Mathematics 9709/31 May/June 2022: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 3 Pure Mathematics 3, May/June 2022. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
Solve the equation 2 32x−1 = 4x+1 , giving your answer correct to 2 decimal places. [4]
−2
(a) Expand 2 − x2
Mark scheme
1 Use law of the logarithm of a product or a quotient or a power *M1
Obtain a correct linear equation in any form A1 e.g. ln2 + ( 2x −1) ln3 = ( x +1) ln4 or
lo g 2 2 + ( 2 x − 1 ) lo g 2 3 = ( 2 x + 2 ) lo g 2 2
Solve for x DM1 Allow for unsimplified expression x = …
Allow M1 M1 for x =1.45 from 6 2 x − 1 = 4 x + 1 .
Obtain answer x = 2.21 A1 The question asks for 2 dp.
Alternative method for question 1
Correct use of indices to obtain 2.25 x = 6 or 1.5 2 x = 6 M1 A1
Correct use of logarithms to solve for x M1 Allow solution of 2.25 x = 6 by trial and
improvement as far as 2.2…
Obtain answer x = 2.21 A1 Need to see an intermediate step / sequence of
iterations.
4
in ascending powers of x, up to and including the term in x4 , simplifying the coefficients. [4]
(b) State the set of values of x for which the expansion is valid. [1]
Mark scheme
2(a) State a correct unsimplified version of the x 2 or the x 4 term of the expansion of M1
1
2
x2 −2. − 3 x2
−2 1 + 2 + ...
1 2 4
( 2 − x ) or 1 − 2 x
2 −2
2 2 2
Symbolic binomial coefficients are not sufficient
for the M1.
1 B1 Accept 2-2 .
State correct first term
4
1 2 3 4 A1 A1 A1 for each one correct ISW.
Obtain the next two terms x + x
4 16 (
Full marks for 14 1 + x 2 + 34 x 4 ISW. )
2 4
SC allow M1 A1 A1 for 4 and 1 + x + 34 x SOI.
1
2 4
SC allow M1 A1 for 1 + x + 34 x
4
2(b) State answer x < 2 B1 Or − 2 < x < 2 .
1
Solve the equation 2 cot 2x + 3 cot x = 5, for 0Å < x < 180Å. [6]
Mark scheme
3 Use correct trigonometric formulae to form an equation in tanx *M1 1 − tan 2 x 3
e.g. + =5
tan x tan x
Obtain a correct linear equation in any form A1 1 − tan 2 x + 3 = 5 tan x
Reduce equation to a 3-term quadratic A1 tan 2 x + 5 tan x − 4 = 0 , or 3-term equivalent
Solve a 3-term quadratic in tan x and obtain a value of x DM1
Obtain answer, e.g. x = 35.1 ° A1
Obtain second answer, e.g. x = 99.9 ° , and no other in ( 0 ° , 180 ° ) A1 Ignore answers outside ( 0 ° , 180 ° ) .
Treat answers in radians ( 0.612,1.74) as a misread.
Alternative method for question 3
Use correct formulae for sin2x and cos2x to form an equation in sinx and *M1
cosx
cos x sin x A1
Obtain 4 − =5
sin x cos x
Reduce equation to a 3-term quadratic A1 tan 2 x + 5 tan x − 4 = 0 , or 3-term equivalent
Solve a 3-term quadratic in tan x and obtain a value of x DM1
Obtain answer, e.g. x = 35.1 ° A1
Obtain second answer, e.g. x = 99.9 ° , and no other in ( 0 ° , 180 ° ) A1 Ignore answers outside ( 0 ° , 180 ° ) .
Treat answers in radians ( 0.612,1.74) as a misread.
6
The variables x and y satisfy the differential equation
= dy xy dx 1 + x2
and y = 2 when x = 0.
Solve the differential equation, obtaining a simplified expression for y in terms of x. [7]
Mark scheme
4 Separate variables correctly B1
x
1+ x
2
dx = dy Accept without integral signs.
1
y
Obtain term ln y B1
(
State term of the form k ln 1 + x 2 ) M1
A1 OE
State correct term
1
2
(
ln 1 + x 2 )
Evaluate a constant, or use limits x = 0, y = 2 in a solution containing terms M1 If they remove logs first the constant must be of the
( )
a lny and b ln 1 + x 2 where ab ≠ 0 correct form.
Obtain correct solution in any form A1 e.g ln y + ln 1 = 1 ln 1 + x 2
2 2 ( )
2 A1 OE The question asks for simplification, so need to
Simplify and obtain y = 2 1 + x
deal with exp(ln(...)) .
7
The polynomial ax3 − 10x2 + bx + 8, where a and b are constants, is denoted by p x. It is given that x − 2 is a factor of both p x and p′ x.
(a) Find the values of a and b. [5]
(b) When a and b have these values, factorise p x completely. [3]
3
Let I = Ô
27
Mark scheme
5(a) Substitute x = 2, equate to zero M1 Or divide by x – 2 and equate constant remainder to
zero.
Obtain a correct equation, e.g. 8a – 40 + 2b + 8 = 0 A1 Seen or implied in subsequent work.
Differentiate p(x), substitute x = 2 and equate result to zero M1 Or divide by x – 2 and equate constant remainder to
zero.
Obtain 12a – 40 + b = 0, or equivalent A1 SOI in subsequent work.
Obtain a = 3 and b = 4 A1
Alternative method for question 5(a)
State or imply ( x − 2) is a factor M1
2
p ( x ) = ( x − 2) ( ax + 2) A1
2
Obtain an equation in b M1
e.g. by comparing coefficients of x: b = 4a − 8 A1
Obtain a = 3 and b = 4 A1
SC If uses x = −2 in both equations allow M1 and
allow A1 for a = −3, b = −4.
5
5(b) Attempt division by (x – 2) M1 The M1 is earned if division reaches a partial
quotient of ax 2 + kx , or if inspection has an
2
unknown factor ax + ex + f and an equation in e
and/or f. Where a has the value found in part 5(a).
Obtain quadratic factor 3 x 2 − 4 x − 4 A1
Obtain factorisation ( 3x + 2)( x − 2)( x − 2) A1
Alternative method for question 5(b)
State or imply ( x − 2) is a factor B1
2
Attempt division by ( x − 2) , reaching a quotient ax + k or use inspection with M1
2
unknown factor cx + d reaching a value for c or for d
Obtain factorisation ( 3x + 2)( x − 2)
2 A1
( 3)
Accept 3 x + 2 ( x − 2 ) .
2
3
2 dx.
9 + x2
0
1π
(a) Using the substitution x = 3 tan 1, show that I = Ó cos2 1 d1.
4
[4]
0
(b) Hence find the exact value of I . [4]
2 − a 2i
The complex number u is defined by u =
1 + 2i
Mark scheme
6(a) State or imply dx = 3sec 2θ dθ B1
Substitute throughout for x and dx M1
Obtain any correct form in terms of θ A1 81sec2 θ
e.g. 9 + 9tan θ dθ
( )
2 2
π
A1 AG
cos θ dθ correctly
4 2
Justify change of limits and obtain
0
4
6(b) Obtain indefinite integral of the form a + b cos 2θ dθ , where ab ≠ 0 *M1
1 1 A1
Obtain θ + sin 2θ
2 4
Use correct limits correctly in an expression containing pθ and q sin 2θ where DM1 + −0
π 1
8 4 ( )
pq ≠ 0
1 A1 1 1
Obtain answer ( π + 2) Or exact equivalent e.g. π+ .
8 8 4
4
, where a is a positive integer.
(a) Express u in terms of a, in the form x + iy, where x and y are real and exact. [3]
It is now given that a = 3.
(b) Express u in the form rei1 , where r > 0 and −π < 1 ≤ π, giving the exact values of r and 1. [2]
(c) Using your answer to part (b), find the two square roots of u. Give your answers in the form rei1 , where r > 0 and −π < 1 ≤ π, giving the exact values of r and 1. [3]
Mark scheme
7(a) Multiply numerator and denominator by 1 – 2i, or equivalent M1 At least one multiplication completed.
Obtain correct numerator (1− 2a) 2 − ( 2 + a) 2i A1 OE
1 − 2a 2+a A1 OE
Obtain final answer 2− 2i
5 5
Alternative method for question 7(a)
Multiply x + iy by 1+2i and compare real and imaginary parts M1
A1
Obtain x − 2y = 2 and 2x + y = a 2
1 − 2a 2+a A1 OE
Obtain final answer 2− 2i
5 5
3
7(b) Obtain r = 2 B1 FT
3 B1
Obtain θ = − π
4
2
7(c) Use correct method to find r or θ M1
3
− πi A1 FT
State answer 2e 8
5
πi A1 FT
State answer 2e 8
3
The equation of a curve is x3 + y3 + 2xy + 8 = 0.
dy
(a) Express in terms of x and y. [4]
dx
The tangent to the curve at the point where x = 0 and the tangent at the point where y = 0 intersect at the acute angle !.
(b) Find the exact value of tan !. [5]
Mark scheme
8(a) 2 dy 3 B1
State or imply 3y as derivative of y
dx
dy B1
State or imply 2 y + 2 x as derivative of 2xy
dx
Complete differentiation and equate attempted derivative to zero and solve for M1
dy
dx
3x2 + 2 y A1
Obtain answer − 2
3 y + 2x
4
8(b) Find gradient at either (0, – 2) or (– 2, 0) M1
1 A1 A1
Obtain answers and 3
3
Use tan ( A±B) formula to find tan α M1
4 A1
Obtain answer tan α =
3
5
G F
M
D k E
O C j
i
A N B
In the diagram, OABCDEFG is a cuboid in which OA = 2 units, OC = 4 units and OG = 2 units. Unit vectors i, j and k are parallel to OA, OC and OG respectively. The point M is the midpoint of DF.
The point N on AB is such that AN = 3NB.
(a) Express the vectors OM and MN in terms of i, j and k. [3]
(b) Find a vector equation for the line through M and N . [2]
?
(c) Show that the length of the perpendicular from O to the line through M and N is 53 . [4]
6
Mark scheme
9(a)
ρ B1
Obtain OM = i + 2j + 2k
ρ ρ ρ ρ ρ ρ
Use a correct method to find MN M1 e.g. MO + OA + AN or MO + ON
ρ A1 Accept any notation.
Obtain MN = i + j − 2k
3
9(b) Use a correct method to form an equation for MN M1 Allow without r = ...
Obtain r = 2i + 3j + 𝜆 (i + j – 2k) A1 FT OE e.g. r = i + 2 j + 2 k + μ ( i + j − 2 k )
Must have r = ... Follow their answers to part 9(a).
2
ρ
9(c) State O P for a general point P on MN in component form, B1
e.g. (2 + λ, 3 + λ , −2λ )
ρ
Equate scalar product of O P and a direction vector for MN to zero and solve M1
for 𝜆
5 A1 OE e.g. μ = 1
Obtain λ = − 6
6
A1
( 76 ) + ( 136 ) + ( 53 )
53
Obtain correctly 2 2 2
6 AG e.g. from
4
y
x
O a π
The curve y = x sin x has one stationary point in the interval 0 < x < π, where x = a (see diagram).
(a) Show that tan a = − 12 a. [4]
(b) Verify by calculation that a lies between 2 and 2.5. [2]
(c) Show that if a sequence of values in the interval 0 < x < π given by the iterative formula xn+1 = π − tan−1 12 xn converges, then it converges to a, the root of the equation in part (a). [2]
(d) Use the iterative formula given in part (c) to determine a correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]
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) Use correct product rule M1 Condone incorrect / missing chain rule
Obtain correct derivative in any form A1 dy x cos x
e.g. = sin x + or
dx 2 sin x
dy
2y = 2 x sin x + x 2 cos x
dx
Equate derivative to zero and obtain an equation in tan x or tan a M1
Obtain tana =− 2 a correctly
1 A1 AG
4
10(b) Calculate the value of a relevant expression or pair of expressions at M1 Must be working in radians
a = 2 and a = 2.5 At least one correct
Complete the argument correctly with correct calculated values A1 e.g. −1 > −2.18 and − 1.25 < −0.747
2
10(c) B1 A correct equation without subscripts
State a suitable equation, e.g. x = π − tan − 1 1 x
2 or quote tan θ = − tan (π − θ )
Using tan( A± B) formula, or otherwise, rearrange this as tan x = − x
1 B1 Complete argument correctly
2
2
10(d) Use the iterative process correctly at least once M1 Must be working in radians
Obtain answer a = 2.29 A1
Show sufficient iterations to 4 dp to justify 2.29 to 2 dp or show there is a sign A1 e.g. 2.25, 2.2974, 2.2871, 2.2893, 2.2888, …
change in the interval (2.285, 2.295)
3
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Mathematics 9709/31 for May/June 2022 is out of 75 marks, with 1 hour 50 minutes to answer it.
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