Mathematics 9709/12 February/March 2025: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 1 Pure Mathematics 1, February/March 2025. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
y
M
P
O x
5
The diagram shows the curve with equation y = 2x 2 - + 3. The curve crosses the x-axis at the point x
P (1, 0) and M is a minimum point.
(a) Find the gradient of the curve at P. [2]
(b) Find the coordinates of M. Give each coordinate correct to 3 significant figures. [3]
34
Mark scheme
2(a) 5 B1 OE
4x +
x2
dy B1 FT Correct use of x = 1 in their two-term differentiated
dx = 9 expression, defined as an expression with one correct
power.
2
2(b) 5 M1 B
Their 4 x + 2 = 0 and valid method as far as ' x = ...' Equate their derivative of the form Ax ± to zero, where
x x2
A, B ≠ 0, and solve. If no working is seen, this can be
implied by a correct answer for x.
x =−1.08 A1 AWRT
y = 9.96 A1 AWRT
3
(a) Find the complete expansion of b2x - l . [4]
x
34
(b) Hence determine the coefficient of x 2 in the expansion of `x 2 + 5jb2x - l . [2]
x
Mark scheme
3(a) ±3
4 B1 May be seen in a full expansion.
( 2x ) and x
4
81
This can be implied by 16x 4 and + 4 unless they are
x
3
clearly using + throughout.
x
3 −3 2 −3
2
−3
3 B1 Correct combination of numerical coefficients for the
4 ( 2x ) + 6 ( 2x ) + 4 ( 2x ) middle three terms.
x x x Can be implied by a correct full expansion.
16 x 4 + k1 x 2 + k2 x0 + k3 x −2 +81x −4 M1 OE
Powers now simplified correctly with their k1 , k2 , k3 ≠ 0.
16 x 4 − 96 x 2 + 216 − 216 x −2 + 81x − 4 A1 OE
4
3(b) Use of (their 216) + 5×(their –96) only, to arrive at the coefficient of x 2 M1 Other terms may be seen.
–264 A1 Accept −264 x 2 as the final answer.
2
A B
4 cm 4 cm
C D
6 cm 0.8 rad 6 cm
O
The diagram shows a triangle OAB where OA = OB = 10 cm and angle AOB = 0.8 radians.
Points C and D on OA and OB respectively are such that the arc CD is part of a circle with centre O and radius 6 cm. The shaded region is bounded by the arc CD and the line segments CA, AB and BD.
(a) Find the perimeter of the shaded region. [3]
(b) Find the area of the shaded region. [3]
Mark scheme
4(a) 6 × 0.8 B1 45.8
Accept × 12π.
360
( )
AB 2 =102 + 102 − 2 × 10 × 10cos 0.8 or AB = 2 10sin 0.4 or
M1 Allow angles correctly converted to degrees for this mark.
ˆ = OBA
0.8 rad = 45.8°, OAB ˆ = 67.1°.
10sin 0.8
AB = π − 0.8
π − 0.8 = 1.17
sin 2
2 This mark can be implied by AWRT 7.8.
20.6 A1 AWRT
3
4(b) 1 2 B1
[Area of sector =] × 6 × 0.8
2
1 M1 OE
[Area of triangle =] × 102 × sin 0.8 or 10sin0.4 ×10cos0.4
Allow use of their value of θ or 2 θ in degrees.
2 1
or other complete method.
21.5 A1 AWRT
3
An arithmetic progression has first term 5 and common difference 6.
For this progression, find the sum of all the terms that lie between 150 and 400. [6]
Mark scheme
5 Attempt to solve either: M1 Attempt to determine positions of first and last terms
5 + ( n − 1) 6 = 150 or 5 + ( n − 1) 6 = 400 involved
151 401 A1 OE
and Can be implied by 25 or 26 and 66 or 67.
6 6
Correct use of Sn formula with their 66 and their 25 M1
S66 or 67 − S25 or 26 M1
S66 – S25 A1
11275 A1
Alternative Method for Question 5:
Attempt to find new a and l for reduced series M1
155 and 395 A1
their 395 = their 155 + ( n − 1) 6 M1 Attempt to find n, which results in 40, 41 or 42.
n = 41 A1 CWO
their 41 M1 OE
Stheir 41 = ( their 155 + their 395)
2
11275 A1
6
y
r
O x
The diagram shows a circle C of radius r, where x 2 0 and y 2 0 for all points on C. The least distance between any point on C and the x-axis is 8 units, and the least distance between any point on C and the y-axis is 5 units.
(a) State the coordinates of the centre of the circle in terms of r. [1]
(b) Given that the distance between the origin and the centre of the circle is 15 units, find the value of r. [3]
(c) The point on the circle furthest from the origin is denoted by P.
Find the gradient of the tangent to the circle at P. [2]
8 sin 2 i - 5 sin 4 i
Mark scheme
6(a) ( r + 5, r + 8 ) B1 OE
Allow x = r + 5, y = r + 8.
If values are stated without reference to x and y, take the
first value to be their x.
1
6(b) B1 FT OE
their ( r + 5) + their ( r + 8) = 152
2 2
Following their answers to (a), which must both contain r.
r 2 + 13r − 68[ = 0] ( r + 17 )( r − 4 ) [ = 0] M1 Or other valid method of solution for their three-term
quadratic.
[ r =] 4 A1 CWO
r = −4 r = 4 scores A0.
Special Case: After B1M0, r = 4 scores SCB1, but after
B1M0, r = −4 r = 4 scores B0.
3
6(c) their ( r + 8 )
from (a) with their r from (b), or
( their r from ( b ) ) + 8 M1 r > 0 only.
their ( r + 5 ) ( their r from ( b ) ) + 5
−
3 A1 FT
OE, i.e. −
their ( r + 5)
or −
( their r from ( b ) ) + 8 .
4 their ( r + 8) ( their r from ( b ) ) + 5
2
(a) Show that 3 tan 2 i + 5 sin 2 i / . [3]
1 - sin 2 i
(b) Hence solve the equation 3 tan 2 i + 5 sin 2 i = 9 for 0° 1 i 1 270° . [4]
Mark scheme
7(a) M1
Use of tan 2 θ = sin 2 θ
2
cos θ
Relevant use of cos2 θ = 1 − sin 2 θ at least once M1
8sin 2 θ − 5sin 4 θ A1 AG
1 − sin 2 θ All necessary detail needed.
3
7(b) Attempt to solve their 5sin 4 θ − 17sin 2 θ + 9 = 0 using a “correct” method *M1 Allow ± errors in arriving at their quadratic in sin 2 θ . This
can be implied by either sin 2 θ = 2.744 or 0.6559.
A1 Condone inclusion of sinθ = 1.65 for this mark.
17 − 109
sin θ = [ ± ] 0.81[ 0 ] or
10 17 ± 109
Allow .
10
This mark can be implied by correct values.
Their 54.1, and 180 − their 54.1 or 180 + their 54.1 DM1 A correct method for obtaining a second angle within the
range 0 < their 54.1 < 90.
54.1, 125.9, 234.1 A1 AWRT
A0 for additional values between 0° and 270°.
4
A geometric progression is such that its second term is - 120 and its sum to infinity is 160.
(a) Find the common ratio. [4]
(b) The first nine terms of the progression are now removed.
Find the sum to infinity of the remaining terms of the progression. [3]
d2y 6 5 1
Mark scheme
8(a) a B1
ar = −120 and = 160
1− r
120 1 a M1 Elimination of either a or r.
− × = 160 or = 160
r 1− r 120 Condone ± errors for this mark.
1+
a
4r 2 − 4r − 3[= 0] or a2 −160a −19200[= 0] A1 OE
Rearrange to arrive at a three-term quadratic.
r = − 12 only A1
4
8(b) [ a =] 240 B1 FT −120 ÷ ( their r ) , where −1 < r < 1, r ≠ 0.
1 M1 With (their 240) and ( their r ) as long as −1 < r < 1, r ≠ 0.
240 1 − (− )9
2 Condone reversed subtraction.
160 −
1
1− −
2
Alternative Method 1 for first two marks of Question 8(b)
[ a =] 240 B1 FT −120 ÷ ( their r ) , where −1 < r < 1, r ≠ 0.
1 M1 Correctly using the 10th term as ‘a’ and the sum to infinity.
240 × (− )9
2 With (their 240) and ( their r ) as long as −1 < r < 1, r ≠ 0.
1
1− −
2
Alternative Method 2 for first two marks of Question 8(b)
B1 FT
15 −120 ÷ ( their r ) , where −1 < r < 1, r ≠ 0.
8
[10th term =] − , 0.46875 OE
32
15 M1 With ( their a and r ) , where −1 < r < 1, r ≠ 0.
−
32
1
1− −
2
5 A1 5
− or − 0.3125 A0 for −0.313 without sight of − or −0.3125.
16 16
3
Condone use of r = to provide a second solution.
2
3
A curve is such that 2 = 4 - 3 . It is given that the curve has a stationary point at b , 9l. dx x x 2 d2y
(a) Use the expression for to determine whether the stationary point is a maximum or a dx 2 minimum point. [2]
(b) Find the equation of the curve. [7]
Mark scheme
9(a) 6 5 M1 Substitute x = 1 and evaluate second derivative.
4
− 3
2
1 1
2 2
96 − 40 [ = 56 ] > 0 Minimum A1 CWO
Evidence and conclusion.
2
SC B1 for d y2 > 0 without sight of 96 − 40 or 56.
dx
2
9(b) 6 −3 5 −2 B1 B1 OE
x − x [ +c1 ]
−3 −2 B1 for each correct { }.
6 1
−3
5 1
−2 M1 Substitute x = 12 into two terms of an integrated expression
0= − + c1 (at least one correct power), now with c1 , and equate to 0 to
−3 2 −2 2
find c1 .
c1 = 6 A1
k1 x −2 + k2 x −1 + k3 x [ +c2 ] M1
Integration of their
dy
to produce at least two terms with
dx
correct powers, k1 , k2 ≠ 0.
1 5 1 M1 OE
9= − + 6 + c2
1 Substitute ( 12 , 9 ) into integrated expression (at least two
2
2
1 2
2 correct powers) to find c2 .
2
5 −1 A1 OE
y = x −2 − x + 6x + 7 Condone their final answer being c2 = 7 if a completely
2
correct simplified expression for the equation containing
c 2 has been stated previously.
7
y
Q
B
A
P
O x
The diagram shows the curve with equation
1 y = 4 (3x + 4) 2 - 2x - 6
for values of x such that 0 G x G 7. The tangent to the curve at the point P (7, 0) meets the y-axis at the point Q. Region A is bounded by the curve and the two axes. Region B is bounded by the curve, the line segment PQ and the y-axis.
(a) Find the area of region A. [4]
(b) Find the area of region B. [5]
Mark scheme
10(a) 2 4 3
2 x2 B1 B1 B1 for each correct { }.
× ( 3x + 4 ) 2 − − 6x
3 3 2
8 3
8 3
M1 Correct use of 7 and 0 in an expression with at least two
[ A =] ( 21 + 4 ) 2 − 72 − 6 × 7 − ( 4 ) 2 terms with two correct powers.
9 9
13 A1
4
10(b) 1
B1
2 ( 3 x + 4 ) −
2 ×3 − 2
dy *M1 dy
y − 0 = their with x = 7 ( x − 7 ) Using x = 7, their value for and then any form of the
dx dx
equation of a straight line using ( 7, 0 ) .
Either:
DM1 Use x = 0 in their equation of PQ.
[Q is ] 0,
28
5
[Area of OPQ =] 1 × 7 × their 28 DM1
2 5
1 28 33 A1 FT Only FT, following B0M1DM1DM1, from their (a) if
2 × 7 × their 5 − ( their 13 from ( a ) ) = 5
(their 13 from ( a )) < 98
5
and the area is > 0.
Or:
4 DM1
their − ( x − 7) dx
5
4 x2 DM1
Evaluating their − – 7 x with limits 7 and 0
5 2
10(b) A1 FT Only FT, following B0M1DM1DM1, from their (a) if
( their area of OPQ ) − ( their 13 from ( a ) ) =
33
5 (their 13 from ( a )) < 98
5
and the area is > 0.
5
Functions f and g are defined for all real values of x by
f (x) = 4x 2 - c and g (x) = 2x + k ,
where c and k are positive constants. It is given that g -1 (3k + 1) = c .
(a) Show that gf (x) = 8x 2 - k - 1. [4]
(b) The curve with equation y = 8x 2 - k - 1 is transformed to the curve with equation y = h (x) by the following sequence of transformations.
2
Translation of e o
3
Stretch in the y-direction by scale factor k
Reflection in the x-axis
Find an expression for h (x) in terms of x and k. [3]
(c) The range of h is given by h (x) G 15.
Find the values of c and k. [3]
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) 1 B1
State or imply g −1 ( x ) = ( x − k ) or equivalent
2
1 B1 OE
Obtain ( 3k + 1 − k ) = c and hence 2k + 1 = 2c
2
Or:
g ( c ) = 3k + 1 B1
[ 2c + k = 3k + 1 ] 2c = 2k + 1 B1 OE
Then:
( )
gf ( x ) = 2 4 x 2 − c + k M1 Allow ± errors only.
8x 2 − 2c + k 8 x 2 − ( 2k + 1) + k gf ( x ) = 8 x 2 − k − 1 A1 AG
All necessary detail needed.
4
11(b) B1 OE
8 ( x − 2) − k − 1 + 3
2
Translation coming before the stretch
( (
k their 8 ( x − 2 ) − k − 1 + 3
2
)) B1 FT OE
Stretch
(( (
h ( x ) = − k their 8 ( x − 2 ) − k − 1 + 3
2
))) B1 FT OE
Reflection
3
11(c) k 2 − 2k = 15 or k 2 − 2k 15 B1 OE
k = 5 only B1
11 B1 OE
c= only
2
3
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Mathematics 9709/12 for February/March 2025 is out of 75 marks, with 1 hour 50 minutes to answer it.
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