Mathematics 9709/13 May/June 2025: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 1 Pure Mathematics 1, May/June 2025. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
A curve has equation y = 2x + . x2
Find the equation of the tangent to the curve at the point (-2, -1). Give your answer in the form y = mx + c . [4]
Mark scheme
1 2 + (12 )( −2 ) x −3 B1 Correct differential but can be unsimplified.
2 + (12 )( −2 )( −2 )
−3
[ = 5] *M1 Substitute x = −2 into their differential, which must contain
x−3 .
y − ( −1) DM1 Attempt to find equation of tangent through ( −2, − 1) with
Either ( their 5 ) =
x − ( −2 ) their numerical gradient obtained as described above.
or −1 = ( their 5) × ( −2 ) + c c =
y = 5x + 9 A1
4
The first two terms of a geometric progression are
4 sin 2 i , 8 sin 3 i ,
where i is an angle such that 0 1 i 1 16 r .
Given that the sum to infinity of the progression is 12 , find the value of i . Give your answer in the form sin -1 k , where k is a rational number. [4]
y e(4x -a 3) + 2o dx = 12, find the value of the constant a.
3
Mark scheme
2 8sin 3 θ B1 A correct unsimplified expression for r, this may only be seen
r= [ = 2 sin θ ] in a correct S∞
4 sin 2 θ
4 sin 2 θ 1 *M1 OE
= Correct S∞ with their expression for r.
1 − 2 sin θ 2
Or
8 sin 3 θ 4 sin 2 θ 1 B1
r= and =
4 sin θ 2
1− r 2
8sin 3 θ *M1
1 − 8sin 2 θ =
4 sin θ
2
Then
Expect 8sin 2 θ + 2 sin θ − 1 [ = 0] DM1 OE and attempt to solve.
Attempt to solve a three-term quadratic to find at least one
value of sin θ .
This can be implied by a final answer of θ = 0.253 c AWRT
1 A1 OE
[θ =] sin −1
, 0.25 Clearly identify one answer only.
4
1 1
Accept sin θ = k = etc.
4 4
4
2 12 160 160 160 may be replaced by 12.65 AWRT and 32 by 5.657
Or AWRT.
1 Working in degrees should give EAF ˆ = 53.13° which
2 EF 32 2 converts to the correct value in radians.
=
−1
2 sin = or other valid method.
AE 160 10
ˆ = 0.927295…=] 0.9273
[ EAF A1 AG
If decimals are used, at least one of 0.32175.., 0.6435 or
0.92729.. must be seen in the first method and 5.6569 and
12.649 or 0.46364 seen in the second method for the correct
level of accuracy.
Note: EAFˆ = 53.14° does not convert to the correct answer,
so A0.
Alternative Method for Question 8(a)
1 M1 Attempt a complete valid method for finding angle EAF.
122 − ( 24 + 24 + 32 ) = ( theirAE × their AF ) sin EAF
ˆ
Area of ABCD – (Area ∆s ADF, ABE &CEF) = Area ∆AEF.
2
2 12 − ( 24 + 24 + 32 )
2
ˆ = sin −1 their 4 .
EAF
ˆ =
sin EAF 5
( theirAE × their AF ) 3
2 122 − ( 24 + 24 + 32 ) Using the cosine rule should lead to cos −1 .
5
ˆ
EAF = sin −1
( theirAE × their AF )
1 A1 AG
EF
ˆ −1 4 2
EAF = sin 5 = AE
2
3 1 −1
B1 OE
a(4 x − 3) + 2 x Do not accept ( −2 + 1) as equivalent to −1.
−1× 4
Apply correct limits, x = 3 & 1, to their integral *M1 Their integral must contain ( 4 x − 3)−1 .
Condone using x = 1 and 3.
−a −a 8a DM1 OE
+6− + 2 = 12 + 4 = 12 Equate their linear unsimplified expression in a to 12.
36 4 36
a = 36 A1
4
(a) Find the first three terms in the expansion of b2 - 32 xl in ascending powers of x. [3]
(b) Use your answer to part (a), with a suitable value of x, to find an approximation to 1.985 5 . [3]
Mark scheme
4(a) 32 x −120 x+180 x
0 2 B1B1B1 Coefficients must be simplified.
Ignore further terms in the expansion.
15
SC B3: 25 1 − x +
45 2
x . can be awarded full marks,
4 8
ISW.
3
4(b) 1 B1 Identify 0.01 as the only value to substitute, condone any
x = 0.01 or method.
100
(Their 32 − 120 x + 180 x ) with x = ( their 0.01)
2 M1 Clear use of their x-value from their solution of
3
2 − x = 1.985 and their answer to (a).
2
Expect 32 – 120(0.01) + 180(0.01)2 = 32 – 1.2+ 0.018.
Condone substitution into further terms in the expansion.
30.8[18] A1 Correct answer to at least 3sf coming from substitution into
three or more terms.
‘Correct’ answers without appropriate working get A0.
3
Solve the equation
4 sin i tan i = 1 + 5 cos i
for - 180° 1 i 1 180° . [6]
Mark scheme
5 sin θ M1
Replace tan θ with
cos θ
Replace sin 2 θ with 1 − cos2 θ leading to a 3-term quadratic *M1 To obtain a 3-term quadratic in cos θ .
Accept a cubic that has a common factor of cos θ .
Condone +/–sign errors during simplification.
9 cos 2 θ + cos θ − 4 [ = 0] A1 OE
Attempt to solve (their quadratic in cos θ ) using a valid method DM1 Only available for solution of a three-term quadratic.
Correct values of cosθ are 0.6134 and – 0.7245.
Any two of ±52.2, ± 136.4 A1 AWRT as final answers.
Any two correct answers between −180° and 180° as their
final answers.
All four values A1 Condone other answers outside the range but no others
between −180° and 180°.
SC after M1 *M1 A1 DM0, correct answers can score B1 B1.
6
An arithmetic progression has first term a and common difference 2. The N th term is 55 and the sum of the first 3N terms is 5760.
Find the values of N and a. [6]
dy
Mark scheme
6 a + 2 ( N − 1) = 55 B1 SOI
1 B1 SOI
× 3N ( 2a + 2 ( 3N − 1) ) = 5760
2
Eliminate either a or N M1 Using a valid substitution, expect either
( 57 − a )
a = 57 − 2 N or N = .
2
Condone +/–sign errors in their substitution or during
simplification.
N 2 + 56N −1920 or a − 226a + 1953 [ = 0 ] A1 All terms need not be collected on one side.
2
N = 24 , a = 9 A1 A1 Ignore presence of extra solutions for N and a.
6
A curve is such that = 3x 2 + 10x - 8 . dx
(a) Find the set of values of x for which y decreases as x increases. [3]
(b) It is given that the maximum point of the curve has y-coordinate 27.
Find the equation of the curve. [4]
Mark scheme
7(a) 2 *B1 OE
5 49
3x + 10x − 8 < 0 or 3 x + −
2
<0 Condone = or ⩽ 0, but use of > or ⩾ can only score B0B1B0.
3 3 If there is no < , = or ⩽ 0, then a correct final answer implies
< 0 here.
2 B1
−4 and
3
2 DB1 FT OE
−4 < x <
3 2
Use of ⩽ sign(s) gets DB0. Condone x > −4, x < ,
3
2 2
x > −4 and x < but not x > −4 or x < .
3 3
FT on their critical values.
3
7(b) Identify x = −4 as the x-value of the maximum point. B1 FT SOI
FT on their lower critical value from part (a).
3 10 B1 Correct integration of the three terms given.
[ y =] x + 3
x 2 − 8 x [ +c ]
Condone f ( x ) = .
3 2
2 M1 2
Use of y = 27, x = their −4 or in their integral. Their integral must be a cubic, their −4 or must come
3 3
from (a) or a restart.
c = −21 y = x + 5 x − 8 x − 21
3 2 A1 Do not ISW if they continue to find a straight line. Condone
omission of final statement if y or f ( x ) = x 3 + 5 x 2 − 8 x + c
has been seen earlier.
4
A 12 cm B
E
12 cm
D F C
The diagram shows a square ABCD where each side has length 12 cm. Points E and F lie on the sides
BC and CD respectively and are such that BE = 13 BC and DF = 13 DC . The arc EF is part of a circle with centre A. The shaded region is bounded by the arc EF and the line segments EC and FC.
(a) Show that the size of angle EAF is 0.9273 radians, correct to 4 significant figures. [2]
(b) Find the perimeter of the shaded region. [3]
(c) Find the area of the shaded region. [3]
Mark scheme
8(a) 1 4 4 12 M1 Attempt a complete valid method for finding angle EAF.
π − 2 tan −1 or sin −1 or cos −1
8(b) 4 B1 OE
[AE or AF =] 160 or Expect AWRT 12.65.
ˆ
sin EAB
[rθ =] (their 12.65) × 0.9273 (
M1 Use of rθ with their 160 .
)
Note: using r = 12 scores M0.
[11.729... + 8 + 8 =] 27.7 AWRT A1
3
8(c) 1 M1 1
( their 12.65) × 0.9273 [= 74.184] r 2θ for area of sector with ( their 12.65 ) .
2
Use
2 2
Note: using r = 12 scores M0.
144 − 24 − 24 − ( 1
2
their12.65 ) × 0.9273
2
M1 Attempt a complete method for finding area of shaded region.
Condone use of r = 12.
[Area =] 21.8 A1 AWRT
Alternative Method for Question 8(c)
1 M1 Area of the segment with ( their 12.65 ) .
( their 12.65) ( 0.9273 − sin 0.9273) [=10.183]
2
2 Note: using r =12 scores M0.
1 M1 Attempt a complete method for finding area of shaded region,
( their12.65) ( 0.9273 − sin 0.9273)
2
32 − condone use of r = 12.
2
[Area =] 21.8 A1 AWRT
3
Three points P, Q and R have coordinates P (-13, 5), Q (5, 1) and R (2, k), where k is a constant. It is given that the angle PRQ is a right angle.
(a) Show that one of the possible values of k is 10, and find the other possible value. [4]
(b) It is now given that k = 10 . A circle passes through the points P, Q and R.
Find the equation of the tangent to the circle at R. Give your answer in the form ax + by + c = 0 , where a, b and c are integers. [5]
9
Mark scheme
9(a) k −5 k −1 B1 Obtain at least one relevant gradient.
Gradient of PR = or gradient of RQ =
2 + 13 2−5
(their gradient of PR) × (their gradient of RQ) = −1 and attempt to simplify M1 Clear of fractions and expand brackets.
Expect k 2 − 6k + 5 = 45 OE.
Condone +/–sign errors during simplification.
k 2 − 6k − 40 [ = 0 ] A1 OE
k = 10, k = −4 A1
Alternative Method for Question 9(a)
PR 2 = (k − 5)2 + ( 2 + 13)
2 B1 Obtain at least one of PR 2 and RQ 2 .
These expressions may be seen under a square root sign.
or RQ 2 = (k − 1) 2 + ( 2 − 5 )
2
( their PR ) + ( their RQ ) = 18 + 4 and attempt to simplify
2 2 2 2 M1 Expect 2k 2 − 12k + 260 = 340 OE.
Condone +/–sign errors.
2k 2 − 12k − 80 [ = 0 ] A1 OE
k = 10, k = −4 A1
9(a) Alternative Method 2 for Question 9(a)
x 2 + y 2 + 8 x − 6 y − 60 = 0 B1 OE
22 + k 2 + 8 × 2 − 6 × k − 60 = 0 M1 Substitution of ( 2, k ) into their circle equation.
k 2 − 6k − 40 [ = 0 ] A1 OE
k = 10, k = −4 A1
4 If none of the above marks are awarded, then SC B1 for using
k = 10 to find the gradients of QR (–3) and PR ( 13 ) and
showing that their product is –1 (or other similar verification),
and then stating that this shows that 10 is a possible value for
k.
9(b) [Centre is] ( −4, 3) B1 SOI
( their 3) − 10 7 *M1 Attempt at finding the gradient of the radius using ( 2, 10 )
Radius gradient = =
( their − 4 ) − 2 6 and their centre, but not P, Q, the mid-point of PR or QR.
−1 y − 10 DM1 OE
Either = Correct method to find the equation of the tangent using
their 7 x−2
−1
6 and ( 2, 10 ) .
their radius gradient
−1
Or 10 = × 2 + c c = ...
their 7
6
A1 OE
( x − 2 ) or y = − x +
6 6 82
y − 10 = − Correct equation but not stated in the required form.
7 7 7
6 x + 7 y − 82 = 0 or 82 − 6 x − 7 y = 0 A1 All correct terms on one side, but condone them being in the
wrong order.
Alternative Method for Question 9(b)
x 2 + y 2 + 8 x − 6 y − 60 = 0 B1 OE
Equation of the circle.
dy dy *M1 Differentiate implicitly to arrive at an expression with two
2x + 2 y +8−6 =0 dy
dx dx terms containing .
dy −2 x − 8 1 dx
( )
−
Or = 85 − x 2 − 8 x − 16 2 Or rearrange to make y the subject and differentiate to arrive
dx 2
at an expression of the form f ′ ( x ) × f ( x ) .
9(b) dy dy DM1 Substitute (2,10) into their implicit differential.
2 ( 2 ) + 2 (10 ) +8−6 =0
dx dx
dy
−2 ( 2 ) − 8 Or substitute x = 2 into their expression for
1
(85 − ( 2) − 8 ( 2) − 16)
dy 2 − .
Or = 2
dx
dx 2
6 6 82 A1 OE
y − 10 = − ( x − 2 ) or y = − x+ Correct equation but not stated in the required form.
7 7 7
6 x + 7 y − 82 = 0 A1 All correct terms on one side, but condone them being in the
wrong order.
5
A curve C has equation y = + 2x - 5.
2x - 5
(a) Find the coordinates of the two stationary points. [4]
d2y
(b) Find and hence determine the nature of each stationary point. [3]
dx 2
-3
(c) The curve C is transformed to the curve C1 using a translation of e o followed by reflection in
7 the x-axis.
(i) State the coordinates of the maximum point of C1 . [1]
a
(ii) Find the equation of C1 in the form y = + dx + e , where a, b, c, d and e are integers. bx + c
[3]
Mark scheme
10(a) −9 × 2(2 x − 5)−2 + 2 B1 Correct differential.
( their − 18(2 x − 5) + 2 ) = 0 and rearrange to form a quadratic.
−2 M1
Equating a two term
dy
to 0 and dealing correctly with the
dx
( 2 x − 5) = 9 or 8 x − 40 x + 32 = 0
2 2
negative power.
dy −2
Their two term must contain (2 x − 5) .
dx
(1, − 6 ) and ( 4, 6 ) A1, A1 A1 for two correct x-values or one correct point, second A1
for all correct.
4
10(b) B1 FT Following through on their first derivative which must
144 x − 360
−18 × −2 × 2(2 x − 5) −3 = 72(2 x − 5) −3 or −2
contain (2 x − 5) .
( 2 x − 5)
4
(
Use ( their x = 1 and x = 4 ) in (their 72 2 x − 5) −3 ) M1 Substitute x-coordinate of each stationary point and determine
their nature. Nature of the turning points must correctly
To determine the nature of both turning points.
d2 y
follow from their values of 2 .
dx
d 2 y 72 A1 CWO
For x = 1, 2
= − or < 0 ⟹ maximum
dx 27
d 2 y 72
For x = 4 , 2 = or > 0 ⟹ minimum
dx 27
3
10(c)(i) (1, − 13) B1
1
10(c)(ii) 9 M1 −3
[ y =] + 2 ( x ± 3) − 5 ± 7 Application of to the original expression for C but
2 ( x ± 3) − 5 7
condone +/ − sign errors.
9 M1 9
[ y =] − + 2 ( x ± 3) − 5 ± 7 SC B1 for [ y =] − + 2x − 5 .
2 ( x ± 3) − 5 2x − 5
9 A1 Answer must be in this format; the ' y = ' can be implied by
y=− − 2x − 8
2x + 1 earlier inclusion.
3
The function f is defined by f (x) = x 2 + 4ax + a for x ! R , where a is a constant.
The function g is such that g -1 (x) = 3 2x - 4 for x ! R .
(a) Given that the range of f is f (x) H - 33, find the possible values of a. [4]
(b) Given instead that fgg (0) = 96 , find the value of a. [6]
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) f ( x ) = ( x + 2a ) 2 − 4a 2 + a B1
(their ( −4a + a ) ) = −33
2 M1 Condone or > .
4a2 − a − 33 [= 0] B1 Condone or < .
11 A1 OE
− , 3 Do not ISW if their final answer is given as a range or if one
4
of the answers is rejected.
Alternative Method for Question 11(a)
x = −2a , y = −4a 2 + a B1 The co-ordinates of the minimum point (likely to be either
from differentiation or completing the square).
( their − 4a + a ) = −33
2 M1 Their y-coordinate equated to −33. Condone or > .
−4a2 + a + 33[ = 0] B1 OE
Condone or > .
11 A1 OE
− , 3 Do not ISW if their final answer is given as a range or if one
4
of the answers is rejected.
11(a) Alternative Method 2 for Question 11(a)
x2 + 4ax + a + 33 [=0] B1 Condone or > .
M1 Condone or > .
( 4a ) − 4 [1] ( a + 33) [ = 0]
2
16a2 − 4a − 132 [=0] B1 OE
Condone or > .
Accept 4a2 − a − 33 or multiples thereof.
11 A1 OE
− , 3 Do not ISW if their final answer is given as a range or if one
4
of the answers is rejected.
4
11(b) 1 B1 Expression for g ( x ) . SOI
[ g ( x ) =] ( x + 4 )
3
2
Either
M1 Replacing x with 0 in their g ( x ) .
[ g ( 0 ) =]
1
2
(( 0) ± 4)
3
[ g ( 0 ) = 2]
gg ( 0 ) = 6 A1
f ( their 6) [ = 36 + 24a + a ] M1
Or
2 M1 Either composite function using their g ( x ) .
x3 + 4 x3 + 4
fg ( x ) = + 4 a +a
2 2
3
x3 + 4
2 +4 ( x 3 + 4 )3
or gg x =
( ) = + 2
2 16
A1 Complete algebraic expression for fgg ( x ) .
( x + 4) ( x + 4)
2
3 3
3 3
fgg ( x ) = + 2 + 4a + 2 + a
16 16
2 M1 Replacing x with 0 in their fgg ( x ) .
( 0 + 4)
(0 + 4)
3 3
fgg ( 0 ) = + 2 + 4a + 2 + a
16
16
11(b) Then
36 + 24a + a = 96 A1 OE
12 A1 OE
[ a =]
5
6
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