Mathematics 9709/13 October/November 2023: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 1 Pure Mathematics 1, October/November 2023. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
A curve is such that its gradient at a point x, y is given by = x − 3x 2 . It is given that the curve dx passes through the point 4, 1.
Find the equation of the curve. [4]
Mark scheme
1 B1 B1 Any unsimplified correct form, ISW for extra terms,
x 2 −3x 2
1
allow lists.
[ y ] = 1 [ +c ]
2
2
1 = 8 ‒ 12 + c M1 Substitute (into an integrated expression)
x = 4, y = 1. c must be present. Expect c = 5.
1 2 1 A1 Condone c = 5 as the final line so long as ‘y =’
y= x − 6 x 2 + 5 , allow f(x) = present.
2
4
The circle with equation x − 32 + y − 52 = 40 intersects the y-axis at points A and B.
(a) Find the y-coordinates of A and B, expressing your answers in terms of surds. [2]
(b) Find the equation of the circle which has AB as its diameter. [2]
Mark scheme
2(a) ( 0 − 3)2 + ( y − 5)2 = 40 M1 OE. Substitute x = 0, may use y 2 − 10 y − 6 = 0 .
y = 5 ± 31 A1 OE. Must be surd form.
2
2(b)
x + ( y − 5) = 31 Allow ( x − 0)
2 2 2 B1FT
B1FT
B1 FT for their 5 and B1 FT for their 31. Don’t allow
surd form.
2
(a) Show that the equation
5 cos 1 − sin 1 tan 1 + 1 = 0 may be expressed in the form a cos2 1 + b cos 1 + c = 0, where a, b and c are constants to be found. [3]
(b) Hence solve the equation 5 cos 1 − sin 1 tan 1 + 1 = 0 for 0 < 1 < 2π. [4]
Mark scheme
3(a) 5cos2θ − sin 2θ + cosθ [= 0] M1
Multiply by cosθ and replace tanθ by
sin θ
.
cos θ
( )
5cos2θ − 1 − cos2θ + cosθ [ = 0] M1
6cos2θ + cosθ − 1 = 0 A1 Missing ‘= 0’ can be condoned if ‘= 0’ appears
earlier.
3
3(b) (3cosθ − 1)(2cosθ + 1) = 0 M1 1 1
Must have 3 term quadratic, expect cosθ = , − .
3 2
Factors (OE) must be shown.
2π 4π A1 A1 For A1 FT is for both 2π ‒ 1st solutions.
θ = 1.23 ; {2.09 or } ; {5.05 and 4.19 (allow )} A1 FT
3 3
4
(a) Expand the following in ascending powers of x up to and including the term in x2 .
(i) 1 + 2x5 . [1]
(ii) 1 − ax6 , where a is a constant. [2]
In the expansion of 1 + 2x5 1 − ax6 , the coefficient of x2 is −5.
(b) Find the possible values of a. [4]
Mark scheme
4(a)(i) 1 + 10 x + 40 x2 B1 Ignore any additional terms (ISW).
Allow x 0 or 1 x 0 for the first term, allow lists.
1
4(a)(ii) 1−6ax +15a2 x2 B2, 1, 0 Ignore any additional terms (ISW).
Allow x 0 or 1 x 0 for the first term, allow lists.
2
4(b) 15a 2 − 60a + 40 = −5 M1 A1 Correct 3 products from their expansions for M1.
Condone inclusion of x 2 for M1.
[15] ( a − 1)( a − 3) [ = 0] OE DM1 OE. Rearranging and solving a quadratic.
a = 1 and 3 A1 Special case: If M1 A1 DM0 scored then SC B1 can
be awarded for correct answers.
4
The first, second and third terms of a geometric progression are 2p + 6, 5p and 8p + 2 respectively.
(a) Find the possible values of the constant p. [3]
(b) One of the values of p found in (a) is a negative fraction.
Use this value of p to find the sum to infinity of this progression. [4]
Mark scheme
5(a) 5p 8p + 2 M1 OE. Setting up a valid relationship in terms of p.
=
2p + 6 5p
9 p 2 − 52 p − 12 [ = 0] DM1 OE. Simplifying to a 3 term quadratic equation, only
condone sign errors.
( 9 p + 2)( p − 6) = 0 leading to p = −92 and 6 A1
3
5(b) 2 50 *M1
FT their −
2
a = 2 − + 6 = , allow any negative non-integer.
9 9 9
10 50 1 *M1 2
r=− ÷ =− Ft their − , allow any negative non-integer.
9 9 5 9
50 1 125 DM1 A1 Can only get DM1 if |r| < 1.
S∞ = ÷ 1 − − = Accept AWRT 4.63 .
9 5 27
4
A line has equation y = 6x − c and a curve has equation y = cx2 + 2x − 3, where c is a constant. The line is a tangent to the curve at point P.
Find the possible values of c and the corresponding coordinates of P. [7]
3
Mark scheme
6 cx 2 + 2 x − 3 = 6 x − c leading to cx2 − 4x + ( c − 3) [= 0] B1 3-term quadratic.
16 − 4c ( c − 3) = 0 *M1 Apply b 2 − 4ac = 0 (‘= 0’ may be implied in
subsequent work). Their coefficients must be
substituted correctly
4c2 −12c −16[= 0] leading to [ 4] ( c − 4 )( c + 1) [ = 0] leading to A1 Dependent on factorisation oe.
c = 4 and −1
When c = 4, 4x2 − 4x + 1[= 0] ( 2 x − 1) = 0 DM1 OE. Substituting their c = 4 into their quadratic
2
equation.
1 A1 Both required.
x= , y = −1
2
When c = −1, x2 + 4x + 4[= 0] ( x + 2 ) = 0 DM1 OE. Substituting their c = -1 into their quadratic
2
equation.
x = −2, y = −11 A1 Both required.
Alternative method for Question 6
dy B1
= 2cx + 2
dx
2cx + 2 = 6 M1 Equating their curve gradient and 6.
2 A1 SOI
c=
x
2x2 + 3x − 2 [= 0] DM1
Substitute c =
2
into cx2 + 2 x − 3 = 6 x − c . Simplify
x
to 3-term quadratic.
6 1 A1 Dependent on factorisation. Both required.
( 2 x − 1)( x + 2 ) [ = 0] → x = or −2
2
c = 4 and −1 A1 Both required, if DM0 given SC B1 for both.
y = −1 and − 11 A1 Both required, if DM0 given SC B1 for both.
SC one correct (x, y). A1 only
7
The function f is defined by f x = 1 + 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 = 2x − 2 for x > 0.
(c) Obtain a simplified expression for gf x. [2]
Mark scheme
7(a) Range is [y] > 1 B1 Allow f, f(x), (1,∞), etc.
1
7(b) 3 3 M1 Clearing the fraction.
y= + 1 leading to y − 1 = leading to ( x − 2 )( y − 1) = 3
x−2 x−2
3 DM1 Reaching a stage which only requires one further
x−2=
y −1 operation.
3 A1 2x + 1
OE. Slightly longer routes lead to f −1 ( x ) =
3
x= + 2 leading to f −1 ( x ) = +2 .
y −1 x −1 x −1
[Domain is] x > 1 B1FT Must use x FT their (a), must be a range.
4
7(c) 3 x +1 M1 Substitute f(x) into g(x).
gf ( x ) = 2 + 1 − 2 or 2 −2
x−2 x−2
6 A1
x−2
2
y
1 y = sin a x + b
0.5
x
− 23 π − 13 π 0 1
3π
2
3π π 4
3π
5
3π 2π
−0.5
The diagram shows part of the graph of y = sin a x + b , where a and b are positive constants.
(a) State the value of a and one possible value of b. [2]
Another curve, with equation y = f x, has a single stationary point at the point p, q, where p and q are constants. This curve is transformed to a curve with equation
y = −3f 14 x + 8 .
(b) For the transformed curve, find the coordinates of the stationary point, giving your answer in terms of p and q. [3]
1
A curve has equation y = 2x 2 − 1.
(a) Find the equation of the normal to the curve at the point A 4, 3, giving your answer in the form y = mx + c. [3]
1
A point is moving along the curve y = 2x 2 − 1 in such a way that at A the rate of increase of the x-coordinate is 3 cm s−1 .
(b) Find the rate of increase of the y-coordinate at A. [2]
At A the moving point suddenly changes direction and speed, and moves down the normal in such a way that the rate of decrease of the y-coordinate is constant at 5 cm s−1 .
(c) As the point moves down the normal, find the rate of change of its x-coordinate. [3]
Mark scheme
9(a) dy −1 1 B1
= x 2 → m = at x = 4
dx 2
Equation of normal is y − 3 = −2 ( x − 4 ) M1
Through (4, 3) with gradient −
1
.
their m
(Dependent on differentiation used).
y = −2 x + 11 A1 Only acceptable answer.
3
9(b) dy 1 M1 Correct use of the chain rule with a numerical
= their × 3 gradient.
dt 2
3 A1
2
2
9(c) dy B1FT SOI. FT from their part (a) if a normal gradient has
Required gradient = = −2
dx been found from m1m2 = −1 and differentiation.
dx 1 M1 Correct use of the chain rule . Allow method mark
= × −5 also for +5, must be numerical. Their normal gradient
dt their normal gradient
must come from m1m2 = −1 and differentiation in
part(a) unless ‘restarted’ here.
5 A1
2
3
A 2.8 rad B r
O
R r
C
The diagram shows points A, B and C lying on a circle with centre O and radius r . Angle AOB is
2.8 radians. The shaded region is bounded by two arcs. The upper arc is part of the circle with centre
O and radius r. The lower arc is part of a circle with centre C and radius R.
(a) State the size of angle ACO in radians. [1]
(b) Find R in terms of r. [1]
(c) Find the area of the shaded region in terms of r. [7]
Mark scheme
10(a) Angle ACO = 0.7 B1 Don’t allow AWRT 0.7 .
1
10(b) [ R = ] 1.53 r B1 Allow AWRT 1.53r.
1
10(c) 1 B1
Sector OAB = r 2 × 2.8 = 1.4r 2
2
1 *M1
Sector CAB = ( their R )2 × 2 × their 0.7
2
1.638 r 2 A1 Allow AWRT 1.64 r 2 .
1 1 *M1
[2] × r 2 sin (π − 1.4) OR [2] × r × theirR sin 0.7
2 2
2 × 0.4927r 2 A1 Allow AWRT 0.98 r 2 to 0.99 r 2 .
(
1.4r 2 − their 1.638r 2 − their 0.985r 2 ) DM1
0.747r 2 to 0.748r 2 A1
7
10(c) General guidance for alternative methods
Finding any useful sector area of the circle radius, r B1 May be ‘nested’ in a segment.
Finding the area of sector CAB *M1A1 May be ‘nested’ in a segment.
Finding the area of one useful triangle *M1 May be ‘nested’ in a segment.
Finding the total area of useful triangles A1 May be ‘nested’ in a segment.
A correct plan for the shaded area DM1
0.747r 2 to 0.748r 2 A1
7
y
P 2 y=x+
2x − 12
Q
R
x
O 1 2
2
The diagram shows part of the curve with equation y = x + . The lines x = 1 and x = 2
2x − 12 intersect the curve at P and Q respectively and R is the stationary point on the curve.
(a) Verify that the x-coordinate of R is 23 and find the y-coordinate of R. [4]
(b) Find the exact value of the area of the shaded region. [6]
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
dx
= −2 × 2 × ( 2x − 1) × 2 + 1
−3 B1B1
Expect
−8
( 2 x − 1)3
+ 1.
3 dy −8 DB1 AG.
Substitute x = leading to = +1 = 0 .
2 dx 8 dy
Or correct solution of =0.
3 dx
Hence x-coordinate of R is
2
3 2 3 B1 Answer only is acceptable.
When x = , y = + = 2
2 4 2
4
11(b) 20 B1 Both required.
y-coordinate of P = 3, y-coordinate of Q =
9
2 ( 2 x − 1)
−1 B1 B1 Area below curve.
1 2
+ x
− ×
1 2 2
1 1 5 1 M1
Apply limits 1→2 to an integral. Expect
13
− 3 + 2 − −1 + 2 = 3 − − 2
.
6
1 20 47 M1 Area of trapezium, only allow errors in y-coordinate
3 + = of Q.
2 9 18
47 13 4 A1 Shaded region.
− =
18 6 9
6
Alternative method 1: Changes the award of the first M1
−7 34 M1 Must be some evidence of use of limits.
Their equation of line PQ:[ y = x + ] . Integrating between 1 and 2.
9 9
Alternative method 2: Changes the award of the first M1, a B1 and the second M1
−16 34 2 M1 34
Combining line and curve: x+ − dx For area under the line if their is seen integrated
9 9 ( 2 x − 1)
2 9
correctly and limits used. Correct first and 3rd terms.
−8 2 34 1 B1 B1
= x + x+
9 9 ( − 1)
2 x
Use of limits on the whole integral M1
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Mathematics 9709/13 for October/November 2023 is out of 75 marks, with 1 hour 50 minutes to answer it.
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