Mathematics 9709/13 October/November 2021: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 1 Pure Mathematics 1, October/November 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
The graph of y = f x is transformed to the graph of y = 3 − f x.
Describe fully, in the correct order, the two transformations that have been combined. [4]
Mark scheme
1 {Reflection} {[in the] x-axis} *B1 DB1 {} indicate how the B1 marks should be awarded
or throughout.
{Stretch of scale factor -1} {parallel to y-axis}
0 B1 B1 Or Translation 3 units in the positive y-direction.
Then {Translation} N.B. If order reversed a maximum of 3 out of 4 marks
3 awarded.
Alternative method for question 1
0 B1 B1 Or Translation 3 units in the negative y-direction.
{Translation}
−3
Then {Reflection} {in the x-axis} *B1 DB1 N.B. If order reversed a maximum of 3 out of 4 marks
or {Stretch of scale factor -1} {parallel to y-axis} awarded.
4
(a) Find the first three terms, in ascending powers of x, in the expansion of 1 + ax6 . [1]
(b) Given that the coefficient of x2 in the expansion of 1 − 3x 1 + ax6 is −3, find the possible values of the constant a. [4]
Mark scheme
2(a) 1 + 6ax + 15a 2 x 2 B1 Terms must be evaluated.
1
2(b) their 15a 2 ± ( 3 × their 6a ) *M1 Expect 15a 2 − 18a .
15a 2 − 18a = −3 A1
( 3)( a − 1)( 5a − 1) [ = 0] DM1 Dependent on 3-term quadratic.
Or solve using formula or completing the square.
1 A1 WWW. If DM0 awarded SC B1 if both answers correct.
a = 1,
5
4
(a) Express 5y2 − 30y + 50 in the form 5 y + a2 + b, where a and b are constants. [2]
(b) The function f is defined by f x = x5 − 10x3 + 50x for x ∈ >.
Determine whether f is an increasing function, a decreasing function or neither. [3]
Mark scheme
3(a)
{5 ( y − 3) } {+5}
2 B1 B1 Accept a = ‒3, b = 5
2
3(b) f ′ ( x ) = 5 x 4 − 30 x 2 + 50 B1
( ) M1
2
5 x 2 − 3 + 5 or b 2 < 4ac and at least one value of f’(x) > 0
> 0 and increasing A1 WWW
3
The first term of an arithmetic progression is 84 and the common difference is −3.
(a) Find the smallest value of n for which the nth term is negative. [2]
It is given that the sum of the first 2k terms of this progression is equal to the sum of the first k terms.
(b) Find the value of k. [3]
Mark scheme
4(a) 84 − 3 ( n − 1) = 0 M1 OE, SOI. Allow either = 0 or < 0 (to -3).
Smallest n is 30 A1 SC B2 for answer only n = 30 WWW.
2
4(b) 2k k M1 A1 M1 for forming an equation using correct formula.
168 + ( 2k − 1)( −3) = 168 + ( k − 1)( −3) A1 for at least one side correct.
2 2
k = 19 A1
3
B
Y
9 cm
11 cm
X C
A
In the diagram, X and Y are points on the line AB such that BX = 9 cm and AY = 11 cm. Arc BC is part of a circle with centre X and radius 9 cm, where CX is perpendicular to AB. Arc AC is part of a circle with centre Y and radius 11 cm.
(a) Show that angle XYC = 0.9582 radians, correct to 4 significant figures. [1]
(b) Find the perimeter of ABC. [6]
Mark scheme
5(a) 9 B1 AG. OE using cosine rule.
Angle XYC = sin −1 = 0.9582
11
9
or sin XYC = leading to XYC = 0.9582
11
1
5(b) XY = 112 − 92 = 40 or using 0.9582 and trigonometry *M1 A1
AB = 9 + 11 − theirXY B1 FT OE e.g. 20 − 2 √ 10 , 2 + 9 − 2 10 + 11 − 2 √ 10
Arc AC = 11 × 0.9582 M1
π M1
Arc BC = 9×
2
Perimeter = [13.6(8) + 10.5(4) +14.1(4) =] 38.4 A1 AWRT. Answer must be evaluated as a single decimal.
6
y
x
O
y = f x
The diagram shows the graph of y = f x.
(a) On this diagram sketch the graph of y = f −1 x. [1]
x
It is now given that f x = − where −2 < x < 2.
4 − x2
(b) Find an expression for f −1 x. [4]
The function g is defined by g x = 2x for −a < x < a, where a is a constant.
(c) State the maximum possible value of a for which fg can be formed. [1]
(d) Assuming that fg can be formed, find and simplify an expression for fg x. [2]
tan x + cos x
Mark scheme
6(a) B1 A reflection of the given curve in y = x (the line y = x can
be implied by position of curve).
1
6(b) −x *M1 Squaring and clearing the fraction. Condone one error in
y= (
leading to x 2 = y 2 4 − x 2 ) squaring – x or y
4 − x2
( )
x2 1 + y 2 = 4 y 2 DM1 OE. Factorisation of the new subject with order of
operations correct. Condone sign errors.
2y DM1 4 y2
x = (±) x = (±) OE is acceptable for this mark.
1+ y 2 2
(1 + y
Isolating the new subject. Order of operations correct.
Condone sign errors.
−2 x A1 Selecting the correct square root.
f −1 ( x ) = Must not have fractions in numerator or denominator.
2
1+ x
4
6(c) 1 or a = 1 B1 Do not allow x = 1 or −1 < x < 1
1
6(d) −2 x B1 −2 x
[ fg ( x ) = f ( 2 x ) =] Allow or any correct unsimplified form.
4 − ( 2x)
2 2
4 − 4x
−x −x x B1 Result of cancelling 2 in numerator and denominator.
fg ( x ) = or 2
1 − x 2 or 2 1 − x2
1− x 2 1− x x −1
2
(a) Show that the equation = k, where k is a constant, can be expressed as tan x − cos x k + 1 sin2 x + k − 1 sin x − k + 1 = 0. [4]
tan x + cos x
(b) Hence solve the equation = 4 for 0Å ≤ x ≤ 360Å. [4]
tan x − cos x
Mark scheme
7(a) tan x + cos x = k(tan x ‒cos x) leading to sin x + cos 2 x = k (sin x − cos 2 x ) M1 sinx
Use tanx = and clear fraction.
cosx
sinx + 1 − sin 2 x = ksinx − k + ksin 2 x *M1 Use cos 2 x = 1 − sin 2 x twice to obtain an equation in sine.
ksin 2 x + sin 2 x + ksinx − sinx − k − 1 = 0 DM1 Gather like terms on one side of the equation.
( k + 1) sin 2 x + ( k − 1) sin x − ( k + 1) = 0 A1 AG. Factorise to obtain answer.
4
7(b) 5sin 2 x + 3sin x − 5 = 0 B1
−3 ± 9 + 100 M1 Use formula or complete the square.
sin x =
10
x = 48.1°, 131.9° A1 AWRT. Maximum A1 if extra solutions in range.
A1 FT FT for 180 ‒ their answer or 540 – their answer if sinx is
negative
If M0 given and correct answers only SCB1B1 available.
If answers in radians; 0.839, 2.30 can score SCB1 for
both.
4
y
A 14 , 2
1 y = 52 − x 2
1 B 4, 21 y = x− 2 x
O
−1 1
The diagram shows the curves with equations y = x 2 and y = 52 − x 2 . The curves intersect at the points
A 14 , 2 and B 4, 21 .
(a) Find the area of the region between the two curves. [6]
−1
(b) The normal to the curve y = x 2 at the point 1, 1 intersects the y-axis at the point 0, p.
Find the value of p. [4]
Mark scheme
8(a) 5 1
−
1 M1 5
− x 2 − x 2 dx ( )
OR as 2 separate integrals − x1/2 dx − x −1/2 dx
2
2
5 2 2
3
1 A1 A1 A1 If two separate integrals with no subtraction SC B1 for
x − x {−} 2 x 2 each correct integral.
2 3
16 5 1 DM1 1
10 − − 4 − − − 1 Substitute limits → 4 at least once, must be seen.
3 8 12 4
9 A1 WWW. Cannot be awarded if π appears in any integral.
or 1.125
8
6
8(b) dy 1 − 2
3 B1
dx = − 2 x
1 M1 Substitute x = 1 into a differential.
When x = 1, m = −
2
[Equation of normal is] y − 1 = 2 ( x − 1) M1 1 1− p
Through (1, 1) with gradient − or =2
m 1
[When x = 0,] p = ‒1 A1 WWW
4
The line y = 2x + 5 intersects the circle with equation x2 + y2 = 20 at A and B.
(a) Find the coordinates of A and B in surd form and hence find the exact length of the chord AB.
[7]
A straight line through the point 10, 0 with gradient m is a tangent to the circle.
(b) Find the two possible values of m. [5]
Mark scheme
9(a) x 2 + ( 2 x + 5) = 20
2
leading to x 2 + 4 x 2 + 20 x + 25 = 20 M1 Substitute y = 2 x + 5 and expand bracket.
( 5 ) ( x 2 + 4 x + 1) [ = 0] A1 3-term quadratic.
−4 ± 16 − 4 M1 OE. Apply formula or complete the square.
x=
2
(
A = −2 + 3,1 + 2 3 ) A1 Or 2 correct x values.
(
B = −2 − 3,1 − 2 3 ) A1 Or all values correct.
SC B1 all 4 values correct in surd form without working.
SC B1 all 4 values correct in decimal form from correct
formula or completion of the square
AB 2 = their ( x2 − x1 ) + their ( y2 − y1 )
2 2 M1 Using their coordinates in a correct distance formula.
Condone one sign error in x2 − x1 or y2 − y1
AB 2 = 48 + 12 leading to AB = 60 A1 OE. CAO. Do not accept decimal answer. Answer must
come from use of surd form in distance formula.
7
9(b) x 2 + m2 ( x − 10 ) = 20
2 *M1 Finding equation of tangent and substituting into circle
equation.
( ) (
x 2 m 2 + 1 − 20m 2 x + 20 5m 2 − 1 ) [ = 0] DM1 OE. Brackets expanded and all terms collected on one side
of the equation.
( )(
[b 2 − 4ac =]400m 4 − 80 m 2 + 1 5m 2 − 1 ) M1 Using correct coefficients from their quadratic equation.
( ) (
400m 4 − 80 5m 4 + 4m 2 − 1 = 0 → ( −80 ) 4m 2 − 1 = 0 ) A1 OE. Must have ‘=0’ for A1.
1 A1
m=±
2
Alternative method for question 9(b)
Length, l of tangent, is given by l 2 = 10 2 − 20 M1
l = 80 A1
20 1 M1 A1 Where α is the angle between the tangent and the x-axis.
tan α = =
80 2
1 A1
m=±
2
5
A curve has equation y = f x and it is given that
−2 f ′ x = 12 x + k − 1 + k−2 ,
where k is a constant. The curve has a minimum point at x = 2.
(a) Find f ′′ x in terms of k and x, and hence find the set of possible values of k. [3]
It is now given that k = −3 and the minimum point is at 2, 3 12 .
(b) Find f x. [4]
(c) Find the coordinates of the other stationary point and determine its nature. [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
10(a) f ′′ ( x ) = − ( 12 x + k )
−3 B1
f ′′ ( 2 ) > 0 − (1 + k ) > 0
−3 M1 Allow for solving their f ′′ ( 2 ) > 0
k < −1 A1 WWW
3
10(b) ( 1 x − 3)−1 B1 B1 1
−1
(( ) x −2
f ( x ) = 1 −2
x−3 − ) ( −2 ) −2
dx = 2 − Allow −2 x + k OE for 1st B1 and − (1 + k ) x OE
2 2
−1 × 2 4
1
for 2nd B1
3 12 = 1 - 12 + c M1 Substitute x = 2, y = 3 1 into their integral with c present.
2
−2 x A1 OE
f ( x) = − +3
( 1
2
x−3 ) 4
4
10(c) M1 Substitute k = −3 and set to zero.
( 12 x − 3) − ( −2 )−2 = 0
−2
leading to ( 12 x − 3) = 4 12 x − 3 = ( ± ) 2 leading to x = 10
2 A1
(10, ‒ 12 ) A1 Or when x = 10, y = ‒1 ‒ 2 1 + 3 = ‒ 1
2 2
f ′′ (10 ) = − ( 5 − 3) → < 0 → MAXIMUM
−3 A1 WWW
4
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On this page, free: the mark scheme for AS & A Level Mathematics 9709/13, October/November 2021 (9709_w21_ms_13), and the question paper (9709_w21_qp_13), to read or download, with each question's marking points written out underneath.
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Mathematics 9709/13 for October/November 2021 is out of 75 marks, with 1 hour 50 minutes to answer it.
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