Mathematics 9709/13 May/June 2023: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 1 Pure Mathematics 1, May/June 2023. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
y
6
B
5
C
4
3
A B′
C′
2
1
A′
0 x
0 1 2 3 4 5 6 7 8 9
The diagram shows the graph of y = f x, which consists of the two straight lines AB and BC. The lines A′ B′ and B′ C′ form the graph of y = g x, which is the result of applying a sequence of two transformations, in either order, to y = f x.
State fully the two transformations. [4]
Mark scheme
1 0 B2, 1, 0 B2 for fully correct, B1 with two elements correct.
{Translation} {} indicates different elements.
−2
{Stretch}{[scale] factor 2} {parallel to x-axis} B2, 1, 0 B2 for fully correct, B1 with two elements correct.
4 Transformations can be in either order.
The function f is defined for x ∈ > by f x = x2 − 6x + c, where c is a constant. It is given that f x > 2 for all values of x.
Find the set of possible values of c. [4]
@ A
2 5
Mark scheme
2 x 2 − 6 x + c > 2 leading to ( x − 3) − 9 + c > 2 M1 A1 M1 for completion of the square with an equation or in
2
equality with the ‘2’.
c > 11 − ( x − 3) ( x − 3)2 ≥ 0 M1 SOI
2
and
c > 11 A1
Alternative Method 1
dy M1 dy
= 2x − 6 = 0 M1 for differentiating and setting = 0.
dx dx
x=3 A1
When x = 3, y = 9 − 18 + c M1
[ −9 + c > 2 ] c > 11 A1
Alternative Method 2
x 2 − 6 x + c > 2 leading to x 2 − 6 x + c − 2[> 0] then use of ' b 2 − 4ac ' M1
36 − 4 (1)( c − 2 ) < 0 M1 A1 OE Must be correct inequality for M1.
c > 11 A1
4
(a) Give the complete expansion of x + . [2]
x
A @
2 5 2 1
(b) In the expansion of a + bx x + , the coefficient of x is zero and the coefficient of is 80. x x
Find the values of the constants a and b. [4]
Mark scheme
3(a) 80 80 32 B2, 1, 0 B2, all terms correct, B1 5 terms correct. Terms must be
x5 + 10 x3 + 40 x + + + or x 5 + 10 x 3 + 40 x + 80 x −1 + 80 x −3 + 32 x −5 simplified. Lists of terms allowed.
x x3 x5
2
3(b) ( )
their 40 × a + their coefficient of x −1 × b = 0 M1 Coefficients of a and b must be non-zero, allow x’s so
long as they are dealt with correctly.
( their coefficient of x ) × a + ( their coefficient of x ) × b = 80
−1 −3 M1 Coefficients of a and b must be non-zero, allow x’s as
long as they are dealt with correctly.
a=2 b = −1 A1 A1 Dependent on both M marks, may be seen without
working.
4
(a) Show that the equation
3 tan2 x − 3 sin2 x − 4 = 0 may be expressed in the form a cos4 x + b cos2 x + c = 0, where a, b and c are constants to be found. [3]
(b) Hence solve the equation 3 tan2 x − 3 sin2 x − 4 = 0 for 0Å ≤ x ≤ 180Å. [4]
Mark scheme
4(a) 3sin 2 x − 3sin 2 xcos 2 x − 4cos 2 x [ = 0] M1 sin 2 x
Replace tan 2 x with and multiply by cos 2 x .
cos 2 x
( ) ( )
3 1 − cos 2 x − 3 1 − cos 2 x cos 2 x − 4cos 2 x [ = 0] M1 Replace sin 2 x by 1 − cos 2 x twice.
3cos 4 x − 10cos 2 x + 3 = 0 or −3cos 4 x + 10cos 2 x − 3 = 0 A1 Or multiple of these equations.
3
4(b) ( 3cos x − 1)( cos x − 3) [ = 0]
2 2 M1 OE, using their equation in the given form.
Allow unusual notation if meaning is clear.
1 A1 SOI Answer only SC B1.
cos x = [ ± ]
3
54.7º, A1
125.3º A1 FT Only other answer and must be from correct
factorisation for A1.
FT for 180° − their first answer .
Answers only SC B1, SC B1 FT.
4
A circle has equation x − 12 + y + 42 = 40. A line with equation y = x − 9 intersects the circle at points A and B.
(a) Find the coordinates of the two points of intersection. [4]
(b) Find an equation of the circle with diameter AB. [3]
Mark scheme
5(a) ( x − 1)2 + ( x − 9 + 4 )2 = 40 M1 Substitute line into circle.
x 2 − 6 x − 7 [ = 0] leading to ( x + 1)( x − 7 ) [ = 0] M1 Simplify to 3-term quadratic and factorise OE.
(‒1, ‒10), (7, ‒2) or x = ‒1 and 7, y = ‒10 and ‒2 A1 A1 Answers only SC B1, SC B1 but must see a correct
quadratic equation.
4
5(b) their x1 + their x2 their y1 + their y2 M1 Expect (3, ‒6).
[C is mid-point =] ( , )
2 2
M1 Expect
( their x − their 3)2 + ( their y − their ( −6 ) ) 32 .
2
Radius = OR
(
their √ ( 7 − ( −1) ) + ( −2 − ( −10 ) )
2 2
)/2
( x − 3)2 + ( y + 6 )2 = 32 A1 OE
3
A r cm
O 1 rad
r cm
B
The diagram shows a sector OAB of a circle with centre O and radius r cm. Angle AOB = 1 radians.
It is given that the length of the arc AB is 9.6 cm and that the area of the sector OAB is 76.8 cm2.
(a) Find the area of the shaded region. [5]
(b) Find the perimeter of the shaded region. [2]
5
Mark scheme
6(a) ½ r 2θ 76.8 1 9.62 M1 Eliminate θ or r using correct formulae SOI.
= or 2 θ = 76.8
rθ 9.6 2 θ
r = 16 A1
θ = 0.6 A1 Accept 34.4o
ΔOAB = ½ × their 162 × sin their 0.6 M1 Allow Segment = 76.8 –½ × their 162 × sin their 0.6.
Expect 72.27 .
[Area = 76.8 ‒ 72.27 =] 4.53 A1 AWRT
5
6(b) AB = 2 × 16 × sin 0.3 OR AB 2 = 16 2 + 16 2 − 2 × 16 2 cos 0.6 M1 Any valid method with their r, θ. Expect AB = 9.46.
Perimeter = 9.6 + 9.46 = 19.1 A1 AWRT
2
The function f is defined by f x = 2 − for x > −2. x+2
(a) State the range of f. [1]
(b) Obtain an expression for f −1 x and state the domain of f −1 . [4]
The function g is defined by g x = x + 3 for x > 0.
ax + b
(c) Obtain an expression for fg x giving your answer in the form , where a, b, c and d are cx + d integers. [3]
a2
Mark scheme
7(a) [ y ] < 2 OR [f(x)] < 2 B1 OE e.g. f < 2, ( −∞, 2 ) , −∞ < f [ x ] < 2 .
Do not accept x < 2 or f(x) ⩽ 2.
1
7(b) 5 M1 5
y = 2− leading to y ( x + 2 ) = 2 ( x + 2 ) − 5 leading to xy + 2 y = 2 x − 1 or = 2 − y (allow sign errors).
x+2 x+2
2 y + 1 = 2 x − xy leading to 2 y + 1 = x ( 2 − y ) DM1 5
or = x + 2 (allow sign errors).
2− y
2y +1 2x + 1 A1 5
x= → f −1 ( x ) = OE or y = − 2.
2− y 2− x 2− x
Domain is x < 2 B1 FT FT on the numerical part of their range from part (a),
including x ≠ 2 not penalized.
No FT for x ,x = k , x ≠ k .
4
7(c) 5 B1
fg ( x ) = 2 −
x +3+ 2
2 ( x + 5) − 5 2 ( x + 5) 5 M1 Use of their common denominator.
= or −
x+5 x+5 x+5
2x + 5 A1
=
x+5
3
A progression has first term a and second term , where a is a positive constant. a+2
(a) For the case where the progression is geometric and the sum to infinity is 264, find the value of a. [5]
(b) For the case where the progression is arithmetic and a = 6, determine the least value of n required for the sum of the first n terms to be less than −480. [5]
2
Mark scheme
8(a) a B1 OE SOI
r=
a+2
a M1 Use of S∞ formula.
= 264
a
1−
a+2
a ( a + 2) a ( a + 2) M1* Process to a 3 term quadratic or a 3 term cubic.
= 264 leading to = 264 leading to a 2 + 2a − 528 [ = 0] May contain terms on LHS and RHS.
a+2−a 2
( a − 22 )( a + 24 ) [ = 0] DM1 Attempt to solve.
a = 22 (only) A1 22 without working SC DB1 (dep on 2nd M1).
5
8(b) 62 3 B1
d= − 6 =−
6+2 2
n −3 M1* Forming an inequation with their numerical d. May use
12 + ( n − 1) [<] − 480 an equality.
2 2
[3] ( n 2 − 9n − 640 ) [> 0] A1 OE May contain terms on LHS and RHS.
9 ± 81 + 2560 DM1 OE. Expect 30.19 . Working for solution must be
[n =] shown.
2
31 only A1 Must come from a correct first inequality (or an
equality).
31 no working SC DB1 (dep on correct quadratic and
correct inequality/equality).
5
A curve which passes through 0, 3 has equation y = f x. It is given that f ′ x = 1 − . x − 13
(a) Find the equation of the curve. [4]
The tangent to the curve at 0, 3 intersects the curve again at one other point, P.
(b) Show that the x-coordinate of P satisfies the equation 2x + 1 x − 12 − 1 = 0. [4]
(c) Verify that x = 32 satisfies this equation and hence find the y-coordinate of P. [2]
Mark scheme
9(a) [ y =] x { + ( x − 1)−2 } [ +c ] B1 B1 May be unsimplified.
Sub x = 0, y = 3 leading to 3 = 0 + 1 + c M1 Substitution into an integral, expect c = 2.
y = x + ( x − 1) + 2 f ( x ) = x + ( x − 1) + 2 A1 −2 ( x − 1)
−2 −2 −2
or −2
or must be simplified.
( −2 )( x − 1) 2
−2
4
9(b) [Gradient of tangent =] f ′(0) = 3 B1
Equation of tangent is y − 3 = their gradient at x = 0 ( x − 0 ) M1* Expect y = 3x + 3 , normal gets M0.
Intersection given by 3 x + 3 = x + ( x − 1) + 2 DM1 FT their equation from part (a).
−2
1 A1 WWW AG
→ ( 2 x + 1)( x − 1) − 1 = 0 or solve equation before given
2
2x + 1 =
( x − 1) 2
form reached and show solution ( x = 3 / 2 ) satisfies given result
4
9(c) 3 B1 1
2
leading to ( 2 x + 1)( x − 1) − 1 leading to 4 × ¼ − 1 = 0 .
2
Substitute x = Evaluation of each bracket must be shown. Allow
2 2
3
for second bracket. Solution of ( 2 x + 1)( x − 1) − 1 = 0
2
Hence x =
2
If shown in (b) must be referenced here (in part (c)) is acceptable.
3 B1
When x = y = 7½
2
2
y
3 y = 9x − 2x + 1 2
A
1 12 , 5 21
B 7 12 , 3 12
x
O
The diagram shows the points A 1 12 , 5 12 and B 7 12 , 3 12 lying on the curve with equation
3 y = 9x − 2x + 1 2 .
(a) Find the coordinates of the maximum point of the curve. [4]
(b) Verify that the line AB is the normal to the curve at A. [3]
(c) Find the area of the shaded region. [5]
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
10(a) dy 3 B1, B1 Including ‘+c’ makes the second term B0.
dx = 9 + − 2 ( 2 x + 1) × 2
1/2
9 − 3 ( 2 x + 1) M1 Set differential to zero and solve by squaring SOI.
1/2
= 0 leading to 2 x + 1 = 9
Beware 92 − 32 ( 2 x + 1) = 0 M0A0.
2 x + 1 = 3 or 2 x + 1 = ±9 get M0.
Max point = (4, 9) A1 WWW y = 9 must come from original equation.
4
10(b) dy M1 dy
When x = 1½, shows substitution or =3 Substituting x = 1½ into their .
dx dx
5½ − 3½ −1 M1 Substituting into a correct expression for mAB.
Gradient of AB is =
1½ − 7½ 3
1 A1
− x3 = −1 . [Hence AB is the normal]
3
Alternative method for Question 10(b)
dy M1
When x = 1½ = 3 ,[ perpendicular gradient is -1/3]
dx
−x M1
Perpendicular through A has equation y = + 6 which contains B(7.5,3.5) A1
3
leading to AB is a normal to the curve at A
3
10(c) 5 B1 B1 Integrating y with respect to x.
9 x 2 − ( 2 x + 1) 2
+
2 5 ×2
2
9 2 1 2.5 9 1 2.5 M1 OE
7.5 − ( 2 × 7.5 + 1) − 1.5 − ( 2 × 1.5 + 1) or
2
Apply limits 1½ to 7½ to an integral. Working must be
2 5 2 5
seen. Expect 44.6 .
9 225 1024 81 32 1933 149
× − − − or − or 48.325 – 3.725
2 4 5 8 5 40 40
15 B1 SOI
2
1 1 1 −1 Area of trapezium. May be seen combined with the area
5
2 2
+ 3
2
× 6 or
3
( x + 6)dx =
3 under the curve integral.
2
−1 15 2 15 −1 3
2
3 285 69
× + 6 × − × + 6 × or − [ = 27]
6 2
2 6 2
2 8 8
[Shaded area = 44.6 – 27 =] 17.6 A1 SC B1 if no substitution of the limits seen.
5
10(c) Alternative method for Question 10(c)
15 15 M1 Finding the equation of AB and subtracting from the
2 2
3
−1 28 3
equation of the curve.
A= ((9 x − (
3
)
2x + 1 2 ) −
3
x + 6 )dx =
3
(( x − ( 2 x + 1) 2 − 6)dx
3
2 2
5 A1 A1
28 2 − ( 2 x + 1) 2
= x − 6x +
3× 2 5×2
2
127 −49 M1 Apply limits 1½ to 7½ to an integral. Working must be
− seen.
10 10
17.6 A1 SC B1 if no substitution of limits seen.
5
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