Mathematics 9709/15 October/November 2025: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 1 Pure Mathematics 1, October/November 2025. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
A circle has centre (2, 6) and radius 10.
(a) State the equation of the circle. [2]
(b) The circle passes through the point (8, k).
Find the two possible values of k. [3]
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Mark scheme
1(a)
{( x − 2 ) } + ( y − 6 )
2 2
= 100 or 10
2 B2,1,0 B2 All 3 components correct,
B1 Any 2 components correct.
or x 2 − 4 x + y 2 − 12 y −60 = 0 oe
2
1(b) (8 − 2)2 + ( k − 6)2 = 100 or k 2 − 12k − 28 = 0 M1* Substituting the point into their equation to obtain
an equation in k.
k − 6 = ±8 or ( k − 14 )( k + 2 ) = 0 oe DM1 Simplify to the point where k can be found (allow
one sign error).
k = 14 , −2 B1 SC B2 only for both answers without working.
Allow ‘ y = ’.
3
A geometric progression has first term 3 + 4 2 and second term 5 - 2 .
(a) Find the common ratio of the geometric progression. Give your answer in the form 2 + p , where p is an integer to be found. [3]
(b) Find the sum to infinity of the geometric progression. [2]
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Mark scheme
2(a) 5− 2 B1 Correct quotient SOI.
r=
3+ 4 2
=
(5 − 2 )(3 − 4 2 ) M1 Rationalise their denominator (must see clear
process), or equate their r to 2 + p and solve.
9 − 32
23 − 23 2 B1
= = 2 − 1 [ p = −1]
−23
3
2(b) 3+ 4 2 M1 Uses S∞ formula.
S∞ =
1− ( 2 − 1) FT a = 3 + 4 2 and their r from 2(a), provided
−1 < r < 1 and r is evaluated.
Do not condone missing brackets unless corrected
in the calculation.
14 + 11 2 11 2 A1 Allow answers from a correct expression which
= or 7 + round to 14.8 to 3sf.
2 2
2
(a) Express 4x 2 + 10x + 6 in the form a (x + b) 2 + c , where a, b and c are rational constants to be determined. [2]
(b) The curve with equation y = 4x 2 + 10x + 6 and the line y = k have exactly one point of
intersection.
Using your answer to part (a) or otherwise, state the value of the constant k. [1]
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Mark scheme
3(a) 5
2 B2,1,0 B2 All 3 components correct,
4 x + 1
− B1 Any 2 components correct.
4
4 5 1
Allow a = 4, b = , c = − unless contradicted
4 4
by the algebraic expression.
Allow decimals.
2
3(b) 1 1 1 B1 FT Follow through their c from 3(a).
k =− or −0.25 allow − or y = − 1
4 4 4 Don’t allow x = − .
4
If solved through restarting answer must be
WWW.
1
A B
r cm
1
3 r rad
O r cm C
The diagram shows the design for a company’s new logo. The sector of the circle, centre O, has radius r cm. The acute angle AOC = 13 r radians. The quadrilateral OABC is a rhombus.
(a) Find an expression for the perimeter of the design. Give your answer in terms of r and r. [2]
(b) It is now given that the perimeter of the design is 200 cm.
Find the area of the design. Give your answer to 3 significant figures. [5]
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Mark scheme
4(a) 1 5 M1 Finding the reflex angle AND using arc length
Arc Length = 2π − π × r = πr
3 3 1
formula or 2 πr − r π .
3
5 1 A1 Answer must be exact.
Perimeter = πr + 2r oe e.g. 2πr − πr + 2r
3 3
2
4(b) 5π M1 Equating their expression to 200 and attempting to
r + 2 = 200 solve (their expression must be linear).
3
200 A1 CAO
r = 27.6 [ 4] allow r = oe
5π
+2
3
1 1 M1 Using 2(0.5ab sin C) with a = b = their r and
Area of Rhombus = 2 ( 27.64 ) sin π ( = 661.6 )
2
2
3 1
C = π to get the area of the rhombus (or suitable
1 2π 1 1 3
Or Area of segment = ( 27.64 ) − ( 27.64 ) sin π
2
other method), with their value of r which may be
2 3
2 3 substituted later.
Expect segment area = 400 – 330.8 = 69.2.
Must be an attempt to find and use r.
1 25 M1 5
Area of Sector = ( 27.64 ) π ( = 2000 ) oe Using sector area formula with their r and π,
2 3 3
Or Total Area = Circle area – Segment area + Triangle area or complete method using the segment.
Must be an attempt to find and use r.
Total Area = 2000 + 661.6 = 2660 (3sf) A1 CAO WWW
AWRT 2660 (3sf).
Or Total area = 2400 – 69.2 + 330.8 = 2660 (3sf)
5
(a) Show that the equation
1 4
6 sin i + = tan i sin i can be written in the form
6 cos 2 i - cos i - 2 = 0 . [3]
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(b) Hence, solve the equation
1 4
6 sin i + = tan i sin i for 0° G i G 360° . [4]
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Mark scheme
6(a) cosθ 4 M1 Using identity for tanθ and multiplying through
6sin θ + = 6sin 2 θ + cosθ = 4 by sinθ .
sin θ sin θ
6(1 − cos 2 θ ) + cos θ = 4 M1 Uses sin 2 θ + cos2 θ = 1.
6cos2 θ − cos θ − 2 = 0 A1 AG
Expands out and simplifies to give the answer.
3
6(b) ( 3cosθ − 2 )( 2cosθ + 1) = 0 M1 Suitable method for solving quadratic.
2 1 B1 Obtaining both values of cosθ .
cosθ = , cosθ = −
3 2
θ = 48.2, 120, 240, 311.8 B1 Any 2 values of θ correct.
B1 All 4 values of θ correct and no other values in
the given range.
SC B1 All four answers in radians: 0.841, 2π/3,
4π/3, 5.44.
4
r cm
h cm
A manufacturer wishes to design an open cylindrical tank, as shown in the diagram. The tank will have a base but no top. The outside of the tank will have a fixed surface area of 600r cm2. The radius r cm
and height h cm of the tank can vary.
(a) Show that the volume, V cm3, of the tank is given by
rr (600 - r 2)
V= .
2
[3]
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(b) Find the exact value of r which corresponds to the maximum value of V. [3]
(c) Hence, find the maximum value of V. [2]
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Mark scheme
7(a) 600π − πr 2 600 − r 2 M1* Uses given info and correct formulae to form
2πrh + πr 2 = 600π oe h = = equation in h and r.
2πr 2r
V=
( 600 − r ) × πr oe
2
2
DM1 Sub their expression for h into V = πr 2 h.
2r
=
(
πr 600 − r 2 ) A1 AG
CAO
2 WWW
3
7(b) 3πr 2 M1* Differentiate given expression for V.
1
(
V = π 600r − r 3
2
dV 1
) ( )
= π 600 − 3r 2 or 300π −
dr 2 2
1
Condone missing π (must be of form a − br 2 ).
2
Alternative: h = r at max V, so 2πr ( r ) + πr 2 = 600π oe
600 − 3r 2 = 0 DM1 Equate their derivative to zero and attempt to
solve as far as an equation of the form ‘ r = ’.
Alternative: Solving 2πr ( r ) + πr 2 = 600π
r = 10 2 A1 CAO (accept 200 ).
3
7(c) π
(
V = 600 10 2 − 10 2 ) ( ) M1 Sub their value of r from 7(b) into the given
3
2 expression for V, providing their value of r > 0
and gives V > 0.
8890 ( 3sf ) Accept exact answer 2000 π 2 A1 AWRT 8890 (3sf).
2
y
O a 2a x
The diagram shows the curve with equation y = 6x + 5 . The shaded region is bounded by the curve, the x-axis and the lines x = a and x = 2a , where a is a positive constant.
The shaded region is rotated through 360° about the x-axis to form a solid. The solid has volume, V, such that V H 46r .
(a) Show that 9a 2 + 5a - 46 H 0 . [4]
(b) Find the range of possible values of a. [3]
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Mark scheme
8(a) 2a B1 CAO SOI
V =π ( 6 x + 5 ) [ dx ]
a Must include π and the correct limits.
M1* Correct integral of 6x + 5 (can be awarded if π is
2a
= [ π ] 3x 2 + 5 x
a missing and the limits are missing or incorrect).
2a
( 6 x + 5)
2
or = [ π ]
12
a
(
= [ π] 12a 2 + 10a − 3a 2 − 5a ) [ 46π] DM1 Correct substitution of correct limits.
9a 2 + 5a − 46 0 A1 AG
Can only be awarded if the argument leading to
the statement is complete and clear.
4
8(b) Solve inequality (or equation) M1 Solve quadratic by suitable method.
23 B1 23
Obtain − , 2 Condone absence of − .
9 9
Condone use of x.
Final answer a 2 B1 WWW
Don’t allow x 2.
3
In the expansion of (p + qx) 4 , the coefficient of x is equal to the coefficient of x 2. The constants p and q are both positive.
(a) Find the ratio p : q. Give your answer in its simplest form. [3]
(b) It is given that the coefficient of x 3 is 486.
Find the values of p and q. [4]
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Mark scheme
9(a) 4 × p 3 × q or 6 × p 2 × q 2 M1 Attempt to find either term in x or x 2 – must use
binomial coefficients.
May see x terms.
p 3 DM1 Equate coefficients and simplify to an equation
= oe equivalent to 2 p = 3q (no x’s should be present).
q 2
p : q = 3: 2 A1 Must be in this form only.
3
9(b) 4 p × q 3 = 486 B1 Set coefficient of the term in x 3 equal to 486.
May include x 3 on both sides of the equation.
3q 3 M1 Substitute their 9(a) equation, providing this is of
4× × q = 486 the form p = aq where a is positive and rational.
2
9 A1 Solve for q (or p).
q 4 = 81 q = 3, p =
2
A1 Solve for p and q with no negative answer.
SC B1 for answers without working.
4
The function f is defined by
7 f (x) = 3 + x-2 for x 2 2 .
(a) It is given that f (a) = 4 .
Find the value of a. [2]
(b) Find an expression for f –1 (x) and state the domain of f –1 . [4]
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(c) The function g is defined by
1 + 4x g (x) =
2x - 3 for x 2 32 .
Show that fg (x) / kx , where k is a constant to be determined. [3]
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Mark scheme
10(a) 7 M1 Forms an equation in a.
4 = 3+
a−2
a =9 A1 Condone x = 9.
2
10(b) Swap x and y and change the subject. This can be viewed as four stages: M1 Any two stages.
7 7 7 7
e.g. x = 3 + , x−3= , y−2= , y =2+ M1 The other two stages.
y−2 y−2 x−3 x−3
Condone one sign error present in the complete
7 2x + 1
or x = 3 + , x ( y − 2 ) = 3 y − 6 + 7, xy − 3 y = 2 x + 1, y = method.
y−2 x−3 One algebraic slip would result in only one M1.
7 A1 2x + 1
f −1 ( x ) = 2 + CAO Could be written as
x−3 x−3
Need to see 'f −1 ( x ) = ' or 'f −1 : x → ', not just ' y = '.
Domain: x > 3 B1 Allow equivalent statement but must involve x.
4
10(c) 7 B1 Substitute g ( x ) into f ( x ) .
fg ( x ) ≡ 3 +
1 + 4x
−2
2x − 3
1 + 4x 7
Alternative 1: g ( x ) = f −1 ( kx ) → =2+
2x − 3 kx − 3
14 x − 21 7 ( 2 x − 3) M1 Simplify to expression with no embedded
3+ or 3 + oe fractions (working may be minimal).
1 + 4x − 4x + 6 1 + 4 x − 2 ( 2 x − 3)
Alternative 1: Clearly amending fractions to a linear equation to solve
7
Alternative 2: Solve 3 + = kx to a linear equation
1 + 4x
−2
2x − 3
14 x − 21 A1 Cannot just state k = 2. Must have supporting
3+ ≡ 2 x or k = 2 algebra.
7
Trying several values to establish k = 2 scores
M0A0.
3
y
P
O x
The diagram shows the curve with equation y = 4x 2 - x 3 and the tangent to the curve at the point P.
The point P has x-coordinate 3.
(a) Find the equation of the tangent to the curve at the point P. Give your answer in the form y = mx + c . [5]
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(b) The shaded region is bounded by the curve, the x-axis and the tangent to the curve at P.
Find the exact area of the shaded region. [6]
The graph of y = 4x 2 - x 3 is transformed by a stretch of scale factor 13 in the x-direction. The point Q is the image of P under this transformation. The transformed shaded region is bounded by the transformed curve, the x-axis and the tangent to the transformed curve at Q.
(c) (i) Find the equation of the transformed curve in the form y = mx 2 + nx 3 , where m and n are integers to be found. [1]
(c) (ii) State the coordinates of Q and the area of the transformed shaded region. [2]
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Additional page
If you use the following page to complete the answer to any question, the question number must be clearly shown.
DFD
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Mark scheme
11(a) dy B1 CAO
= 8 x − 3x 2
dx
dy B1
= 24 − 27 = −3 [ when x = 3]
dx
y = 9 [when x = 3] B1 SOI
y − 9 = −3 ( x − 3) or y = −3x + c → 9 = −9 + c → c = 18 oe M1
Uses their y and their numerical
dy
to find
dx
equation of the tangent; condone one sign error.
y = −3x + 18 A1
5
11(b) 4 M1* Must obtain ax3 + bx 4 and indicate the limits 3
( )
Area between curve and x-axis = 4 x 2 − x3 [ dx ] and attempt to integrate
and 4.
3
4 x3 x 4
4 A1 SC B1 for use of wrong or no limits (only for
= − correct integral).
3 4 3
256 81 DM1 Correct sub of correct limits (allow one slip).
− 64 − 36 − 64 63
3 4 Minimum acceptable: − .
3 4
67 A1 SOI
= May be implied by a correct final answer if the
12
two areas are combined.
SC B1 if substitution of the limits is not seen.
9 67 DM1 27
Shaded region = ( 6 − 3) × − their Expect − ‘their integral’, but must be ‘area
2 12 2
−3 x3
6
67 under their line’ minus ‘their area under the
or + 18 x – their curve’, where ‘their integral’ is an attempt at the
2 3 12 area under the curve between x = 3 and x = 4.
May use the lengths from their tangent equation.
95 A1 Calculating area of triangle – (correct) area under
= any equivalent exact answer the curve.
12
11(b) Alternative Method for Question 11(b):
Finds area between curve and tangent between x = 3 and x = 4 M1* Integrate at least two of the four terms correctly.
4 Area under the line could be found from the
( −3x + 18) − ( 4x − x ) dx
2 3
( 4 − 3) 9 + 6 .
3 trapezium area ( )
2
3x 2 4 x3 x 4
4 A1 Integrating all four terms correctly.
= − + 18 x − + 4 x3 x 4
2 3 4 3 SC B1 for the correct integral ± − + .
3 4
256 27 81 DM1 Correct sub of limits (allow one slip).
= −24 + 72 − + 64 − − + 54 − 36 +
3 2 4
23 A1 SOI
= SC B1 if substitution of the limits is not seen.
12
1 23 DM1 Calculating area of triangle between x = 4 and
Shaded region = ( 6 − 4 ) × 6 × +
2 12 x = 6 + their area, providing limits of 3 and 4 are
used to find the area between the curve and the
tangent.
95 A1 Must be exact.
=
12
6
11(c)(i) y = 36 x 2 − 27 x 3 or state m = 36, n = −27 B1 CAO (must be expanded)
1
11(c)(ii) Q(1, 9) B1 CAO coordinates of Q.
95 B1 FT 1
Area = of their area from 11(b).
36 3
Allow 2.64.
2
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Mathematics 9709/15 for October/November 2025 is out of 75 marks, with 1 hour 50 minutes to answer it.
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