Mathematics 9709/12 February/March 2023: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 1 Pure Mathematics 1, February/March 2023. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
A line has equation y = 3x − 2k and a curve has equation y = x2 − kx + 2, where k is a constant.
Show that the line and the curve meet for all values of k. [4]
Mark scheme
1 x2 − kx + 2 = 3x − 2k leading to x2 − x ( k + 3) + ( 2 + 2k ) [= 0] M1 3-term quadratic, may be implied in the
discriminant.
b2 − 4ac = ( k + 3) − 8 (1 + k ) (ignore ‘= 0’ at this stage) DM1 Cannot just be seen in the quadratic formula.
2
= ( k − 1) accept ( k − 1)( k − 1) A1 Or use of calculus to show minimum of zero at
2
k = 1 or sketch of f ( k ) = k 2 − 2k + 1.
0 Hence will meet for all values of k A1 Clear conclusion.
4
A function f is defined by f x = x2 − 2x + 5 for x ∈ >. A sequence of transformations is applied in the following order to the graph of y = f x to give the graph of y = g x.
Stretch parallel to the x-axis with scale factor 21
Reflection in the y-axis
Stretch parallel to the y-axis with scale factor 3
Find g x, giving your answer in the form ax2 + bx + c, where a, b and c are constants. [4]
Mark scheme
2 Stretch: ( 2 x ) − 2 ( 2 x ) + 5 or ( x − 1) + 4 leading to ( 2 x − 1) + 4 M1 Replacing x by 2x.
2 2 2
Reflection: ( −2 x ) − 2 ( −2 x ) + 5 or ( −2 x − 1) + 4 M1 Replacing x by – x. FT on their stretch.
2 2
Stretch: 3 ( −2 x ) − 2 ( −2 x ) + 5 or 3 ( −2 x − 1) + 4
2 2
M1 Multiplying the whole function by 3. FT on their
(stretch plus reflection).
12 x 2 + 12 x + 15 A1
4
1 3x + 12 and a point is moving along the curve.
A curve has equation y = 60
Find the x-coordinate of the point on the curve at which the x- and y-coordinates are increasing at the same rate. [4]
Mark scheme
3 dy 1 B1 B1 1
= ( 3x + 1) × 2 ×3 May see (18x + 6 ) .
dx 60 60
1 M1 dy
( 3x + 1) = 1 Equate their to 1.
10 dx
x=3 A1
4
The circumference round the trunk of a large tree is measured and found to be 5.00 m. After one year the circumference is measured again and found to be 5.02 m.
(a) Given that the circumferences at yearly intervals form an arithmetic progression, find the circumference 20 years after the first measurement. [2]
(b) Given instead that the circumferences at yearly intervals form a geometric progression, find the circumference 20 years after the first measurement. [3]
Mark scheme
4(a) 5.00 + 20 × 0.02 or 5.02 + 19 × 0.02 M1 Allow for a = 5, n = 20 with d = 0.02 only.
a = 5, n = 21(OE) with d = 0.2 gets M1 only.
5.40 A1
2
4(b) 5.02 251 B1
r= = 1.004 or
5 250
5.00 × ( their1.004) or 5.02 × ( their1.004 ) M1 Allow a = 5, n = 20.
20 19
5.42 A1 Any correct rounding of 5.41557108 .
3
Points A 7, 12 and B lie on a circle with centre −2, 5. The line AB has equation y = −2x + 26.
Find the coordinates of B. [6]
A @ x a 7
Mark scheme
5 r 2 = ( 7 + 2 ) + (12 − 5) B1 Expect 130, may use AC rather than r.
2 2
Equation of circle is ( x + 2 ) + ( y − 5) = 130 B1 FT OE FT their 130, may use distance BC rather than
2 2
circle.
M1 Substitute y = −2 x + 26 into a circle equation.
( x + 2) + ( −2 x + 21) = 130
2 2
5x2 − 80 x + 315 [= 0] leading to [5] ( x − 9)( x − 7 ) M1 Factorisation OE must be seen.
x=9 A1 With or without x = 7.
y = 8 OR (9, 8) A1 y = 8 or ( 9,8) only. Both A1’s dependent on the
first M1.
6
In the expansion of + , it is given that a x2 the coefficient of x4
= 3. the coefficient of x
Find the possible values of the constant a. [6]
Mark scheme
6 6 6 5 2 5 2 B1 B1 Coefficients x 4 & x . Can be seen in an
x a x a x a x a
7C1 2 or 7C6 2 7C2 2 or 7C5 2 expansion.
a x a x a x a x
7 M1 OE. Allow extraneous x 4 and x at this stage;
5
a =3 numerator and denominator must be functions of
a.
21
3 Allow errors in evaluation of the combinations.
a
A1 Completely correct.
1 A1 1
a2 = SOI (implied by a = ).
9 3
1 A1 Allow ± 0.333 .
a=±
3
6
(a) By first obtaining a quadratic equation in cos 1, solve the equation tan 1 sin 1 = 1 for 0Å < 1 < 360Å. [5]
tan 1 sin 1
(b) Show that − tan 1 sin 1. [3]
sin 1 tan 1
Mark scheme
7(a) tan θ sin θ = 1 leading to sin 2θ = cos θ M1 sin θ
Use of tan θ = and multiplication by
cos θ
cos θ .
1 − cos2θ = cosθ or cos2θ + cosθ −1 [= 0] M1 Use of trig identity to form a 3-term quadratic.
−1 ± 5 M1 Use of formula or completion of the square must
[cos θ =] be seen on a 3-term quadratic. Expect 0.6180 .
2
51.8º, A1 Both A marks dependent on the 2nd M1.
308.2º A1 FT FT for (360º ‒ 1st soln), A0 if extra solutions in
range.
Radians 0.905 and 5.38, A1 only for both.
5
7(b) tan θ sin θ sin θ sin θ cos θ 1 M1 sin θ
− = − = − cos θ Use tan θ = twice with correct use of
sin θ tan θ sin θ cos θ sin θ cos θ cos θ
fractions.
1 − cos 2θ sin 2θ M1 Use 1 − cos 2θ = sin 2θ with correct use of
= =
cos θ cos θ fractions.
= tan θ sin θ A1 WWW
3
C
4 cm
A 5 cm D 3 cm B
The diagram shows triangle ABC in which angle B is a right angle. The length of AB is 8 cm and the length of BC is 4 cm. The point D on AB is such that AD = 5 cm. The sector DAC is part of a circle with centre D.
(a) Find the perimeter of the shaded region. [5]
(b) Find the area of the shaded region. [3]
Mark scheme
8(a) 4 4 3 M1 −1 4
tan BDC = or sin BDC = or cos BDC = used to find ADC May use cosine rule or CAD = tan .
3 5 5 8
BDC = 0.927 [3] → ADC = π − 0.927 [3] [= 2.214 to 2.215 ] A1 Allow degrees, 126.87, and 0.7048 π or 0.705 π .
Arc AC = 5 × their 2.214 M1 θ
Use of rθ or .2πr Expect 11.07 .
360
M1 Expect 8.94 .
AC = 82 + 42 or 2 × 5 × sin1.107
[Perimeter =11.07 + 8.94 =] 20.0 A1 Accept AWRT [20.01, 20.02].
5
8(b) Sector ACD = ½ × 52 × their 2.214 M1 1 2 θ
See use of r θ or .πr 2 . Expect 27.7 .
2 360
1 2 M1 Subtracting the area of Δ ADC, expect − 10.
Subtracting the area of ΔADC = ½ × 5 × 4 or 5 sin their 2.214 or
2
1 1
×8 × 4 − × 3 × 4
2 2
Shaded area = 27.7 −10 =17.7 A1 Accept AWRT [17.67, 17.68]. Correct answer
cannot come from an angle of 2.215 .
3
The function f is defined by f x = −3x2 + 2 for x ≤ −1.
(a) State the range of f. [1]
(b) Find an expression for f −1 x. [3]
The function g is defined by g x = −x2 − 1 for x ≤ −1.
(c) Solve the equation fg x − gf x + 8 = 0. [5]
dy −1
Mark scheme
9(a) [ y] −1 B1 Accept f or f ( x ) − 1 , −∞ < y −1, ( −∞, −1] .
Do not accept x −1 .
1
9(b) 2− y 2− x M1
y = −3x 2 + 2 rearranged to 3 x 2 = 2 − y , leading to x 2 = or y =
2
3 3
2− y 2 − x A1 A1 2− x x−2
x = [±] → [f −1 ( x )] = − A1 for minus, A1 for , allow − .
3 3 3 −3
3
9(c)
( )
fg ( x ) = −3 − x 2 − 1 + 2
M1 SOI expect −3x 4 − 6 x 2 − 1 .
2
( )
gf ( x ) = − −3 x 2 + 2 − 1
M1 SOI expect −9 x 4 + 12 x 2 − 5 .
2
fg ( x ) − gf ( x ) + 8 = 0 leading to 6x4 −18x2 + 12 [= 0] A1 OE
[6] ( x2 −1)( x2 − 2) [= 0] or formula or completion of the square M1 Solving a 3-term quadratic equation in x 2 must be
seen.
x = −1 , − 2 only these two solutions A1 Allow − √ 1, − 1.41[4] Answers only SC B1.
5
At the point 4, −1 on a curve, the gradient of the curve is − 32 . It is given that = x 2 + k, where k dx is a constant.
(a) Show that k = −2. [1]
(b) Find the equation of the curve. [4]
(c) Find the coordinates of the stationary point. [3]
(d) Determine the nature of the stationary point. [2]
Mark scheme
10(a) 3 1 B1 −
1
1
− = + k leading to k = −2 AG Need to see 4 2
evaluated as 1
or better.
2 2
2
4
1
10(b) 1 M1 A1 1
[ y ] = 2 x 2 − 2 x [ +c ] Allow
x2
− 2x .
1
2
−1 = 4 − 8 + c M1 Substitute x = 4, y = −1 (c present) Expect c = 3.
1 A1 Allow if f(x) = or y = anywhere in the solution.
y = 2 x − 2 x + 3 or y = 2 x − 2x + 3
2
4
10(c) x −1/2 − 2 = 0 M1 dy
Set their to zero.
dx
1 A1 2
x= 1 1 1 1
If = ± max of M1A1 if ,3 seen.
4 2 4 4 2
(¼, 3½) A1
3
10(d) d2 y 1 − 32 B1
=− x
dx 2 2
< 0 (or −4 ) hence Maximum DB1 1
WWW Ignore extra solutions from x = − .
4
2
y
A 5, 2
x
O
B 2, −1
x = y2 + 1
The diagram shows the curve with equation x = y2 + 1. The points A 5, 2 and B 2, −1 lie on the curve.
(a) Find an equation of the line AB. [2]
(b) Find the volume of revolution when the region between the curve and the line AB is rotated through 360Å about the y-axis. [9]
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) 2 − ( −1) M1 Expect 1, must be from Δy / Δx .
Gradient of AB =
5−2
Equation of AB is y − 2 = 1( x − 5) or y + 1 = 1( x − 2 ) A1 OE. Expect y = x − 3 .
2
11(b)
[ π ] x 2dy = [ π ] ( y 2 + 1) dy = [ π ] ( y 4 + 2 y 2 + 1) dy M1 For curve: Attempt to square y 2 + 1 and attempt
2
integration.
Subtracting curve equation from line equation
before squaring is M0.
Integration before squaring M0.
y5 2 y3 A2, 1, 0
[ ] + + y
π
5 3
[ π] ( y + 3) dy = [ π] ( y 2 + 6 y + 9 ) dy M1 For line: Attempt to square their y + 3 and attempt
2
integration.
y3 ( y + 3 )3 A2, 1, 0 Not available for incorrect line equations.
[ π ] + 3 y + 9 y or [ π]
2
3 3
8 32 16 1 2 DM1 Apply limits −1 → 2 to either integral providing
[ π] + 12 + 18 − − + 3 − 9 or [ π]
1
+ + 2 − − − − 1 3
3 3 5 3 5 3 they have been awarded M1. Expect 15 [ π]
5
and/or 39[ π] . Some evidence of substitution of
both − 1 and 2 must be seen. Dependent on at least
one of the first 2 M1 marks.
3 DM1 Appropriate subtraction. Dependent on at least one
Volume = [ π] (39 ‒ 15 ) of the first 2 M1 marks.
5
2 117 A1
= 23 π or π or awrt 73.5[1327]
5 5
9
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Mathematics 9709/12 for February/March 2023 is out of 75 marks, with 1 hour 50 minutes to answer it.
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