 12 × = π
5 5
May be done using 1. This would be the only mark
1
available if candidate integrates y.
1 M1 No further marks available if y .
Volume under curve = [ π ] (5x − 4) 2
dx
3
3 1
B1 B1 Calculator used for integration scores no further marks.
= [ π] ( 5x − 4) 3
5
3
1
M1 Uses limits 1, 2.4 in an integral of y2.
= [ π ] 8 3 − 1 = π
3
5
5
7 3 4 A1 SC B1 if the only error is not showing substitution.
Volume = π− π = π
5 5 5
6
10
( x − 3)2 + y2 = 18 y = mx − 9 leading to ( x − 3)2 + ( mx − 9)2 = 18 M1 Finding equation of tangent and substituting into circle
equation.
Must be mx −9 .
x 2 − 6 x + 9 + m 2 x 2 − 18 mx + 81 = 18 leading to M1 Brackets expanded and all terms collected on one side of
( m + 1) x − ( 6 + 18m ) x + 72[= 0]
2 2 the equation. May be implied in the discriminant. m
cannot be numeric.
( 6 + 18m ) − 4 ( m 2 + 1) × 72 [ = 0] *M1 Use of b 2 − 4 ac . Not in quadratic formula. m cannot be
2
numeric, c must be numeric.
36m2 + 216m − 252[ = 0] leading to m 2 + 6m − 7 = 0 DM1 Simplifies to 3 term quadratic.
m = 1 or m =−7 A1 Condone no method for solving quadratic shown.
m = 1 leading to 2 x 2 − 24 x + 72 = 0 leading to x = 6 DM1 Must be correct x for their quadratic.
6 DM1 Must be correct x for their quadratic.
m = −7 leading to 50 x 2 + 120 x + 72 = 0 leading to x = −
5
A1
( 6, − 3 ) , −
6 3
,−
5 5
The equation of a circle is (x - 3) 2 + y 2 = 18 . The line with equation y = mx + c passes through the point (0, - 9) and is a tangent to the circle.
Find the two possible values of m and, for each value of m, find the coordinates of the point at which the tangent touches the circle. [8]
Mark scheme
10 Alternative Method 1 for first 4 marks of Question 10
3m − 1( 0 ) − 9 (M1) Use of the formula for the length of a perpendicular from
a point to a line.
m2 + 1
3m − 1( 0 ) − 9 (M1) Equates length of a perpendicular from a point to a line to
= 18 the radius.
m2 + 1
(3 m – 9)2 = 18( m 2 + 1) (M1) Squares and clears the fraction.
9 m 2 - 54 m + 81 = 0 leading to m 2 + 6 m − 7 = 0 (M1)
Alternative Method 2 for first 3 marks of Question 10
(3 – x )(9 + 6 x - x 2 )-1/2 = m (M1) OE
dy
Differentiates implicitly or otherwise and equates to
dx
m.
2 (M1) Brackets expanded and all terms collected on one side of
( 1 + m 2 ) x 2 – 6(1 + m ) x + 9(1 – m 2 )[ = 0]
the equation. May be implied in the discriminant.
36( 1 + m 2 )2 – 4( 1 + m 2 ) × 9(1 – m 2 )[ = 0] (M1) Use of b 2 − 4 ac .
8
y
O x
4 3
A function is defined by f (x) = - + 2 for x ! 0 . The graph of y = f (x) is shown in the diagram. x3 x
(a) Find the set of values of x for which f (x) is decreasing. [5]
(b) A triangle is bounded by the y-axis, the normal to the curve at the point where x = 1 and the tangent to the curve at the point where x =- 1.
Find the area of the triangle. Give your answer correct to 3 significant figures. [8]
Additional page
If you use the following page to complete the answer to any question, the question number must be clearly shown.
Mark scheme
11(a) dy 12 3 B1
=− 4 + 2
dx x x
dy 12 3 M1 Set = 0 or uses < , ⩽ and simplifies.
= − 4 + 2 = 0 leading to 3 x 4 − 12 x 2 = 0 or -12 + 3x2 = 0
dx x x dy A B
Must be from = + .
dx x 4 x 2
( )
3 x 2 x 2 − 4 = 0 leading to x =±2 only A1 SC B1 for x =±2 if M0 scored.
−2 < x < 0 and 0 < x < 2 or (-2, 0) and (0, 2) or −2 < x < 2 and x ≠ 0 B1FT Allow and/or.
B1FT Allow −2 x < 0 and / or 0 < x 2 but only B1B0 if 0
included in either or both.
Allow [–2, 0) and (0, 2].
Allow B1B0 for −2 < x < 2 or (–2, 2).
dy A B
Must be from = 4 + 2 .
dx x x
5 dy A B
B marks only available if = 4+ 2.
dx x x
11(b) [At x = 1] y = 3 and m tan = − 9 *M1 dy
Using their .
dx
1 1 DM1
m norm = − =
−9 9
A1
Equation of normal is y − 3 = 1 ( x − 1) leading to y = 1 x + 26
9 9 9
At x = −1, y = 1, m = −9 M1
[
Equation of tangent is y −1 =−9( x +1) leading to y =−9x − 8 ] A1
M1 Equates their tangent and their normal.
Meet when 1 x + 26 = − 9 x − 8 leading to x = − 1.19512 , − 49
9 9 41
M1 If y 2 − y1 is used integration must be correct and
Area = 1 × their 1 .19512 × their 26 + 8
2 9 substitution shown.
6.51 A1 AWRT
Accept fraction wrt 6.51
8
Checked your answers against these? Let AI mark the whole paper, on your own working.
Examiners follow the generic marking principles at the front of the scheme, then each question's points: method (M), accuracy (A) and independent (B) marks, error carried forward where the scheme allows it, and positive marking, so you're credited for what you got right.
Check your answers against it
Done this paper? Upload your script and CIE Buddy's AI marker marks it question by question against this exact scheme, puts the ticks on your own working and says what each lost mark needed.
Where can I find the 9709/11 May/June 2024 mark scheme?+
On this page, free: the mark scheme for AS & A Level Mathematics 9709/11, May/June 2024 (9709_s24_ms_11), and the question paper (9709_s24_qp_11), to read or download, with each question's marking points written out underneath.
How many marks is 9709/11 out of?+
Mathematics 9709/11 for May/June 2024 is out of 75 marks, with 1 hour 50 minutes to answer it.
Can I get my 9709/11 answers marked?+
Yes. Upload your answers to 9709/11/M/J/24 as a PDF or photos, and CIE Buddy marks every question against this paper's official mark scheme, using Cambridge's marking rules. You get your score, each lost mark explained on your own script, and the topics to revise. It's a guide, not an official Cambridge grade.