Mathematics 9709/15 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.
The equation of a curve is such that = 12 (2x - 5) 2 + 8x . It is given that the curve passes through the dx point (2, 4).
Find an equation of the curve. [4]
Mark scheme
1 B1 B1 OE
[ y =] ( 2 x − 5)3
12 8 2
+ x [ +c ] Terms may be unsimplified.
3 × 2 2
May see [ y =] 16 x3 −116 x 2 +300 x [ +c ].
4 = 2×(2×2 – 5)3 + 4×22 + c [⇒ c = –10] M1 Sub (2, 4) correctly into their integrated expression
with a ‘+ c’.
y or f or f ( x ) = 2 ( 2 x − 5) + 4 x2 − 10 A1 May see y = 16 x3 − 116 x 2 + 300 x − 260.
3
' y =′ or 'f ( x ) =′ or 'f = ' can be implied if seen in
working.
4
In the expansion of (3 + ax) 5 + (6 - x) 4 , the coefficient of x 2 is six times the coefficient of x.
Find the possible values of the constant a. [5]
Mark scheme
2 x2 coeff = 10 × 33 × a2 and 6 × 62 × ( −1) allow correct terms with x2 B1 Expect 270a 2 and 216.
2
Terms may be seen separately.
Allow if seen in an expansion. Combinations must be
evaluated.
x coeff = 5 × 34 × a and 4 × 63 × ( −1) allow correct terms with x B1 Expect 405a and − 864.
Terms may be seen separately.
Allow if seen in an expansion. Combinations must be
evaluated.
270a 2 + 216 = 6 ( 405a − 864 ) M1 OE
For forming a correct quadratic equation using their 4
terms only.
45a2 − 405a + 900 = 0 ⇒ 45 ( a − 4 )( a − 5) = 0 DM1 For evidence of a correct method of solving.
If quadratic formula used a full substitution must be
seen.
a = 4 or a = 5 A1 Only dependent on the first M1.
5
(a) Use completing the square to find the exact solutions of the equation 4x 2 - 4x - 1 = 0 . [2]
1
(b) Hence solve the equation 4 tan i = 4 + for 0° 1 i 1 180° . [3]
tan i
Mark scheme
3(a) 1
2
2
1 1 M1 OE
4 x − [ −2 = 0] or ( 2 x − 1) [ −2 = 0] or x − − = 0
2
Must deal with the coefficient of x2 and x correctly to
2 2 2
produce an ( ax + b ) term.
2
( )
1 1 A1 1
x= ± OE, e.g. x = 1± 2 .
2 2 2
SC B1 only for correct solutions from another method.
2
M1 Setting tanθ = their x for at least one value of their x.
( )
3(b) 1
tanθ = 1± 2 May restart and solve the quadratic in tanθ .
2
[θ =] 50.4, 168.3 AWRT and no other answers in the range 0 < θ < 180 A1 A1 SC A1 only for AWRT 0.879 and 2.94 radians.
SC M0 B1 B1 for answers only.
SC M0 B1(only) for both answers only in radians.
3
y
1 y = x2 - x2
O 2 x
1
The diagram shows part of the curve y = x 2 - 2 . The shaded region is bounded by the curve, the line x = 2 and the x-axis. x
Find the volume formed when the shaded region is rotated through 360° about the x-axis, giving your answer correct to 2 decimal places. [5]
Mark scheme
4 Intersects x-axis when x4 = 1 x = 1 B1 WWW
May be seen without working.
May be seen as a limit in the integration.
Allow ±1 if x = 1 is seen as the lower limit in the
integration.
1
2
1 *M1 1
V = [ π ] y dx = [ π ] x 2 − 2 dx = [ π ] x 4 − 2 + 4 dx with attempt at
2
With some attempt at squaring (accept x 4 ± as
x x x4
integration evidence).
1 5 1 A1
x − 2x − 3
5 3x
1 1 DM1 Use of limits, x = 2 and their x = 1, min evidence if
[ π] × 25 − 2 × 2 −
1 1
− − 2 −
3 283 32
5 3× 2 5 3 correct answer: + . If answer incorrect,
120 15
substitution of limits must be clear.
DM0 for use of the limit x = 0.
π × 4.49… = 14.11 A1 AWRT, WWW
539
Allow π or 14.1, dependent on the first M1 and
120
A1.
5
D
A 5 cm C 5 cm B
The diagram shows a sector ABD of a circle with centre A and radius 10 cm. The perpendicular bisector of AB passes through D.
(a) Find the perimeter of the shaded region BCD, giving your answer correct to 1 decimal place. [4]
(b) Find the area of the shaded region BCD, giving your answer correct to 1 decimal place. [2]
Mark scheme
5(a) π B1 OE
Angle CAD = or 60o, CD = 75 or 5 3 ˆ or CD. Allow CAD
ˆ = 1.05, CD = 8.7
3 For either CAD
or 8.66.
B1 B1 OE
π
60
[Perimeter =] 10 × + 75 + 5 or
3 360
× 2π (10 ) + 75 + 5
B1 for the two sides and B1 for arc length (allow 10.5).
= 24.1 cm B1
4
5(b) 1 π 1 60 1 M1 OE
Area = × 102 × − × 5 × 75 or × π × 102 − × 5 × 75 Use of sector area formula minus triangle area
2 3 2 360 2
formulae, with their angle CAD and their side CD.
= 30.7 cm2 A1
2
Each year, on her birthday, Ananya receives some money from each of her parents.
On Ananya’s first birthday, her father gives her $10. Every subsequent year, her father gives her $5 more than he gave her the previous year.
On Ananya’s first birthday, her mother also gives her $10. Every subsequent year, her mother gives her
20% more than she gave her the previous year.
(a) Show that on Ananya’s eleventh birthday she receives more from her mother than from her father.
[3]
(b) Find the total amount of money Ananya receives up to and including her eighteenth birthday. [5]
Mark scheme
6(a) Eleventh birthday: Father 10+10×5 [= 60] M1 For correct use of AP formula OE.
Mother 10×1.210 [= 61.9(2)] M1 For correct use of GP formula OE.
60 and 61.9 A1 Both answers. Accept 2 sf answers.
3
6(b) 18 M1A1 A1 may be implied by a correct final answer.
Father ( 2 × 10 + 17 × 5) = 945
2
Mother
(
10 1 − 1.218 ) = 1281.17 M1A1 A1 may be implied by a correct final answer.
(1 − 1.2 )
Total = 2226.17 A1 Accept 3 or more sf answers.
Ignore S17 ( = 1909.3) as an extra answer.
5
In the parallelogram ABCD, the coordinates of A are (3, 7), the coordinates of B are (6, p) and the coordinates of D are (1, p). It is given that the gradient of AB is - 23 .
(a) Find the value of p. [2]
(b) Find the coordinates of C. [2]
(c) Find the area of the triangle formed by the perpendicular bisector of AB and the x- and y-axes. [5]
Mark scheme
7(a) 7− p 2 2 M1 OE
= − or use of straight line equations with m = − , ( x, y ) = ( 3,7 )
3−6 3 3 2
Expect y − 7 = − ( x − 3) with (6, p) substituted or
with (6, p) substituted. 3
−2
y= x + 9 with (6, p) substituted.
3
p=5 A1
2
7(b) ({4}, {their 2p-7}) B1 WWW
B1FT Extra solutions lose both marks.
2
7(c) Mid-point AB is (4.5, 6) B1
B1
−1 3
Gradient of perp bisector = =
−2 2
3
3 M1 OE. Must be using their mid-point and their
Equation y − 6 = ( x − 4.5) perpendicular gradient.
2
Crosses axes at (0, − 0.75), (0.5, 0) DM1 Correct use of their perpendicular bisector equation to
find the x- and y-intercept.
Area = 0.1875 (accept 3 sf accuracy) A1 3
OE, e.g. . Last 2 marks can be gained by integrating
16
the line equation between zero and 0.5.
5
The equation of a curve is y = x 3 + ax 2 + bx + 5 . The curve has a stationary point at (1, 9).
(a) Find the values of the constants a and b. [5]
(b) Find the coordinates of the other stationary point. [3]
(c) A point P is moving along part of the curve in such a way that the y-coordinate of P is increasing at a constant rate of 6 units per second.
Find the rate at which the x-coordinate of P is increasing when x = 5. [3]
Mark scheme
8(a) 9 =1+ a + b + 5 B1
dy B1
= 3x 2 + 2ax + b
dx
Gradient = 0 at (1, 9) so 0 = 3 + 2a + b M1 dy
Setting their to zero and substituting x = 1.
dx
Attempt to solve their linear equations simultaneously DM1 Can be implied by their answers.
a = −6, b = 9 A1 WWW
5
8(b) dy M1 dy
= 3x 2 − 12 x + 9 = 0 Setting their to zero.
dx dx
Solution [leading to x = 1 or x = 3] DM1 Solving their 3-term quadratic
(3, 5) or x = 3, y = 5 A1 WWW
Ignore (1, 9 ) if given as a second answer.
Only dependent on the first M1.
3
8(c) dy M1 dy
At x = 5, = 3 × 52 − 12 × 5 + 9 Substituting x = 5 into their . May be implied.
dx dx
dx dt M1 OE
6 = their 24 × or their 24 = 6 ×
dt dx dy dy dx
Use of chain rule SOI = × .
dt dx dt
dx dt
Linking correctly (or ), their 24 and 6.
dt dx
dx 1 A1 OE
=
dt 4
3
Functions f and g are defined as follows.
f (x) = cos x for 0 G x G r g (x) = 3 cos (x - r) + 2 for r G x G 2r
(a) Describe fully the transformations that have been combined to transform the graph of y = f (x) to the graph of y = g (x) . [4]
(b) On the given axes, sketch the graphs of y = f (x) and y = g (x) . [4]
y
O r 2r x
(c) Find g -1 f `13 rj. [4]
(d) Explain why the composite function fg cannot be formed. [1]
Mark scheme
9(a) {Stretch}{factor 3} {in y-direction} B2,1,0 If 2 or more stretches or extra transformations, give B0
for stretches.
π B2,1,0 π
{Translation or shift} Translation may be split to before the stretch and
2 0
0
after the stretch.
2
If both vectors correct but ‘translation’/’shift’ not
stated, then B1 only for the translations.
Alternative Method for Question 9(a)
π 0 B2,1,0 If two or more stretches or any extra incorrect
π
{Translation or shift} 2 or and 2 transformation is given, then B0 for the stretches.
0 If order incorrect, maximum of 3/4.
3 3
followed by {stretch} {factor 3} {in y direction} B2,1,0
4
9(b) 7 y
B1† Graph of f(x) with correct domain.
6
5
4 B1† For g(x) being a decreasing function in the domain π to
3 2π.
2
1
x B1 Domain π to 2π for g(x).
π/2 π 3π/2 2π
−1
−2
B1 Range for g approximately correct (should be from 5 to
−1).
−3
4 †Sketches must be curves and have zero gradient at the
ends of the given domains.
9(c) π 1 B1 May be implied by correct substitution later.
[g-1] f = [g-1 ]
3
2
x−2 −1 y − 2
M1 Finding inverse of g.
[g −1 ( x ) or y =] cos −1 + π or [ x = ] cos +π Allow one sign error.
3 3
1
Alt: 3cos ( x − π ) + 2 =
2
1 DM1 1
−2 Substituting their x = into their expression for
1
[g −1 =] cos −1 2 +π
2
2 3 g −1 ( x ) .
1 2π
Alt: Use of arccos for cos ( x − π ) = − ⇒ x−π=
2 3
2π 5π A1
[g −1 ( x ) =] +π= or 5.24 AWRT
3 3
4
9(d) The domain of f does not include the whole of the range of g B1 OE, but must mention range of g and domain of f.
Alternative Method for Question 9(d)
Show clearly that a particular value or set of values in the domain of g gives a B1 Value of x for substitution into g ( x ) must be in the
value of g ( x ) which is outside the domain of f ranges: π x < 4.322 or 5.447 < x 2π.
1
The equation of a circle is x 2 + y 2 + 4x - 8y - 12 = 0 .
(a) Find an equation of the tangent to the circle at the point (2, 8), giving your answer in the form ax + by + c = 0 . [4]
(b) Given that the line x + 3y = k does not intersect the circle, show that k 2 - 20k - 220 2 0 . [5]
Additional page
If you use the following page to complete the answer to any question, the question number must be clearly shown.
Mark scheme
10(a) Centre is (–2, 4) B1
8−4 M1 OE, e.g. using equation of the radius.
Gradient of perpendicular = [ = 1]
2 − ( −2 )
Equation of tangent is y – 8 = –1(x – 2) M1 Using a correct point and their tangent gradient from
dy
an attempt to find , or the negative reciprocal of
dx
their perpendicular gradient.
x + y − 10 = 0 or –x – y + 10 = 0 A1 CAO – this form is required.
Answer without working, allow maximum of 2/4.
Alternative Method for Question 10(a)
1 1 B1 OE; a correct expression for y.
( ) (
y = [ ± ] 28 − x 2 − 4 x 2 + 4 or y = [ ± ] 32 − ( x + 2 ) )
2 2
+4
−1 M1 OE
dy
dx
1
( )
= [ ± ] 28 − x 2 − 4 x 2 ( −2 x − 4 ) [= − 1 at x = 2]
2
Differentiating y with no more than one sign error.
Equation of tangent is y – 8 = –1(x – 2) M1 Using a correct point and their tangent gradient from
dy
an attempt to find , or the negative reciprocal of
dx
their perpendicular gradient.
x + y − 10 = 0 or –x – y + 10 = 0 A1 CAO – this form is required.
Answer without working, allow maximum of 2/4.
Alternative Method 2 for Question 10(a)
dy dy B1 Differentiating implicitly.
2x + 2 y + 4−8 = 0
dx dx
dy ( 2 x + 4 ) M1 dy
−1 at ( 2, 8)
= = Rearranging to make the subject,
dx ( 8 − 2 y ) dx
dy
May substitute (2, 8) and then rearrange to find .
dx
Equation of tangent is y – 8 = –1(x – 2) M1 Using a correct point and their tangent gradient from
dy
an attempt to find , or the negative reciprocal of
dx
their perpendicular gradient.
x + y − 10 = 0 or –x – y + 10 = 0 A1 CAO – this form is required.
Answer without working, allow maximum of 2/4.
4
10(b) ( k − 3 y )2 + y 2 + 4 ( k − 3 y ) − 8 y − 12 = 0 M1* k−x
Sub x = k − 3 y or y = into circle equation.
or ( k − 3 y + 2) + ( y − 4) = 32
2 2 3
y=
k−x ( k − x ) 2 − 8 k − x + x 2 + 4 x − 12 = 0
gives:
3 9 3
2
k−x
or ( x + 2 ) +
2
− 4 = 32.
3
9 y 2 − 6ky − 20 y + y 2 + k 2 + 4k − 12 = 0 DM1 OE
All squared brackets expanded to give a quadratic
equation in y (or x, which gives:
9x2 + 60x − 2kx + k 2 + x2 − 24k −108 = 0 or
x 2 + k 2 − 2kx 8k − 8 x
− + 16 + x 2 + 4 x + 4 = 32. )
9 3
(
b 2 − 4ac = ( 6k + 20 ) − 4 × 10 × k 2 + 4k − 12
2
) M1** OE
Factorising out y (or x) from their quadratic (this is the
quadratic where the previous DM1 or possibly DM0
was awarded) to identify ‘b’ and then finding the
discriminant. From equation in x, this is:
(
( 60 − 2k )2 − ( 4 )(10 ) k 2 − 24k − 108 . )
−4k 2 + 80k + 880 < 0 DM1 OE
Simplify the discriminant and setting it to less than
zero. From eliminating y this is:
−36k 2 + 720k + 7920 < 0
Only dependent on the previous M1.
k 2 − 20k − 220 > 0 A1 AG
WWW
10(b) Alternative Method for Question 10(b)
Centre of circle (–2, 4) and radius of circle is 32 B1
1( −2 ) + 3( 4 ) − k M1* Correct use of ‘distance of a line from a point formula’
Distance of x + 3 y − k from ( −2,4 ) = with their centre coordinates.
(12 + 32
1( −2 ) + 3 ( 4 ) − k DM1 Setting distance to be greater than their radius.
Setting > 32
(12 + 32 )
Squaring both sides DM1 Removing the square roots.
Rearranging to k 2 − 20k − 220 > 0 A1 AG
5
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Mathematics 9709/15 for May/June 2025 is out of 75 marks, with 1 hour 50 minutes to answer it.
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