Mathematics 9709/11 October/November 2024: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 1 Pure Mathematics 1, October/November 2024. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
In the expansion of bkx + l , where k is a positive constant, the term independent of x is equal to 150. x
Find the value of k and hence determine the coefficient of x 2 in the expansion. [4]
a
Mark scheme
1 2
2 M1 4!
Identify correct term and obtain 6(kx) ·
2 Needs numerical coefficient or , not 4 C 2 .
x 2!2!
5 A1 5
Equate to 150 and obtain k = Ignore –
2 2
3 2 M1 4!
Identify correct term 4 ( kx ) · with their value of k Needs numerical coefficient or .
x 3!1!
Obtain coefficient 125 A1 Accept 125x 2 as final answer.
4
The curve y = x 2 - has a stationary point at (-3, b). x
Find the values of the constants a and b. [4]
Mark scheme
2 Differentiate to obtain 2 x + ax −2 or equivalent B1
Equate first derivative to zero, substitute x = −3 and attempt value of a M1 Must be an attempt at differentiation.
Obtain a = 54 A1
Obtain b = 27 A1
4
B C
15 cm 15 cm
A D
O
The diagram shows a sector of a circle, centre O, where OB = OC = 15 cm . The size of angle BOC is
2
5 r radians. Points A and D on the lines OB and OC respectively are joined by an arc AD of a circle with centre O. The shaded region is bounded by the arcs AD and BC and by the straight lines AB and
DC. It is given that the area of the shaded region is 209 2
5 r cm .
Find the perimeter of the shaded region. Give your answer in terms of r . [5]
Mark scheme
3 Use correct sector area formula M1
1 2 1 2 209 A1
Obtain × 152 × π − × x 2 × π = π or equivalent
2 5 2 5 5
Obtain [ x ] = 4 A1 AWRT 4.00.
Use correct arc length formula twice M1
38 A1 OE. Must be in terms of π.
Obtain 22 + π Like terms must be collected.
5
Not from a rounded value of x.
5
Show that the curve with equation x 2 - 3xy - 40 = 0 and the line with equation 3x + y + k = 0 meet for all values of the constant k. [5]
dy
Mark scheme
4 Substitute for y (or x) in first equation and simplify *M1 All terms to one side and brackets expanded.
Obtain 10 x 2 + 3kx − 40 [= 0] (or 10 y2 + 11ky + k 2 − 360[= 0] ) A1
Attempt b2 − 4ac for 3-term quadratic involving k DM1 Not in quadratic formula unless b2 − 4ac is
isolated.
Obtain 9k 2 + 1600 (or 81k 2 + 14400 ) A1
9k 2 + 1600 > 0 A1 FT FT for ak2 + b > 0 with a, b > 0.
5
The equation of a curve is such that = 4x - 3 x + 1. dx
(a) Find the x-coordinate of the point on the curve at which the gradient is 11
2 . [3]
(b) Given that the curve passes through the point (4, 11), find the equation of the curve. [4]
Mark scheme
5(a) Attempt correct process for solving 3-term quadratic equation in x M1 Accept 8y2 – 6y – 9 → (2y – 3)(4y + 3), if y = √x
specified.
Obtain at least 2 x − 3 = 0 or equivalent A1 Ignore 4 x + 3 = 0.
3
SC B1 for x = with no method shown for
2
solving the 3-term quadratic.
9 9 A1 SC B1 if no method shown for solving the 3-term
Conclude x = ignore quadratic.
4 16
Alternative Method for Q5(a)
9 81 M1
3 x = 4x – → 16x2 – 45x + o.e and attempt correct process to solve
2 4
9 9 A1 SC B1 if no method shown for solving the 3-term
Obtain x = or quadratic.
4 16
9 9 A1 SC B1 if no method shown for solving the 3-term
x= ignore quadratic.
4 16
3
5(b) 3 M1
Integrate to obtain form k1 x 2 + k2 x 2 + k3 x where k1k2 k3 ≠ 0
3 A1 Allow unsimplified.
Obtain correct 2 x 2 − 2 x 2 + x or equivalent
Substitute x = 4 and y = 11 to attempt value of c M1 Dependent on at least 2 correct terms involving x.
3 A1 Must be simplified.
Obtain y = 2 x 2 − 2 x 2 + x − 9 Allow ‘f(x) =’.
Allow y missing if y appears previously.
4
Circles C1 and C2 have equations
x 2 + y 2 + 6x - 10y + 18 = 0 and (x - 9) 2 + (y + 4) 2 - 64 = 0
respectively.
(a) Find the distance between the centres of the circles. [4]
P and Q are points on C1 and C2 respectively. The distance between P and Q is denoted by d.
(b) Find the greatest and least possible values of d. [3]
Mark scheme
6(a) State or imply centre of C1 is ( −3, 5 ) B1
State or imply centre of C2 is ( 9, − 4 ) B1
Attempt correct process for finding distance between centres M1
Obtain 15 A1
4
6(b) R = 4 and R = 8 B1
Obtain least or greatest distance B1 FT ‘15’ – R1 – R2 or ‘15’ + R1 + R2.
Obtain 3 and 27 B1 FT ‘15’ – R1 – R2 and ‘15’ + R1 + R2.
3
y
A
O 7 x
2
12
The diagram shows part of the curve with equation y = 3 . The point A on the curve has
2x + 1 coordinates b72 , 6l.
(a) Find the equation of the tangent to the curve at A. Give your answer in the form y = mx + c . [4]
(b) Find the area of the region bounded by the curve and the lines x = 0 , x = 72 and y = 0 . [4]
Mark scheme
7(a) −
4 M1
Differentiate to obtain form k1 (2 x + 1) 3
−
4 A1
Obtain correct −8(2 x + 1) 3 or unsimplified equivalent
7 M1 Gradient must come from a differentiated
Attempt equation of tangent at , 6 with numerical gradient expression.
2
1 31 A1
Obtain y = − x + or equivalent of requested form
2 4
4
7(b) 2 M1
Integrate to obtain form k2 (2 x + 1) 3
2 A1
Obtain correct 9(2 x + 1) 3 or unsimplified equivalent
Use correct limits correctly to find area M1 Substitute correct limits into an integrated
expression.
36 – 9 minimum working required.
Obtain 27 A1 SC B1 if M1 A1 M0 scored.
4
(a) It is given that b is an angle between 90° and 180° such that sin b = a .
Express tan 2 b - 3 sin b cos b in terms of a. [3]
(b) Solve the equation sin 2 i + 2 cos 2 i = 4 sin i + 3 for 0° 1 i 1 360° . [5]
Mark scheme
8(a) sin 2 β B1 sin 2 β
Use tan 2 β = E.g. tan 2 β = and then replaces sin 2 β
cos2 β cos β
2
with a2 or cos 2 β with 1 – a2.
cos β = − 1 − a2 B1
a2 B1
Obtain + 3a 1 − a 2
1− a 2
3
8(b) Use correct identity to obtain 3-term quadratic equation in sinθ *M1
Obtain sin 2 θ + 4sinθ + 1[= 0] A1
Attempt to solve quadratic DM1 −4 ± 12
At least as far as .
2
–15.5o implies attempt at solving quadratic.
Obtain 195.5 A1
Obtain 344.5 A1FT Following first answer; and no others for
0° < θ < 360° but must be in 4th quadrant.
SC B1 for 3.41c and 6.01c.
5
The equation of a curve is y = 4 + 5x + 6x 2 - 3x 3.
(a) Find the set of values of x for which y decreases as x increases. [4]
(b) It is given that y = 9x + k is a tangent to the curve.
Find the value of the constant k. [4]
Mark scheme
9(a) Differentiate to obtain 5 + 12 x − 9 x 2 B1
Attempt to find two critical values by solving quadratic equation or inequality M1
1 5 A1 SC B1 if no method for solving the quadratic.
Obtain values − and
3 3
1 5 A1FT SC B1 if no method for solving the quadratic.
Conclude x < − , x >
3 3
4
9(b) Equate first derivative to 9 and simplify to 3 term quadratic *M1
2 A1 SC B1 for solving 5 + 12 x − 9 x 2 = 9 without
Obtain x =
3 simplifying to a 3-term quadratic.
Use x-value and corresponding y-value to determine value of k DM1
28 A1 28
Obtain k = SC B1 for k = from solving 5 + 12 x − 9 x 2 = 9
9 9
without simplifying to a 3-term quadratic.
4
An arithmetic progression has first term 5 and common difference d, where d 2 0 . The second, fifth and eleventh terms of the arithmetic progression, in that order, are the first three terms of a geometric progression.
(a) Find the value of d. [3]
(b) The sum of the first 77 terms of the arithmetic progression is denoted by S77 . The sum of the first
10 terms of the geometric progression is denoted by G 10 .
Find the value of S77 - G 10 . [5]
Mark scheme
10(a) State or imply that first 3 terms of GP are 5 + d , 5 + 4d , 5 + 10d B1
Form equation (5 + 4d )2 = (5 + d )(5 + 10d ) or equivalent M1
Obtain d = 2.5 A1 Ignore 0 as a solution.
SC B1 Obtain d = 2.5 and 7.5, 15, 30 by trial and
improvement www.
Alternative Method for Question 10(a):
State or imply that first 3 terms of GP are 5 + d , 5 + 4d , 5 + 10d B1
5 − 5R M1 OE
(5 + d )R = 5 + 4d → d = , ( 5 + d ) R 2 = 5 + 10 d → R2 – 3R + 2 [= 0] Eliminates d.
R−4
Obtain d = 2.5 A1
3
10(b) Use correct formula for sum of AP with their value of d M1
Obtain or imply 7700 A1
State or imply GP is 7.5, 15, 30, ... B1
Use correct formula for sum of GP with their common ratio M1
Obtain S77 − G10 = 27.5 A1
5
The function f is defined by f (x) = 3 + 6x - 2x 2 for x ! R .
(a) Express f (x) in the form a - b (x - c) 2 , where a, b and c are constants, and state the range of f. [3]
(b) The graph of y = f (x) is transformed to the graph of y = h (x) by a reflection in one of the axes followed by a translation. It is given that the graph of y = h (x) has a minimum point at the origin.
Give details of the reflection and translation involved. [2]
The function g is defined by g (x) = 3 + 6x - 2x 2 for x G 0 .
(c) Sketch the graph of y = g (x) and explain why g is a one-one function. You are not required to find the coordinates of any intersections with the axes. [2]
(d) Sketch the graph of y = g -1 (x) on your diagram in (c), and find an expression for g -1 (x) .
You should label the two graphs in your diagram appropriately and show any relevant mirror line.
[4]
Additional page
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
Mark scheme
11(a) 3 B1
Obtain b = 2 and c =
2
15 3
2 B1
Obtain − 2 x −
2 2
15 15 B1 FT Following their value of a.
State range is y or f ( x ) with ⩽ given or clearly implied (not <)
2 2
3
11(b) State that reflection is in x-axis B1 Accept transformations in any order.
3 B1 FT Following their values of a and c in part (a).
− 2 Accept transformations in any order.
State or imply that translation is by or equivalent
15
2
2
11(c) Sketch the correct graph appearing in second and third quadrants only B1
State that each y-value is associated with a single x-value or equivalent B1 Accept passes the horizontal line test.
Ignore passes the vertical line test.
2
11(d) Sketch the correct graph with suitable labelling to distinguish the two curves B1 Appearing in third and fourth quadrants only.
Draw the line y = x B1 See above; no need to label the line.
Attempt correct process for finding the inverse function M1 Allowing use of ± and y so far.
3 15 1 A1 Must involve x at the conclusion.
Obtain − − x or equivalent
2 4 2
4
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On this page, free: the mark scheme for AS & A Level Mathematics 9709/11, October/November 2024 (9709_w24_ms_11), and the question paper (9709_w24_qp_11), to read or download, with each question's marking points written out underneath.
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Mathematics 9709/11 for October/November 2024 is out of 75 marks, with 1 hour 50 minutes to answer it.
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