Mathematics 9709/13 May/June 2020: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 1 Pure Mathematics 1, May/June 2020. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
Find the set of values of m for which the line with equation y = mx + 1 and the curve with equation y = 3x2 + 2x + 4 intersect at two distinct points. [4]
dy 1 −1
Mark scheme
1 3x 2 + 2 x + 4 = mx + 1 → 3 x 2 + x ( 2 − m ) + 3 ( = 0 ) B1
( 2 − m )2 − 36 SOI M1
(m + 4)(m ‒ 8) (>/= 0) or 2 ‒ m >/= 6 and 2 ‒ m </= ‒6 OE A1
m < ‒4, m > 8 WWW A1
Alternative method for question 1
dy M1
= 6x + 2 → m = 6x + 2 → 3x 2 + 2 x + 4 = ( 6 x + 2 ) x + 1
dx
x = ±1 A1
m = ±6 + 2 → m = 8 or − 4 A1
m < ‒4, m > 8 WWW A1
4
In each of parts (a), (b) and (c), the graph shown with solid lines has equation y = f x. The graph shown with broken lines is a transformation of y = f x.
(a) y
3
2 y = f x
1
x
−3 −2 −1 0 1 2 3
State, in terms of f, the equation of the graph shown with broken lines. [1]
(b) y
6
5
4
3
2 y = f x
1
x
0 1 2 3
State, in terms of f, the equation of the graph shown with broken lines. [1]
(c) y
3
2 y = f x
1
x
−3 −2 −1 0 1 2 3
−1
−2
State, in terms of f, the equation of the graph shown with broken lines. [2]
Mark scheme
3(a) ( y ) = f ( −x) B1
1
3(b) ( y ) = 2f ( x ) B1
1
3(c) ( y ) = f ( x + 4) − 3 B1 B1
2
(a) Expand 1 + a5 in ascending powers of a up to and including the term in a3 . [1]
(b) Hence expand 1 + x + x2 5 in ascending powers of x up to and including the term in x3 , simplifying your answer. [3]
Mark scheme
4(a) 1 + 5a + 10 a 2 + 10 a 3 + ... B1
1
4(b)
( ) ( ) + 10 ( x + x ) + ... SOI M1
2 2 3
1 + 5 x + x 2 + 10 x + x 2
( ) ( ) ( )
1 + 5 x + x 2 + 10 x 2 + 2 x 3 + ... + 10 x 3 + ... + ... SOI A1
1 + 5 x + 15 x 2 + 30 x 3 + ... A1
3
O
5
A B
13
P
The diagram shows a cord going around a pulley and a pin. The pulley is modelled as a circle with centre O and radius 5 cm. The thickness of the cord and the size of the pin P can be neglected. The pin is situated 13 cm vertically below O. Points A and B are on the circumference of the circle such that AP and BP are tangents to the circle. The cord passes over the major arc AB of the circle and under the pin such that the cord is taut.
Calculate the length of the cord. [6]
Mark scheme
5 5 M1 A1
cos POA = → POA = 1.17 ( 6 ) Allow 67.4°
13
12 12
or sin = or tan =
13 5
Reflex AOB = 2π − 2 ×their 1.17 ( 6) OE in degrees M1
or minor arc AB = 5×2×their1.17(6)
Major arc = 5 × their 3.93(1) M1
or 2π × 5 - their 11.7(6)
AP (or BP) = 132 − 52 = 12 B1
Cord length = 43.7 A1
6
A point P is moving along a curve in such a way that the x-coordinate of P is increasing at a constant
1 rate of 2 units per minute. The equation of the curve is y = 5x − 1 2 .
(a) Find the rate at which the y-coordinate is increasing when x = 1. [4]
(b) Find the value of x when the y-coordinate is increasing at 85 units per minute. [3]
tan 1 tan 1 2
Mark scheme
6(a) dy 1 −1/2 B1 B1
= ( 5x − 1) × [5]
dx 2
dy dy M1
Use = 2 × their when x = 1
dt dx
5 A1
2
4
6(b) 5 5 M1
2 × their ( 5 x − 1)−1/2 = oe
2 8
( 5 x − 1)1/2 = 8 A1
x = 13 A1
3
(a) Show that + . [4]
1 + cos 1 1 − cos 1 sin 1 cos 1
tan 1 tan 1 6
(b) Hence solve the equation + = for 0Å < 1 < 180Å. [4]
1 + cos 1 1 − cos 1 tan 1
Mark scheme
7(a) tan θ tan θ tan θ (1 − cos θ ) + tan θ (1 + cos θ ) M1
+ =
1 + cos θ 1 − cos θ 1 − cos 2θ
2 tan θ M1
=
sin 2θ
2sin θ M1
=
cos θ sin 2θ
2 A1
= AG
sin θ cosθ
4
7(b) 2 6cos θ M1
=
sin θ cosθ sin θ
1 A1
cos 2θ = → cos θ = ( ± ) 0.5774
3
54.7º, 125.3º A1
(FT for 180º ‒ 1st solution) A1FT
4
The first term of a progression is sin2 1, where 0 < 1 < 12 π. The second term of the progression is sin2 1 cos2 1.
(a) Given that the progression is geometric, find the sum to infinity. [3]
It is now given instead that the progression is arithmetic.
(b) (i) Find the common difference of the progression in terms of sin 1. [3]
(ii) Find the sum of the first 16 terms when 1 = 13 π. [3]
Mark scheme
8(a) r = cos 2θ SOI M1
sin 2θ M1
S∞ =
1 − cos2θ
1 A1
3
8(b)(i) d = sin 2θ cos 2θ − sin 2θ M1
(
sin 2θ cos 2θ − 1 ) M1
−sin 4θ A1
3
8(b)(ii) 16 M1
Use of S16 = [ 2a + 15d ]
2
3 9 A1
With both a = and d = −
4 16
1 A1
S16 = −55
2
3
The functions f and g are defined by
f x = x2 − 4x + 3 for x > c, where c is a constant,
1 g x = for x > −1. x+1
(a) Express f x in the form x − a2 + b. [2]
It is given that f is a one-one function.
(b) State the smallest possible value of c. [1]
It is now given that c = 5.
(c) Find an expression for f −1 x and state the domain of f −1 . [3]
(d) Find an expression for gf x and state the range of gf. [3]
Mark scheme
9(a) ( x − 2 )2 [ −1] B1 B1
2
9(b) Smallest c = 2 B1FT
(FT on their part (a))
1
9(c) y = ( x − 2) − 1 → ( x − 2) = y + 1
2 2 *M1
x = 2(±) y +1 DM1
(f −1 ( x )) = 2 + x + 1 for x > 8 A1
3
9(d) 1 1 B1
gf ( x ) = = OE
( x − 2) − 1 + 1 ( x − 2)
2 2
1 B1 B1
Range of gf is 0 < gf(x) <
9
3
(a) The coordinates of two points A and B are −7, 3 and 5, 11 respectively.
Show that the equation of the perpendicular bisector of AB is 3x + 2y = 11. [4]
(b) A circle passes through A and B and its centre lies on the line 12x − 5y = 70.
Find an equation of the circle. [5]
Mark scheme
10(a) Mid-point is (‒1, 7) B1
Gradient, m, of AB is 8/12 OE B1
12 M1
y−7 =− ( x + 1)
8
3 x + 2 y = 11 AG A1
4
10(b) Solve simultaneously 12 x − 5 y = 70 and their 3 x + 2 y = 11 M1
x = 5, y = −2 A1
Attempt to find distance between their (5, ‒2) and either (‒7,3) or (5, 11) M1
A1
( r ) = 122 + 52 or 132 + 0 = 13
Equation of circle is ( x − 5) + ( y + 2 ) = 169
2 2 A1
5
y
A y = x3 − 2bx2 + b2 x
x
O a b
The diagram shows part of the curve with equation y = x3 − 2bx2 + b2 x and the line OA, where A is the maximum point on the curve. The x-coordinate of A is a and the curve has a minimum point at b, 0, where a and b are positive constants.
(a) Show that b = 3a. [4]
(b) Show that the area of the shaded region between the line and the curve is ka4 , where k is a fraction to be found. [7]
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) dy B1
= 3x 2 − 4bx + b 2
dx
3x 2 − 4bx + b 2 = 0 → ( 3 x − b )( x − b ) ( = 0 ) M1
b A1
x= or b
3
b A1
a= → b = 3a AG
3
Alternative method for question 11(a)
dy B1
= 3x 2 − 4bx + b 2
dx
dy M1
Sub b = 3a & obtain = 0 when x = a and when x = 3a
dx
d2 y A1
= 6 x − 12 a
dx 2
< 0 Max at x = a and > 0 Min at x = 3a. Hence ܾ = 3ܽ AG A1
4
11(b) (
Area under curve = x 3 − 6ax 2 + 9a 2 x dx ) M1
x4 9a 2 x 2 B2,1,0
− 2ax 3 +
4 2
a4 9a 4 11a 4 M1
− 2a 4 + =
4 2 4
(M1 for applying limits 0 → a)
3 3
When x = a, y = a − 6a + 9a = 4a
3 3 B1
1 M1
Area under line = a × their 4a3
2
11a 4 3 A1
Shaded area = − 2a 4 = a 4
4 4
7
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On this page, free: the mark scheme for AS & A Level Mathematics 9709/13, May/June 2020 (9709_s20_ms_13), and the question paper (9709_s20_qp_13), to read or download, with each question's marking points written out underneath.
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Mathematics 9709/13 for May/June 2020 is out of 75 marks, with 1 hour 50 minutes to answer it.
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