Further Math P1
Topical
2012-2021
Compiled By A Fellow Student
Roots Of Polonomial
Rational Functions
Summation Of Series
Matrices 1
Polar Coordinates
Vectors
Proof By Induction
1-68
69-152
153-226
227-252
253-340
341-439
440-500
Assalam-o-Alaikum,
So basically while doing AS I was unable to find any topical for FM
that was not behind a paywall and easily accessible and due to
that I was unable to practice much for the paper. Therefore, after
my P1 paper I felt like I should compile this workbook so that
others may have an easier time practicing and exploring this
interesting/unique/difficult/depression inducing subject.
Oh and as only a student compiled this workbook there might be
some errors. I have tried my best to eliminate them but sorry if
you find any.
A fellow student,
Craftear/AN
Craftear
Table Of Contents:
1
Craftear
ROOTS OF POLYNOMIAL
2
Q1
3
6
The equation x 4 - 2x 3 - 1 = 0 has roots a , b , c , d .
(a) Find a quartic equation whose roots are a 3 , b 3 , c 3 , d 3 and state the value of a 3 + b 3 + c 3 + d 3 .
[4]
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(b) Find the value of
1
1
1
a
b
c
3+
3+
3+
1
d3
.
[3]
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(c) Find the value of a 4 + b 4 + c 4 + d 4 .
[2]
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4
Q2
2
4
The cubic equation 2x 3 - 4x 2 + 3 = 0 has roots a , b , c . Let Sn = a n + b n + c n .
(a) State the value of S1 and find the value of S2 .
[3]
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(b) (i) Express Sn +3 in terms of Sn + 2 and Sn .
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(ii) Hence, or otherwise, find the value of S4 .
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(c) Use the substitution y = S 1 - x , where S1 is the numerical value found in part (a), to find and
simplify an equation whose roots are a + b , b + c , c + a .
[3]
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(d) Find the value of
1
a+b
+
1
b+c
+
1
c+a
.
[2]
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6
Q3
1
It is given that
a + b + c = 3,
a 2 + b 2 + c 2 = 5,
a3 + b3 + c3 = 6.
The cubic equation x 3 + bx 2 + cx + d = 0 has roots a , b , c .
Find the values of b, c and d.
[6]
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Q4
4
6
The cubic equation x 3 + 2x 2 + 3x + 3 = 0 has roots a , b , c .
(a) Find the value of a 2 + b 2 + c 2 .
[2]
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(b) Show that a 3 + b 3 + c 3 = 1.
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Q5
3
6
The cubic equation x 3 + cx + 1 = 0 , where c is a constant, has roots a, b, c.
(a) Find a cubic equation whose roots are a 3 , b 3 , c 3 .
[3]
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(b) Show that a 6 + b 6 + c 6 = 3 - 2c 3 .
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f
p
1 a3 b3
(c) Find the real value of c for which the matrix a 3 1 c 3 is singular.
b3 c3 1
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10
Q6
1
The cubic equation x 3 + bx 2 + cx + d = 0 , where b, c and d are constants, has roots a, b, c. It is given
that abc = -1.
(a) State the value of d.
[1]
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(b) Find a cubic equation, with coefficients in terms of b and c, whose roots are a + 1, b + 1, c + 1.
[3]
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(c) Given also that c + 1 = - a - 1, deduce that (c - 2b + 3) (b - 3) = b - c .
[4]
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Q7
2
4
The cubic equation 6x 3 + px 2 - 3x - 5 = 0 , where p is a constant, has roots a, b, c.
(a) Find a cubic equation whose roots are a 2 , b 2 , c 2 .
[3]
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(b) It is given that a 2 + b 2 + c 2 = 2 (a + b + c) .
(i) Find the value of p.
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(ii) Find the value of a 3 + b 3 + c 3 .
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Q8
1
2
The cubic equation 7x 3 + 3x 2 + 5x + 1 = 0 has roots a, b, c.
(a) Find a cubic equation whose roots are a -1, b -1, c -1 .
[3]
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(b) Find the value of a -2 + b -2 + c -2 .
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(c) Find the value of a -3 + b -3 + c -3 .
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Q9
7
The equation x3 + 2x2 + x + 7 = 0 has roots !, ", ' .
(i) Use the relation x2 = −7y to show that the equation
has roots
! " '
,
,
.
"' '! !"
49y3 + 14y2 − 27y + 7 = 0
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(ii) Show that
!2
"2
'2
+
+
= 58 .
"2 ' 2 ' 2 !2 !2 "2 49
[3]
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(iii) Find the exact value of
"3
'3
!3
+
+
.
"3 ' 3 ' 3 ! 3 !3 " 3
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Q10
6
The equation
x3 − x + 1 = 0
has roots !, ", ' .
(i) Use the relation x = y 3 to show that the equation
1
y3 + 3y2 + 2y + 1 = 0
has roots !3 , "3 , ' 3 . Hence write down the value of !3 + "3 + ' 3 .
[3]
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Let Sn = !n + "n + ' n .
(ii) Find the value of S−3 .
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(iii) Show that S6 = 5 and find the value of S9 .
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Q11
9
A cubic equation x3 + bx2 + cx + d = 0 has real roots !, " and ' such that
1 1 1
+ + =−5,
12
! " '
!"' = −12,
!3 + "3 + ' 3 = 90.
(i) Find the values of c and d .
[3]
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(ii) Express !2 + "2 + ' 2 in terms of b.
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(iii) Show that b3 − 15b + 126 = 0.
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(iv) Given that 3 + iï 12 is a root of y3 − 15y + 126 = 0, deduce the value of b.
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Q12
4
It is given that the equation
x3 − 21x2 + kx − 216 = 0,
where k is a constant, has real roots a, ar and ar−1 .
(i) Find the numerical values of the roots.
[6]
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(ii) Deduce the value of k.
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Q13
6
The equation
has roots !, ", ' .
9x3 − 9x2 + x − 2 = 0
(i) Use the substitution y = 3x − 1 to show that 3! − 1, 3" − 1, 3' − 1 are the roots of the equation
y3 − 2y − 7 = 0.
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The sum 3! − 1n + 3" − 1n + 3' − 1n is denoted by Sn .
(ii) Find the value of S3 .
[2]
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(iii) Find the value of S−2 .
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Q14
2
The roots of the equation
x3 + px2 + qx + r = 0
are !, 2!, 4!, where p, q, r and ! are non-zero real constants.
(i) Show that
2p! + q = 0.
[4]
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(ii) Show that
p3 r − q3 = 0.
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Q15
1
The roots of the cubic equation
are !, ", ' .
x3 − 5x2 + 13x − 4 = 0
(i) Find the value of !2 + "2 + ' 2 .
[3]
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(ii) Find the value of !3 + "3 + ' 3 .
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Q16
7
By finding a cubic equation whose roots are !, " and ' , solve the set of simultaneous equations
! + " + ' = −1,
! + "2 + ' 2 = 29,
1 1 1
+ + = −1.
! " '
2
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© UCLES 2017
9231/11/M/J/17
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Craftear
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28
Q17
1
The roots of the cubic equation x3 + 2x2 − 3 = 0 are !, " and ' .
(i) By using the substitution y =
1
1 1
1
, find the cubic equation with roots 2 , 2 and 2 .
2
! "
'
x
[3]
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(ii) Hence find the value of
1
1
1
+ 2 + 2.
2
!
"
'
[1]
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
(iii) Find also the value of
1
1
1
+ 2 2 + 2 2.
! "
" '
' !
2 2
[1]
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© UCLES 2017
9231/13/M/J/17
Craftear
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29
Q18
4
The cubic equation 2x3 − 3x2 + 4x − 10 = 0 has roots !, " and ' .
(i) Find the value of ! + 1 " + 1 ' + 1.
[4]
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(ii) Find the value of " + ' ' + ! ! + ".
[4]
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© UCLES 2017
9231/11/O/N/17
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Craftear
........................................................................................................................................................
Q19
1
, or
x
1
1
1
otherwise, find the cubic equation whose roots are 1 + , 1 + and 1 + , giving your answer in the
!
"
'
form ay3 + by2 + cy + d = 0, where a, b, c and d are constants to be found.
[4]
The roots of the cubic equation 2x3 + x2 − 7 = 0 are !, " and ' . Using the substitution y = 1 +
Craftear
1
30
© UCLES 2016
9231/11/M/J/16
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31
Using the cartesian equation of the curve, find the area of this region.
Q20
8
[3]
The cubic equation
z3 − z2 − z − 5 = 0
has roots !, " and ' . Show that the value of !3 + "3 + ' 3 is 19.
[4]
Find the value of !4 + "4 + ' 4 .
[2]
Show that the cubic equation with roots
'−1
!−1 "−1
,
and
may be found using the substitution
!
"
'
Craftear
1
z=
, and find this equation, giving your answer in the form px3 + qx2 + rx + s = 0, where p, q, r
1−x
and s are constants to be determined.
[4]
© UCLES 2016
9231/13/M/J/16
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Q21
2
32
Find the cubic equation with roots !, " and ' such that
! + " + ' = 3,
!2 + "2 + ' 2 = 1,
!3 + "3 + ' 3 = −30,
[6]
Craftear
giving your answer in the form x3 + px2 + qx + r = 0, where p, q and r are integers to be found.
© UCLES 2016
9231/11/O/N/16
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Q22
4
33
The roots of the cubic equation x3 − 7x2 + 2x − 3 = 0 are !, " and ' . Find the values of
(i)
1
,
!" "' '!
(ii)
1
1
1
+
+
,
!" "' ' !
(iii)
1
1
1
.
+ 2 +
! "' !" ' !"' 2
2
[6]
1
1 1
,
and
.
!" "'
'!
[2]
Craftear
Deduce a cubic equation, with integer coefficients, having roots
© UCLES 2015
9231/11/M/J/15
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Q23
1
The quartic equation x4 − px2 + qx − r = 0, where p, q and r are real constants, has two pairs of equal
roots. Show that p2 + 4r = 0 and state the value of q.
[6]
The curve C has polar equation r = e41 for 0 ≤ 1 ≤ !, where ! is measured in radians. The length of
C is 2015. Find the value of !.
[6]
Craftear
2
34
© UCLES 2015
9231/13/M/J/15
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Q24
5
35
The cubic equation x3 + px2 + qx + r = 0, where p, q and r are integers, has roots !, " and ' , such that
! + " + ' = 15,
! + "2 + ' 2 = 83.
2
[3]
Given that !, " and ' are all real and that !" + !' = 36, find ! and hence find the value of r.
[5]
Craftear
Write down the value of p and find the value of q.
© UCLES 2015
9231/11/O/N/15
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Q25
The equation x3 + px + q = 0, where p and q are constants, with q ≠ 0, has one root which is the
[5]
reciprocal of another root. Prove that p + q2 = 1.
Craftear
1
36
© UCLES 2014
9231/11/M/J/14
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37
Q26
11
4
Answer only one of the following two alternatives.
EITHER
The roots of the quartic equation x4 + 4x3 + 2x2 − 4x + 1 = 0 are , , and . Find the values of
(i)
+ + + ,
[1]
(ii)
2
+ 2 + 2 + 2 ,
[2]
1 1 1
+ + ,
[2]
(iii)
(iv)
1
+
+
+
+
.
[2]
Using the substitution y = x + 1, find a quartic equation in y. Solve this quartic equation and hence
[7]
find the roots of the equation x4 + 4x3 + 2x2 − 4x + 1 = 0.
OR
Craftear
The square matrix A has as an eigenvalue with e as a corresponding eigenvector. Show that if A is
non-singular then
© UCLES 2014
9231/11/O/N/14
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Q27
3
38
The cubic equation x3 − 2x2 − 3x + 4 = 0 has roots !, ", ' . Given that c = ! + " + ' , state the value
of c.
[1]
Use the substitution y = c − x to find a cubic equation whose roots are ! + ", " + ' , ' + !.
Find a cubic equation whose roots are
1
1
1
+
+
.
2
2
' + !2
! + "
" + '
[2]
[2]
Craftear
Hence evaluate
1
1
1
,
,
.
!+" "+' '+!
[3]
© UCLES 2013
9231/11/M/J/13
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Q28
2
39
The roots of the equation x4 − 4x2 + 3x − 2 = 0 are !, ", ' and $ ; the sum !n + "n + ' n + $ n is denoted
by Sn . By using the relation y = x2 , or otherwise, show that !2 , "2 , ' 2 and $ 2 are the roots of the
equation
y4 − 8y3 + 12y2 + 7y + 4 = 0.
3
State the value of S2 and hence show that
3
Craftear
S8 = 8S6 − 12S4 − 72.
© UCLES 2013
9231/13/M/J/13
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Q29
2
40
The cubic equation x3 − px − q = 0, where p and q are constants, has roots !, ", ' . Show that
(i) !2 + "2 + ' 2 = 2p,
[1]
(ii) !3 + "3 + ' 3 = 3q,
[2]
[3]
Craftear
(iii) 6 !5 + "5 + ' 5 = 5 !3 + "3 + ' 3 !2 + "2 + ' 2 .
© UCLES 2013
9231/11/O/N/13
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Q30
5
41
The equation
8x3 + 36x2 + kx − 21 = 0,
Craftear
where k is a constant, has roots a − d, a, a + d. Find the numerical values of the roots and determine
the value of k.
[8]
© UCLES 2013
9231/13/O/N/13
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Q31
1
42
The roots of the cubic equation x3 − 7x2 + 2x − 3 = 0 are α , β , γ . Find the values of
(i) α 2 + β 2 + γ 2 ,
[2]
(ii) α 3 + β 3 + γ 3 .
Craftear
[3]
© UCLES 2012
9231/11/M/J/12
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Q32
43
11 EITHER
The roots of the equation x4 − 3x2 + 5x − 2 = 0 are α , β , γ , δ , and α n + β n + γ n + δ n is denoted by Sn .
Show that
Sn+4 − 3Sn+2 + 5Sn+1 − 2Sn = 0.
[2]
Find the values of
(i) S2 and S4 ,
[3]
(ii) S3 and S5 .
[6]
Hence find the value of
[3]
Craftear
α 2 (β 3 + γ 3 + δ 3 ) + β 2 (γ 3 + δ 3 + α 3 ) + γ 2 (δ 3 + α 3 + β 3 ) + δ 2 (α 3 + β 3 + γ 3 ).
© UCLES 2012
9231/12/O/N/12
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Q33
7
44
A cubic equation has roots α , β and γ such that
α + β + γ = 4,
Find the value of αβ + βγ + γ α .
α 2 + β 2 + γ 2 = 14,
α 3 + β 3 + γ 3 = 34.
[2]
Show that the cubic equation is
x3 − 4x2 + x + 6 = 0,
[6]
Craftear
and solve this equation.
© UCLES 2012
9231/13/O/N/12
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Q34
8
45
The cubic equation x3 − x2 − 3x − 10 = 0 has roots α , β , γ .
(i) Let u = −α + β + γ . Show that u + 2α = 1, and hence find a cubic equation having roots −α + β + γ ,
α − β + γ, α + β − γ.
[5]
1 1
1
,
,
.
βγ γ α αβ
[5]
Craftear
(ii) State the value of αβγ and hence find a cubic equation having roots
© UCLES 2012
9231/13/M/J/12
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ROOTS OF POLYNOMIAL
MS
Craftear
46
Q1
3(c)
3(b)
Question
3(a)
Question
Craftear
α 3 β 3δ 3 + α 3 β 3γ 3 + β 3γ 3δ 3 + α 3γ 3δ 3 6
=
−1
α 3 β 3γ 3δ 3
Guidance
3
A1
M1
1
1
1
1
A1 FT Relates α 3 + β 3 + γ 3 + δ 3 to coefficients.
2
A1
δ
=
3
1
= 20
γ
+
3
1
M1 Uses original equation.
β
+
3
1
Marks
4
α 4 + β 4 + γ 4 + δ 4 = 2(α 3 + β 3 + γ 3 + δ 3 ) + 4
−6
α
+
3
1
α3 + β3 + γ 3 +δ 3 = 8
Answer
A1
y 4 − 8 y 3 − 12 y 2 − 6 y − 1 = 0
B1 FT
M1 Obtains an equation not involving radicals.
4
y 3 − 2 y −1 = 0 y 4 = ( 2 y + 1)3 = 8 y 3 + 12 y 2 + 6 y + 1
Guidance
B1 Correct substitution.
Marks
y = x3
Answer
47
Q2
M1 Uses formula for sum of squares.
A1 Correct answer implies M1A1.
S2 = S12 − 2(0)
=4
A1
=4
2(d)
2(c)
M1 Uses their recursive formula from part (i) to find
S4 S3 = 72 .
S4 = 2S3 − 32 S1 = 2(2S2 − 23 S0 ) − 32 S1
2(b)(ii)
=
8
3
Craftear
OR
Or use 2 S 2 − 8S1 + 8S0 − 3S −1 = 0 with substitution of their values
α'
1
+
2
+
1
=
α ' β '+ β ' γ '+ γ 'α '
.
β' γ'
α ' β 'γ '
1
A1 FT FT from 2(c).
M1
Uses
A1 OE but must be an equation.
2 y3 − 8 y 2 + 8y − 3 = 0
8
2
−3
2
M1 Makes their substitution.
2(2 − y )3 − 4(2 − y )2 + 3 = 0
3
B1 SOI
x =2− y
2
1
B1 CAO or as a single fraction.
Sn+3 = 2Sn+ 2 − 32 Sn
3
B1
S1 = 2
2(b)(i)
2(a)
48
Q4
Q3
4(b)
4(a)
Question
1
Question
A1
−2
A1 AG
1
2
M1 Uses original equation or formula for sum of cubes.
α 3 + β 3 + γ 3 = −2(−2) − 3(−2) − 3(3)
2
M1 Uses formula for sum of squares.
Guidance
(−2)2 − 2(3)
Marks
A1
[Equation is x 3 − 3 x 2 + 2 x + 1 = 0 ]
d =1
Answer
M1 Uses original equation or formula for sum of cubes.
6 − 3(5) + 2(3) + 3d = 0
6
A1
Craftear
Guidance
M1 A1 Uses formula for sum of squares.
B1
Marks
c=2
5 = 32 − 2(αβ + βγ + γα )
b = − (α + β + γ ) = −3
Answer
49
Q5
3(a)
Question
4(c)
n
r =1
r =1
)
Craftear
3
A1
(
y 3 + 3 y 2 + 3 + c3 y + 1 = 0
Guidance
M1 Correct attempt to eliminate cube root.
1
3
Marks
y + cy 3 + 1 = 0 −c3 y = ( y + 1) = y 3 + 3 y 2 + 3 y + 1
Answer
6
M1 A1 Simplifies.
M1 Applies formulae from MF19.
B1 Substitutes.
)
Expands.
M1 A1 Collects like terms and uses results from parts (a) and
(b).
B1
y = x3 x = y 3
1
n + 14 n(n + 1) 3n 2 − 5n −16
(
n + 14 n(n + 1) ( −12 − 4(2n +1) + 3n(n +1) )
n − 3n(n + 1) − n( n + 1)(2n + 1) + 34 n 2 ( n + 1) 2
n − 6 ( 12 n( n + 1) ) − 6 ( 16 n( n + 1)( 2n + 1) ) + 34 n 2 ( n + 1) 2
3
3
3
( (α + r ) + ( β + r ) + (γ + r ) ) = (1 + 3(−2)r + 3(−2)r 2 + 3r 3 )
n
(α + r )3 = α 3 + 3α 2 r + 3α r 2 + r 3
50
3(c)
3(b)
Craftear
A1
c=32
5
M1 Sets determinant equal to zero.
1
M1 A1 Evaluates determinant.
B1 If using their answer to (a) FT
2c 3 − 4 = 0
β3 γ 3
1
α3
γ 3 = 1 − (α 6 + β 6 + γ 6 ) + 2α 3 β 3γ 3 = 2c 3 − 4
α3 β3
1
α 3 β 3γ 3 = −1
3
A1 AG
α 6 + β 6 + γ 6 = (α 3 + β 3 + γ 3 )2 − 2(α 3 β 3 + β 3γ 3 + γ 3α 3 )
= 3 − 2c 3
2
M1
B1 FT Using their answer to (a).
α 6 + β 6 + γ 6 = ( −3 ) − 2 ( 3 + c 3 )
α 3 + β 3 + γ 3 = −3 α 3 β 3 + β 3γ 3 + γ 3α 3 = 3 + c3
51
Q6
1(c)
y = x +1 x = y −1
1(b)
Answer
A1
y 3 + (b − 3) y 2 + (c − 2b + 3) y + b − c = 0)
B1 Uses sum of products in pairs.
M1 Applies product of roots.
A1 AG
−(α +1)(α + 1) = c − 2b + 3
− (α + 1) ( β + 1) (α + 1) = − ( b − c )
( c − 2b + 3) ( b − 3) = b − c
4
B1 Uses sum of roots.
β + 1 = − ( b − 3)
3
M1 Using these relationships.
(α + 1 + β + 1 + γ +1) = 3 − b, (α + 1)( β + 1)(γ +1) = c − b
M1 A1 Substitutes and expands.
B1
Craftear
Guidance
B1 Uses correct substitution.
1
B1
Marks
(α + 1)( β +1) + (α + 1)(γ + 1) + ( β + 1)(γ + 1) = c − 2b + 3
Alternative method for question 1(b)
y 3 + (b − 3) y 2 + (c − 2b + 3) y + b − c = 0
d =1
1(a)
Question
52
Q8
Q7
1
1
Answer
A1
p = −6
α −2 + β −2 + γ −2 = ( −5) 2 − 2 ( 3 ) = 19
α −3 + β −3 + γ −3 = −5(19) − 3(−5) − 21 = −101
1(c)
7 y −3 + 3 y −2 + 5y −1 +1 = 0 y 3 + 5 y 2 + 3y + 7 = 0
y = x−1
Craftear
A1
α3 + β3 +γ 3 = 5
4
M1 A1
M1 A1
3
M1 A1
B1
Marks
2
M1
6(α 3 + β 3 + γ 3 ) = 6(α 2 + β 2 + γ 2 ) + 3(α + β + γ ) + 15
3
M1
p 2 + 36
2p
=−
p 2 + 12 p + 36 = 0
36
6
2
B1
2
3
p 2 + 36
α + β +γ =
36
2
36 y 3 − ( p 2 + 36) y 2 + (10 p + 9) y − 25 = 0
y (6 y − 3) 2 = ( − py + 5) 2 y (36 y 2 − 36 y + 9) = p 2 y 2 −10 py + 25
A1
M1
3
6 y 2 + py − 3 y 2 − 5 = 0 y 2 ( 6 y − 3) = − py + 5
Marks
B1
Answer
y = x2
1(b)
1(a)
Question
2(b)(ii)
2(b)(i)
2(a)
Question
53
Q9
7(iii)
7(ii)
7(i)
Question
2
x2
x2
=
−7 αβγ
⇒
α3
β3
γ3
317
+
+
=−
3 3
3 3
3 3
βγ αγ α β
343
α3
β3
γ3
58
2
49 3 3 + 3 3 + 3 3 = −14 + 27 − − 21
αγ α β
49
7
β γ
α2
β2
γ2
2
27 58
+
+
= − − 2 − =
2 2
2 2
2 2
β γ
α γ
α β
7
49 49
2
α
β
γ
2 1
1
1
27
+
+
=− , 2 + 2 + 2 =−
7 γ
49
βγ αγ αβ
β α
α2
α β2
β γ2
γ
=
,
=
,
=
So roots are
αβγ βγ αβγ αγ αβγ αβ
y=
⇒ 49 y 3 +14 y 2 − 27 y + 7 = 0
⇒ −7 y ( −7 y + 1) = 14 y − 7 ⇒ −7 y ( −7 y + 1) = (14 y − 7 )
−7 y ( −7 y ) + 2 ( −7 y ) + −7 y + 7 = 0
Answer
Craftear
2
Guidance
AG
2
A1
M1 Uses 49α ' 3 = −14α ′ 2 + 27α ′ − 7 .
3
2
α ' 2 + β ' 2 + γ ' 2 = (α '+ β '+ γ ') − 2 (α ' β '+ α ' γ '+ β ' γ ')
M1 A1 Uses
B1 States sum of roots and α ' β '+ α ' γ '+ β ' γ ' .
4
A1 AG
M1 Uses αβγ = −7 .
A1 AG
M1 Uses given substitution and eliminates radical.
Marks
54
Q11
Q10
9(i)
Question
6(iii)
6(ii)
6(i)
Question
+
β
1
d = 12
α
1
+
γ
1
=
αβ + β γ + αγ
⇒ c = αβ + βγ + αγ = 5
αβγ
Answer
S9 = −3S6 − 2S3 − 3 = −3 ( 5 ) − 2 ( −3) − 3 = −12
2
S6 = ( −3) − 2 ( 2 ) = 5
α 3 β 3 + β 3γ 3 + α 3γ 3 2
=
= −2
−1
α 3 β 3γ 3
Craftear
∑αβ . AG.
3
B1
M1 A1
Marks
4
Uses
α
1
+
β
1
+
γ
1
=
αβ + βγ + αγ
.
αβγ
Guidance
M1 A1 Sums y 3 − 3 y 2 + 2 y − 1 = 0 for y = α 3 , β 3 , γ 3 .
α ≠β
M1 A1 Uses ( ∑ α )2 = ∑ α 2 + 2
2
M1 A1
3
B1
S3 = −3
S −3 =
Guidance
M1 Obtains an equation in y not involving radicals.
A1 AG
3
Marks
y = y 3 + 3 y 2 + 3y + 1 ⇒ y 3 + 3y 2 + 2 y +1 = 0
y = ( y +1)
Answer
55
9(iv)
9(iii)
9(ii)
)
(−b = α + β + γ )
Real root is b = −6
Craftear
M1 A1 Gives the real root for b .
4
M1 A1 Equates expressions.
5b − 126
= b2 − 10 ⇒ b3 −15b + 126 = 0
b
2
M1 A1 Finds sum of squares in terms of b.
A1
(M1) Substitutes known values.
M1 A1 Uses expansion of (α + β + γ )3
A1
M1 A1 Sums over roots and uses 9(ii)
M1 Formulates equation.
2
M1 A1 Sums over roots
α 2 + β 2 + γ 2 = (α + β + γ ) − 2 (αβ + βγ + γα ) = b 2 − 10
Alternative method 2 for 9(iii)
b3 −15b + 126 = 0
−b3 = 90 −15b + 36
3
(α + β + γ ) = α 3 + β 3 + γ 3 + 3(α + β + γ ) (αβ + βγ + γα ) − 3αβγ
Alternative method 1 for 9(iii)
b3 −15b + 126 = 0
90 + b b 2 − 10 − 5b + 36 = 0
(
x 3 + bx 2 + 5 x + 12 = 0
90 + bS2 − 5b + 36 = 0 ⇒ S2 =
Alternative method for 9(ii)
5b −126
b
A1
(α + β + γ ) −10 = b2 −10
2
M1
2
α 2 + β 2 + γ 2 = (α + β + γ ) − 2 (αβ + βγ + γα )
56
Q13
Q12
6(ii)
6(i)
Question
4(ii)
4(i)
Question
)
M1 Uses y 3 = 2 y + 7 . Or uses formula for Σ (3α – 1)3
A1
S3 = 2 S1 + 7 × 3
= 21
Craftear
Guidance
M1 Accept substitution of y = 3x − 1 into given equation
and derivation of equation in x .
Marks
2
M1 A1 Or finds coefficient of x in ( x − 3)( x − 6 )( x − 12 ) .
Or substitutes root into equation
6
A1
A1 AG.
y +1
3
Guidance
M1 A1 Substitutes for a and solves quadratic
M1 Uses sum of roots
M1 A1 Uses product of roots
Marks
Obtains the given result
Substitutes x =
Answer
k = αβ + αγ + βγ = 6 (12 ) + 6 ( 3) + 12 ( 3) =126
Roots are 6, 12, 3
2 r 2 − 5r + 2 = 0 ⇒ r = 2 or r = 0.5
6 1 + r + r −1 = 21
(
a + ar + ar −1 = 21
αβγ = a3 = 216 ⇒ a = 6
Answer
57
Q14
2(i)
Question
6(iii)
4
49
1
=
∑(3α −1) 2 (3β − 1) 2
Craftear
A1 Verifies result (AG).
⇒ 2 pα + q = 0
4
M1 Combines equations.
14α 2
q
=−
p
7α
Guidance
B1 Sum of products in pairs.
Marks
7z3 + 2z2 – 1 = 0 A1
Uses S2 = (S1)2 – 2xΣαβ M1
4
A1
=
49
1
, etc. M1
2
written down directly.
7
Alt method: Finds cubic with roots
2α 2 + 4α 2 + 8α 2 = q
Answer
8
A1
M1
M1 A1
A1
B1 Sum of roots.
4
( − 2)2 – 2(7)(0)
=
2
49
7
Award M1A1 if S−1 = −
M1 Uses 7 y −2 = y − 2 y −1 .
M1 A1
α + 2α + 4α = − p
=
2
( 3α − 1) (3β − 1)2 (3γ − 1)2
(∑ ( 3α − 1)( 3β − 1))2 − 2 ( 3α −1) ( 3β −1)( 3γ −1) (∑ ( 3α −1))
s−2 =
∑
( 3α − 1)( 3β −1) + ( 3α − 1)( 3γ − 1) + ( 3β − 1) ( 3γ − 1) = −2
.
7
( 3α − 1)( 3β −1) ( 3γ − 1)
7 S −2 = S1 − 2 S −1
S −1 =
58
Q15
1(ii)
1(i)
Question
2(ii)
)
Or ( ∑α ) 3 – 3 ( ∑α ) ( ∑ αβ ) +3 αβγ
Craftear
Alt method: Use formula e.g. ∑ α 3 = ( ∑ α ) ∑ α 2 − ∑ αβ + 3αβγ
2
A1
= 5( −1) −13( 5) + 12 = −58
(
M1 Uses α 3 = 5α 2 − 13α + 4.
α 3 + β 3 + γ 3 = 5 (α 2 + β 2 + γ 2 ) − 13 (α + β + γ ) + 12
3
www
A1
= ‒1
Guidance
M1 Uses ∑ α 2 = ( ∑ α )2 − 2 ( ∑ αβ )
Marks
α 2 + β 2 + γ 2 = 52 − 2 (13)
Answer
2
B1 Verifies result (AG).
B1 Sum of roots and αβ + αγ + βγ . SOI
q3
⇒ p3r − q3 = 0
3
p
B1 Product of roots.
α + β + γ = 5 , αβ + αγ + βγ = 13
⇒r=
8α 3 = −r
59
Q17
Q16
1(iii)
1(ii)
1(i)
Question
7
Question
)
2
1
2
−3=0⇒
= 3−
y
y
y y
1
2
2
+
α β
α
2
1
+
β
2
1
β γ
2 2
2
1
γ
1
+
+
=
γ α
2
1
2
=
12
4
or
9
3
4
9
⇒ 9 y 3 − 12 y 2 + 4 y − 1 = 0
SR B1 for finding cubic by manipulating roots
+
1
1
⇒x=
2
x
y
y y
1
y=
Craftear
Answer
⇒Solution is –1, in ± 14 any order. Accept ±3.74 (awrt)
SR B1 for correct roots without working
⇒ ( x + 1) x 2 − 14
(
⇒ x 3 + x 2 − 14 x − 14 = 0
αβγ
Total:
Guidance
8
A1
Total:
Total:
Total:
Guidance
1
B1FT
1
B1FT
3
A1 OE
M1 Substituting and
squaring
M1 Rearranging to make
x the subject
Marks
M1A1 Attempt to factorise cubic
A1
M1A1 FT
∑αβ = −14 = −1 ⇒ αβγ = 14
αβγ
M1A1
Marks
∑αβ = 1 − 29 ⇒ ∑αβ = −14
2
Answer
60
Q18
4(ii)
4(i)
Question
3
2
αβ + βγ + γα = 2
1
2
27 9 3 3
− × + × 2 − 5 = −2
8 4 2 2
=
Guidance
M1, A1
A1
M1
4
M1A1
A1
M1 Alt methods: = (∑ α ) (∑ αβ ) − αβγ or ∑ α 2 ∑ α + 2αβγ − ∑ α 3
4
A1FT Alt method:
Let x= y–1
Sub and expand
2y3 – 9y2 16y – 19 = 0
Product of roots = 19/2
M1A1 Multiply out and group for M1
B1 (Can be awarded in (ii) if not seen here)
SOI
Marks
Craftear
27 9
3
− (α + β + γ ) + (αβ + βγ + γα ) − αβγ
8 4
2
1
2
=
1
2
( β + γ )( γ + α ) (α + β ) = 1 − α 1 − β
1 − γ
1
1
= 5 + 2 +1 +1 = 9
2
2
αβγ = 5 + β + γ =
(α + 1)( β + 1) (γ + 1) = αβγ + (αβ + βγ + γα ) +
(α + β + γ ) + 1
3
αβ + βγ + γα = 2αβγ = 5
2
α + β +γ =
Answer
61
62
y =1+
1
1
⇒x=
x
y −1
2
( y − 1)
(
3
+
1
( y −1)
M1
3
2
− 7 = 0 ⇒ 2 + ( y − 1) − 7 ( y − 1) = 0
)
⇒ 7 y 3 − 3 y 2 + 3y −1 − y + 1 − 2 = 0 ⇒ 7 y 3 − 21y 2 + 20 y − 8 = 0
Q20
8
A1
M1A1
[4]
ALT METHOD: ∑α, ∑αβ, αβγM1 A1, ∑(1+1/α) etc M1 A1
Let Sn denote αn + βn +γn.
S 2 = S12 − 2 ∑αβ = 12 − 2 × ( −1) = 3
S3 = S2 + S1 +15 = 3 + 1 + 15 = 19 (AG)
Alt method: ∑ α = 1, Σαβ= -1, and αβγ = 5 M1, A1
S3 = S13- 3Σα Σαβ + 3αβγ =19 M1, A1
S 4 = S3 + S 2 + 5S1 = 19 + 3 + 5 = 27
z −1
1
1
1
=1− ⇒ =1− x ⇒ z =
z
z
z
1− x
1
1
1
−
−
− 5 = 0 is the required equation.
Hence
3
2
(1 − x ) (1 − x ) (1 − x )
x=
2
3
⇒ 1 − (1− x ) − (1− x ) − 5 (1− x ) = 0 ⇒ … ⇒ 5 x 3 − 16 x 2 + 18 x − 6 = 0
M1A1
M1A1
[4]
M1A1
[2]
B1
M1
M1A1
[4]
Craftear
Q19 1
Q21
Q22
2
(
2
4
B1
(iii)
⇒ x3 −
Craftear
M1A1√
(2)
Total: 8
M1A1
(6)
1 1
1 1
1 αβ + βγ + γα 2
1
1
1
=
+
+
+ + =
=
2
2
αβγ α β γ αβγ
αβγ
α βγ αβ γ αβγ
9
(ii)
7 2 2
1
x + x − = 0 ⇒ 9 x 3 − 21x 2 + 2 x −1 = 0
3
9
9
M1A1
1
1
1
α + β +γ
7
+
+
=
=
αβ
βγ
γα
αβγ
3
(i)
2
B1
α + β + γ = 7 , αβ + βγ + γα = 2 , αβγ = 3 (Stated or implied by working.)
1
1
1
=
=
2
(αβ )( β γ )(γα ) (αβγ )
9
ALT METHOD: S3 – 3S2+4S1 + 3r = 0 M1 3r = 30 +3 × 1 – 4 × 3 A1 r=7 A1
3
A1
⇒ αβγ = −7
Required cubic equation is x − 3x + 4 x + 7 = 0 must see final equation
M1
A1
A1
3
( ∑ α ) = ∑ α 3 + 3 ∑ α ∑ αβ − 3αβγ
)
M1A1
Correct substitution in formula
or
Use of, e.g.: ∑ α 3 − 3αβγ = ∑ α ∑ α 2 − ∑ αβ
2 ∑ αβ = 9 − 1 ⇒ ∑ αβ = 4
[6]
63
64
Q23
2α + 2β = 0
ii α 2 + 4αβ + β 2 = − p
iii 2α 2 β + 2αβ 2 = − q
iv α 2 β 2 = −r
1
i
All four correct
(Any two correct)
Use of β = −α in ii, iii or iv
B2
(B1)
M1
⇒ p = 2α 2 and α 4 = − r ⇒ p 2 + 4r = 0 (AG)
M1A1
B1
(6)
Total
6
and q = 0 (CAO)
Q24
α + β + γ = − p = 15 ⇒ p = −15
B1
(
)
2(αβ + βγ + γα) = (α + β + γ ) 2 − α 2 + β 2 + γ 2 = 2q
1
⇒ q = (225 − 83) = 71
2
M1
A1
[3]
36
= 15 − α (= [β + γ ])
α
⇒ a 2 − 15α + 36 = 0 ⇒ α = 3 , α ≠ 12 , e.g. since 122 > 83 or other reason
βγ = 71− 36 = 35
⇒ r = −αβγ = −3× 35 = −105 (extra answer penalised)
M1
M1A1
B1
A1
[5]
Total
8
Q25
1
Let roots be α, α–1 and β ⇒ α + α–1 + β = 0
(1) Any of three for M1
Product of roots
⇒ β = –q
(2) A1 for another.
Sum of products in pairs ⇒ 1+ β (α + α–1) = p (3) A1 for a third
M1
A1
A1
From (1) and (3) 1 – β2 = p
Using (2)
1 – q2 = p or p + q2 = 1
M1
A1
[5]
Wrong sign in (2) scores
M1A0A1M1A0
Craftear
5
Q26
11E (i)
65
α + β + γ + δ = −4
B1
(1)
(ii)
α 2 + β 2 + γ 2 + δ 2 = (− 4) − 2 × 2 = 12
(iii)
1 1 1 1 − (− 4)
+ + + =
=4
α β γ δ
1
(iv)
α
β
γ
δ
α 2 + β 2 + γ 2 + δ 2 12
+
+
+
=
=
= 12
βγδ αγδ αβδ αβγ
αβγδ
1
2
M1A1
(2)
M1A1
(2)
M1A1
(2)
y = x +1 ⇒ x = y −1
( y −1)4 + 4( y −1)3 = y 4 − 6 y 2 + 8y − 3
2
2( y − 1) − 4( y − 1) +1 = 2 y 2 − 8y + 7
4
3
2
4
M1A1
A1
2
⇒ x + 4 x + 2 x − 4x +1 = y − 4 y + 4 = 0
A1
(y − 2) = 0 ⇒ y = ± 2 (twice).
A1
2
2
⇒ x = ± 2 −1 (twice).
(Some indication of four roots for fina
M1A1
(7)
[14]
Q27
Uses
−b
∑α = a .
Uses substitution
c=2
B1
( α + β = c − γ etc.) ⇒ y = c − x ⇒ x = c − y
M1
M1
(2 − y ) 3 − 2(2 − y ) 2 − 3(2 − y) + 4 = 0 …(their c)
to obtain required cubic
equation.
⇒ y3 − 4y2 + y + 2 = 0
A1
Obtains equation whose roots
are reciprocals of those in
previous cubic equation.
Uses z = y −1 to obtain 2 z 3 + z 2 − 4 z + 1 = 0
M1A
M1A1
Uses
∑ α = (∑ α ) − 2∑ αβ
2
2
∑
1
Craftear
3
2
1
1
1
= − 2(−2) = 4
2
4
(α + β )
2
M1A1
2
[8]
Q28
2
1
Makes substitution.
Squares.
Obtains result.
y 2 − 4 y + 3y 2 − 2 = 0
4
2
2
3
⇒ 9 y = 4 + y + 16 y − 4 y + 16 y − 8y
(N.B. Must see both terms in y2.)
⇒ y 4 − 8y 3 + 12 y 2 + 7 y + 4 = 0 (AG)
S 2 = 0 2 − 2 × (−4) = 8
S 8 = 8S 6 − 12 S 4 − 7 S 2 − 16
⇒ S 8 = 8S 6 − 12 S 4 − 56 − 16 = 8S 6 − 12 S 4 − 72 (AG)
Alternatively – for final two marks.
S2 = 8 , S3 = –9 , S4 = 40 , S5 = –60 , S6 =203 , S7 = –378
S8 = 1072 (generated by substitution of roots in equations
and summing.)
Then 8S 6 −12S 4 − 72 = 1624 − 480 − 72 = 1072 = S 8
M1 requires a complete method, A1 if all correct.
M1
M1
A1
3
B1
M1
A1
3
[6]
66
Q29
2 (i)
α 2 + β 2 + γ 2 = (α + β + γ ) − 2(αβ + βγ + γα )
= 0 − 2(− p ) = 2 p (AG)
B1
1
2
2
Finds S2.
(ii)
Finds S3.
α 3 + β 3 + γ 3 = p ∑α + 3q = 0 + 3q = 3q (AG)
M1A1
(iii)
Finds S5.
α5 + β5 +γ 5 = p
∑α + q∑α
M1
3
2
= p.3q + q.2 p = 5 pq
5
3
⇒ 6 ∑α = 30pq = 5∑α ∑α
2
A1
(AG)
A1
[6]
3
Q30
36
9
3
Uses sum of roots
∑ α = 3a = − 8 = − 2 ⇒ a = − 2
Uses product of roots
αβγ = a a 2 − d 2 =
) 218
M1
Substitutes for a,
9
2 21
7
−d2 = − ×
=−
4
3 8
4
M1
and solves
⇒ d 2 = 4 ⇒ d = ±2
A1
7 3 1
Roots are − , − ,
2 2 2
A1
Uses sum of products in
pairs.
(Or expands
(2x + 7)(2x + 3)(2x + 1)
(
21
7
3
k
M1A1
∑ αβ = 4 − 4 − 4 = 8
M1
⇒ k = 22
A1
Craftear
5
(8)
and equates
coefficients)
[8]
Q31
1
States
∑ α and ∑ αβ
Uses formula for
correctly.
Uses formula for
to obtain result.
∑α
3
∑ α = 7 ∑ αβ = 2
∑ α = 7 − 2 × 2 = 45
2
2
∑ α = 7 ∑ α − 2∑ α + 9
3
2
= 315-14+9 = 310
B1
B1
2
M1
A1A1
3
[5]
67
Q32
(i)
(ii)
∑ α = (∑ α ) − 2∑ αβ
2
2
α is a root ⇒ α 4 − 3α 2 + 5α − 2 = 0
⇒ α n+ 4 − 3α n+ 2 + 5α n+1 − 2α n = 0
Repeat for β , γ , δ and sum
⇒ S n+ 4 − 3S n+ 2 + 5S n+1 − 2 S n = 0 (AG)
Finds S 4 from formula.
S 2 = 0 − 2 × ( −3) = 6
S 4 = 3 × 6 − 5 × 0 + 2 × 4 = 26
∑ αβγ
S =
S −1 =
Uses
−1
αβγδ
Finds S 3 from formula.
Finds S 5 from formula.
2
B1
M1A1
3
−5 5
=
−2 2
M1A1
S3 = 3 × 0 − 5 × 4 + 2 ×
M1A1
5
= −15
2
S 5 = 3 × (−15) − 5 × 6 + 2 × 0 = −75
∑α β = S S − S
2
3
2
3
5
= 6 × (−15) − (−75) = −15
Q33
7
M1
A1
(α + β + γ ) 2 − 2(αβ + βγ + γα ) = α 2 + β 2 + γ 2 ⇒
Either
Required equation is x 3 − 4 x 2 + x + c = 0
⇒ ∑ α 3 − 4∑ a 2 + 4 + 3c = 0
⇒ 3c = 56 − 34 − 4 = 18 ⇒ c = 6
∑ αβ = 1
(AG)
Or
α 3 + β 3 + γ 3 − 3αβγ = (α + β + γ )(α 2 + β 2 + γ 2 − αβ − βγ − γα )
(or some other appropriate identity, e.g.
(α + β + γ ) 3 = α 3 + β 3 + γ 3 + 3(α + β + γ )(αβ + βγ + γα ) − 3αβγ )
⇒ ... ⇒ αβγ = −6
(AG)
⇒ x3 − 4x 2 + x + 6 = 0
⇒ (x + 1)(x − 2)(x − 3) = 0 ⇒ x = –1,2,3.
M1A1
6
M1
M1A1
3
M1A1
[14]
2
M1
M1
A1
Craftear
EITHER
Substitute α into equation.
Multiply by α n .
Obtain result.
11
(M1)
(M1A1)
A1
M1A1
6
[8]
Q34
8 (i) Deduces initial result.
Substitutes into cubic
equation.
68
u = −α + β + γ ⇒ u + 2α = α + β + γ = 1
1− u
⇒α =
2
3
Deduces new cubic
equation.
(ii) Deduces initial result.
Substitutes into cubic
equation.
Deduces new cubic
equation.
M1
2
1− u 1− u
1− u
⇒
−
− 3
−10 = 0
2 2
2
⇒ … ⇒ u 3 − u 2 − 13u + 93 = 0
αβγ = 10
⇒v=
M1A1
A1
A1
5
B1
1
v
1
1
⇒ =
= ⇒ α = 10v
βγ
α αβγ 10
M1A1
(10v )3 − (10v )2 − 3(10v ) − 10 = 0
M1
⇒ 100v 3 − 10v 2 − 3v − 1 = 0
A1
5
[10]
Alternatively:
Award M1 for an
attempt at formulae for
all three coefficients.
A1 for any two correct.
A1 for completion
(ii) For final 4 marks in
(ii):
Award M1 for an
attempt at formulae for
all three coefficients.
A1 for any one correct.
A1 for a second one
correct.
A1 for completion.
–b = ∑α = 1 ⇒ b = –1
c = 4∑αβ – (∑α)2
= 4 × (–3) – 12 = –13
−d =4
∑α ∑αβ − (∑α ) − 8αβγ
3
= 4 × 1 × (–3) – 13 – 8 × 10 = –93
So u3 – u2 – 13u + 93 = 0
(5)
Let equation be v3 + bv2 + cv + d = 0.
Craftear
8 (i) For final 3 marks in (i): Let equation be u3 + bu2 + cu + d = 0.
Σα
1
1
= ⇒b=−
αβγ 10
10
3
Σαβ
−3
c=
=
=−
100
(αβγ ) 2 10 2
−b =
1
1
1
=
⇒d =−
100
(αβγ ) 2 10 2
1
3
1
v−
=0
So v 3 − v 2 −
10
100
100
or 100v3 – 10v2 – 3v – 1 = 0.
−d =
(5)
[10]
69
Craftear
RATIONAL FUNCTIONS
70
Q1
7
The curve C has equation y =
x2 + x + 9
.
x+1
(a) Find the equations of the asymptotes of C.
[3]
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(b) Find the coordinates of the stationary points on C.
[4]
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71
(c) Sketch C, stating the coordinates of any intersections with the axes.
[3]
(d) Sketch the curve with equation y =
x2 + x + 9
and find the set of values of x for which
x+1
2 x 2 + x + 9 2 13 x + 1 .
© UCLES 2021
[5]
9231/11/M/J/21
Craftear
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72
Q22
7
x2 - x - 3
.
1 + x - x2
(a) Find the equations of the asymptotes of C.
The curve C has equation y =
[2]
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[3]
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(b) Find the coordinates of any stationary points on C.
73
[3]
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(d) Sketch the curve with equation y =
which
© UCLES 2021
x2 - x - 3
1 3.
1 + x - x2
x2 - x - 3
and find in exact form the set of values of x for
1 + x - x2
[6]
9231/13/M/J/21
Craftear
(c) Sketch C, stating the coordinates of the intersections with the axes.
74
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75
Q3
7
The curve C has equation y =
4x + 5
.
4 - 4x 2
(a) Find the equations of the asymptotes of C.
[2]
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(b) Find the coordinates of any stationary points on C.
[4]
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76
(d) Sketch the curve with equation y =
which 4 4x + 5 2 5 4 - 4x 2 .
[3]
4x + 5
and find in exact form the set of values of x for
4 - 4x 2
[6]
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(c) Sketch C, stating the coordinates of the intersections with the axes.
77
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78
Q4
6
The curve C has equation y =
x2
.
x-3
(a) Find the equations of the asymptotes of C.
[3]
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(b) Show that there is no point on C for which 0 1 y 1 12 .
[4]
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79
(c) Sketch C.
x2
and y = x - 3 on a single diagram, stating the coordinates
x-3
of the intersections with the axes.
[4]
(d) (i) Sketch the graphs of y =
(ii) Use your sketch to find the set of values of c for which
x2
G x + c has no solution. [1]
x-3
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© UCLES 2021
9231/12/O/N/21
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[2]
80
Q55
6
The curve C has equation y =
x2 + x - 1
.
x-1
(a) Find the equations of the asymptotes of C.
[3]
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(b) Show that there is no point on C for which 1 1 y 1 5.
[4]
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81
(c) Find the coordinates of the intersections of C with the axes, and sketch C.
[3]
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Craftear
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(d) Sketch the curve with equation y =
© UCLES 2020
x2 + x - 1
.
x-1
9231/11/O/N/20
[2]
[Turn over
82
Q66
6
Let a be a positive constant.
(a) The curve C1 has equation y =
x-a
.
x - 2a
[2]
Sketch C1 .
x-a 2
l . The curve C3 has equation y = x - a .
x - 2a
x - 2a
(b) (i) Find the coordinates of any stationary points of C2 .
[3]
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© UCLES 2020
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The curve C2 has equation y = b
83
(ii) Find also the coordinates of any points of intersection of C2 and C3 .
[3]
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(c) Sketch C2 and C3 on a single diagram, clearly identifying each curve. Hence find the set of values
of x for which b
© UCLES 2020
x-a 2
l G x-a .
x - 2a
x - 2a
[5]
9231/12/O/N/20
[Turn over
Craftear
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84
Q7
1
Let a be a positive constant.
ax
.
x+7
[2]
Craftear
(a) Sketch the curve with equation y =
© UCLES 2020
9231/11/M/J/20
85
(b) Sketch the curve with equation y =
ax
ax
a
and find the set of values of x for which
2 .
x+7
x+7
2
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[4]
86
Q88
3
The curve C has equation y =
x2
.
2x + 1
(a) Find the equations of the asymptotes of C.
[3]
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(b) Find the coordinates of the stationary points on C.
[3]
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87
(c) Sketch C.
Craftear
[3]
© UCLES 2020
9231/11/M/J/20
[Turn over
88
Q9
6
The curve C has equation y =
10 + x - 2x 2
.
2x - 3
(a) Find the equations of the asymptotes of C.
[3]
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(b) Show that C has no turning points.
[3]
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89
[3]
Craftear
(c) Sketch C, stating the coordinates of the intersections with the axes.
© UCLES 2020
9231/13/M/J/20
90
which
10 + x - 2x 2
1 4.
2x - 3
10 + x - 2x 2
and find in exact form the set of values of x for
2x - 3
[6]
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© UCLES 2020
9231/13/M/J/20
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Craftear
(d) Sketch the curve with equation y =
91
Q10
4
The line y = 2x + 1 is an asymptote of the curve C with equation
y=
x2 + 1
.
ax + b
(i) Find the values of the constants a and b.
[3]
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[1]
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(iii) Sketch C. [Your sketch should indicate the coordinates of any points of intersection with the
y-axis. You do not need to find the coordinates of any stationary points.]
[3]
© UCLES 2019
9231/11/O/N/19
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Craftear
(ii) State the equation of the other asymptote of C.
92
Q11
10
The curves C1 and C2 have equations
y=
ax
x+5
and
y=
respectively, where a is a constant and a > 2.
x2 + a + 10x + 5a + 26
x+5
(i) Find the equations of the asymptotes of C1 .
[2]
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(ii) Find the equation of the oblique asymptote of C2 .
[2]
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(iii) Show that C1 and C2 do not intersect.
[2]
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93
(iv) Find the coordinates of the stationary points of C2 .
[3]
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(v) Sketch C1 and C2 on a single diagram. [You do not need to calculate the coordinates of any
[3]
points where C2 crosses the axes.]
© UCLES 2019
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Craftear
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94
Q12
6
The curve C has equation
y=
where k is a positive constant.
x2
,
kx − 1
(i) Obtain the equations of the asymptotes of C.
[3]
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(ii) Find the coordinates of the stationary points of C.
[3]
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95
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(iii) Sketch C.
© UCLES 2019
[3]
9231/13/M/J/19
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Craftear
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96
Q13
6
The curve C has equation
y=
where b is a positive constant.
x2 + b
,
x+b
(i) Find the equations of the asymptotes of C.
[3]
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(ii) Show that C does not intersect the x-axis.
[1]
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97
(iii) Justifying your answer, find the number of stationary points on C.
[2]
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(iv) Sketch C. Your sketch should indicate the coordinates of any points of intersection with the
y-axis. You do not need to find the coordinates of any stationary points.
[3]
© UCLES 2018
9231/11/M/J/18
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Craftear
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98
Q14
4
The curve C has equation
y=
x2 + 7x + 6
.
x−2
(i) Find the coordinates of the points of intersection of C with the axes.
[2]
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(ii) Find the equation of each of the asymptotes of C.
99
(iii) Sketch C.
Craftear
[3]
© UCLES 2018
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[Turn over
100
Q15
6
The curve C has equation
where a is constant and a > 1.
y=
x2 + ax − 1
,
x+1
(i) Find the equations of the asymptotes of C.
[3]
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(ii) Show that C intersects the x-axis twice.
[1]
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101
(iii) Justifying your answer, find the number of stationary points on C.
[2]
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(iv) Sketch C, stating the coordinates of its point of intersection with the y-axis.
© UCLES 2018
9231/11/O/N/18
[3]
[Turn over
Craftear
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102
Q16
9
16
The curve C has equation
y=
5x2 + 5x + 1
.
x2 + x + 1
(i) Find the equation of the asymptote of C.
[2]
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(ii) Show that, for all real values of x, − 13 ≤ y < 5.
[4]
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103
17
(iii) Find the coordinates of any stationary points of C.
[2]
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(iv) Sketch C, stating the coordinates of any intersections with the y-axis.
© UCLES 2018
9231/12/O/N/18
[2]
[Turn over
Craftear
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104
Q17
9
The curve C has equation y =
x2 − 3x + 6
.
1−x
(i) Find the equations of the asymptotes of C.
[3]
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© UCLES 2017
9231/11/M/J/17
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(ii) Find the coordinates of the turning points of C.
105
(iii) Find the coordinates of any intersections with the coordinate axes.
[2]
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(iv) Sketch C.
Craftear
[3]
© UCLES 2017
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106
Q18
9
The curve C has equation
y=
3x − 9
.
x − 2 x + 1
(i) Find the equations of the asymptotes of C.
[2]
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(ii) Show that there is no point on C for which 13 < y < 3.
[4]
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(iii) Find the coordinates of the turning points of C.
[3]
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(iv) Sketch C.
© UCLES 2017
[3]
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Craftear
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Q19
A curve C has equation y =
x2
. Find the equations of the asymptotes of C.
x−2
[3]
Show that there are no points on C for which 0 < y < 8.
[4]
Sketch C, giving the coordinates of the turning points.
[3]
Craftear
7
108
© UCLES 2016
9231/11/M/J/16
[Turn over
Q20
The curve C has equation y =
dy
x+2
. Show that
< 0 at all points on C.
2
dx
x −9
[3]
State the equations of the asymptotes of C.
[2]
Sketch C, showing the coordinates of any points of intersection with the coordinate axes.
[3]
Craftear
5
109
© UCLES 2016
9231/13/M/J/16
[Turn over
Q21
10
110
4x2 − 3x
. Verify that the equation of C may be written in the form
x2 + 1
1
3x − 12
9
x + 32
y=− +
and
also
in
the
form
y
=
−
.
[3]
2 2 x2 + 1
2 2 x2 + 1
The curve C has equation y =
Hence show that − 12 ≤ y ≤ 92 .
[2]
Without differentiating, write down the coordinates of the turning points of C.
[2]
State the equation of the asymptote of C.
[1]
Craftear
Sketch the graph of C, stating the coordinates of the intersections with the coordinate axes and the
asymptote.
[3]
© UCLES 2015
9231/13/M/J/15
[Turn over
Q22
8
The curve C has equation y =
C has no stationary points.
111
2x2 + kx
, where k is a constant. Find the set of values of k for which
x+1
[5]
Craftear
For the case k = 4, find the equations of the asymptotes of C and sketch C, indicating the coordinates
of the points where C intersects the coordinate axes.
[6]
© UCLES 2015
9231/11/O/N/15
[Turn over
Q23
112
12 OR
The curve C has equation
ax2 + bx + c
,
x+d
where a, b, c and d are constants. The curve cuts the y-axis at 0, −2 and has asymptotes x = 2 and
y = x + 1.
y=
[1]
(ii) Determine the values of a, b and c.
[6]
(iii) Show that, at all points on C, either y ≤ 3 − 26 or y ≥ 3 + 26.
[7]
Craftear
(i) Write down the value of d.
© UCLES 2014
9231/11/M/J/14
[Turn over
Q24
113
11: EITHER
Express
2x2 − x + 5
A
B
in the form 2 +
+
, where A and B are integers to be found.
2
x−1 x+1
x −1
The curve C has equation y =
dy
= 0.
dx
[3]
2x2 − x + 5
. Show that there are two distinct values of x for which
x2 − 1
[4]
Craftear
Sketch C, stating the equations of the asymptotes and giving the coordinates of any points of
intersection with the coordinate axes and with the asymptotes. You do not need to find the coordinates
of the turning points.
[7]
© UCLES 2014
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[Turn over
Q25
4
A curve C has equation y =
114
2x2 + x − 1
. Find the equations of the asymptotes of C.
x−1
[3]
Show that there is no point on C for which 1 < y < 9.
Craftear
[4]
© UCLES 2014
9231/11/O/N/14
[Turn over
Q26
10
The curve C has equation y =
115
2x2 − 3x − 2
. State the equations of the asymptotes of C.
x2 − 2x + 1
[2]
at all points of C.
Show that y ≤ 25
12
[4]
Find the coordinates of any stationary points of C.
[3]
Craftear
Sketch C, stating the coordinates of any intersections of C with the coordinate axes and the asymptotes.
[4]
© UCLES 2013
9231/11/M/J/13
[Turn over
Q27
10
116
The curve C has equation
y=
where p is a positive constant and p ≠ 3.
px2 + 4x + 1
,
x+1
(i) Obtain the equations of the asymptotes of C.
[3]
(ii) Find the value of p for which the x-axis is a tangent to C, and sketch C in this case.
[4]
Craftear
(iii) For the case p = 1, show that C has no turning points, and sketch C, giving the exact coordinates
of the points of intersection of C with the x-axis.
[5]
© UCLES 2013
9231/11/O/N/13
[Turn over
Q28
7
117
The curve C has equation
y=
2x2 + 5x − 1
.
x+2
Find the equations of the asymptotes of C.
Show that
[3]
dy
> 2 at all points on C.
dx
[3]
Sketch C.
8
[3]
The points A, B, C have position vectors
4i + 5j + 6k,
5i + 7j + 8k,
2i + 6j + 4k,
[4]
Craftear
respectively, relative to the origin O. Find a cartesian equation of the plane ABC.
© UCLES 2013
9231/13/O/N/13
[Turn over
Q29
9
118
The curve C has equation
y=
Show that, for all x, 1 ≤ y ≤ 52 .
2x2 + 2x + 3
.
x2 + 2
[4]
Find the coordinates of the turning points on C.
[3]
Find the equation of the asymptote of C.
[2]
Craftear
Sketch the graph of C, stating the coordinates of any intersections with the y-axis and the asymptote.
[2]
© UCLES 2012
9231/11/M/J/12
[Turn over
Q30
7
119
The curve C has equation
x
,
x−2
where λ is a non-zero constant. Find the equations of the asymptotes of C.
y = λx +
[3]
[3]
Sketch C in the case λ = −1, stating the coordinates of the intersections with the axes.
[3]
Craftear
Show that C has no turning points if λ < 0.
© UCLES 2012
9231/12/O/N/12
[Turn over
Q31
The curve C has equation y =
x2 − 3x + 3
. Find the equations of the asymptotes of C .
x−2
[3]
Show that there are no points on C for which −1 < y < 3.
[4]
Find the coordinates of the turning points of C.
[3]
Sketch C.
[2]
Craftear
9
120
© UCLES 2012
9231/13/O/N/12
[Turn over
Q32
The curve C has equation y =
x2
. Find the equations of the asymptotes of C.
x−2
[3]
Find the coordinates of the turning points on C.
[3]
Draw a sketch of C.
[3]
Craftear
6
121
© UCLES 2012
9231/13/M/J/12
[Turn over
RATIONAL FUNCTIONS
MS
Craftear
122
Q1
7(c)
7(b)
7(a)
Question
x( x + 1) + 9
9
=x+
x +1
x +1
9
x
Craftear
3
B1 Lower branch correct and good approach to asymptotes
throughout, no extra branches.
B1 Upper branch with (0, 9) stated or shown on diagram.
B1 Axes labelled and correct asymptotes drawn.
4
A1
( −4, − 7 )
M1 A1 Differentiates and sets derivative equal to 0.
3
A1
M1 Finds oblique asymptote.
A1
y
Guidance
B1 States vertical asymptote.
Marks
( 2, 5)
dy
= 1 − 9( x +1)−2 = 0 ( x + 1)2 = 9
dx
y=x
y=
x = −1
Answer
123
7(d)
Question
Craftear
5
A1
+ 152 x + 312 = 0
x < 12 and x > 5.
or x
2
solutions since 7 > 132 .
M1 M1 Finds critical points, award M1 for each case.
May state that x 2 + x + 9 = − 132 (x + 1) has no real
A1
− 112 x + 52 = 0
13
13
(x +1) or x 2 + x + 9 = − (x + 1)
2
2
x
Guidance
B1 FT FT from sketch in (c) with asymptotes shown.
Marks
x = 12 , 5
x
2
x2 + x + 9 =
9
y
Answer
124
Q2
7(c)
7(b)
7(a)
Question
)
1
2
( +
1
2
13,0 ,
) ( −
1
2
1
2
13,0 , (0, −3)
)
x
Craftear
dy
equal to 0 and forms equation.
dx
3
B1 States exact coordinates of intersections with axes.
B1 Correct shape and position, with all asymptotic
behaviour clear.
B1 Both axes labelled and correct asymptotes shown.
3
A1 WWW
dy
.
dx
Sets their
Finds
( 12 , − 135 )
M1
2
B1 Horizontal asymptote.
M1
y
Guidance
B1 Vertical asymptotes. Must be exact.
Marks
(2 x − 1)(−2) = 0
dy (1 + x − x 2 )(2 x −1) − ( x 2 − x − 3)(1 − 2 x)
=
2
dx
1 + x − x2
(
1− 5
1+ 5
, x=
2
2
y = −1
x=
Answer
125
Q3
7(b)
7(a)
Question
7(d)
(
)
( −2, 14 ) , ( − 12 ,1)
16 x 2 + 40 x + 16 = 0
dy (4 − 4 x 2 )(4) − (4x + 5)(−8 x)
=
2
dx
4 − 4 x2
Craftear
dy
.
dx
Guidance
4
A1 A1
M1 Sets equal to 0 and forms equation.
M1
Finds
B1 Horizontal asymptote.
y=0
2
B1 Vertical asymptotes.
Marks
x = 1, x = −1
Answer
6
A1 FT Must be three distinct regions and strict
inequalities.
A1 Must be exact.
x = 12 + 12 7, x = 12 − 12 7 , x = 0, x = 1
x < 12 − 12 7, 0 < x < 1, x > 12 + 12 7
M2 Finds critical points, award M1 for each case.
x2 − x − 3
x2 − x − 3
or
=
3
= −3
1 + x − x2
1 + x − x2
4 x 2 − 4 x − 6 = 0 or −2 x 2 + 2 x = 0
B1 Correct shape as x tends to infinity and
intersections with x axis.
B1 FT FT from sketch in (c).
126
7(d)
4x5
4-4x ଶ
x
B1 Correct shape at infinity.
B1 FT FT from sketch in part (c).
3
B1 States coordinates of intersections with axes.
2
− 53
5
6 < x < − 54 , 0 < x < 52 + 53 6 , x ≠ ±1
Craftear
6
A1 FT Condone exclusion of x = ±1 from the range.
A1 A0 if –1.07, 1.87
4x5
ฬ
yൌ ฬ
4-4x ଶ
yൌ
x
B1 Correct shape and position.
B1 Axes and asymptotes.
x = − 54 , x = 0 or x = 52 − 53 6, x = 52 + 53 6
xൌ-1
M2 Finds critical points, award M1 for each case.
( 0, 54 )
xൌ-1
4x + 5 5
4x + 5
5
= or
=−
2
2
4
4
4 − 4x
4 − 4x
2
2
5 x + 4 x = 0 or 5 x − 4 x − 1 0 = 0
(− 54 ,0),
xൌ1
xൌ1
7(c)
127
6(c)
6(b)
6(a)
Question
9
leading to y = x + 3
x −3
Answer
𝑦
0 < y < 12
𝑥
𝑥
3
y 2 − 4(3 y) < 0 leading to y 2 − 12 y < 0
yx − 3y = x2 leading to x2 − yx + 3 y = 0
y = x +3+
x=3
3
𝑥
Q4
Craftear
Guidance
2
B1 Branches correct.
B1 Axes and asymptotes.
4
A1 AG
M1 Uses that discriminant is negative.
M1 A1 Forms quadratic in x.
3
M1 A1 Finds oblique asymptote.
B1 States vertical asymptote.
Marks
128
Q5
6(b)
6(a)
Question
6(d)(ii)
Answer
B1 States vertical asymptote.
Guidance
M1 A1 Forms quadratic in x.
3
M1 A1 Finds oblique asymptote.
Marks
1
B1
4
4
A1 AG
Craftear
x
B1 Correct intercepts with axes (may be seen on graph).
B1 Correct shape of y = x − 3.
B1 Correct shape at infinity.
1< y < 5
3
B1 FT FT from sketch in (c).
M1 Uses that discriminant is negative if there are no values
of x.
𝑥
𝑥
(1 − y )2 − 4( y −1) < 0 y 2 − 6 y + 5 < 0
yx − y = x 2 + x − 1 x 2 + (1− y)x + y −1 = 0
x 2 + x −1 = ( x −1)(x + 2) +1 y = x + 2
x =1
c ⩽ –3
𝑦
3
𝑥
6(d)(i)
129
6(d)
6(c)
(0,1),
1
2
1
2
1
2
1
2
( − + √ 5,0 ) , ( − − √ 5,0 )
y
x
Craftear
x
2
B1 Correct shape at extremities.
B1 FT FT from sketch in (c) with both branches.
3
B1 States coordinates of intersections with axes, can be
labelled on graph. Accept ( 0.618,0 ) and ( −1.62,0 ) .
B1 Branches correct.
B1 Axes and asymptotes.
130
Q6
6(b)(ii)
6(b)(i)
6(a)
Question
Craftear
A1
Since x ≠ 2a, stationary point is ( a,0 ) .
A1
(a, 0) and ( 32 a , 1)
3
A1 Both x values or one correct point.
x = a and 32 a or (a, 0) or ( 32 a , 1)
dy
= 0.
dx
M1 Finds a critical point.
Sets
( x − a )2 = ( x − a)( x − 2a) or ( x − a )2 = −( x − a)( x − 2a)
3
M1
2
Solves for x.
x
M1
2a
Guidance
B1 Correct position and shape. Not too truncated.
B1 Asymptotes labelled.
Marks
2a ( x − a )
dy
=−
=0
3
dx
( x − 2a )
1
y
Answer
131
Q7
1(a)
C2
1
y
C3
C2
2a
C3
C3
C2
x
Answer
Craftear
(B1 for axes and asymptotes correct, B1 for branches correct)
x ⩽ 32 a
Question
6(c)
5
B1 Accept algebraic method.
B1 Relative positions and intersections correct.
Marks
2
B1
B1
C3 FT from their sketch in part (a). Clearly identified.
B1 Correct shape of C2 . Clearly identified.
B1 FT
B1 Axes and asymptotes.
132
Q8
3(b)
3(a)
Question
1(b)
7
3
Craftear
A1
(0,0),(−1, −1)
3
A1
3
A1
M1
B1
x = 0, −1
1
4( 2 x + 1)
Marks
M1
= 12 x − 14 +
Answer
4
A1
A1
M1
dy 1
1
= −
= 0 (2 x + 1) 2 = 1
dx 2 2 ( 2x +1)2
2x + 1
( 2 x + 1) ( 12 x − 14 ) + 14
y = 12 x − 14
y=
x = − 12
7
x < −7 , −7 < x < − , x > 7
3
x = 7 and x = −
ax
ax
a
a
= or
=−
x+7 2
x+7
2
(B1 FT from sketch in part (a))
B1 FT
133
Q9
6(b)
6(a)
Question
3(c)
)
dy
14
or
= −1 −
2
dx
( 2 x − 3)
2
Craftear
122 − 4(4)(23) = −224 < 0 (or y ' < 0 ) No turning points.
4 x 2 − 12 x + 23 = 0
dy ( 2 x − 3)(1 − 4 x ) − 2 10 + x − 2x
=
dx
( 2 x − 3) 2
(
−2 x 2 + x + 10 = ( 2 x − 3 ) ( − x − 1) + 7 y = − x − 1
x = 32
Answer
(B1 for axes and asymptotes correct, B1 for upper branch correct, B1 for lower branch correct)
3
A1
A1
M1
3
M1 A1
B1
Marks
3
B1
B1
B1
134
6(c)
(0, − 103 ), ( −2,0 ) ,( 52 ,0)
Craftear
(B1 for axes and asymptotes correct, B1 for branches correct)
3
B1
B1
B1
135
4(ii)
4(i)
Q10 Question
6(d)
1
2
(
Answer
− 112 < x < 14 9 − √ 65 and 2 < x < 14 9 + √ 65
)
)
Marks
Guidance
Craftear
1
B1 FT
3
A1 A1
M1 Uses that 2 x + 1 is the quotient.
6
A1 FT
A1
x = − 112 , 2 or x = 14 ( 9 ± √ 65 )
(
M2
B1
B1 FT
Marks
10 + x − 2 x 2
10 + x − 2 x 2
= −4
= 4 or
2x − 3
2x − 3
2 x 2 + 7 x − 22 = 0 or 2 x 2 − 9 x + 2 = 0
1
1
⇒a= , b=−
2
4
x=
Answer
(B1 FT for their sketch in (c), B1 for correct shape at infinity)
x 2 + 1 = ( ax + b )( 2x + 1) + c
Question
136
Q11
M1 By inspection or long division.
A1
oblique asymptote is y = x + a + 5
Guidance
x 2 + ( a + 10 ) x + 5a + 26 = ( x + 5 )( x + a + 5 ) + 1
2
2
B1 B1
Marks
3
Deduct at most one mark for poor forms at infinity.
B1 FT Right branch correct.
B1 Left branch correct.
10(ii)
Craftear
Guidance
B1 Intersection (0,-4) given and asymptotes drawn.
Marks
x = −5 and y = a
Answer
Answer
10(i)
Question
4(iii)
Question
137
10(v)
10(iv)
10(iii)
Craftear
A1 Correct discriminant and conclusion.
102 − 4 ( 5a + 26 ) = −4 − 20a < 0 so no intersection point
A1
A1 Must have both points.
x 2 + 10 x + 24 = 0
Stationary points are ( −4,a + 2 ) and ( −6,a − 2 )
C2 correct.
B1
3
C1 correct.
B1
B1 Asymptotes drawn, intersection correct.
3
M1 Differentiates and forms quadratic equation.
( x + 5 )( 2 x + a + 10 ) − x 2 − ax − 10 x − 5a − 26 = 0
2
M1 Puts y –values equal and forms quadratic equation.
x 2 + 10 x + 5a + 26 = 0
138
Q12
6(iii)
6(ii)
6(i)
Question
kx − 1
Craftear
3
B1 Lower branch correct.
Deduct at most 1 mark for poor forms at infinity.
B1 Upper branch correct.
B1 Axes and asymptotes correct.
3
A1 Finds y -coordinates
M1 Differentiates and equates to 0.
( 0,0 ) , ( 2k −1 , 4k −2 )
2
= 0 ⇒ kx 2 − 2 x = 0
A1 Finds x -coordinates.
( kx − 1)
2 x ( kx − 1) − kx 2
x = 0, 2k −1
y′ =
3
A1
M1 Finds oblique asymptote.
( kx − 1) ( k −1 x + k −2 ) + k −2
Guidance
B1 States vertical asymptote.
Marks
1
k
Oblique asymptote is y = k −1 x + k −2
y=
x=
Answer
139
Q13
6(iii)
6(ii)
6(i)
Question
2
)
( x + b )2
b2 + b
=0.
Craftear
dy
b2 + b
= 0 gives
and setting
dx
x+b
b2 + b > 0
Therefore there are two stationary points on C
1−
Or differentiating y = x − b +
dy 2 x ( x + b ) − x + b
=
= 0 ⇒ x 2 + 2bx − b = 0
2
dx
( x + b)
(
If y = 0 then x 2 + b = 0 which has no real root.
Thus the oblique asymptote is y = x − b
x−b
x + b x2 + 0x + b
Find
dy
and set = 0
dx
2
A1 Use discriminant or (x + b)2 to show two stationary
points
M1
1
B1 Must refer to b > 0 OE
3
A1
M1 By inspection or long division.
x 2 + b = ( x + b ) ( x − b ) + b 2 + b or
Guidance
B1
Marks
Vertical asymptote is x = −b .
Answer
140
Q14
4(iii)
4(ii)
4(i)
Question
6(iv)
24
⇒ other asymptote is y = x + 9 .
x−2
Craftear
B1 Lower branch correctly located and orientated.
Penalise at most 1 mark for poor forms at infinity
8
B1 Upper branch correctly located and orientated.
B1 Sketches axes and asymptotes, labelled or to scale
M1 A1 By inspection or long division. A0 if error in division
B1
One asymptote is x = 2 .
y = x+9+
B1 States y -intercept
( 0, −3)
Guidance
B1 States points of intersection with x -axis.
Marks
( −6,0 ) , ( −1,0 )
Answer
3
B1 B1 Each branch correct
Penalise at most one mark for poor forms at infinitty
1) g ven and asym
mptotes drawn
B1 Intersection (0,1
141
Q15
Craftear
A1
Therefore there are no stationary points on C .
6(iv)
M1
( x + 1)( 2 x + a ) − ( x 2 + ax − 1)
= 0 ⇒ x2 + 2 x + a + 1 = 0
2
( x + 1)
Discriminant = 4 − 4 ( a + 1) < 0
6(iii)
3
B1 B1 Each branch correect
2
pt and asymptotess drawn.
B1 Correct y-intercep
1
B1
3
A1
a2 + 4 > 0
Thus the oblique asymptote is y = x + a −1
2
x + a −1
or x +1 x + ax − 1
M1 By inspection or long division.
x 2 + ax − 1 = ( x + 1)( x + a − 1) − a
Guidance
B1
Marks
Vertical asymptote is x = −1 .
Answer
6(ii)
6(i)
Question
142
Q16
9(iii)
9(ii)
9(i)
Question
4
x + x +1
2
2
A1
)
1
1
⇒ 4 ( 2 x + 1) = 0 ⇒ x = − , y = −
2
3
(
M1 Differentiates and equates to 0.
)
y ′ = 0 ⇒ x 2 + x + 1 (10 x + 5 ) − 5 x 2 + 5 x + 1 ( 2 x + 1) = 0
(
A1 Explaining strict upper inequality (AG)
1
⇒ − - y < 5, because y = 5 is an asymptote (www)
3
4
M1 Factorising
2
⇒ ( y − 5)( 3 y + 1) - 0
B1 Forms quadratic equation in x .
2
A1
M1 Uses discriminant
Craftear
Guidance
M1 Alt method: Finding limit
Marks
For real x , ( y − 5) − 4 ( y − 5) ( y − 1) . 0 (condone >)
⇒ ( y − 5) x2 + ( y − 5) x + ( y − 1) = 0
yx2 + yx + y = 5x2 + 5x + 1
As x → ±∞ , y → 5 ∴ y = 5 CAO
y =5−
Answer
143
Q17
9(iii)
9(ii)
9(i)
Question
9(iv)
Total:
Craftear
2
B1
y = 0 ⇒ x 2 − 3x + 6 ; ∆ = 9 − 24 ⇒ No intersection with x-axis
Total:
B1
(0 , 6)
3
A1
M1
Turning points are ( −1,5) and ( 3,−3)
=0
3
A1
−2
Total:
M1A1
B1
Marks
⇒ x = −1,3
y ' = −1+ 4 (1 − x )
y =2− x
x =1
Answer
2
Guidance
orrect asymptote and completely corrrect
B1 Co
graaph.
B1FT Possitive y-intercept at (0,1), FT dep on
min
nimum point from (iii).
144
Q18
9(ii)
9(i)
Question
9(iv)
= 0 is
1
< y<3
3
Craftear
4
A1 AG
A1
⇒ 9 y 2 − 30 y + 9 < 0 ⇒ 3 y 2 − 10 y + 3 < 0
⇒ ( 3 y −1)( y − 3) < 0 ⇒
M1
No points on C if ( y + 3) − 4 y ( 9 − 2 y ) < 0
2
M1
2
B1
B1
Marks
yx 2 − ( y + 3) x + 9 − 2 y = 0
( x + 1)( x − 2 ) = 0 ⇒ x = −1 and ⇒ x = 2 are vertical asymptotes.
Degree of numerator < degree of denominator ⇒
horizontal asymptote.
Answer
Total:
3
Guidance
B1B1 Each brancch.
B1
1 FT Asymptotees correct.
145
9(iv)
9(iii)
Craftear
RH branch ; Other two branchhes
Axes, asympptotes and points on axes
(0, 4.5) (3,0).
3
B1
1B1
B1
3
B1
1
⇒ Turning points are (1,3) and 5, .
3
B1
B1
)
⇒ … ⇒ ( x − 1) ( x − 5) = 0
(
dy
= 0 ⇒ 3 x 2 − x − 2 − ( 3 x − 9 )( 2 x −1) = 0
dx
146
147
Q19 7
B1
Vertical asymptote is x = 2.
4
y = x+2+
⇒ Oblique asymptote is y = x + 2 .
x−2
M1A1
[3]
x2
⇒ x 2 − yx + 2 y = 0
x−2
Quadratic has no real roots (i.e. no points on C) if ∆ < 0 ⇒ y 2 − 8 y < 0
y=
B1
M1
⇒ y ( y − 8) < 0 ⇒ 0 < y < 8 . (AG) Correct inequality
M1A1
[4]
B1
B1B1
[3]
Axes and asymptotes.
Each branch, showing (0,0) and (4,8).
(Deduct at most 1 mark for poor forms at infinity and/or missing coordinates.)
Q20
(
)
x2 − 9 − 2x ( x + 2)
dy
=
2
dx
x2 − 9
=
(
− x2 − 4 x − 9
( x − 9)
2
2
B1
)
2
=
− ( x + 2) − 5
( x − 9)
2
2
⇒
dy
<0
dx
M1A1
[3]
Asymptotes: x = ±3 ; y = 0
B1B1
[2]
Sketch: Axes and asymptotes ; Outside branches
2
Middle branch, showing 0, − and ( −2,0 ) .
9
(Deduct at most 1 mark for poor forms at infinity.)
B1B1
B1
[3]
1 (3 x − 1) 2 − x 2 − 1+ 9 x 2 − 6x +1 8 x 2 − 6 x 4 x 2 − 3 x
+
=
=
= 2
2 2( x 2 + 1)
x +1
2( x 2 + 1)
2( x 2 + 1)
M1A1
Craftear
5
Q21
10
−
9 ( x + 3) 2 9 x 2 + 9 − x 2 − 6 x − 9 8 x 2 − 6 x 4 x 2 −3x
−
=
=
= 2
2 2( x 2 + 1)
2( x 2 + 1)
2(
1)
x +1
A1
(3)
(3 x − 1) 2
1
>0⇒ y>−
2
2
2( x + 1)
B1
(x + 3) 2
9
1
9
>0⇒ y< ⇒ − < y<
(Not hence gets B1 only.)
2
2
2
2
2( x + 1)
Turning points occur at equality, so −3,
B1
(2)
9
1
1
and , −
2
2
3
B1B1
(2)
Asymptote is y = 4 .
B1
(1)
3
4
4
3
Intersection with axes at (0, 0) and , 0 , Intersection with asymptote at − , 4
B1B1
Correct graph.
B1
(3)
Total
11
Q22
(
)
y′ = 0 ⇒ (x + 1)(4 x + k) − 2x 2 + kx ×1 = 0
2
2
2
M1
⇒ 4 x + (4 + k ) x + k − 2 x − kx = 0 ⇒ 2 x + 4 x + k = 0
A1
B 2 − 4 AC < 0 ⇒ no stationary points ⇒ 16 − 8k < 0
⇒ k > 2 for no stationary points.
M1A1
A1
[5]
When k = 4:
Vertical asymptote: x = –1
Oblique asymptote: y = 2 x + 2 −
2
⇒ y = 2x + 2
x +1
Axes and asymptotes
Each branch.
B1
M1A1
B1
B1B1
[6]
Total
11
Q23
12O (i) d = –2
c
⇒c=4
d
4 + 4a + 2b
y = ax + (b + 2a) +
x−2
Oblique asymptote y = x + 1 ⇒ a = 1, b + 2a = 1 ⇒ b = –1
(ii) x = 0 ⇒ y = –2 =
(iii)
x2 − x + 4
⇒ x 2 − ( y +1) x + 2 y + 4 = 0
x−2
For real x B2 – 4AC > 0 ⇒ (y + 1)2 – 4(2y + 4) > 0
⇒ y2 – y – 15 > 0
y=
B1
[1]
M1A1
M1A1
A1A1
[6]
M1
M1
A1
Craftear
8
148
⇒ (y – 3)2 > 24 (or for solving y2 – 6y – 15 = 0)
⇒ y – 3 < – 2 6 or y – 3 > – 2 6 (or for thumbnail sketch)
M1A1
M1
⇒ y < 3 – 2 6 or y > 3 + 2 6 (AG)
(Note: if done by differentiation, max and min must be demonstrated for full marks.)
A1
[7]
EITH
A
B
2( x 2 − 1) + A( x +1) + B( x −1) 2 x 2 + ( A + B ) x + A − B − 2 2 x 2 − x + 5
2+
+
=
=
=
x − 1 x +1
x2 −1
x2 −1
x2 −1
⇒ A = 3 , B = –4
M1
Q24
11
1
y′ = 0 ⇒ (x2 – 1)(4x – 1) – (2x2 – x + 5).2x = 0 ⇒ x2 – 14x + 1 = 0
B2 – 4AC = (–14)2 – 4 × 1 × 1 = 192 > 0 ⇒ two distinct turning points.
Asymptotes are x = 1 , x = –1 : y = 2
y = 2 ⇒ 2x2 – x + 5 = 2x2 – 2 ⇒ x = 7 ⇒ (7, 2) (Accept if labelled on graph.)
Sketch: Middle branch crossing y-axis at (0, –5) and left branch.
Right branch.
Working to show no intersections with x-axis.
A1A1
[3]
M1A1
M1A1
[4]
B1B1
M1A1
B1
B1
B1
[7]
Q25
4
149
Vertical asymptote is x = 1
−1
y = 2 x + 3 + 2( x − 1) ⇒ y = 2 x + 3 is the oblique asymptote.
B1
M1A1
(3)
2 x 2 + (1 − y )x + ( y −1) = 0 has real roots
⇔ (1 − y ) − 8( y − 1) > 0
2
⇔ y 2 − 10 y + 9 > 0
⇔ ( y −1)( y − 9 ) > 0
Hence ( y − 1)( y − 9 ) < 0 ⇒ no real roots.
i.e. 1 < y < 9 ⇒ no points on C. (AG) Thumbnail sketch, or similar, required.
M1A1
M1
A1
(4)
[7]
Q26
States asymptotes.
Vertical: x = 1 and Horizontal: y = 2
B1B1
Obtains quadratic form in x.
yx 2 − 2 yx + y = 2x 2 − 3x − 2
M1A1
2
⇒ ( y − 2) x 2 − (2 y − 3) x + ( y + 2) = 0
Uses B 2 − 4 AC ≥ 0 for real
roots.
Finds condition for y ′ = 0 .
For real x (2 y − 3) 2 − 4( y − 2)( y + 2) ≥ 0
25
.
⇒ 12 y ≤ 25 ⇒ y ≤
12
M1
y′ = 0 ⇒
M1
A1
4
( x 2 − 2 x + 1)(4 x − 3) − (2 x 2 − 3x − 2)(2 x − 2) = 0
Solves
⇒ x 2 − 8 x + 7 = 0 ⇒ ( x − 7)( x − 1) = 0
⇒ x = 7 , (since x = 1 is vertical asymptote).
A1
Obtains stationary point.
25
Stationary point is 7 ,
12
A1
Sketch showing:
Axes and asymptotes
(-0.5,0) , (2,0) , (0,–2) and (4,2)
Left hand branch.
Right hand branch.
B1
B1
B1
B1
Craftear
10
4
[13]
Q27
10 (i)
(ii)
(iii)
Vertical asymptote.
Oblique asymptote.
Asymptotes : x = –1
−1
y = px + 4 − p + ( p − 3)( x +1) ⇒ y = px + 4 − p
B1
M1A1
3
Obtains value of p.
Sketches curve.
p = 4 ⇒ x-axis is a tangent
Correct location of turning points and asymptotes.
Each branch.
B1
B1
B1B1
4
Proves required result.
Sketches graph.
p = 1 ⇒ y = x + 3 − 2( x + 1) ⇒ y′ = 1 + 2( x + 1)
−1
(
Intersections on x-axis at − 2 ± 3 , 0
Each branch.
−2
)
([ 1)
M1A1
B1
B1B1
5
[12]
150
Q28
7
States vertical
asymptote
Vertical asymptote is x = −2
B1
Expresses y in a suitable
form
y = 2x + 1 − 3(x + 2)
M1
States oblique
asymptote
Oblique asymptote is y = 2x + 1
A1
Differentiates wrt x
y ′ = 2 + 3(x + 2)
M1A1
and explains result
(x + 2)−2 > 0 ⇒ y ′ > 2 (AG)
A1
Sketches graph.
Axes and asymptotes
B1
(Deduct 1 mark for poor Each branch, showing intersection with axes.
forms at infinity)
B1B1
−1
−2
(3)
(3)
(3)
[9]
Q29
Possible approach for
first two parts
together.
2x2 + 2x + 3
(x + 1) 2
=
+
1
x2 + 2
x2 + 2
(x +1) 2
States 2
≥ 0 ⇒ y ≥1
x +2
From this it is clear that (–1, 1) is a turning point.
Writes y =
Writes y =
States
2 x 2 + 2 x + 3 5 (x − 2) 2
= −
2 2( x 2 + 2)
x2 + 2
(B1)
(M1A1)
(B1)
5
(x − 2) 2
≥0⇒ y ≤
2
2
2( x + 2)
From this it is clear that (2, 2
(i) can come after
finding turning points:
Continuous function
(implied by graph)
⇒ (2,2.5) Max and
(–1,1) Min
5
(AG)
⇒1≤ y ≤
2
N.B. Award B1 if Max
and Min assumed
without proof. i.e. 1/4 .
(B1)
Craftear
9(i)
and
(ii)
1
) is the other turning point.
2
(B1)
(A1)
(7)
(M1)
(M1A1)
(A1)
(4)
151
OR
Forms quadratic
equation in x.
yx 2 + 2 y = 2x 2 + 2x + 3
⇒ ( y − 2) x 2 − 2 x + (2 y − 3) = 0
M1
A1
Uses discriminant to
obtain condition for
real roots.
For real x 4 – 4(y – 2)(2y – 3) ≥ 0
M1
5
(AG)
2
A1
Differentiates and
equates to zero.
Solves equation.
⇒ (2 y − 5)( y −1) ≤ 0
⇒1≤ y ≤
y′ = 0
⇒ (x 2 + 2)(4x + 2) − 2x(2x 2 + 2x + 3) = 0
Expresses y in an
appropriate form. (May
alternatively divide
numerator and
denominator by x2.
Finds y-intercept and
intersection with
y = 2.
Completes graph.
M1
⇒ (x –2)(x + 1) = 0 ⇒ x = –1 or x = 2
(Or substitutes y =1 and
States coordinates of
turning points.
4
5
in equation of C.)
2
1
Turning points are (–1, 1) and 2, 2
2
y = 2+
2x − 1
x2 + 2
A1A1
3
M1
As x → ±∞ y → 2 ∴ y = 2
A1
1
1
Shows 0, 1 and , 2
2
2
B1
Completely correct graph.
B1
2
2
Craftear
9
[11]
Q30
7
Finds asymptotes to C.
Differentiates and
equates to 0.
Sketch of graph.
Deduct 1 mark
for poor forms at
infinity.
Deduct 1 mark if
intersections
with axes not shown.
Vertical asymptote x = 2.
2
y = λx + 1 +
⇒ y = λx + 1 is oblique asymptote.
x−2
y ′ = λ − 2( x − 2) −2 = 0 for turning points.
M1A1
2
> 0 ⇒ no turning points if λ < 0 .
(x − 2) 2
A1
λ=
B1
3
M1A1
Or y ′ = 0 ⇒ λx 2 − 4λx + 4λ − 2 = 0
Uses discriminant to show 8λ < 0 ⇒ no T.P.s.
(M1A1)
(A1)
Axes and asymptotes.
LH branch. RH branch. (Indicating intersections with
axes at (0,0) and (3,0).)
B1
B1B1
3
3
[9]
152
9
States vertical
asymptote.
Vertical asymptote is x = 2 .
⇒ oblique asymptote is y = x − 1 .
Divides and
states oblique
asymptote.
Rearranges as
quadratic in x.
xy − 2 y = x 2 − 3x + 3 ⇒ x 2 − ( y + 3) x + (3 + 2y) = 0
Uses discriminant
to obtain
condition stated.
For real x, B 2 − 4AC ≥ 0
∴( y + 3) 2 − 4(3 + 2 y) ≥ 0
⇒ ... ⇒ ( y − 3)( y + 1)) ≥ 0
⇒ y ≤ −1 or y ≥ 3
∴ no points for −1 < y < 3 (AG)
B1
M1
A1
3
B1
M1
A1
A1
4
Differentiates,
puts = 0 and
obtains x-values.
y′ =1 − (x − 2) −2 = 0 ⇒ x = 1 or 3
(Or uses y = –1 and 3 to obtain x-values.)
M1
States stationary
points.
Stationary points are (1,–1) and (3,3)
A1A1
3
Sketch.
One mark for each branch correctly placed.
B1B1
2
Vertical asymptote is x = 2.
B1
y = x+2+
M1
A1
Q32
6
States vertical
asymptote.
Finds oblique asymptote.
Oblique asymptote is y = x + 2.
y′ = 1 −
Differentiates and
equates to zero.
Finds x coordinates.
4
x−2
4
= 0 ⇒ (x − 2) 2 = 4
2
(x − 2)
3
[12]
Craftear
Q31
M1
x = 0, 4.
A1
Turning points are (0,0) and (4,8)
A1
3
Axes and both asymptotes correct.
Upper branch correct.
Lower branch correct.
B1
B1
B1
3
States coordinates of
turning points.
Deduct at most 1 mark
for poor forms at
infinity.
[9]
153
Craftear
SUMMATION OF SERIES
154
Q1
2
n
/
`1 - r - r 2j in terms of n,
(a) Use standard results from the List of formulae (MF19) to find
simplifying your answer.
[3]
r =1
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155
(b) Show that
r+1
r
1 - r - r2
=
2
2
2
2
`r + 2r + 2j`r + 1j `r + 1j + 1 r + 1
and hence use the method of differences to find
n
/ `r + 12r-+r2-j`rr + 1j .
r =1
2
2
2
[5]
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3
(c) Deduce the value of
/
1 - r - r2
.
2
2
`
j
`
j
+
+
+
r
2
r
2
r
1
r =1
[1]
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156
Q2
1
(a) Show that
tan (r + 1) - tan r =
sin 1
.
cos (r + 1) cos r
[2]
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Let ur =
1
.
cos (r + 1) cos r
(b) Use the method of differences to find
n
/u .
r =1
r
[3]
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157
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(c) Explain why the infinite series u1 + u2 + u3 + f does not converge.
[1]
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158
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2
Q3
(a) Use standard results from the list of formulae (MF19) to find
n
/ r (r + 1) (r + 2) in terms of n,
r =1
fully factorising your answer.
[3]
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159
(b) Express
1
in partial fractions and hence use the method of differences to find
r (r + 1) (r + 2)
n
/ r (r + 11) (r + 2) .
[5]
r =1
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/ r (r + 11) (r + 2) .
3
(c) Deduce the value of
[1]
r =1
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160
Q4
n
3
Let Sn =
/ ln r(r(r++12) ) .
r =1
2
(a) Using the method of differences, or otherwise, show that S n = ln
n+2
.
2 (n + 1)
[4]
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161
/ ln r(r(r++12) ) .
3
Let S =
r =1
2
(b) Find the least value of n such that Sn - S 1 0.01.
[3]
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162
(c) Use standard results from the list of formulae (MF19) to show that
n
/`(a + r) + (b + r) + (c + r) j = n + n (n + 1)`an + bn + cj,
3
3
3
r =1
where a, b and c are constants to be determined.
1
4
2
[6]
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163
Q5
2
4
(a) Use standard results from the List of Formulae (MF19) to show that
n
/ (7r + 1) (7r + 8) = an 3 + bn 2 + cn ,
r =1
where a, b and c are constants to be determined.
[3]
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164
n
(b) Use the method of differences to find
/ (7r + 1)1(7r + 8) in terms of n.
[4]
r=1
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/ (7r + 1)1(7r + 8) .
3
(c) Deduce the value of
[1]
r=1
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165
Q6
3
(a) By simplifying `x n - x 2n + 1j`x n + x 2n + 1j , show that
1
= - x n - x 2n + 1 .
2n
x - x +1
n
[1]
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Let un = x n+1 + x 2n+2 + 1 +
1
.
x - x 2n + 1
n
N
(b) Use the method of differences to find / un in terms of N and x.
n =1
[3]
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(c) Deduce the set of values of x for which the infinite series
u1 + u2 + u3 + f
is convergent and give the sum to infinity when this exists.
[3]
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166
Q7
4
8
(a) By first expressing
1
in partial fractions, show that
r -1
2
n
/ r 1- 1 = 34 - 2nan(n++b1) ,
r =2
2
where a and b are integers to be found.
[5]
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167
3
(b) Deduce the value of
/ r 1- 1 .
r =2
[1]
2
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2n
/ r n- 1 .
r = n +1
2
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(c) Find the limit, as n " 3, of
168
Q8
3
Let Sn = 2 2 + 6 2 + 10 2 + f + `4n - 2j .
2
(a) Use standard results from the List of Formulae (MF19) to show that S n = 43 n (4n 2 - 1) .
[4]
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169
n
(b) Express
in partial fractions and find
Sn
N
/ Sn in terms of N.
n =1
n
[4]
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3
(c) Deduce the value of
/ Sn .
n =1
[1]
n
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170
Q9
Let SN = Ð 5r + 1 5r + 6 and TN =
N
5
r=1
Ð 5r + 1 5r + 6 .
N
1
r=1
(i) Use standard results from the List of Formulae (MF10) to show that
SN = 13 N 25N 2 + 90N + 83.
[3]
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(ii) Use the method of differences to express TN in terms of N .
[4]
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171
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(iii) Find lim N −3 SN TN .
[2]
N →∞
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172
Q10
2
Let un =
4 sin n − 12 sin 12
cos 2n − 1 + cos 1
.
(i) Using the formulae for cos P ± cos Q given in the List of Formulae MF10, show that
un =
1
1
−
.
cos n cos n − 1
[2]
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(ii) Use the method of differences to find Ð un .
N
[2]
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(iii) Explain why the infinite series u1 + u2 + u3 + à does not converge.
[1]
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Craftear
n=1
173
Q11
4
Ð 3r + 1 3r − 2 = 3 − 3 3N + 1 .
N
(i) Use the method of differences to show that
1
1
1
[4]
r=1
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174
(ii) Find the limit, as N → ∞, of
Ð 3r + 1 3r − 2 .
N2
N
[4]
r=N +1
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175
Q12
Let Sn = Ð −1r−1 r2 .
n
5
r=1
(i) Use the standard result for Ð r 2 given in the List of Formulae (MF10) to show that
n
r=1
S2n = −n 2n + 1.
[4]
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176
S2n
S
and find lim 2n2+1 .
2
n→∞ n
n→∞ n
(ii) State the value of lim
[4]
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177
Q13
2
(i) Verify that
n e − 1 + e
1
1
.
= n−
n+1
ne
n n + 1e
n + 1en+1
[1]
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Ð n n + 1e
n=1
n e − 1 + e
n+1
.
(ii) Express SN in terms of N and e.
[2]
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Let SN =
N
178
Let S = lim SN .
N →∞
(iii) Find the least value of N such that N + 1 S − SN < 10−3 .
[3]
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179
Q14
11
20
Answer only one of the following two alternatives.
EITHER
(i) By considering 2r + 12 − 2r − 12 , use the method of differences to prove that
Ð r = 12 n n + 1.
n
r=1
[3]
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180
(ii) By considering 2r + 14 − 2r − 14 , use the method of differences and the result given in part (i)
to prove that
Ð r3 = 14 n2 n + 12 .
n
r=1
[5]
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181
The sums S and T are defined as follows:
S = 13 + 23 + 33 + 43 + à + 2N 3 + 2N + 13 ,
T = 13 + 33 + 53 + 73 + à + 2N − 13 + 2N + 13 .
(iii) Use the result given in part (ii) to show that S = 2N + 12 N + 12 .
[1]
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(iv) Hence, or otherwise, find an expression in terms of N for T , factorising your answer as far as
possible.
[2]
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(v) Deduce the value of
S
as N → ∞.
T
[2]
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182
Q15
7
Let
SN = Ð 3r + 1 3r + 4
N
r=1
and
TN =
Ð 3r + 1 3r + 4 .
N
1
r=1
(i) Use standard results from the List of Formulae (MF10) to show that
SN = N 3N 2 + 12N + 13.
[3]
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(ii) Use the method of differences to show that
TN =
1
1
−
.
12 3 3N + 4
[3]
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183
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(iii) Deduce that
SN
is an integer.
TN
[2]
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SN
.
N →∞ N 3 T
N
(iv) Find lim
[2]
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© UCLES 2018
9231/12/O/N/18
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184
Q16
It is given that Ð ur = n2 2n + 3, where n is a positive integer.
n
1
r=1
Ð ur .
2n
(i) Find
[2]
r=n+1
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(ii) Find ur .
[3]
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185
Q17
2
(i) Verify that
2r + 1
1
=
r r + 1 r + 2 2
D
E
2r + 1 2r + 3
2r − 1 2r + 1
.
−
r + 1 r + 2
r r + 1
[2]
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Ð
n
(ii) Hence show that
r=1
2r + 1
1
=
r r + 1 r + 2 2
D
E
2n + 1 2n + 3 3
.
−
n + 1 n + 2
2
[2]
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∞
(iii) Deduce the value of
Ð r r + 1 r + 2 .
2r + 1
[2]
r=1
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186
Q18
Find Ð 4r − 3 4r + 1, giving your answer in its simplest form.
n
1
r=1
[4]
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187
Q19
2
4
Express
in partial fractions and hence find
r r + 1 r + 2
∞
Deduce the value of
Ð r r + 1 r + 2 .
4
Ð r r + 1 r + 2 .
n
4
[5]
r =1
[1]
Craftear
r =1
© UCLES 2016
9231/11/M/J/16
[Turn over
188
Q20
1
Verify that
1
1
1
1
−
=
.
3r + 1 3r + 4 3 3r + 1 3r + 4
Ð
N
Let SN denote
r=1
1
and let S denote
3r + 1 3r + 4
∞
Ð 3r + 1 3r + 4 . Find the least value of N
r=1
1
[5]
Craftear
such that S − SN < 10 1000 .
[1]
© UCLES 2016
9231/13/M/J/16
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189
Q21
1
Use the method of differences to find
∞
Deduce the value of
1
.
2r2 − 1
Ð 2r − 1 .
[1]
2
Craftear
r =1
1
[4]
© UCLES 2016
9231/11/O/N/16
[Turn over
Q22
1
190
Use the List of Formulae (MF10) to show that Ð 3r2 − 5r + 1 and Ð r3 − 1 have the same
9
r=1
r=0
[4]
Craftear
numerical value.
13
© UCLES 2015
9231/11/M/J/15
[Turn over
Q23
4
191
The sequence a1 , a2 , a3 , … is such that, for all positive integers n,
an = n+5
n+6
− 2
.
n − n + 1
n + n + 1
2
The sum Ð an is denoted by SN . Find
N
n=1
[3]
(ii) the least value of N for which SN > 4.9.
[4]
Craftear
(i) the value of S30 correct to 3 decimal places,
© UCLES 2015
9231/11/O/N/15
[Turn over
Q24
2
192
Expand and simplify r + 14 − r4 .
[1]
n
n
r=1
r=1
Use the method of differences together with the standard results for r and r2 to show that
n
r3 = 14 n2 n + 12 .
4
Craftear
r=1
© UCLES 2014
9231/11/M/J/14
[Turn over
Show that a, b, c is a basis for >3 .
193
[3]
Express d in terms of a, b and c.
Q25
[2]
2
Show that the difference between the squares of consecutive integers is an odd integer.
[1]
Find the sum to n terms of the series
5
7
2r + 1
3
+ 2
+ 2
+à+ 2
+à
2
2
2
1 ×2
2 ×3
3 ×4
r r + 12
2
[5]
Craftear
and deduce the sum to infinity of the series.
© UCLES 2014
9231/13/M/J/14
[Turn over
Q26
1
194
Given that
uk = 1
1
,
−
2k − 1
2k + 1
n
express uk in terms of n.
[4]
k=13
∞
Deduce the value of uk .
[1]
Craftear
k=13
© UCLES 2014
9231/11/O/N/14
[Turn over
Q27
5
195
Ð 2r + 1 2r + 3 = 6 − 2 2N + 3 .
N
Use the method of differences to show that
1
1
1
[5]
r=1
Ð 2r + 1 2r + 3 < 8N .
2N
Deduce that
1
1
[4]
Craftear
r=N +1
© UCLES 2013
9231/11/M/J/13
[Turn over
Q28
Let f r = r! r − 1. Simplify f r + 1 − f r and hence find Ð r! r2 + 1.
2n
r=n+1
[5]
Craftear
1
196
© UCLES 2013
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Q29
3
197
It is given that
Sn = Ð ur = 2n2 + n.
n
r=1
Write down the values of S1 , S2 , S3 , S4 . Express ur in terms of r, justifying your answer.
[4]
Find
Ð ur .
3
9231/11/O/N/13
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2n
Craftear
r=n+1
© UCLES 2013
Q30
1
Express
198
1
in partial fractions.
r r + 1 r − 1
Find
[1]
Ð r r + 1 r − 1 .
4
Ð r r + 1 r − 1 .
1
n
1
r=2
State the value of
∞
1
Craftear
r=2
© UCLES 2013
9231/13/O/N/13
[Turn over
Q31
3
1
Given that f (r) =
, show that
(r + 1)(r + 2)
199
f (r − 1) − f (r) =
n
Hence find
2
.
r(r + 1)(r + 2)
1
∑ r(r + 1)(r + 2) .
[2]
[3]
r=1
∞
Deduce the value of
1
∑ r(r + 1)(r + 2) .
[1]
Craftear
r=1
© UCLES 2012
9231/11/M/J/12
[Turn over
200
Q32
4
Let f (r) = r(r + 1)(r + 2). Show that
f (r) − f (r − 1) = 3r(r + 1).
[1]
n
Hence show that ∑ r(r + 1) = 13 n(n + 1)(n + 2).
r=1
n
n
r =1
r=1
Using the standard result for ∑ r, deduce that ∑ r2 = 16 n(n + 1)(2n + 1).
[2]
[2]
Find the sum of the series
12 + 2 × 22 + 32 + 2 × 42 + 52 + 2 × 62 + . . . + 2(n − 1)2 + n2 ,
where n is odd.
Craftear
[3]
© UCLES 2012
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Q33
201
2n
Show that ∑ r2 = 16 n(2n + 1)(7n + 1).
[4]
r=n+1
Craftear
1
© UCLES 2012
9231/13/O/N/12
[Turn over
Q34
1
202
Find the sum of the first n terms of the series
1
1
1
+
+
+ ...
1×3 2×4 3×5
[5]
Craftear
and deduce the sum to infinity.
© UCLES 2012
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SUMMATION OF SERIES
MS
Craftear
203
2(c)
2(b)
2(a)
Question
Q1
(
(
) (
)(
)
)
(
)(
)
2
1− r − r2
2
=
n
r +1
2
r
r =1
n +1
n
2 1 3 2 4 3
− + − + − + +
− 2
2
5 2 10 5 17 10
(n + 1) +1 n +1
− 12
1
n +1
=− +
2 ( n + 1)2 +1
=
r =1
2
( r + 2r + 2)( r + 1) (r + 1) +1 − r +1
n
Craftear
(r + 1) r 2 + 1 − r r 2 + 2r + 2
r +1
r
1− r − r2
=
−
=
(r + 1) 2 +1 r 2 +1
r 2 + 2r + 2 r 2 + 1
r 2 + 2r + 2 r 2 + 1
1
n − n 2 − 13 n 3
3
n − 12 n(n + 1) − 16 n(n + 1)(2n + 1)
Answer
Guidance
1
B1 FT FT from their answer to part (b).
5
A1 ISW
M1 A1 Shows at least three complete terms including first and
last. Cancellation may be implicit.
M1 A1 Puts over a common denominator and expands, AG.
3
A1 Simplifies by collecting terms.
M1A1 Substitutes correct formulae from MF19.
Expanding brackets correctly.
Marks
204
1(c)
1(b)
1(a)
Question
Q2
sin1
cos(r +1) cos r
=
ur =
tan(n + 1) − tan1
sin1
r =1
r
u = tan 2 − tan1 + tan 3 − tan 2 + ... + tan(n + 1) − tan n
n
Craftear
tan(n +1) oscillates as n → ∞ so u1 + u2 + u3 + does not converge.
r =1
n
[sin1]
tan r + tan1
tan r + tan1 − tan r + tan 2 r tan 1
− tan r =
1 − tan r tan1
1 − tan r tan1
2
tan1sec r
tan1
sin1
=
=
=
1 − tan r tan1 cos r(cos r − sin r tan1) cos r (cos r cos1 − sin r sin1)
sin1
=
cos r cos(r +1)
sin r cos r cos1+ cos 2 r sin1 − sin r cos r cos1 + sin 2 r sin1
cos(r +1)cos r
=
sin( r + 1) sin r sin(r + 1) cos r − cos( r + 1) sin r
sin(r + 1 − r)
−
=
=
cos(r + 1) cos r
cos(r + 1) cos r
cos(r +1) cos r
sin( r + 1) sin r sin r cos1 + cos r sin1 sin r
OR
=
−
−
cos(r + 1) cos r
cos( r + 1)
cos r
Answer
Guidance
1
B1 States ‘oscillates’ or refers to diverging values of
tan(n +1) , or states that tan(n +1) does not tend to
a limit.
3
A1 OE and ISW.
M1 A1 Shows enough terms for cancellation to be clear.
There must be three complete terms including first
and last. Cancelling may be implied.
2
A1 SC B2 for this route completely correct to AG.
M1 Applies all relevant correct addition formulae
from MF19 for the route chosen and combines
into a single fraction.
Marks
205
2(c)
2(b)
2(a)
Question
Q3
3
2
1
4
2
2
1
2
(
)
r =1
1
f (1) − f (2) + 12 f (3)
2
1
4
=
1
1
1
−
+
4 2 ( n + 1) 2 ( n + 2 )
+ 12 f (n) − f (n + 1) + 12 f (n + 2)
+ 12 f (n − 1) − f (n) + 12 f (n +1)
+ 12 f (3) − f (4) + 12 f (5)
+ 12 f (2) − f (3) + 12 f (4)
1
1
r
1
B1 FT FT from their answer to part (b).
5
A1 ISW
Where f ( x) =
M1 A1 Shows enough terms for cancelation to be clear.
n
r (r + 1)(r + 2) =
3
A1
M1 A1 Finds partial fractions.
Craftear
Guidance
M1 A1 Expands and substitutes formulae.
At least two formulae used for M1.
Marks
1
1
1
1
+
=
−
r (r + 1)(r + 2) 2r r + 1 2(r + 2)
= 14 n( n + 1) n 2 + n + 4n + 2 + 4 = 14 n( n + 1)( n + 2)( n + 3)
r =1
( r + 3r + 2r ) = n (n + 1) + n(n + 1)(2n + 1) + n(n + 1)
n
Answer
206
3(b)
3(a)
Question
Q4
Craftear
A1 CAO
Least value of n is 99
3
M1 Forms inequality
n+2
0.01
S n − S = ln
< 0.01 leading to n + 2 < e (n +1)
n +1
4
A1 AG
M1 A1 Shows enough terms to make cancellation clear.
B1 States sum to infinity. AEF
n+2
.
2(n + 1)
Guidance
B1 Separates logarithms into correct form using a
difference.
Or as logarithm of product.
Marks
S = − ln 2
[ln1] − ln 2 − ln( n + 1) + ln( n + 2) = ln
ln n − 2ln(n + 1) + ln(n + 2)
ln(n − 1) − 2ln n + ln(n +1)
ln 3 − 2 ln 4 + ln 5
ln 2 − 2ln 3 + ln 4
ln1 − 2ln 2 + ln 3
r =1
( ln r − 2ln(r + 1) + ln(r + 2) )
n
Answer
207
2(c)
2(b)
2(a)
Question
Q5
2
1
56
11
1
1
1
1
Craftear
1
B1 FT
4
A1
1
1
11
= −
7 8 7n + 8
r =1
n
M1 Writes at least three correct terms, including first and
last.
1
M1 A1 Finds partial fractions.
2
r and r .
(7r + 1)(7r + 8) = 7 8 − 15 + 15 − 22 + + 7n + 1 − 7n + 8
1
1 1
1
=
−
(7 r + 1)(7 r + 8) 7 7r + 1 7r + 8
3
A1
M1 Expands.
= 493 n3 + 56n 2 + 143
n
3
r =1
49r + 63r + 8
n
Guidance
M1 Substitutes formulae for
(7 r + 1)(7 r + 8) =
Marks
= 49 ( 16 n(n + 1)(2n + 1) ) + 63 ( 12 n(n + 1) ) + 8n
r =1
n
Answer
208
3(c)
3(b)
3(a)
Question
Q6
n
2n
n
2n
(
)
(
)
2n
n
2
4
N
2
2N
3
6
2
4
2 N +2
)
N +1
(
(
)
(
)
x N +1 → 0 as n → ∞ u1 + u2 + u3 + = 1 − x − √ x 2 + 1
−1 < x < 1
= x N +1 + √ x 2 N + 2 + 1 − x − √ x 2 + 1
n =1
Craftear
u = x + √ ( x + 1) − x − √ ( x + 1) + x + √ ( x + 1) − x − √ ( x + 1)
+ + x
+ √(x
+ 1) − x − √ ( x + 1)
N
un = x n +1 + √ x 2 n + 2 + 1 − x n − √ x 2 n + 1
2n
( x − √ ( x + 1)) ( x + √ ( x + 1)) = x − ( x + 1) = −1
Answer
Guidance
3
M1 A1 Finds sum to infinity.
B1
3
A1 (Don’t allow in terms of n .)
M1 Writes at least three complete correct terms, including the
first and last.
B1 Uses the identity given in part (a).
1
B1 Uses the difference of two squares and obtains the given
answer. AG
Marks
209
2
1
1 1 1
1
1
1
1
1
1
1
1
Answer
1
→ 12 as n → ∞
Craftear
3
3
4n + 1
2n + 1 n ( 2n +1) n ( 4n + 1)
−
= n −
− n −
=
4 2n ( n + 1) 2n ( n + 1) 4n ( 2n +1)
4 4n ( 2n + 1)
2n
n
n
n
n
=
−
2
2
2
r = n +1 r − 1
r =2 r − 1
r =2 r −1
2n
4(c)
3
2n + 1
−
4 2n ( n + 1)
3
4
=
1 1 1
1
= 1+ − −
2 2 n n +1
r =2
r − 1 = 2 1 − 3 + 2 − 4 + 3 − 5 + 4 − 6 + + n − 2 − n + n − 1 − n + 1
n
1
1
1 1
1
=
=
−
r − 1 ( r −1)( r + 1) 2 r − 1 r + 1
2
4(b)
4(a)
Question
Q7
4
A1
M1A1
M1
1
B1
5
A1
A1
M1
M1 A1
Marks
210
3(c)
3(b)
3(a)
Question
Q8
r =1
n
(4r − 2) 2 = 16
r =1
n
r 2 − 16
r =1
( 4n − 1) AG
n
n
n =1
3
1
1
1
1
1
1
1
3
8
3
1
= 1−
8 2N + 1
S = 8 1 − 3 + 3 − 5 + 5 − 7 + 2 N − 1 − 2 N + 1
N
n
3
3 1
1
=
=
−
S n 4 ( 2n − 1) (2n + 1) 8 2n − 1 2n + 1
4
n(4n 2 + 6n + 2 − 6n − 6 + 3) = 34 n
3
2
r + 4n
n
= 83 n(n + 1)(2n + 1) − 8n(n + 1) + 4n
Sn =
Craftear
Answer
1
B1
4
A1
M1
M1 A1
4
A1
M1
M1 A1
Marks
211
5(iii)
5(ii)
5(i)
Question
Q9
2
)
SN
25 1
5
T → ×
=
3 N
3 30 18
N
Craftear
2
M1 A1 Divides S by N 3 and takes limits as N → ∞
N
4
A1 At least 3 terms including last.
11
1 1
1
−
−
=
5 6 5 N + 6 30 5 ( 5 N + 6 )
M1 A1 Finds partial fractions.
3
M1 Expresses terms as differences.
(
1 1 1 1 1
1
1
TN = − + − + " +
−
5 6 11 11 16
5 N +1 5 N + 6
1 1
1
=
−
( 5r + 1)( 5r + 6 ) 5 5r + 1 5r + 6
1
)
A1 Simplifies to the given answer (AG).
(
35
35
25
1
= N
2 N 2 + 3N + 1 + N + + 6 = N 25 N 2 + 90 N + 83
2
2
6
3
r =1
M1 Expands.
Guidance
M1 Substitutes formulae for ∑ r and ∑ r 2 .
r =1
r =1
N
Marks
1
1
25 N ( N +1)( 2 N + 1) + 35 N ( N + 1) + 6 N
6
2
N
N
∑ ( 5r + 1)( 5r + 6 ) = 25∑r + 35∑r + 6N
Answer
212
2(iii)
2(ii)
2(i)
Question
Q10
∑
Craftear
cos N oscillates as N → ∞ so u1 + u 2 + u 3 + … does not converge.
1
cos N
1
1
1
1
1
1
− +
−
−
+ …+
cos1 1 cos 2 cos1
cos N cos ( N − 1)
−1 +
=
∑
1 1
4sin n − sin
N
1
1
2 2
=
−
cos ( 2n − 1) + cos1 n =1 cos n cos ( n − 1)
n =1
N
1
B1 States “oscillates” or refers to diverging values of cos N .
2
A1
M1 Applies (i), shows enough terms and cancelation.
2
A1 AG
1
1
−
cos n cos ( n −1)
Guidance
M1 Uses formulae for cos P ± cos Q.
Marks
1 1
4sin n − sin
2 2 2 ( cos ( n − 1) − cos n )
=
2cos n cos ( n −1)
cos ( 2n −1) + cos1
Answer
213
4(ii)
4(i)
Question
Q11
=
N2
N
−
→
1
as N → ∞
9
(
)
(
N
r =1
)
N
N
N3 − N2
−
=
3 ( 3N +1) 3 3 N 2 + 1 ( 3N +1) 3 N 2 + 1
)
=
(
N
N
N
N
− −
−
2
3 3 3N + 1 3 3 ( 3N + 1)
r =1
=
N
N
Craftear
∑ (3r + 1)( 3r − 2 ) ∑ ( 3r + 1)( 3r − 2) ∑ ( 3r +1)( 3r − 2 )
r = N +1
N2
1
Uses
r =1
−
r =1
N
4
B1
A1 Allow simplification to common denominator.
=
N2
∑ ∑ ∑
r = N +1
N2
M1 Applies (i)
M1
4
A1 AG
∑
N
1
1
= 1 −
3 3N + 1
Guidance
M1 At least 3 term including final term.
M1 A1 Finds partial fractions.
Marks
11 1 1 1
1
1
= − + − +…
−
3r + 1)( 3r − 2 ) 3 1 4 4 7
3N − 2 3 N +1
r =1 (
1
1 1
=
−
( 3r +1)( 3r − 2 ) 3 3r − 2 3r + 1
1
Answer
214
5(ii)
5(
5(i)
Question
Q12
2
2n 2
r =1
n 2
r =1
Thus lim n →∞
S2 n+1
=2
n2
Craftear
n
1
2
AG
2
correct sign
4
S 2n
and
n2
M1
Uses the result given in (i) or using lim n→∞
A1
n
1
A1 Alt: Find limit from previous line directly
M1
So, S2n+1 = −n ( 2n + 1) + ( 2n + 1) = ( 2n + 1) ( n + 1)
2
M1
( 2n + 1)2
S2 n +1 = S2n + ( −1)
2n
B1
4
n
1
= −n ( 2n + 1)
2
A1
n
1
∑ 4 r − 4∑ ( r ) + n − 4∑ ( r )
n
1
∑ ( 2r − 1) − ∑ ( 2r ) =A1
M1
A1 Alt method: Use
2
Guidance
M1 Uses correct difference.
Marks
S2 n
= −2
n →∞ n 2
lim
1
Factorising, S 2n = n ( 2n + 1) ( 4n + 1 − 4n − 4 ) = − n ( 2n +1)
3
1
8
( 2n ) ( 2n + 1) ( 4n + 1) − n ( n +1)( 2n +1)
6
6
Thus S 2n =
n
r =1
2n 2
r =1
so S 2 n =
∑ r − 2∑ ( 2r ) = ∑ r − 8∑ r
2
2
2
2
2n = 1 – 2 + 3 –4 …..
Answer
215
11E(i)
Question
Q14
2(iii)
2(ii)
2(i)
Question
Q13
1
e
1
1
⇒
r =1
1
r = −12 + ( 2n + 1)
r =1
∑r = 2 n ( n + 1)
n
⇒ 8
∑
n
2
( 2r + 1) − ( 2r −1) = 8r
2
⇒ e N +1 > 103
⇒ least such N is 6 .
Answer
1
< 10−3 ⇒ N +1 < 10−3
e
( N +1) ( S − S N )
S=
N
∑ ne − ( n + 1) e
Craftear
Marks
Guidance
3
A1 AG.
M1 Sums both sides and uses method of differences.
B1
6
A1
M1 Attempts to find difference between sum and sum to
infinity.
B1 Finds S
A1
SN =
1
1
−
.
e ( N +1) e N +1
Guidance
M1 Uses difference method to sum.
B1 Verifies result (AG).
Marks
1
1
1
1
=( −
+
+
−
n
n +1
2
2
3
e 2e
2e
3e
n =1
1
1
−
………….
) SOI =
N
Ne
( N + 1) e N +1
( n + 1) e − n = n ( e − 1) + e
1
1
−
=
n
n +1
n ( n + 1) en +1 n ( n 1) e
ne
( n + 1) e
Answer
216
11E(v)
11E(iv)
11E(iii)
11E(ii)
(
)
(
r3 +
2
M1 Sums both sides and uses method of differences.
A1
M1 Uses difference of squares or expands.
4
4
2
Craftear
2
A1 Accept ( N + 1)2 ( 2N ( N + 2 ) + 1) .
M1 Eliminates even terms from S .
2
2
A1
2
Converges to 2 as N → ∞ .
2
M1 Writes fraction as quadratic in N divided by quadratic
in N .
2
1
B1 Uses formula for ∑ r 3 . AG.
5
( 2 N +1) = 4 N 2 + 4 N + 1
S
=
T 2N 2 + 4N + 1 2N 2 + 4N + 1
2
( N + 1) ( 2 N 2 + 4 N + 1)
r =1
3
∑ ( 2r ) = ( 2 N +1) ( N +1) − 2 N ( N + 1)
N
1
2
2
2
2
( 2 N + 1) ( 2 N + 2 ) = ( 2 N + 1) ( N + 1)
4
3
T =S−
S=
∑
n
A1 AG
4
2
2
2
2
2
1 1
1 1 1
2
= n + − n + + − n ( n +1) = n 2 ( n + 1)
2 2
2 2 2
∑
4
1
1
r = − 1 − + n +
2
2
r =1
n
2
M1 Uses formula for ∑ r
r =1
∑
n
)
2
1 1 1
4 r = n + − − n ( n +1)
2 2 2
r =1
⇒4
= 8r 2 + 2 ( 8r ) = 16 4r 3 + r
4
( 2r + 1) − ( 2r − 1) = ( ( 2r + 1) + ( 2r − 1) )( ( 2r + 1) − ( 2r − 1) )
217
7(iii)
7(ii)
7(i)
Question
Q15
N
So
SN
is an integer because all terms are integers
TN
SN
N
⇒
= 4 ( 3 N + 4 ) 3N 2 + 12 N + 13
4 ( 3N + 4 )
TN
(
)
Craftear
SN
as a polynomial
TN
2
A1 Justifies expression being integer
M1
Writes
A1 Cancels to given answer (AG).
1 1
1 1
1
−
= −
3 4 3N + 4 12 3 ( 3N + 4 )
TN =
M1 Expresses terms as differences.
1 1 1 1 1
1
1
TN = − + − + " +
−
3 4 7 7 10
3 ( N − 1) + 1 3 N + 4
3
B1 Finds partial fractions.
1
1 1
1
=
−
( 3r + 1) ( 3r + 4 ) 3 3r + 1 3r + 4
2
3
A1 Shows simplification to the given answer
(AG).
)
= N ( 3 N + 12N + 13)
(
15
15
9
= N 2 N 2 + 3N + 1 + N + + 4
6
2
2
r =1
Guidance
M1 Substitutes formulae for ∑ r and ∑ r 2 .
r =1
r =1
M1 Expands
Marks
1
1
9 N ( N + 1)( 2 N + 1) + 15 N ( N + 1) + 4 N
6
2
N
N
∑ ( 3r + 1) ( 3r + 4) = 9∑r 2 +15∑r + 4N
Answer
218
2(i)
Question
Q17
1(ii)
1(i)
Question
Q16
7(iv)
(
2
2n
n +1 ur =
2
Answer
)
(
)
Total:
Craftear
2
A1 AG
2
Marks
3
2
3
2
1 4r + 8r + 3r − 4r + 8r − r − 2 1 ( 4r + 2 )
( 2r +1)
=
=
=
r ( r + 1) ( r + 2 )
2
2 r ( r +1) ( r + 2 ) r ( r +1) ( r + 2 )
(
3
Guidance
A1 SR: CAO B1 without wrong working
M1A1 Method mark for using Sr – Sr–1 OE
2
A1
M1
2
Answer
Total:
Total:
Guidance
M1 Method mark for using S2n – Sn
Marks
2
A1
M1 Divides expression in (iii) by N 3 and
takes limit
1 4r + 8r + 3r − 4r −1 ( r + 2 )
RHS =
2
r ( r +1) ( r + 2 )
3
ur = r 2 ( 2r + 3) − ( r − 1) ( 2r +1)
= 6r 2 − 1
)
( 2n )2 ( 4n + 3) − n 2 ( 2n + 3)
= 14n3 + 9n 2
∑
→ 4 ( 3)( 3) = 36
4 ( 3 N + 4 ) 3 N + 16 N + 9
SN
=
N3
N 3TN
219
1
Question
Q18
2(iii)
2(ii
2(ii)
r =1
∑
n
r =1
n
=… =
(
)
2
− 3n
n
16n 2 +12n − 13 (3 terms)
3
6
n ( n + 1)
Answer
−8
∑ r − 3n
r2 − 8
n ( n + 1)( 2n + 1)
r =1
n
∑
= 16
ur = 16
1
3
1
S∞ = × 4 − = 1 .
2
4
4
1 ( 2n +1)( 2n + 3) 3
=
−
2
2 ( n +1)( n + 2 )
Craftear
Guidance
4
M1A1
A1 AG
M1
4
A1 OE
M1 For using formulae correctly in their expression
M1A1 M1 for split into 3 parts
Marks
Total:
( 2n − 1)( 2n +1) ( 2n − 3)( 2n − 1) ( 2n + 1)( 2n + 3) ( 2n − 1)( 2n +1)
1 3.5 1.3 5.7 3.5
−
−
+
−
+…+
−
+
n ( n − 1)
n ( n + 1)
n ( n +1)
2 2.3 1.2 3.4 2.3
( n +1)( n + 2 )
to n terms is:
220
221
Q19
2
2
4
2
−
+
(Award B2 if written down by cover up rule.)
r r +1 r + 2
2 4 1
4
2 2
4
2
2
− +
+
2 − 2 + + 1 − + + … +
+ −
+
3
3
2
n
−
1
n
n
+
1
n
n
+
1
n
2
2
2
+
(AEF)
= 1−
n +1 n + 2
Sum to infinity = 1
M1A1
M1A1
A1
[5]
B1
[1]
Q20
1 1
1 1 ( 3r + 4 ) − ( 3r + 1)
1
AG
−
=
=
3 3r + 1 3r + 4 3 ( 3r + 1)( 3r + 4 ) ( 3r + 1)( 3r + 4 )
1
1 1 1 1 1
1 1 1
1
S N = − + − + … +
= −
−
3 4 7 7 10
3N +1 3N + 4 3 4 3N + 4
1
⇒S =
12
1
1
<
⇒ S – SN =
3 ( 3N + 4 ) 10000
⇒ 3N + 4 >
10000
7
⇒ N > 1109
. Thus least N is 1110.
3
9
B1
[1]
M1 A1
A1
M1
A1
[5]
Craftear
1
Q21
1
1 1 1
1
1
1
1
M1A1
Craftear
B1
[1]
1
∞
n
1
1
1
1
1
−
=
⇒
=
∑
2
2 2n + 1 2n +1
2
r =1 ( 2r ) − 1
r =1
2
[4]
1
M1A1
∑ ( 2r ) − 1 = 2 1 − 3 + 3 − 5 +…+ 2n − 1 − 2n +1 = 2 1 − 2n + 1 ( OE )
n
1 1
1
=
−
( 2r − 1) ( 2r + 1) 2 2r − 1 2r + 1
1
222
223
Q22
3 ×
1
13 × 14 × 27
13 × 14
− 5×
+13 = 2015
6
2
M1A1
2
9 × 10
2 −10 = 2015 (Award M1 for subtracting 9 or 10 here.)
M1A1
(4)
Total: 4
Q23
(i)
7 7
8
36
36
35
6
= 4.820
−
−
= 6 −
−
+ K +
+
3 3
7
931
931
871
1
(ii)
6−
n+6
2
> 4.9 ⇒ 0.21n 2 − 10.79n − 34.79(> 0)
n + n +1
⇒ n > 54.42 K so 55 terms required.
M1A1
A1
[3]
M1A1
dM1A1
[4]
Total
7
Q24
(r + 1)4 – r4 = 4r3 + 6r2 + 4r + 1
2
(n + 1)4 – 14 = 4Σ nr=1 r3 + 6Σ nr=1 r2 + 4Σ nr=1 r + n
B1
[1]
M1
n4 +4n3 + 6n2 + 4n = 4Σ nr=1 r3 + n(2n2 + 3n + 1) + 2n2 + 2n + n
A1A1
⇒ … ⇒ Σ nr=1 r3 = 1 n2(n + 1)2.
4
A1
[4]
(AG)
Craftear
4
Q25
(n + 1)2 – n2 = n2 + 2n + 1 – n2 = 2n + 1 ⇒ odd.
3
5
7
2n +1
2 2 − 12 32 − 2 2 4 2 − 32
(n +1) 2 − n 2
+
+
+
K
=
+
+
+
K
12.2 2 2 2.32 32.4 2
n 2 (n +1) 2
12.2 2
2 2.32
32.4 2
n 2 (n +1) 2
1
1
1
1
1
1
1
= 1− 2 + 2 − 2 + 2 − 2 +K+ 2 −
2
2
3
3
4
(n +1) 2
n
1
=1−
(n +1) 2
Sum to infinity =1.
2
Q26
1
(
1
−
25
1
27
)+ (
∑
n
1
−
27
u = 15 −
r =13 k
∑
∞
u = 15
r =13 k
1
2 n+1
1
29
)+ ... + (
1
−
2 n −1
1
2 n +1
)
B1
[1]
M1A1
M1
A1
A1
[5]
M1A1
M1A1
(4)
B1
(1)
[5]
224
Q27
5
1
1 1
1
=
−
(`2r + 1)(2r + 3) 2 2r + 1 2r + 3
Finds partial fractions.
N
M1A1
1
∑ (2r + 1)(2r + 3)
r =1
Expresses terms as differences.
Shows cancellation.
2N
Uses
∑
N +1
=
2N
N
1
1 1 1
1 1
1
−
− + ... +
23 5
2 2N + 1 2N + 3
=
1
1
(AG)
−
6 2(2 N + 3)
M1A1
A
A11
2N
∑ ∑
−
=
1
1
1
1
− −
= −
2(4 N + 3) 6 2(2 N + 3)
N +1 6
M1
∑
.
1
Applies result
and simplifies.
Deduces inequality.
=
1 1
1
−
2 2N + 3 4N + 3
A1
=
N
(2 N + 3)(4 N + 3)
M1
<
N
1
(AG)
=
2 N.4N 8 N
A1
[9]
1
Simplifies.
f(r + 1) – f(r) = r(r + 1)!−(r −1)r!
2
2
= r!(r + r − r + 1) = r!(r + 1)
Uses difference
method.
Obtains result.
n
∑ = f(2) – f(1) + f(3) – f(2) +…f(n + 1) – f(n)
1
∴
Craftear
Q28
M1
A1
M1
A1
= n(n + 1)!−0 = n(n + 1)!
2n
∑ = 2n(2n + 1)!−n(n + 1)!
A1
[5]
5
n+1
2n
(Or directly using
∑ = f (2n + 1) − f (n + 1) from the
n +1
method of differences.)
Q29
3
Writes first four
sums.
S1…S4 ~ 3, 10, 21, 36
Deduces first four
terms, conjectures
and justifies result.
u1…u4 ~ 3, 7, 11, 15 ⇒ ur = 4r ‒ 1
n
since S n = {6 + 4(n −1)} = 2n 2 + n as given.
2
B1B1
Or ur = S r − S r −1 = 2r 2 + r − 2(r −1) − (r −1)
= 4r – 1
B1B1
B1
B1
2
Obtains required
sum.
∑ 2n n+1 (4r − 1) = 4.
2n(2n +1)
n(n +1)
− 2n − 4.
− n
2
2
(
)
= 8n 2 + 2n − 2n 2 + n = 6n 2 + n
Or Sum of AP =
n
(4n + 3 + 8n − 1) = 6n 2 + n
2
B1
4
M1A1
A1
3
[7]
225
Q30
1
Finds partial fractions.
1
1
1
1
=
− +
r (r -1) (r + 1) 2(r − 1) r 2(r + 1)
Expresses each term in
fractions
1
1
1 1 1 1 1 1 1
− +
− + + − + ...
2 2 6 4 3 8 2(n − 1) n 2(n + 1)
Cancels terms and sums
=
Find sums to infinity
S∞ =
1 1
1
−
+
4 2n 2(n + 1)
(OE)
1
4
B1
(1)
M1A1
M1A1
(4)
B1
(1)
[6]
Q31
Proves initial result.
1
1
−
r (r + 1) (r +1)(r + 2)
r +2−r
2
=
=
(AG)
r (r +1)(r + 2) r (r +1)(r + 2)
f(r – 1) – f(r) =
1
n
Sets up method of
differences.
Shows cancellation to
get result.
States sum to infinity.
‘Non hence’ method
for last two parts
i.e. penalty of 1 mar
i.
1 1
1
∑ r (r + 1)(r + 2) = 2 1 × 2 − 2 × 3 ……
1
+
1 1
1
−
2 n(n + 1) (n + 1)(n + 2)
=
1 1
1
−
(OE)
4 2 (n + 1)(n + 2)
1
1
∴∑
=
4
1 r ( r + 1)( r + 2)
∞
1
1
1
1
=
−
+
r (r + 1)(r − 2) 2r (r +1) 2(r + 2)
⇒ ⋅⋅⋅ ⇒
1 1 1
1
1
1
− + ... +
−
+
2 2 4
2(n + 1) (n +1) 2(n + 2)
=
M1
A1
2
M1A1
A1
3
A1√
1
(M1)
(A1)
1 1
1
−
(OE)
4 2 (n +1)(n + 2)
(A1)
(3)
1
1
=
4
1 r ( r + 1)( r + 2)
(A1√)
(1)
∞
∴∑
Craftear
3
[6]
226
Q32
4
r(r + 1)(r + 2) − (r − 1)r(r + 1) = r(r + 1)(r + 2 − r + 1)
= 3r(r + 1) (AG)
Verifies given result.
n
∑
Uses method of
differences to
sum first series.
1
r =1
1
{[f (n) − f (n −1)] + [f (n −1) − f (n − 2)] + ... + [f (1) − f (0]}
3
1
= n(n + 1)(n + 2) (AG) (Award B1 if ‘not
3
hence’.)
n
n
Subtracts
B1
r ( r + 1) =
∑
∑r
r =1
r =1
to obtain sum of second
series.
n
∑
r ( r + 1) −
∑r =
(12 + 2 2 + ... + n 2 ) + (2 2 + 4 2 + ... + (n −1) 2 ) =
Splits series into two
series.
Applies sum of squares
formula to obtain result.
A1
2
n
n(n + 1)(n + 2) n(n + 1)
−
3
2
r =1
r =1
1
1
= n(n + 1)(2n + 4 − 3) = n(n + 1)(2n + 1) (AG)
6
6
r2 =
M1
n − 1 n + 1
4
n
1
n(n + 1)(2n + 1)
2 2
=…= n 2 (n + 1)
+
2
6
6
M1
A1
2
M1
M1A1
3
[8]
4
[4]
5
[5]
1
Use of:
Use of:
Obtains result.
2n
2n
n
n+1
n
1
1
∑ = ∑ −∑
Craftear
Q33
M1
n( n + 1)(2n + 1)
6
1
2n(2n + 1)(4n + 1) n( n +1)(2n + 1)
−
6
6
1
1
= n(2n + 1)(8n + 2 − n − 1) = n(2n + 1)(7 n + 1) (AG)
6
6
∑r =
M1
2
A1
A1
Q34
1
Finds partial fractions.
1
1 1
1
= −
r (r + 2) 2 r r + 2
n
1
=
∑
r =1 r ( r + 2)
Use method of
differences.
1 1
1 1
1
1 1 1
+
−
+ ... + − + 1 −
−
2 n n + 2 n − 1 n + 1
2 4 3
Obtains results.
=
3
1 3
1
1
−
−
(acf) ⇒ S ∞ =
4
2 2 n + 1 n + 2
M1A1
M1
A1A1√
227
Craftear
MATRICES 1
228
Q1
4
8
The matrix M represents the sequence of two transformations in the x-y plane given by a rotation of 60°
anticlockwise about the origin followed by a one-way stretch in the x-direction, scale factor d (d ! 0) .
(a) Find M in terms of d.
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(b) The unit square in the x-y plane is transformed by M onto a parallelogram of area 12 d 2 units2.
Show that d = 2 .
[2]
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J1 1N
The matrix N is such that MN = KK1 1OO .
L2 2P
(c) Find N.
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(d) Find the equations of the invariant lines, through the origin, of the transformation in the x-y plane
represented by MN.
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Q2
4
The matrices A, B and C are given by
2 k k
A = f5 - 1 3p,
1 0 1
1 0
B = f0 1p and
1 0
0 1 1
o,
C =e
-1 2 0
where k is a real constant.
(a) Find CAB.
[3]
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(b) Given that A is singular, find the value of k.
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(c) Using the value of k from part (b), find the equations of the invariant lines, through the origin, of
the transformation in the x-y plane represented by CAB.
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Q3
4
cos i - sin i 3 0
oe
o.
The matrix M is given by M = e
sin i cos i 0 1
(a) The matrix M represents a sequence of two geometrical transformations.
State the type of each transformation, and make clear the order in which they are applied.
[2]
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(b) Find the values of i, for 0 G i G r , for which the transformation represented by M has exactly
one invariant line through the origin, giving your answers in terms of r .
[9]
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Q4
1
(a) Give full details of the geometrical transformation in the x-y plane represented by the matrix
6 0
e
o.
[1]
0 6
............................................................................................................................................................
............................................................................................................................................................
3 4
o.
Let A = e
2 2
(b) The triangle DEF in the x-y plane is transformed by A onto triangle PQR.
Given that the area of triangle DEF is 13 cm2, find the area of triangle PQR.
[2]
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6 0
o.
(c) Find the matrix B such that AB = e
0 6
[2]
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(d) Show that the origin is the only invariant point of the transformation in the x-y plane represented
by A.
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Q5
1
2
1 b a 0
oe
o, where a and b are positive constants.
The matrix M is given by M = e
0 1 0 1
(a) The matrix M represents a sequence of two geometrical transformations.
State the type of each transformation, and make clear the order in which they are applied.
[2]
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The unit square in the x-y plane is transformed by M onto parallelogram OPQR.
(b) Find, in terms of a and b, the matrix which transforms parallelogram OPQR onto the unit square.
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It is given that the area of OPQR is 2 cm 2 and that the line x + 3y = 0 is invariant under the
transformation represented by M.
(c) Find the values of a and b.
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Q6
4
The matrices A and B are given by
1
1
-2 3
0 1
2
o and B = f 1
A =e
p.
1
1 0
3
2
2
(a) Give full details of the geometrical transformation in the x-y plane represented by A.
[1]
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(b) Give full details of the geometrical transformation in the x-y plane represented by B.
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The triangle DEF in the x-y plane is transformed by AB onto triangle PQR.
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(c) Show that the triangles DEF and PQR have the same area.
238
(d) Find the matrix which transforms triangle PQR onto triangle DEF.
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(e) Find the equations of the invariant lines, through the origin, of the transformation in the x-y plane
represented by AB.
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Q7
6
2 0
o.
Let A = e
1 1
(a) The transformation in the x-y plane represented by A -1 transforms a triangle of area 30 cm 2 into a
triangle of area d cm 2 .
Find the value of d.
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(b) Prove by mathematical induction that, for all positive integers n,
2n 0
o.
An = e n
2 -1 1
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(c) The line y = 2x is invariant under the transformation in the x-y plane represented by A n B , where
1 0
o.
B=e
33 0
Find the value of n.
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Q8
4
The matrix A is given by
k
0
2
A = f0 - 1 - 1p,
1
1 -k
where k is a real constant.
(a) Show that A is non-singular.
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The matrices B and C are given by
0 -3
-3 -1 1
o.
B = f- 1
3p and C = e
1
1 2
0
0
- 2 - 32
It is given that CAB = f
p.
- 1 - 32
(b) Find the value of k.
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(c) Find the equations of the invariant lines, through the origin, of the transformation in the x-y plane
represented by CAB.
[5]
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MATRICES 1
MS
Craftear
243
Q1
4(c)
4(b)
4(a)
Question
1
1
= 2
2− 3
2
3
1
1 1+ 3 1+ 3
=
4 1 − 3 1− 3
1 1 1 1
N = M −1 1 1 = 2
3
2 2 2 − 2
M
−1
3 1 1
1 1
1 2 2
Craftear
A1 CAO
M1 Multiplies on the left by the inverse of their M.
B1 FT Inverse of their M.
2
A1 AG
4
M1 A1 Correct order.
B1 One-way stretch in the x-direction, scale factor d.
d ≠0 d =2
− d 23
1
2
Guidance
B1 Rotation 60° anticlockwise about the origin.
M1 Uses value of det M .
− 23 d2
=
1
3
2
2
− 23
1
2
Marks
d = 12 d 2
d 0 12
M=
0 1 23
d 0
0 1
cos 60 −sin 60 12
=
sin 60 cos 60 23
Answer
244
Q2
4(b)
4(a)
Question
4(d)
−1 3
5 3
5 −1
−k
+k
= 0 leading to −2 − 2k + k = 0
0 1
1 1
1 0
k = −2
2
−1
10
−k +14 −k − 2
Craftear
2 k k 1 0
2 + k k
0 1 1
0 1 1
−1
5 −1 3 0 1 =
8
−1 2 0 1 0
−1 2 0 2
0
1 0
1 0
2 k k 1 0
4
−1
0 1 1
6
Or
5 −1 3 0 1 =
0 1
k
k
1
2
0
8
−
−
−
2
−
+
6
1 0 1 1 0
1 0
Answer
Guidance
3
A1
M1 A1 Sets determinant equal to zero and forms linear
equation.
3
A1
M1 A1 Multiplies two matrices correctly.
Marks
5
A1 WWW
y = 12 x and y = −x
M1 A1 Uses y = mx and Y = mX .
x
X
Transforms to .
y
Y
A1
( x + mx )
B1
1 + m = 2m + 2 m 2 2m 2 + m − 1 = 0
1
x + 12 mx = m
2
1 1 x x + y
1 1 = 1
1
2 2 y 2 x + 2 y
245
Q3
4(a)
Question
4(c)
Craftear
One-way stretch [scale factor 3 parallel to the x-axis] followed by a rotation,
[anticlockwise centred at the origin, through an angle θ . ]
Answer
[m2 − 10m + 16 = 0]
y = 2 x and y = 8 x
16 = 10m − m2
10x − mx = X and 16 x = mX
10 −1 x 10 x − y
=
16 0 y 16 x
x
Transforms .
y
1
Allow q where q is x, t or a nonzero number.
m
Guidance
2
B2 Award B1 if given in the wrong order.
Marks
5
A1
A1 OE
Expect 16x = m (10 x − mx ) .
M1 A1 Uses y = mx and Y = mX .
M1
246
4(b)
1
or
3
tan θ = ±
θ = 16 π and θ = 56 π
cosθ = ± 23
−sin θ x 3x cosθ − y sin θ
=
cosθ y 3x sin θ + y cosθ
−sin θ
cosθ
3cosθ
3sin θ
3cosθ
M =
3sin θ
Craftear
x
X
Transforms to .
y
Y
9
A1 Allow 30° and 150°
M1 Uses an appropriate trigonometric identity.
M1 A1 Sets discriminant equal to 0.
A1
M1 A1 Uses y = mx and Y = mX .
B1
B1
247
Q4
Craftear
A1
Area = 2 × 13 = 26 cm2
1(d)
1(c)
M1 Finds det A.
det A = 6 − 8 = −2
A1 AEF
Could be solved by equations.
2 −4 −6 12
B = −3
=
−2 3 6 −9
4
M1 A1
2 4
Solves equations or states that det
≠ 0 , AG.
2 1
x X
Uses = to form simultaneous equations.
y Y
M1
3x + 4 y = x
2 x + 4 y = 0
leading to
2x + 2 y = y
2 x + y = 0
y = −2 x leading to −6 x = 0 leading to x = 0, y = 0
X
Finds
Y
B1
3 4 x 3x + 4 y
=
2 2 y 2 x + 2 y
2
M1 Finds A −1 .
Guidance
2 −4
A −1 = − 12
−2 3
2
1
1(b)
B1
Marks
Enlargement, scale factor 6.
Answer
1(a)
Question
248
Q5
1(c)
1(b)
1(a)
Question
A1
b=3
Craftear
M1 Uses that line is invariant (or X + 3Y = 0 ).
x = ax − 13 bx 1 = a − 13 b
5
M1 Uses x + 3 y = 0 .
ax − 13 bx
=
−1 x
3
x
X
Transforms to .
y
Y
B1
a b x ax + by
=
0 1 y y
2
M1 A1
2
B2 Both named correctly. Award B1 if given in the wrong
order.
Guidance
B1
1 1 −b
a 0 a
Marks
a=2
M −1 =
One-way stretch followed by a shear.
Answer
249
Q6
4(e)
4(d)
4(c)
B1 Or 60°
1
π
3
1
− 12 12 √ 3
2
=
1
1
1
√ 3 2
− 2 √ 3
2
x
1
X
x
Transforms to . Accept t instead of .
m
Y
y
y
A1 OE
y = (2 − √ 3) x and y + (2 + √ 3) x = 0
5
A1
M1A1 Uses Y = mX . OE. Allow working with y = mx + c as long
as c = 0 is stated explicitly.
B1
2
M1 A1 or using (AB)–1 = B–1A–1
1 − m √ 3 = m √ 3 + m 2 m 2 + 2m √ 3 −1 = 0 m = ±2 − √ 3
1
− 12 m √ 3 = 12 m √ 3 + 12 m 2
2
1
12 √ 3
x 12 x √ 3 + 12 y
2
1
=
− 12 √ 3 y 12 x − 12 y √ 3
2
( AB )
− 1 √3
= − 2 1
−
2
A1
Area of PQR = |-1| Area of DEF
−1
B1
det AB = −1
3
M1 Finds AB or uses product of determinants.
Full marks here may be obtained by arguing that both
reflection and rotation preserve [absolute] value of area and
so also does their combination.
2
1
1 √3
2
AB = 2 1
or det AB = det A det B
1
− 2 √ 3
2
anticlockwise about the origin.
B1 Writes ‘rotation’ or ‘rotate’ (and no other transformation)
Rotation
1
4(b)
Craftear
Guidance
B1 Only mention reflection (and no other transformation).
Marks
Reflection in the line y = x.
Answer
4(a)
Question
250
Q7
6(c)
6(b)
6(a)
Question
(
)
2k +1
0 2k +1
2k
0 2 0
0
= k
=
=
2 − 1 1 1 1 2 2k − 1 +1 1 2k +1 − 1 1
(2n + 32) x = 2n+1 x 2n = 32 n = 5
2n
0 x 2n x
n
= n
2
+
32
0
y (2 + 32) x
2n
0 1 0 2n
0
AnB = n
=
2 − 1 1 33 0 2n + 32 0
Craftear
So, it is also true for n = k +1 . Hence, by induction, true for all positive integers.
Then A
k +1
5
M1 A1
B1
M1A1
5
A1
M1A1
B1
2k
0
Assume that it is true for n = k , so A = k
.
2 − 1 1
k
B1
3
A1
M1 A1
Marks
1
0
2 0 2
A=
=
so true when n = 1.
1
1 1 2 − 1 1
d = 15
−1
det A −1 = ( det A ) = 12
Answer
251
Q8
4(c)
4(b)
4(a)
Question
−1 −1
0 −1
= 0 k2 + k + 2 0
+2
1 −k
1 1
Craftear
A1
y = −x and 3 y − 2 x = 0
5
A1
M1 A1
x + 32 mx = m ( 2x + 32 mx )
1 + 32 m = 2 m + 32 m 2 3m 2 + m − 2 = 0
B1
3
A1
M1 A1
3
A1
M1 A1
Marks
3
2 x + 32 y
x
2
=
3
3
y
2 x + 2 y
2
1
−2 9k + 3 −2 − 32
=
k = − 12
=
3
−
1
−
1
−3
3
−
−
k
2
k 0 2 0 −3
0 −3k
−3 −1 1
−3 −1 1
0 −1 −1 −1 3 =
1 −3
1
1
2
1
1
2
1 1 −k 0 0
−1 0
1 − 4(2) = −7 < 0 Non-singular
k
Answer
252
253
Craftear
POLAR COORDINATES
254
Q1
5
The curve C has polar equation r = a cot b13 r - il, where a is a positive constant and 0 G i G 16 r .
It is given that the greatest distance of a point on C from the pole is 2 3.
(a) Sketch C and show that a = 2 .
[3]
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(b) Find the exact value of the area of the region bounded by C, the initial line and the half-line
1
[4]
i = 6 r.
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(c) Show that C has Cartesian equation 2 `x + y 3j = `x 3 - yj x 2 + y 2 .
[3]
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256
Q2
The curve C has polar equation r =
1
r-i
-
1
r
, where 0 G i G 12 r .
(a) Sketch C.
[3]
(b) Show that the area of the region bounded by the half-line i = 12 r and C is
3 - 4 ln 2
.
4r
[6]
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258
Q3
6
The curve C has polar equation r = 2 cos i (1 + sin i) , for 0 G i G 12 r .
(a) Find the polar coordinates of the point on C that is furthest from the pole.
[5]
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259
(b) Sketch C.
[2]
(c) Find the area of the region bounded by C and the initial line, giving your answer in exact form. [6]
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260
Q4
The curve C has polar equation r = 3 + 2 sin i , for - r 1 i G r .
(a) The diagram shows part of C. Sketch the rest of C on the diagram.
O
[1]
i=0
The straight line l has polar equation r sin i = 2 .
(b) Add l to the diagram in part (a) and find the polar coordinates of the points of intersection of C
and l.
[5]
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8
261
(c) The region R is enclosed by C and l, and contains the pole.
Find the area of R, giving your answer in exact form.
[6]
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262
Q5
7
14
(a) Show that the curve with Cartesian equation
5
(x 2 + y 2) 2 = 4xy (x 2 - y 2)
has polar equation r = sin 4i .
[4]
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263
The curve C has polar equation r = sin 4i , for 0 G i G 14 r .
(b) Sketch C and state the equation of the line of symmetry.
[3]
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[4]
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(c) Find the exact value of the area of the region enclosed by C.
264
(d) Using the identity sin 4i / 4 sin i cos 3 i - 4 sin 3 i cos i , find the maximum distance of C from the
[6]
line i = 12 r . Give your answer correct to 2 decimal places.
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265
Q6
5
8
The curve C has polar equation r = ln (1 + r - i) , for 0 G i G r .
(a) Sketch C and state the polar coordinates of the point of C furthest from the pole.
[3]
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1
`
j
2 (1 + r) ln (1 + r) ln (1 + r) - 2 + r .
[6]
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(b) Using the substitution u = 1 + r - i , or otherwise, show that the area of the region enclosed by C
and the initial line is
266
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(c) Show that, at the point of C furthest from the initial line,
(1 + r - i) ln (1 + r - i) - tan i = 0
and verify that this equation has a root between 1.2 and 1.3.
[5]
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267
Q7
7
The curve C1 has polar equation r = i cos i , for 0 G i G 12 r .
(a) The point on C1 furthest from the line i = 12 r is denoted by P. Show that, at P,
2i tan i - 1 = 0
and verify that this equation has a root between 0.6 and 0.7.
[5]
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The curve C2 has polar equation r = i sin i , for 0 G i G 12 r . The curves C1 and C2 intersect at the
pole, denoted by O, and at another point Q.
(b) Find the polar coordinates of Q, giving your answers in exact form.
[2]
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(c) Sketch C1 and C2 on the same diagram.
[3]
(d) Find, in terms of r, the area of the region bounded by the arc OQ of C1 and the arc OQ of C2 . [7]
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269
Q8
The curve C has polar equation r = a tan i , where a is a positive constant and 0 G i G 14 r .
(a) Sketch C and state the greatest distance of a point on C from the pole.
[2]
(b) Find the exact value of the area of the region bounded by C and the half-line i = 14 r .
[4]
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8
270
x2
.
[3]
a2 - x2
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(c) Show that C has Cartesian equation y =
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1
2a
2
x2
dx.
[2]
a2 - x2
0
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(d) Using your answer to part (b), deduce the exact value of
y
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271
Q9
OR
The curves C1 and C2 have polar equations, for 0 ≤ 1 ≤ 12 0, as follows:
C1 : r = 2 e1 + e−1 ,
C2 : r = e21 − e−21 .
The curves intersect at the point P where 1 = !.
(i) Show that e2! − 2e! − 1 = 0. Hence find the exact value of ! and show that the value of r at P is
4ï2.
[6]
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(ii) Sketch C1 and C2 on the same diagram.
[3]
(iii) Find the area of the region enclosed by C1 , C2 and the initial line, giving your answer correct to
[5]
3 significant figures.
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273
Q10
11
20
Answer only one of the following two alternatives.
EITHER
The curve C1 has polar equation r 2 = 21, for 0 ≤ 1 ≤ 12 0.
(i) The point on C1 furthest from the line 1 = 12 0 is denoted by P. Show that, at P,
21 tan 1 = 1
and verify that this equation has a root between 0.6 and 0.7.
[5]
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274
The curve C2 has polar equation r 2 = 1 sec2 1, for 0 ≤ 1 < 12 0. The curves C1 and C2 intersect at the
pole, denoted by O, and at another point Q.
(ii) Find the exact value of 1 at Q.
[2]
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(iii) The diagram below shows the curve C2 . Sketch C1 on this diagram.
[2]
1 = 21 0
C2
1=0
O
© UCLES 2019
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275
(iv) Find, in exact form, the area of the region OPQ enclosed by C1 and C2 .
[5]
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276
Q11
The curve C has polar equation r2 = ln 1 + 1, for 0 ≤ 1 ≤ 20.
(i) Sketch C.
[2]
(ii) Using the substitution u = 1 + 1, or otherwise, find the area of the region bounded by C and the
initial line, leaving your answer in an exact form.
[5]
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2
277
Q12
The curve C has polar equation r = cos 21, for − 14 0 ≤ 1 ≤ 14 0.
(i) Sketch C.
[2]
(ii) Find the area of the region enclosed by C, showing full working.
[3]
Craftear
3
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(iii) Find a cartesian equation of C.
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[3]
279
Q13
8
The curves C1 and C2 have polar equations, for 0 ≤ 1 ≤ 0, as follows:
C1 : r = a,
C2 : r = 2a cos 1 ,
where a is a positive constant. The curves intersect at the points P1 and P2 .
(i) Find the polar coordinates of P1 and P2 .
[2]
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(ii) In a single diagram, sketch C1 , C2 and their line of symmetry.
© UCLES 2018
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[3]
Craftear
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280
(iii) The region R enclosed by C1 and C2 is bounded by the arcs OP1 , P1 P2 and P2 O, where O is the
pole. Find the area of R, giving your answer in exact form.
[5]
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281
Q14
9
The curve C has polar equation
where 0.01 ≤ 1 ≤ 12 0.
r = 5 cot 1,
(i) Find the area of the finite region bounded by C and the line 1 = 0.01, showing full working. Give
your answer correct to 1 decimal place.
[3]
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Let P be the point on C where 1 = 0.01.
(ii) Find the distance of P from the initial line, giving your answer correct to 1 decimal place.
[2]
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282
(iii) Find the maximum distance of C from the initial line.
[3]
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(iv) Sketch C.
© UCLES 2018
[2]
9231/11/O/N/18
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Craftear
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283
Q15
The curve C has polar equation r = a cos 31, for − 16 0 ≤ 1 ≤ 16 0 , where a is a positive constant.
(i) Sketch C.
[2]
(ii) Find the area of the region enclosed by C, showing full working.
[3]
Craftear
3
4
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© UCLES 2018
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Craftear
(iii) Using the identity cos 31 4 cos3 1 − 3 cos 1, find a cartesian equation of C.
285
Q16
11
The curve C has polar equation r = a 1 + sin 1 for −0 < 1 ≤ 0, where a is a positive constant.
(i) Sketch C.
[2]
(ii) Find the area of the region enclosed by C.
[4]
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286
(iii) Show that the length of the arc of C from the pole to the point furthest from the pole is given by
s = ï2a Ô
10
2
− 12 0
1 + sin 1 d1.
[3]
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287
2
(iv) Show that the substitution u = 1 + sin 1 reduces this integral for s to ï2aÔ 0
evaluate s.
1
du. Hence
2 − u
[4]
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288
Q17
11
19
Answer only one of the following two alternatives.
EITHER
A curve C has polar equation r = 2a cos 21 + 12 0 for 0 ≤ 1 < 20, where a is a positive constant.
(i) Show that r = −2a sin 21 and sketch C.
[4]
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Craftear
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(ii) Deduce that the cartesian equation of C is
x2 + y2 2 = −4axy.
3
[2]
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289
(iii) Find the area of one loop of C.
[5]
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290
(iv) Show that, at the points (other than the pole) at which a tangent to C is parallel to the initial line,
2 tan 1 = − tan 21.
[3]
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Craftear
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291
Q18
OR
The polar equation of a curve C is r = a 1 + cos 1 for 0 ≤ 1 < 20, where a is a positive constant.
(i) Sketch C.
(ii) Show that the cartesian equation of C is
x2 + y2 = a x + x2 + y2 .
[2]
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[2]
292
(iii) Find the area of the sector of C between 1 = 0 and 1 = 13 0 .
[4]
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293
(iv) Find the arc length of C between the point where 1 = 0 and the point where 1 = 13 0 .
[5]
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Q19
4
294
A curve C has polar equation r2 = 8 cosec 21 for 0 < 1 < 12 0. Find a cartesian equation of C.
[3]
Sketch C.
[2]
Determine the exact area of the sector bounded by the arc of C between 1 = 16 0 and 1 = 13 0, the
half-line 1 = 16 0 and the half-line 1 = 13 0.
[3]
Craftear
[It is given that Ó cosec x dx = ln tan 12 x + c.]
© UCLES 2016
9231/11/M/J/16
[Turn over
Q20
7
A curve has polar equation r =
equation of the curve.
295
1
, for 0 < 1 < 20. Find, in the form y2 = f x, the cartesian
1 − cos 1
[3]
Hence sketch the curve, and shade the region whose area is given by 1 Ô
2
Using the cartesian equation of the curve, find the area of this region.
8
30
2
10
2
1
d1.
1 − cos 12
[3]
[3]
The cubic equation
z3 − z2 − z − 5 = 0
[4]
Find the value of !4 + "4 + ' 4 .
[2]
Craftear
has roots !, " and ' . Show that the value of !3 + "3 + ' 3 is 19.
© UCLES 2016
9231/13/M/J/16
[Turn over
Q21
5
296
The curves C1 and C2 have polar equations
C1 :
C2 :
1
, for 0 ≤ 1 < 20,
ï2
r = sin 12 1 , for 0 ≤ 1 ≤ 0.
r=
[2]
Sketch C1 and C2 on the same diagram.
[3]
Find the exact value of the area of the region enclosed by C1 , C2 and the half-line 1 = 0.
[4]
Craftear
Find the polar coordinates of the point of intersection of C1 and C2 .
© UCLES 2015
9231/11/M/J/15
[Turn over
1
297 p, q and r are real constants, has two pairs of equal
The quartic equation x4 − px2 + qx − r = 0, where
2
roots. Show that p + 4r = 0 and state the value of q.
[6]
Q22
The curve C has polar equation r = e41 for 0 ≤ 1 ≤ !, where ! is measured in radians. The length of
C is 2015. Find the value of !.
[6]
Craftear
2
© UCLES 2015
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[Turn over
Q23
298
11 OR
The curve C has polar equation r = a 1 − cos 1 for 0 ≤ 1 < 20. Sketch C.
[2]
Find the area of the region enclosed by the arc of C for which 21 0 ≤ 1 ≤ 32 0, the half-line 1 = 12 0 and
the half-line 1 = 32 0.
[5]
Show that
@
A
ds 2
= 4a2 sin2 12 1 ,
d1
[7]
Craftear
where s denotes arc length, and find the length of the arc of C for which 21 0 ≤ 1 ≤ 32 0.
© UCLES 2015
9231/11/O/N/15
[Turn over
Q24
5
299
The curve C has polar equation r = a1 + sin , where a is a positive constant and 0 ≤ < 2. Draw
a sketch of C.
[2]
Find the exact value of the area of the region enclosed by C and the half-lines = 13 and = 23 . [4]
The linear transformation T : 4 → 4 is represented by the matrix M, where
Craftear
6
© UCLES 2014
9231/11/M/J/14
[Turn over
Q25
The curve C has cartesian equation x2 + y2
polar equation of C is r2 = a2 sin 21.
2
= 2a2 xy, where a is a positive constant. Show that the
[3]
Sketch C for −0 < 1 ≤ 0.
[2]
Find the area enclosed by one loop of C.
[2]
Craftear
4
300
© UCLES 2014
9231/13/M/J/14
[Turn over
Q26
A circle has polar equation r = a, for 0 ≤ < 2, and a cardioid has polar equation r = a1 − cos ,
for 0 ≤ < 2, where a is a positive constant. Draw sketches of the circle and the cardioid on the
same diagram.
[3]
Write down the polar coordinates of the points of intersection of the circle and the cardioid.
[2]
Show that the area of the region that is both inside the circle and inside the cardioid is
5
− 2 a2 .
4
6
Craftear
8
301
© UCLES 2014
9231/11/O/N/14
[Turn over
Q27
Find the area of the region enclosed by the curve with polar equation r = 2 1 + cos 1, for 0 ≤ 1 < 20.
[4]
Craftear
1
302
© UCLES 2013
9231/11/M/J/13
[Turn over
Q28
10
303
Use the identity 2 sin P cos Q sin P + Q + sin P − Q to show that
2 sin 1 cos 1 − 14 0 cos 21 − 34 0 +
1
.
ï2
3
A curve has polar equation r = 2 sin 1 cos 1 − 14 0, for 0 ≤ 1 ≤ 43 0. Sketch the curve and state the polar
equation of its line of symmetry, justifying your answer.
[3]
[6]
Craftear
Show that the area of the region enclosed by the curve is 83 0 + 1.
© UCLES 2013
9231/13/M/J/13
[Turn over
Q29
The curve C has polar equation r = 2e1 , for 61 0 ≤ 1 ≤ 12 0. Find
(i) the area of the region bounded by the half-lines 1 = 16 0, 1 = 12 0 and C,
[2]
(ii) the length of C.
[3]
Craftear
1
304
© UCLES 2013
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[Turn over
Q30
10
305
The curve C has polar equation r = 2 sin 1 1 − cos 1, for 0 ≤ 1 ≤ 0. Find
coordinates of the point of C that is furthest from the pole.
dr
and hence find the polar
d1
[5]
[2]
Find the exact area of the sector from 1 = 0 to 1 = 14 0.
[6]
Craftear
Sketch C.
© UCLES 2013
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[Turn over
Q31
4
306
The curve C has polar equation r = 2 + 2 cos θ , for 0 ≤ θ ≤ π . Sketch the graph of C.
[2]
Find the area of the region R enclosed by C and the initial line.
[4]
Craftear
The half-line θ = 15 π divides R into two parts. Find the area of each part, correct to 3 decimal places.
[3]
© UCLES 2012
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307
Q32
1
√
Find the cartesian equation corresponding to the polar equation r = ( 2) sec(θ − 14 π ).
[3]
Craftear
√
Sketch the the graph of r = ( 2) sec(θ − 14 π ), for − 14 π < θ < 34 π , indicating clearly the polar coordinates
of the intersection with the initial line.
[2]
© UCLES 2012
9231/12/O/N/12
[Turn over
Q33
The curve C has polar equation r = 1 + 2 cos θ . Sketch the curve for − 23 π ≤ θ < 23 π .
[2]
Find the area bounded by C and the half-lines θ = − 13 π , θ = 13 π .
[4]
Craftear
5
308
© UCLES 2012
9231/13/O/N/12
[Turn over
Q34
309
11 EITHER
The curve C has cartesian equation
(x2 + y2 ) = a2 (x2 − y2 ),
2
where a is a positive constant. Show that C has polar equation
r2 = a2 cos 2θ .
[2]
[2]
Find the area of the sector between θ = − 14 π and θ = 14 π .
[3]
Find the polar coordinates of all points of C where the tangent is parallel to the initial line.
[7]
Craftear
Sketch C for −π < θ ≤ π .
© UCLES 2012
9231/13/M/J/12
[Turn over
POLAR COORDINATES
MS
Craftear
310
Q1
5(c)
5(b)
5(a)
Question
1
3
)
3 = 34 3 − 13 π
) (
2
) x +y
r cos θ + 3r sin θ
3r cos θ − r sin θ
2 x+ y 3 = x 3− y
(
1
3
cos π3 cos θ + sin π3 sin θ
sin π3 cos θ − cos π3 sin θ
1
6
x2 + y 2 = 2
r=2
0
2
( 3− π−
1
3
2
2
cosec ( π − θ ) −1 dθ = 2 cot ( π − θ ) − θ
1π
6
4cot 2 ( 13 π − θ ) dθ
2
1π
1 6
2 0
a cot 16 π = 2 3 a = 2
θ =0
Answer
0
1π
6
Craftear
Guidance
3
A1 AG.
M1 Applies r = x 2 + y 2 , x = r cos θ and y = r sin θ .
M1 Uses the identities for cos( A − B) and sin( A − B). Or
identity for tan (A – B).
4
A1 OE, must be exact.
M1 A1 Uses cot 2 ( 1 π − θ ) = cosec2 ( 1 π − θ ) − 1 and integrates.
3
3
M1 Uses 1 r 2 dθ with correct limits.
2
3
B1 Substitutes θ = 1 π . AG.
6
B1 B1 Initial line with correct position and shape of curve.
SCB1 if correct shape with wrong starting position.
Pole position must be clear.
Marks
311
Q2
5(b)
5(a)
Question
1π
1
2
2
2
1
1
− dθ
π −θ π
1
1π
2
θ = 0]
=
3 − 4ln 2
4π
1 2 2 π 1 1 2
1 3 2 1
− + ln π =
+ ln
+ ln +
2 π π 2 2π π π
2 2π π 2
2
1 1
θ 2
+
ln(
+
θ
π
−
)
2 π − θ π
π 2 0
1 2
2 0
( π − θ ) − π ( π − θ ) + π dθ
1π
1 2
2 0
[O]
Answer
Craftear
B1 Approximately correct curve passing through the
pole, O, with correct domain.
Guidance
6
A1 AG
M1 Substitute limits into an expression of the correct
form and simplify the log terms.
M1 A1 Integrates all terms to obtain correct form.
M1 Expands.
M1 Uses 1 r 2 dθ with correct limits.
2
3
B1 Correct form at O.
Approx tangential to initial line at O.
B1 r strictly increasing over the domain 0 to π .
2
Marks
312
Q3
6(c)
Question
6(b)
6(a)
6
)
4c 2 (1 + s)2 dθ = 2
2
2
0
2
1
4
2
2 2
c + 2sc + s c dθ
1π
2
Answer
θ =0
1π
2
0
2
2
1
4
1
4
( cos 2θ + 1) + 2 sin θ cos θ + sin 2θ dθ
1π
2 1
1π
3, 16 π
1π
5
π + 43
8
3
1
4
1
5
2
4 θ + 2 sin 2θ − 3 cos θ − 16 sin 4θ 0
0
cos 2θ + 1 + 4 sin θ cos θ + − cos 4θ dθ
2
3
2
1 2
2 0
(
Craftear
Finds
dr
and sets equal to 0.
dθ
increasing to
π
then decreasing.
6
Correct shape. Tangential to θ =
π
at pole, r strictly
2
Guidance
6
A1
M1 Integrates.
A1 Obtains integrable form.
M1 Uses double angle formulae on first and last terms.
M1 A1 Uses 1 r 2 dθ with correct limits.
2
Marks
2
B1 Correct position. All in first quadrant.
B1
5
A1 Allow ( 2.60, 1 π ) 3sf
6
A1
1 − s − 2s 2 = 0
sin θ = 12 leading to θ
M1
2 ( c(c) − (1 + s)s ) = 0
313
Q4
Answer
Line l parallel to initial line and correct side of pole.
5
A1
A1 SC1 For finding both angles correctly
Forms quadratic in sin θ . Or in r r = 3 +
( 4, 16 π )
( 4, 56 π )
M1
B1
1
M1 Solves for sin θ .
Craftear
Guidance
B1 Correct symmetrical shape, closed loop.
Marks
sin θ = 12
2 = 3sin θ + 2sin 2 θ
5(b)
5(a)
Question
4
r
314
Q5
7(a)
Question
5(c)
− π2
2
( 3 + 2 sin θ ) dθ
π
6
π
6
π
2
(
(
Craftear
4
A1 Applies both double angle formulae, AG.
r = sin 4θ
B1 Uses x = r cos θ and y = r sin θ .
B1 Uses r 2 = x 2 + y 2 .
Guidance
M1 Applies at least one double angle formula.
Marks
6
A1
M1 Adds area of triangle.
A1 A1
M1 Uses double angle formula and integrates.
M1 Finds the part of the required area enclosed by the
curved outer edge and two line segments from the pole.
Limits must be correct.
r = 2sin 2θ cos 2θ
)
Answer
r 5 = 4r 4 cos θ sin θ cos 2 θ − sin 2 θ
)
3
22
π − 52
3
r 5 = 4 xy x 2 − y 2
3 + (4cos 6π ) × 2
22
π − 132
3
[11θ − 12cosθ − sin 2θ ]− = 223 π − 132 3
− π2
9 +12sin θ + 2 (1 − cos 2θ ) dθ
π
6
2 × 12
315
7(d)
7(c)
7(b)
2
= 14
0
Craftear
A1 Substituting θ = ±0.369 gives maximum value of | x | .
θ = 0.369 x = 0.93
(or θ = −0.369 | x |= 0.93 )
6
B1 Allow use of decimals.
)
tan 2 θ = 14 7 ± √ 41 θ = ±0.369, ± 1.071
(
M1 Forms quadratic in tan 2 θ . Must see consideration of
cos θ = 0.
cos θ = 0 or 2 tan 4 θ − 7 tan 2 θ + 1 = 0
)
A1
(
cos θ 2sin 4 θ − 7 sin 2 θ cos 2 θ + cos 4 θ = 0
B1 Uses x = r cos θ .
4
A1
M1 A1 Applies double angle formula and integrates.
M1 Differentiates and sets equal to 0.
)
M1 Uses 1 r 2 dθ with correct limits.
2
3
B1 States the equation of the line of symmetry.
−4sin 2 θ cos3 θ + cos5 θ + 2sin 4 θ cos θ − 3sin 2 θ cos3 θ = 0
x = 4 sin θ cos 4 θ − sin 3 θ cos 2 θ
(
1π
1 − cos8θ dθ = 14 θ − 18 sin 8θ 04
1π
4
= 161 π
0
sin 4θ dθ
1
2
1π
4
θ = 18 π
B1 Correct shape at extremities.
B1 Initial line drawn and one loop in the first quadrant.
316
Q6
5(c)
5(b)
5(a)
Question
ln u du
1+π
−
1
1+ π
1
(1 + π)ln(1 + π)(ln(1 + π) − 2) + π
2
Craftear
5
B1 Shows sign change. Values correct to 2sf should be shown.
(1 + π −1.2) ln(1 + π −1.2) − tan1.2 = 0.602
and (1 + π −1.3) ln(1 + π −1.3) − tan1.3 = −0.634
M1 A1 Sets derivative equal to zero.
B1 Uses y = r sinθ
6
A1 AG
A1
M1 Integrates by parts again (or uses known result for integral
of ln u)
M1 A1 Integrates by parts once.
M1 Uses correct formula with correct limits.
3
B1 May be seen on their diagram. Allow (1.42, 0)
B1 Section 12 π ≤ θ ≤ π correct, becoming tangential at θ = π.
A1 AG
=
Guidance
B1 Correct shape, r decreasing.
Marks
cos θ ≠ 0 (1 + π − θ ) ln(1+ π − θ ) − tan θ = 0
dy
sin θ
= ln(1 + π − θ ) cosθ −
=0
dθ
1+ π − θ
y = ln(1 + π −θ )sinθ
1du
ln 2 u du
1
(1 + π) ln 2 (1 + π) − (1 + π) ln(1 + π) + π
2
1
= 12 u ln 2 u − u ln u + u
1
1+π
− [u ln u ]1
1
1+π
= 12 u ln 2 u
1
−
1+π
1+π
1
θ )dθ = 12
1+π
= 12 u ln 2 u
π
1
ln 2 (1 + π −
2 0
( ln(1+ π),0 )
θ =0
Answer
317
Q7
7(c)
7(b)
7(a)
Question
B1
B1
2(0.6) tan 0.6 − 1 = −0.179 and 2(0.7) tan 0.7 − 1 = 0.179
Craftear
(B1 for initial line drawn and C1 correct, B1 for C2 correct, B1 for intersection correct)
3
B1
B1
B1
2
A1
1
4
( π √ 2, π )
1
8
M1
θ cos θ = θ sin θ tan θ = 1
5
A1
M1 A1
Marks
cos θ ≠ 0 2θ tan θ − 1 = 0
dx
= −2θ cosθ sin θ + cos 2 θ = 0
dθ
x = θ cos 2 θ
Answer
318
Q8
5(a)
Question
7(d)
0
2
θ cos 2θ dθ
1π
4
2
2
Maximum distance of C from the pole is a.
= 641 π 2 − 18
0
4
= 14 θ 2 sin 2θ + θ cos 2θ − 12 sin 2θ
1π
1π
1π
1π
4
4
cos 2θ dθ
= 12 12 θ 2 sin 2θ − − 12 θ cos 2θ 04 − 12
0
0
1π
1π
4
4
= 12 12 θ 2 sin 2θ − θ sin 2θ dθ
0
0
= 12
2
θ ( cos θ − sin θ ) dθ
1π
1 4
2 0
Craftear
Answer
2
B1
B1
Marks
7
A1
A1
M1
M1 A1
M1
M1
319
5(d)
5(c)
5(b)
0
2
2
y
x
(
)
1π
4
0
(
)
AG
( a cos 14 π ) ( a sin 14 π ) − 12 a 2 (1 − 14 π ) = 14 a 2 ( 12 π − 1)
√ a2 − x2
x2
(A1 FT their (b))
1
2
y=
x2 x2 + y 2 = a2 y 2 y 2 a 2 − x2 = x4
)
x2 + y 2 = a
(
1
2
sec θ −1dθ = a [ tan θ − θ ]
1π
4
= 12 a 2 (1 − 14 π )
= 12 a 2
2
tan θ dθ
1π
1 2 4
a
2
0
Craftear
11
M1 A1 FT
3
A1
M1
B1
4
A1
M1 A1
M1
320
Q9
11O(ii)
11O(i)
Question
(
)
)
)
r = 2 1 + 2 + 2 −1 = 4 2
(
α = ln 1 + 2
(
e 2α − 2eα − 1 = 0 ⇒ eα = 1 + 2
e 2α − e −2α = 2 eα + e −α ⇒ eα − e −α = 2
Answer
Craftear
Guidance
C2 has correct shape.
B1
3
B1 Intersection points positioned correctly.
C1 has correct shape.
B1
6
M1 A1 Substitutes to find r .
A1 Must be exact.
M1 A1 Forms quadratic in eα , AG.
M1 Sets equations equal and divides by eα + e −α .
Marks
321
Q10
11E(i)
Question
11O(iii)
Question
)
−θ
2θ
e +e
θ
)
2
0
−2θ
(
)
0
∫ (
ln 1+√2
e −e
2θ
(
θ
Answer
−2
)
dθ
4
−4
Uses
5
B1 Shows sign change.
2 ( 0.6 ) tan ( 0.6 ) −1 = −0.179 and 2 ( 0.7 ) tan ( 0.7 ) −1 = 0.179
1
A1 AG
1
Guidance
M1 A1 Sets derivative of r cosθ equal to zero.
M1 Uses x = r cosθ .
Marks
5
A1
Guidance
1 2
∫ r dθ to formulate correct area.
2
M1 A1 Expands and integrates.
M1 A1
Marks
1 −
−θ 2 sin θ + θ 2 cos θ = 0 ⇒ cos θ = 2θ sin θ ⇒ 2θ tan θ = 1
2
Craftear
) − (1 + 2 ) − 81 (1 + 2 ) − (1 + 2 ) = 5.82
2
(
ln 1+√2
)
2
1
1
1
d
1 −
2θ 2 cosθ = 2 −θ 2 sin θ + θ 2 cosθ = 0
dθ
2
x = 2θ
1
2 cos
= 5ln(1 + √ 2) + 1 + 2
−2θ
1
1
− e 4θ − e −4θ dθ
2
2
1
dθ −
2
∫ 5 + 2e + 2e
ln 1+√2
(
0
∫ (
)
1
1
= 5θ + e 2θ − e−2θ − e4θ + e−4θ
8
8
0
=
2
(
ln 1+√2
Answer
322
11E(iv)
11E(iii)
11E(ii)
∫
∫
π
4
1
2
∫(
π
4
0
)
( secθ − 2 ) = 0 ⇒ θ = π4
Craftear
5
A1 AEF, must be exact.
2
1
π π 1 π 1
ππ
1
−
−
+ ln 2 = ln 2 + − 1 .
4
8 2 2 4 2
84
π
A1
∫ ( 2θ − tan θ ) dθ
M1 A1 Integrates by parts.
M1 Forms correct integral.
2
B1 Intersection correct.
B1 Correct shape.
2
M1 A1 Finds value of θ .
π
1
1
θ ( 2θ − tan θ ) 04 − θ 2 + ln cos θ 4
0
2
2
1
θ ( 2θ − tan θ )
2
4
π
1
04 −
20
π
1
1
1
2θ θ − θ sec 2 θ dθ = θ 2 − sec 2 θ dθ
20
20
20
π
4
2θ = θ sec θ ⇒ θ
2
323
Q11
2(ii)
2(i)
Question
(
∫
2π
)
1
∫
2π+1
1
ln (1 + θ ) dθ
20
2π+1
2 π +1
− [u ]1
1
A = π + ln ( 2π + 1) − π
2
1
1
∫ ln u du = [u ln u ]
2π+1
0
2π
A=
Answer
Craftear
θ =0
Guidance
States
1 2
∫ r dθ with correct expression and limits.
2
5
A1 AEF, must be exact.
M1 A1 Integrates ln u (by parts or otherwise).
M1 Applies given substitution correctly, changes their limits.
M1
2
B1 The initial line is tangential to C at the pole.
B1 Correct shape, domain and orientation.
Marks
324
Q12
3(iii)
3(ii)
3(i)
Question
π
∫
π
4
1 4
cos 4θ + 1dθ =
4 − π4
)
3
Thus x 2 + y 2 2 = x 2 − y 2
(
r = cos 2 θ − sin 2 θ OE
π
1 1
4
= 0.393
sin 4θ + θ =
π
8
4 4
−
=
∫
π
1 4
cos 2 2θ dθ
2 − π4
Answer
1
Craftear
Guidance
2
B1 Single correct loop.
B1 Correct position including label of 1 on initial line,
and symmetric about initial
3
M1 A1 Uses x = r cos θ or y = r sin θ or both , AEF
B1 Uses trig identity
3
A1
M1 Correct use of double angle formula
M1 Correct integral
Marks
325
Q13
8(iii)
8(ii)
8(i)
Qu
uestion
3
π
∫ cos 2θ + 1dθ
2
π
π
1
2
Area =
π
π a2 a2 3
3
− a 2 −
+
= −
6
2
6
2
3
π a2
2 1
Craftear
π 3 π
π
3
2
+ = a 2 −
= 2a sin 2θ + θ = 2a 2 −
2
4 3
3
2
2
π
3
= 2a
2
3
π
4a 2 cos 2 θ dθ
∫
2
π
π
2π
a, and a,
3
3
a = 2a cos θ ⇒ cos θ = ±
Answ
wer
Guidance
Uses cos 2 θ =
1
( cos 2θ + 1)
2
10
M1
π a2
A1FT M1 for subtracting ‘their’ OP1P2 from
6
A1 Integrates correctly.
M1
M1 Finds area of segment OP1 and OP2 of C2
2 and line of symm
metry.
B1 Other half of C2
uding r = 2a.
B1 Half of C2 inclu
B1 Semicircle for C1 including r = a..
ded for A1.
A1 Both points need
M1 Eliminates r .
Marks
326
9(iii)
9(ii)
9(i)
Q14 Question
2
π
π
1
⇒y=
5 2
2
( = 3.54 )
1
2
5
1
sin 2 2θ
−
dy
5
sin 2 2θ cos 2θ = 0 or max ( sin 2θ ) = 1
=
dθ
2
θ = 0.01 ⇒ y ≈ 0.5
1
y = 5cos 2 θ sin 2 θ =
25
ln sin 0.01 ≈ 57.6
2
25
[ln sin θ ]0.201
2
=−
=
∫
25
cot θ dθ
2 0.01
Answer
Craftear
Uses
1 2
∫ r dθ .
2
3
A1
M1 A1
2
A1
Sets
Guidance
dy
= 0 or considers max (AEF).
dθ
M1 Uses y = r sin θ .
3
A1
A1
M1
Marks
327
Q15
3(i)
Qu
uestion
9(iv)
Answer
Craftear
a
2
Guiidance
2
orrect position including (a, 0) labellled
B1 Co
or in table.
mities
B1 Jusst one loop, correcct shape at extrem
Marks
B1 Correct shape.
B1 Intersecting the initial line only when x = 0 and y = 0.
328
Q16
11(i)
Question
3(iii)
π
6
−
6
π
∫π ( cos 6θ +1) dθ
6
) (
(
⇒ x2 + y 2
(
2
2
2
) = ax ( x − 3y )
⇒ r 4 = ax 4 x 2 − 3r 2 ⇒ x 2 + y 2
)
(
2
Answer
2
2
2
Craftear
) = ax ( 4 x − 3 ( x + y ) )
2
x x
r = a cos θ 4cos 2 θ − 3 ⇒ r = a 4 − 3
r r
a2 1
π a2
6
sin
6
θ
θ
+
=
=
π
4 6
12
−
a
4
2 6
−
∫
1
a 2 cos 2 3θ dθ
2 π
6
π
Total:
Guidance
2
B1 Correct oriientation and labeelled
B1 Sketch of a cardioid
Marks
3
A1 Any equivalent cartesian form without
fractions.
M1 For eliminating θ
B1 Uses x = r cosθ and x2 + y 2 = r 2 .
3
A1
M1 Using double angle formula correctly
329
11(iv)
11(iii)
11(ii)
∫ (
)
a2 π
1 + 2sin θ + sin 2 θ dθ
2 −π
2
2
∫−π
π
(
)
, when r=2a θ =
2
π
2
1
∫0 2 −u
2
du AG
1
2
= 2a −2 ( 2 − u ) 2 = 4a
0
so s = 2a
Total:
1
2
− 12
Total:
Total:
π
1 + sin θ dθ
π
dr
= a cosθ
dθ
Craftear
∫
and that
a 2 1 + 2sin θ + a 2 sin 2 θ + a 2 cos 2 θ dθ = 2a
2
π
u = 1 + sin θ ⇒ dduθ = cos θ = 1 − ( u −1) = 2u − u 2
s=
Show that when r=0, θ = −
2
a 2 3θ
1
π 3π a
= − 2cos θ − sin 2θ
=
4
2
−π
2 2
∫
a2 π
1 1
1 + 2sin θ + − cos 2θ dθ
=
−
π
2
2
2
A=
4
M1A1
A1 Including limits
M1
3
M1A1 Uses correct formula for arc length;
AG
B1
4
M1A1
M1
M1
330
Q17
11E
E(iii)
11E
E(ii)
11E
E(i)
Quesstion
2
1
π
2
π
∫ (1 − cos 4θ ) dθ
π
2
1
= a θ − sin 4θ
4
1π
= a2
Area of one loop is
π
π
)
1
= π a2
2
π
2
π
2
∫
∫
1
4a 2sin 2 2θ dθ = a 2 2sin 2 2θ dθ
21
1
(
3
Craftear
y x
r = −4a sin θ cosθ = −4a. . ⇒ r 3 = −4axy ⇒ ( x 2 + y 2 2 = −4axy )
r r
1
1
r = 2a cos 2θ cos π − sinn 2θ sin π = 2a ( 0 − sin 2θ .1) = −2a sin 2θ
2
2
Answeer
To
otal:
To
otal:
To
otal:
Guidancee
5
M1A1
M1A1
M1 OE
2
M1A1 AG
4
B1 Symmetry and shape
correct
B1 Loops in 2nd and 4th
quadrant
M1A1 AG
Marks
331
Q18
11OR(ii)
11OR
Question
11E(iv)
x 2 + y 2 = a ( x + (x 2 + y 2 ))
r = a (1 + cosθ ) ⇒ x 2 + y 2 = a 1 +
Answer
x 2 + y 2
x
(⇒ 2 tan θ = − tan 2θ )
2
B1 Cardioid wiith indication of correct scale.
2
A1 Substitutes for r and cos(θ)
M1
Craftear
Guidance
3
ding at pole, in ap
pproximately correect
B1 Closed curve starting and end
location.
Marks
Total:
A1 AG
M1
dy
= −2a ( 2cos 2θ sin θ + sin 2θ cosθ ) = 0
dθ
2cos 2θ sin θ = − sin 2θ cos θ
B1
y = r sin θ = −2a sin 2θ sin θ
332
11OR(iv)
11OR(iii)
1
π
)
π
1
(
)
2
π
2
2
2
0
1
π
∫ 2 + 2 cosθ dθ
1
π
3
θ 3
= a 4sin = 2a
2 0
=a
0
∫ a (1 + 2cosθ + cos θ ) + a ( −sin θ ) dθ
1
π
3
a2
4π + 9 3
16
Arc length =
=
sin 2θ 3
a 2 3θ
= + 2sin θ +
2 2
4 0
∫
a
cos 2θ
3
=
dθ
+ 2cosθ +
2 0 2
2
2 3
1
a
1 + 2 cosθ + cos 2θ dθ
Sector area =
2 0
∫(
2 3
Craftear
5
M1A1
A1
M1A1
4
A1
M1
M1A1
333
334
Using x = r cos θ and y = r sin θ
4
r 2 = 8 cosec 2θ ⇒ r 2 =
sinθ cos θ
⇒ r cos θ .r sin θ = 4 ⇒ xy = 4
(in simple form)
Sketch: Curve in 1st quadrant with correct concavity, asymptotic to both axes.
1
π
3
∫
π
1
= 2 ln 3 − ln
=2 ln3orln9
3
Q20 7
M1
A1
[3]
B1B1
[2]
1
π
1
8 cosec 2θ dθ = 2 ln tan θ 31
π
21
6
6
B1
x2 + y 2 =
1
1−
x
⇒ x2 + y 2 − x = 1
M1A1
A1
[3]
M1A1
x2 + y2
⇒ y 2 = 1 + 2 x or equivalent RHS
1
Sketch: Parabola symmetrical about x-axis. Intercepts at − , 0 and ( 0 , ±1) .
2
Shading correct area
A1
[3]
Craftear
Q19 4
B1 B1
B1
[3]
3π
2
Recognises
1
1
dθ as the area of the region between parabola and y-axis.
2 π (1 − cos θ )2
∫
2
0
= 2×
1
2
∫ (1 + 2 x ) dx
−
1
2
0
3
2
1
= 2 (1 + 2 x ) 2 =
3
−1 3
2
M1
M1A1
[4]
335
Q21 5
1 1
1
1
1
1
, π (Accept 1.05 for π .)
sin θ = ⇒ θ = π ; Intersection at
2
2
3
3
2 3
C1: Circle centre at pole and radius 1/√2
C2: Curve, approx. correct orientation, from (0,0) to (1,π).
Completely correct correct shape.
1
1 1 13π
1
× π × − ∫ sin θ dθ
0
6
2 2
2
1
1
1 1π 1
3
π × − 2cos θ 3 = π +
=
−1
12
2
2 0
12
2
M1A1
(4)
Total: 9
α
e8θ + 16e8θ dθ = ∫
0
α
0
17e 4θ dθ = (AEF) (LNR)
M1A1
α
17 4θ
17 4α
17
=
e ⇒
e −
= 2015
4
4
4
0
⇒ e4α = 1955.837 K ⇒ α = 1.89 CAO
dM1A1
dM1A1
(6)
Total
6
Craftear
s=∫
B1
B1
B1
(3)
M1A1
Q22
2
M1A1
(2)
Q23
11O
1
1
]
Closed curve through pole with correct orientation.
Completely correct.
π
π 3
1
2 × a 2 1 (1− 2cosθ + cos2 θ )dθ = a 2 1 − 2
π
π 2
2
2
2
∫
∫
sθ +
1
2
s 2θ dθ
1
π
1
3
= a2 θ − 2 nθ +
n 2θ
2
4
1π
1
2
3
= a2 π + 2
4
ds
dθ
2
= 2×
∫
= a 2 (1 − 2
sθ +
= 2a 2 (1−
1
1
sθ ) = 2a 2 .2 n 2 θ = 4a 2 n 2 θ (AG)
2
2
π
1 2a
π
2
= 4a − 2
= 4 2a
1
n θ dθ
2
π
1
s θ
2 1π
s2 θ + n 2 θ )
1
]
1
1
1
1
2
1
]
4
l
336
Q24
1
3
Closed curve through (a, 0), 2a, π , (a, π), 0, π .
2
2
Completely correct shape for cardioid.
5
a2
Area =
2
a2
=
2
∫
2
π
a2
2
3
1 (1+ sinθ ) dθ =
π
2
3
∫
2
π
3 3 + 2sin θ − 1 cos 2θ dθ
1
π
2
2
3
B1
[2]
M1
(LNR)
M1
(LR)
M1
2
π
3
2
=
∫
2
π
2
3
1 (1+ 2sin θ + sin θ )dθ
π
3
B1
a 3
1
θ − 2 cos θ − sin 2θ
2 2
4
1π
A1
[4]
3
1
1
= a 2 π +1 +
3 (CAO)
8
4
Q25
Use of r2 = x2 + y2
Use of x = r cos θ and y = r sin θ (both).
Obtains r2 = a2 sin 2θ (AG)
B1
B1
B1
[3]
1
3
Sketch with two loops, approximately symmetrical about θ = π and θ = − π .
4
4
1
2
∫
1
π
1
2
2
π
2 a 2 sin 2θ dθ = − a cos 2θ
0
4
0
=
B1B1
[2]
M1
(LNR)
1 2
a
2
A1
(2)
Q26
8
Circle sketched
Cardioid – correct location and orientation – correct indentation near pole.
(a, π2 ) and (a, 32π ) (B1 for reverse, or a = a(1− cosθ ) ⇒ θ = π2 , 32π seen.)
π
2 a 2 1 − cos θ 2 dθ
0
)
∫ (
= πa + a ∫ (1 − 2 cos θ + cos θ ) dθ
Area = 12 πa 2 + 2 × 12
π
2
2
0
π
= 12 πa 2 + a 2 2 32 − 2 cos θ + 12 cos 2θ
0
1
2
2
(Half circle + Area of sector)
2
∫ (
= 12 πa 2 + a 2 [32θ − 2 sin θ + 14 sin 2θ ] 02
= 12 πa 2 + a 2 ( 34 π − 2 ) = ( 54 π − 2 ) a 2
π
) dθ
B1
B1B1
(3)
B1B1
(2)
B1M1
A1
(Use of double angle formula.)
M1
(Integration)
M1
(AG)
A1
(6)
[11]
Craftear
4
Q27
1
337
1 2
r dθ
2
A=
Use of double angle formula
and attempt to integrate.
=
Use of
∫
1
2
∫
2π
0
M1
4(1 + 2 cos θ + cos 2 θ ) dθ
2π
M1
∫ (3 + 4 cos θ + cos 2θ ) dθ
0
Integrates correctly.
sin 2θ
= 3θ + 4sin θ +
2 0
2π
A1
Finds value.
= 6π (CWO) Accept 18.8
A1
[4]
Q28
Uses identity.
1
1
1
4
4
1
4
1
1
2
4
√2
M1
2sin θ cos θ – π = sin 2θ – π + sin π
Uses sine-cosine
link.
Obtains result.
= cos π – 2θ+ π +
Sketches graph.
Closed loop through origin, in correct position.
Obtains line of
symmetry.
For line of symmetry 2θ – π = 0 ⇒ θ = π.
Uses area of sector
formula.
Rearranges.
= cos 2θ – π +
4
3
1
π
3
3
4
8
3
1
(AG)
√2
3
π
4
1
3
3
1
2
2
4
= 0 cos 4θ – π +√2 cos 2θ – π +1 dθ
2
1
=
16
= –
Substitutes limits.
=
3
3
8
sin 4θ – π +
1
16
1
3
4
8
2
1
+ + π –
π+1
16
1
2√2
1
–
3
3
θ 4π
4
2 0
sin 2θ – π +
3
B1B1
3
1
M1
A1
B1
A = 04 cos2 2θ – π +√2 cos 2θ – π + dθ
2
4
4
2
Integrates correctly.
Obtains given
answer.
3
3
M1
A1
Craftear
10
dM1A1
dM1
4
(AG)
A1
[12]
6
N.B Method marks are dependent in final par
N.
If factor missing throughout – award M’s (Max 3)
1
3
3
1
1
If 2 × 08 r2 dθ , penultimate line is = π – –
8
8 2
2
Q29
1 (i)
Uses area formula.
π
π
[ ]
M1
= eπ − e 3 ( = 20.3)
A1
Area =
1 2 2θ
4e dθ = e 2θ π2
2 π6
6
∫
π
Obtains result.
(ii)
Uses arc length
formula.
Arc length =
π
2
π
6
∫
4e 2θ + 4e 2θ dθ = 2
=2
π
2 π2 eθ dθ
M1A1
6
π
Obtains result.
∫
2
[ ]
2 eθ π2 = 2
6
π
π
2 e 2 − e 6
(= 8.83)
A1
3
[5]
Q30
10
338
Differentiates wrt θ
dr
= 2 cos θ − 2 cos 2θ
dθ
Equates to zero
2c − 2 2c 2 − 1 = 0 ⇒ 2c 2 − c − 1 = 0
A1
And solves equation
⇒ (2c +1) (c − 1) = 0
M1
1
⇒ c = − or 1
2
A1
2
3
3, π
3
2
A1
(5)
Approximate shape and location
Accurate scaling.
B1
B1
(2)
States required points
on C
Sketches C.
(
1 2
Uses
r dθ
2
1
Area =
2
Obtains an integrable
form
=
∫
∫
M1
)
∫
π
4 4sin 2 θ 1 − 2cosθ + cos 2 θ
0
(
)dθ
π
1
4 1 − cos 2θ − 4cosθ sin 2 θ +
1 − cos 4θ
0
4
[
M1
M1
] dθ
A1A1
π
Obtains result
=
5
1
2
or 0.0103
π− −
16
2
3
M1
A1
(6)
Craftear
Integrates
5θ sin 2θ
sin 3 θ sin 4θ 4
= −
−4
−
2
3
16 0
4
[13]
Q31
4
Draws sketch of C.
Uses
1 β 2
r dθ
2 α
∫
Uses double angle
formula.
Shows (4,0) and (0,π) lie on C.
Correct shape.
(Full cardioid is B1 unless clear evidence of plotting
up to 2π or –π to π.)
1 π
(4 + 8cos θ + 4 cos 2 θ )dθ
2 0
∫
π
∫ (2 + 4 cosθ + 2 cos θ )dθ
= ∫ (3 + 4 cos θ + cos 2θ )dθ
=
B1
B1
2
M1
2
0
π
0
M1
π
Integrates and obtains
area.
Finds areas.
sin 2θ
= 3θ + 4sinθ +
= 3π
2 0
(A1 for correct integral)
3π
π
π
π
+ 4sin + sin cos = 4.712
5
5
5
5
3π − 4.712 = 4.713
A1A1
CWO
4
M1A1
A1
3
[9]
339
Q32
1
and uses compound
angle formula.
π
r cosθ − = 2
4
cos θ sin θ
= 2
r
+
2
2
Changes to cartesian.
⇒ r cosθ + r sinθ = 2 ⇒ x + y = 2 or y = 2 − x .
A1
Sketches graph.
π
to the initial line.
4
Point (2,0) clearly indicated.
B1
Re-writes equation
Straight line at –
M1
A1
3
B1
2
B1
B1
2
[5]
Q33
Sketches graph.
Uses area of
sector formula.
Uses double
angle formula
Integrates.
Correct shape and orientation.
Passing through (0,0) and (3,0)
1
Area = 2 ×
2
=
=
∫
π
3 (1 + 2 cosθ ) 2 dθ
0
∫
π
3 (1 + 4 cos θ + 4 cos 2 θ )dθ
0
∫
π
3 (3 + 4 cos θ + 2 cos 2θ )
0
= [3θ + 4 sin θ + sin 2θ ]
Obtains result.
M1
5
= π +
3
2
π
3
0
dθ
M1
A1
A1
4
[6]
Craftear
5
340
Q34
EITHE
Uses x2 + y2 = r2,
(x2 + y2)2 = a2(x2 – y2)
x = r cos θ and y = r sin θ.
x2 y2
M1
⇒ r 2 = a 2 2 − 2
r
r
= a2 (cos2 θ – sin2 θ) = a2 cos 2θ (AG)
One mark for each loop,
or half of whole curve.
Uses sector area
formula.
Sketches C.
Area =
π
1 4 2
π a cos 2θ dθ =
2 −4
∫
∫
π
4 a 2 cos 2θ dθ
0
A1
2
B2,1,0
2
M1
π
S. Omission of
S.C.
factor, but correct
integration gets B1.
Differentiates.
Puts y′ = 0 .
Obtains coordinates.
sin 2θ 4
a2
= a2
=
2 0
2
2( x 2 + y 2 )(2 x + 2 yy ′) = a 2 (2 x − 2 yy ′)
y ′ = 0 ⇒ 2 x( x 2 + y 2 ) = a 2 x
a
⇒ 2r 2 = a 2 ⇒ r =
(r ≥ 0)
2
1
⇒ cos 2θ =
2
B1B1
M1
⇒θ = ±
A1A1
i.
Alternatively for last 7
marks:
Obtains condition for
tangent parallel to initial
line.
Differentiates equation
of C.
Forms equation.
Solves for tanθ,
and θ.
Writes coordinates of
points.
A1A1
π
5π
, ±
6
6
A1
M1
7
π
5π
a
a
,± and
,±
6
6
2
2
[14]
dy
dy
dr
=0⇒
= 0 ⇒ r cos θ +
sin θ = 0
dx
dθ
dθ
(M1A1)
a2
dr
dr
= −2a 2 sin 2θ ⇒
= − sin 2θ
r
dθ
dθ
(A1)
2r
3
a2
sin 2θ sin θ = 0
r
∴ a 2 cos 2θ cosθ = a 2 sin 2θ sinθ
1
∴
= tan θ
tan 2θ
1− t2
∴
= t ⇒ 2t 2 = 1 − t 2 ⇒ 3t 2 = 1
2t
1
∴t = ±
3
π
5π
∴θ = ± , ±
6
6
5π
π a
a
,±
,±
6
2 6 , 2
∴ r cos θ −
(Award final A1 if r found but result not written
out.)
Craftear
11
(M1)
(M1)
(A1)
(A1)
(7)
[14]
341
Craftear
VECTORS
342
Q1
6
Let t be a positive constant.
The line l1 passes through the point with position vector ti + j and is parallel to the vector - 2i - j. The
line l2 passes through the point with position vector j + tk and is parallel to the vector - 2j + k .
It is given that the shortest distance between the lines l1 and l2 is 21.
(a) Find the value of t.
[5]
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The plane P 1 contains l1 and is parallel to l2 .
(b) Write down an equation of P 1 , giving your answer in the form r = a + mb + nc .
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343
13
The plane P 2 has Cartesian equation 5x - 6y + 7z = 0 .
(c) Find the acute angle between l2 and P 2 .
[3]
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(d) Find the acute angle between P 1 and P 2 .
[3]
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344
Q2
6
The lines l1 and l2 have equations r =- i - 2j + k + s (2i - 3j) and r = 3i - 2k + t (3i - j + 3k)
respectively.
The plane P 1 contains l1 and the point P with position vector - 2i - 2j + 4k .
(a) Find an equation of P 1 , giving your answer in the form r = a + mb + nc .
[2]
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The plane P 2 contains l2 and is parallel to l1 .
[4]
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(b) Find an equation of P 2 , giving your answer in the form ax + by + cz = d .
345
(c) Find the acute angle between P 1 and P 2 .
[5]
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346
(d) The point Q is such that OQ =- 5 OP .
Find the position vector of the foot of the perpendicular from the point Q to P 2 .
[4]
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347
Q3
5
8
The plane P has equation r = - 2i + 3j + 3k + m (i + k) + n (2i + 3j) .
(a) Find a Cartesian equation of P , giving your answer in the form ax + by + cz = d .
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The line l passes through the point P with position vector 2i - 3j + 5k and is parallel to the vector k.
(b) Find the position vector of the point where l meets P .
[3]
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(c) Find the acute angle between l and P .
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(d) Find the perpendicular distance from P to P .
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349
Q4
7
The points A, B, C have position vectors
2i + 2j, - j + k and 2i + j - 7k
respectively, relative to the origin O.
(a) Find an equation of the plane OAB, giving your answer in the form r.n = p .
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The plane P has equation x - 3y - 2z = 1.
(b) Find the perpendicular distance of P from the origin.
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(c) Find the acute angle between the planes OAB and P .
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(d) Find an equation for the common perpendicular to the lines OC and AB.
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Q5
4
The points A, B, C have position vectors
- i + j + 2k ,
- 2i - j,
2i + 2k ,
respectively, relative to the origin O.
(a) Find the equation of the plane ABC, giving your answer in the form ax + by + cz = d.
[5]
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(b) Find the perpendicular distance from O to the plane ABC.
[2]
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(c) Find the acute angle between the planes OAB and ABC.
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354
Q6
7
The points A, B, C have position vectors
- 2i + 2j - k ,
- 2i + j + 2k ,
- 2j + k ,
respectively, relative to the origin O.
(a) Find the equation of the plane ABC, giving your answer in the form ax + by + cz = d.
[5]
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(b) Find the acute angle between the planes OBC and ABC.
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The point D has position vector ti - j.
(c) Given that the shortest distance between the lines AB and CD is 10, find the value of t.
[6]
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356
Q7
5
The lines l1 and l2 have equations r = 3i + 3k + m (i + 4 j + 4k) and r = 3i - 5j - 6k + n (5j + 6k)
respectively.
(a) Find the shortest distance between l1 and l2 .
[5]
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11
The plane P contains l1 and is parallel to the vector i + k .
(b) Find the equation of P, giving your answer in the form ax + by + cz = d .
[4]
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(c) Find the acute angle between l2 and P.
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Q8
7
The lines l1 and l2 have equations r = - 5j + m (5i + 2k) and r = 4i + 2j - 2k + n (j + k) respectively.
The plane P contains l1 and is parallel to l2 .
(a) Find the equation of P, giving your answer in the form ax + by + cz = d .
[4]
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(b) Find the distance between l2 and P.
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15
The point P on l1 and the point Q on l2 are such that PQ is perpendicular to both l1 and l2 .
22
(c) Show that P has position vector 55
27 i - 5j + 27 k and state a vector equation for PQ.
[8]
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360
Q9
6
With O as the origin, the points A, B, C have position vectors
i − j,
2i + j + 7k,
i−j+k
respectively.
(i) Find the shortest distance between the lines OC and AB.
[5]
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(ii) Find the cartesian equation of the plane containing the line OC and the common perpendicular
of the lines OC and AB.
[4]
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Q10
3
The lines l1 and l2 have equations r = 6i + 2j + 7k + , i + j and r = 4i + 4j + - −6j + k respectively.
The point P on l1 and the point Q on l2 are such that PQ is perpendicular to both l1 and l2 . Find the
position vectors of P and Q.
[8]
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364
Q11
7
The line l1 passes through the points A −3, 1, 4 and B −1, 5, 9. The line l2 passes through the
points C −2, 6, 5 and D −1, 7, 5.
(i) Find the shortest distance between the lines l1 and l2 .
[5]
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(ii) Find the acute angle between the line l2 and the plane containing A, B and D.
[5]
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366
Q12
10
The line l1 is parallel to the vector ai − j + k, where a is a constant, and passes through the point
whose position vector is 9j + 2k. The line l2 is parallel to the vector −ai + 2j + 4k and passes through
the point whose position vector is −6i − 5j + 10k.
(i) It is given that l1 and l2 intersect.
6.
(a) Show that a = − 13
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(b) Find a cartesian equation of the plane containing l1 and l2 .
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(ii) Given instead that the perpendicular distance between l1 and l2 is 3 30, find the value of a.
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Q13
7
The lines l1 and l2 have vector equations
r = ai + 9j + 13k + , i + 2j + 3k
and
r = −3i + 7j − 2k + - −i + 2j − 3k
respectively. It is given that l1 and l2 intersect.
(i) Find the value of the constant a.
[3]
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The point P has position vector 3i + j + 6k.
(ii) Find the perpendicular distance from P to the plane containing l1 and l2 .
[4]
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(iii) Find the perpendicular distance from P to l2 .
371
Q14
8
The plane 1 has equation
` a
` a
` a
5
−4
0
r= 1 +s
1 +t 1 .
0
3
2
(i) Find a cartesian equation of 1 .
[3]
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The plane 2 has equation 3x + y − z = 3.
(ii) Find the acute angle between 1 and 2 , giving your answer in degrees.
[2]
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(iii) Find an equation of the line of intersection of 1 and 2 , giving your answer in the form r = a + ,b.
[5]
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Q15
10
18
The position vectors of the points A, B, C, D are
i + j + 3k,
3i + 4j + 5k,
−i + 3k,
mj + 4k,
respectively, where m is a constant.
(i) Show that the lines AB and CD are parallel when m = 32 .
[1]
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(ii) Given that m ≠ 32 , find the shortest distance between the lines AB and CD.
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(iii) When m = 2, find the acute angle between the planes ABC and ABD, giving your answer in
degrees.
[6]
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Q16
OR
The position vectors of the points A, B, C, D are
i + j + 3k,
3i − j + 5k,
3i − j + k,
5i − 5j + !k,
respectively, where ! is a positive integer. It is given that the shortest distance between the line AB
and the line CD is equal to 2ï2.
(i) Show that the possible values of ! are 3 and 5.
[7]
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(ii) Using ! = 3, find the shortest distance of the point D from the line AC, giving your answer correct
to 3 significant figures.
[3]
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(iii) Using ! = 3, find the acute angle between the planes ABC and ABD, giving your answer in
degrees.
[4]
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Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable
effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will
be pleased to make amends at the earliest possible opportunity.
To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge International
Examinations Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cie.org.uk after
the live examination series.
Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local
Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.
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Q17
9
The plane 1 passes through the points 1, 2, 1 and 5, −2, 9 and is parallel to the vector i + 2j + 3k.
(i) Find the cartesian equation of 1 .
[4]
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The plane 2 contains the lines
r = 2i − 3j + k + , i − 2j − k
and
r = 2i − 3j + k + - 2i + 3j − k.
(ii) Find the cartesian equation of 2 .
[4]
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© UCLES 2017
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(iii) Find the acute angle between 1 and 2 .
[3]
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© UCLES 2017
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Craftear
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380
Q18
6
The points A, B and C have position vectors 2i − j + k, 3i + 4j − k and −i + 2j + 4k respectively.
(i) Find the area of the triangle ABC.
[4]
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381
(ii) Find the perpendicular distance of the point A from the line BC.
[3]
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(iii) Find the cartesian equation of the plane through A, B and C.
[2]
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© UCLES 2017
9231/11/O/N/17
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Craftear
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Q19
Find a cartesian equation of the plane 1 passing through the points with coordinates 2, −1, 3,
4, 2, −5 and −1, 3, −2.
[4]
The plane 2 has cartesian equation 3x − y + 2z = 5. Find the acute angle between 1 and 2 .
[3]
Find a vector equation of the line of intersection of the planes 1 and 2 .
[4]
Craftear
8
382
© UCLES 2016
9231/11/M/J/16
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383
Write down matrices P and D such that P−1AP = D, where D is a diagonal matrix, and find P−1 . [5]
Write down a matrix C such that C2 = D, and deduce a matrix B such that B2 = A.
[4]
OR
Q20
11:
The position vectors of the points A, B, C, D are
a = 2i + ,j − 3k,
b = 6i + 3j − 2k,
c = i + 2j − k,
d = i + 7j + 4k
respectively. It is given that the shortest distance between the lines AB and CD is 3.
(i) Show that ,2 + , − 20 = 0.
[7]
Craftear
(ii) The planes p1 and p2 are the planes through A, B and D corresponding to the two values of ,
satisfying the equation in part (i). Find the acute angle between p1 and p2 .
[7]
© UCLES 2016
9231/13/M/J/16
[Turn over
Q21
384
11: EITHER
The lines l1 and l2 have equations
r = 6i − 3j + s 3i − 4j − 2k
and
r = 2i − j − 4k + t i − 3j − k
respectively. The point P on l1 and the point Q on l2 are such that PQ is perpendicular to both l1 and
[7]
l2 . Show that the position vector of P is 3i + j + 2k and find the position vector of Q.
Find, in the form r = a + ,b + -c, an equation of the plane which passes through P and is
perpendicular to l1 .
[3]
The plane meets the plane r = p i + q j in the line l3 . Find a vector equation of l3 .
[4]
OR
Craftear
A curve C has parametric equations
© UCLES 2016
9231/11/O/N/16
[Turn over
385
Q22
11: The lines l1 and l2 have equations r = 8i + 2j + 3k + , i − 2j and r = 5i + 3j − 14k + - 2j − 3k
respectively. The point P on l1 and the point Q on l2 are such that PQ is perpendicular to both l1 and
l2 . Find the position vector of the point P and the position vector of the point Q.
[8]
The points with position vectors 8i + 2j + 3k and 5i + 3j − 14k are denoted by A and B respectively.
Find
−−→ −−→
(i) AP × AQ and hence the area of the triangle APQ,
Craftear
(ii) the volume of the tetrahedron APQB. (You are given that the volume of a tetrahedron is
1 × area of base × perpendicular height.)
3
[6]
© UCLES 2015
9231/11/M/J/15
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Q23
8
386
A line, passing through the point A 3, 0, 2, has vector equation r = 3i + 2k + , 2i + j − 2k. It meets
the plane , which has equation r. i + 2j + k = 3, at the point P. Find the coordinates of P.
[3]
Write down a vector n which is perpendicular to , and calculate the vector w, where
w = n × 2i + j − 2k.
3
Craftear
The point Q lies in and is the foot of the perpendicular from A to . Use the vector w to determine
[4]
an equation of the line PQ in the form r = u + -v.
© UCLES 2015
9231/13/M/J/15
[Turn over
Q24
387
11 EITHER
The points A, B and C have position vectors i, 2j and 4k respectively, relative to an origin O. The
point N is the foot of the perpendicular from O to the plane ABC. The point P on the line-segment
ON is such that OP = 34 ON . The line AP meets the plane OBC at Q. Find a vector perpendicular to
4
.
[4]
the plane ABC and show that the length of ON is 21
[5]
Show that the acute angle between the planes ABC and ABQ is cos−1 23 .
[5]
Craftear
Find the position vector of the point Q.
© UCLES 2015
9231/11/O/N/15
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Q25
The line l1 passes through the points A 2, 3, −5 and B 8, 7, −13. The line l2 passes through the
points C −2, 1, 8 and D 3, −1, 4. Find the shortest distance between the lines l1 and l2 .
[5]
The plane 1 passes through the points A, B and D. The plane 2 passes though the points A, C
and D. Find the acute angle between 1 and 2 , giving your answer in degrees.
[6]
Craftear
11
388
© UCLES 2014
9231/11/M/J/14
[Turn over
Q26
389
11: OR
With respect to an origin O, the point A has position vector 4i − 2j + 2k and the plane 1 has equation
r = 4 + , + 3-i + −2 + 7, + -j + 2 + , − -k,
−−→
−−→
where , and - are real. The point L is such that OL = 3OA and 2 is the plane through L which is
−−→
−−→
parallel to 1 . The point M is such that AM = 3ML.
(i) Show that A is in 1 .
[1]
(ii) Find a vector perpendicular to 2 .
[2]
(iii) Find the position vector of the point N in 2 such that ON is perpendicular to 2 .
[5]
Craftear
(iv) Show that the position vector of M is 10i − 5j + 5k and find the perpendicular distance of M
from the line through O and N , giving your answer correct to 3 significant figures.
[6]
© UCLES 2014
9231/13/M/J/14
[Turn over
Q27
10
390
The line l1 is parallel to the vector i − 2j − 3k and passes through the point A, whose position vector
is 3i + 3j − 4k. The line l2 is parallel to the vector −2i + j + 3k and passes through the point B,
whose position vector is −3i − j + 2k. The point P on l1 and the point Q on l2 are such that PQ is
perpendicular to both l1 and l2 . Find
(i) the length PQ,
(ii) the cartesian equation of the plane
containing PQ and l2 ,
.
[4]
[3]
Craftear
(iii) the perpendicular distance of A from
[5]
© UCLES 2014
9231/11/O/N/14
[Turn over
Hence find the exact value of S.
Q28
391
[2]
11: OR
The points A, B, C and D have coordinates as follows:
A 2, 1, −2,
B 4, 1, −1,
C 3, −2, −1
and
D 3, 6, 2.
The plane 1 passes through the points A, B and C. Find a cartesian equation of 1 .
[4]
Find the area of triangle ABC and hence, or otherwise, find the volume of the tetrahedron ABCD.
[6]
[The volume of a tetrahedron is 31 × area of base × perpendicular height.]
[4]
Craftear
The plane 2 passes through the points A, B and D. Find the acute angle between 1 and 2 .
© UCLES 2013
9231/11/M/J/13
[Turn over
Q29
392
11 EITHER
The line l1 passes through the point A whose position vector is 4i + 7j − k and is parallel to the vector
3i + 2j − k. The line l2 passes through the point B whose position vector is i + 7j + 11k and is parallel
to the vector i − 6j − 2k. The points P on l1 and Q on l2 are such that PQ is perpendicular to both l1
and l2 . Find the position vectors of P and Q.
[8]
Craftear
Find the shortest distance between the line through A and B and the line through P and Q, giving your
answer correct to 3 significant figures.
[6]
© UCLES 2013
9231/13/M/J/13
[Turn over
Q30
8
393
a
` a
` a
2
1
1
The plane 1 has equation r = 3 + s 0 + t −1 . Find a cartesian equation of 1 .
−1
1
−2
`
The plane 2 has equation 2x − y + Ï = 10. Find the acute angle between 1 and 2 .
[3]
[2]
Find an equation of the line of intersection of 1 and 2 , giving your answer in the form r = a + ,b.
[5]
The curve C has parametric equations
Craftear
9
© UCLES 2013
9231/11/O/N/13
[Turn over
Sketch C.
[3]
Q31
8
394
The points A, B, C have position vectors
4i + 5j + 6k,
5i + 7j + 8k,
2i + 6j + 4k,
respectively, relative to the origin O. Find a cartesian equation of the plane ABC.
[4]
The point D has position vector 6i + 3j + 6k. Find the coordinates of E, the point of intersection of
the line OD with the plane ABC.
[4]
[3]
Craftear
Find the acute angle between the line ED and the plane ABC.
© UCLES 2013
9231/13/O/N/13
[Turn over
Q32
395
11 OR
The position vectors of the points A, B, C, D are
2i + 4j − 3k,
−2i + 5j − 4k,
i + 4j + k,
i + 5j + mk,
respectively, where m is an integer. It is given that the shortest distance between the line through
A and B and the line through C and D is 3. Show that the only possible value of m is 2.
[7]
[3]
1
Show that the acute angle between the planes ACD and BCD is cos−1 √ .
3
[4]
Craftear
Find the shortest distance of D from the line through A and C.
© UCLES 2012
9231/11/M/J/12
[Turn over
Q33
9
396
The plane Π has equation
r = 2i + 3j − k + λ (i − 2j + 2k) + µ (3i + j − 2k).
The line l, which does not lie in Π , has equation
r = 3i + 6j + 12k + t(8i + 5j − 8k).
Show that l is parallel to Π .
[4]
Find the position vector of the point at which the line with equation r = 5i − 4j + 7k + s(2i − j + k)
meets Π .
[4]
[4]
Craftear
Find the perpendicular distance from the point with position vector 9i + 11j + 2k to l.
© UCLES 2012
9231/12/O/N/12
[Turn over
Q34
4
397
The points A, B and C have position vectors i + 2j + 2k, 2i + 4j + 5k and 2i + 3j + 4k respectively.
−−→ −−→
Find AB × AC.
[3]
Deduce, in either order, the exact value of
(i) the area of the triangle ABC,
(ii) the perpendicular distance from C to AB.
Craftear
[3]
© UCLES 2012
9231/13/O/N/12
[Turn over
398
Q35
9
The plane Π1 has parametric equation
r = 2i − 3j + k + λ (i − 2j − k) + µ (i + 2j − 2k).
[4]
The plane Π2 has cartesian equation 3x − 2y − 3ß = 4. Find the acute angle between Π1 and Π2 .
[3]
Find a vector equation of the line of intersection of Π1 and Π2 .
[4]
Craftear
Find a cartesian equation of Π1 .
© UCLES 2012
9231/13/M/J/12
[Turn over
VECTORS
MS
Craftear
399
6(d)
0 5
19
−2 −6 = 5 110 cos α leading to cos α = 5 110
1 7
6(c)
76.8°
−1 5
11
2 −6 = 21 110 cos α leading to cos α = 21 110
47
Acute angle between l2 and Π 2 is 90 − α = 54.1°
r = 21
i + j + λ ( 2i + j) + μ ( −2 j + k )
5
−1 −1
t
0 2 = 21 t = 21
= 4.2
5
21
14
i
j k −1
−2 −1 0 = 2
0 −2 1 4
0 t −1
1 − 1 = t 0
t 0 1
Answer
6(b)
6(a)
Question
Q1
Craftear
Guidance
3
A1
M1 Dot product using their normal to Π1 .
A1 FT
3
A1
M1 A1 Uses dot product of −2j + k and normal.
1
B1 FT Using their value of t.
5
M1 A1 Uses formula for shortest distance.
M1 A1 Finds common perpendicular.
B1 Finds vector from any point on l1 to any point on l2.
Marks
400
6(c)
6(b)
6(a)
Question
Q2
5
A1 0.845 rad
DM1 Dot product using their normal vectors.
A1 FT
3 9
32
2 6 = 14 166 cos α leading to cos α = 14 166
1 −7
48.4°
M1 A1 Finds vector perpendicular to Π1 .
i j k 9 3
2 −3 0 = 6 ~ 2
1 0 −3 3 1
4
A1
9 x + 6 y − 7 z = 41
M1 A1 OE. Finds vector perpendicular to Π 2 .
2
B1 FT OE. FT their i – 3k
M1 Uses point on Π 2 .
Craftear
Guidance
B1 OE. Finds direction vector from P to a point of l1.
Marks
−9(3) − 6(0) + 7( −2) = −41
i j k −9
2 −3 0 = −6
3 −1 3 7
r = −i − 2 j + k + λ ( 2i − 3 j ) + μ ( i − 3k )
r = −2i − 2 j + 4k + λ ( 2i − 3j) + μ ( i − 3k )
− ( −2i − 2 j + 4k ) + ( −i − 2 j + k ) = i − 3k
Answer
401
5(a)
Question
Q3
6(d)
4
A1
−3 x + 2 y + 3z = 21
Craftear
Guidance
M1 A1 Finds vector perpendicular to Π.
Marks
M1 Substitutes point on Π.
Answer
4
A1
DM1 Substitutes into the equation of Π 2 .
M1 A1 FT Scales OP by a factor of ±5 and uses multiple of
their normal to Π 2 .
−3(−2) + 2(3) + 3(3) = 21
i j k −3
1 0 1 = 2
2 3 0 3
−7
2
t = − 32 leading to OF = 1
− 19
2
9 (10 + 9t ) + 6 (10 + 6t ) − 7 ( −20 − 7t ) = 41 leading to 290 + 166t = 41
−2 9 10 + 9t
OF = OQ + QF = −5 −2 + t 6 = 10 + 6t
4 −7 −20 − 7t
402
5(d)
5(c)
5(b)
Craftear
A1
2
−3
11
−2 −3
1 18
3 2 =
= 3.84
22
22
2 3
0 2 −2
0 − −3 = 3
7 5 2
Acute angle between l and Π is 90 − α = 39.8°
0 −3
3
0 2 = 1 22 cos α leading to cos α = 22
1 3
M1 Substitutes into the equation for Π.
−3(2) + 2(−3) + 3(5 + t ) = 21 leading to t = 6
3
M1 A1 Uses dot product of their direction and normal vectors.
B1 Finds direction vector from P to plane.
3
A1
M1 Uses dot product of k and their normal.
A1 FT
3
B1 Forms general point on line (given as a single vector).
2
−3
5 + t
403
7(c)
7(b)
7(a)
Question
Q4
=
14
1
22.2°
1 1
6
−1 −3 = 3 14 cos α leading to cos α = 3 14
−1 − 2
12 + 32 + 22
1
1
r. −1 = 0
−1
i j k 2 1
n = 2 2 0 = −2 ~ −1
0 −1 1 −2 −1
Answer
Craftear
Guidance
3
A1 Accept 0.388 radians. Mark final answer.
M1 Takes dot product of normal vectors.
A1 FT
1
B1 Divides by magnitude of the normal to Π
0.267
3
A1
M1 A1 Finds common perpendicular.
Marks
404
7(d)
−0.2
5
r = −0.1 + k −3
1
0.7
2
1
1
λ = − leading to OP = − 1
10
10
−7
2 − 2 μ − 2λ 2
2 − 3μ − λ 1 = 0 14 μ + 54λ = 6
μ + 7 λ −7
Craftear
10
B1 FT FT using their common perpendicular.
M1 A1 Solves for λ and substitutes into OP .
A1 Deduces second equation.
A1 Deduces one equation.
M1 Uses that dot product of PQ with line direction is zero,
or, alternatively, PQ is a multiple of the common
perpendicular (parameter k not 1).
M1 A1 Finds PQ .
2
2 − 2μ
2 − 2 μ − 2λ
leading to
OP = λ 1 , OQ = 2 − 3μ
PQ = 2 − 3μ − λ
−7
μ
μ + 7λ
2 − 2 μ − 2 λ −2
2 − 2 μ − 2λ
5
2 − 3μ − λ −3 = 0 or 2 − 3μ − λ = k −3
1
μ + 7λ 1
μ + 7λ
M1 A1 Finds direction of common perpendicular.
i
j k −20 5
2 1 −7 = 12 ~ −3
−2 −3 1 −4 1
405
4(c)
4(b)
4(a)
Question
Q5
2
=
89
10
Craftear
4
A1
36.2
M1 A1 Finds normal to the plane OAB.
2
M1 A1 Divides by magnitude of normal vector.
1.06…
M1 Uses dot product correctly.
OE
5
A1 A1 A1 A1 OE
M1
−2 2
−6 −4 = 89 29 cos θ
7 3
i
j k 2
−1 1 2 = −4
−2 −1 0 3
22 + 62 +
10
−2 x − 6 y + 7 z = 10
Setting up 3 equations using points given.
Alternative method for question 4(a)
M1 A1 Substitutes point.
−2(−1) − 6(1) + 7(2) = 10 −2 x − 6 y + 7z = 10
B1 Finds direction vectors of two lines in the plane.
BC = 4i + j + 2k
Guidance
M1 A1 Finds normal to the plane ABC.
AC = 3i − j
Marks
i j k −2
1 2 2 = −6
3 −1 0 7
AB = −i − 2 j − 2k
Answer
406
Q6
7(c)
7(b)
7(a)
Question
t = 103
4 + 10t 2
−4 −10t
(
= 10 (4 + 10t) 2 = 10 4 + 10t 2
2 −2
−4 −10t
−4 3t =
2
4 +10t 2 t
4 +10t 2
1
)
Craftear
4
Finds common perpendicular.
6
A1
M1 Sets equal to 10 and solves for t.
M1 A1 Uses formula for perpendicular distance.
M1 A1
A1 0.49(0) rad
28.1°
i j k −2
0 −1 3 = 3t
t 1 −1 t
M1 Uses dot product of normal vectors.
M1 A1 Finds normal to the plane OBC.
5 5
3 2 = 35 45 cosθ
1 4
i
j k 5
−2 1 2 = 2
0 −2 1 4
5
M1 A1 Substitutes point. OE
5(0) + 3(−2) +1(1) = −5 5 x + 3 y + z = −5
B1 Finds
F
direction vectors of two lines in the plane.
BC = 2i − 3j − k
Guidance
M1 A1 Finds normal to the plane ABC. OE
AC = 2i − 4 j + 2k
Marks
i j k 5
0 −1 3 = 3
1 −2 1 1
AB = −j + 3k
Answer
407
Q7
5(c)
5(b)
5(a)
Question
A1
Acute angle between l2 and Π is α − 90 = 10.4°
3
M1 A1FT
0 4
−9
5 3 = 61 41cos α cos α = 61 41
6 −4
(A1 FT their normal)
4
M1 A1
4(3) + 3(0) − 4(3) = 0 4 x + 3 y − 4z = 0
5
M1 A1
M1 A1
Craftear
B1
M1 A1
Marks
i j k 4
1 4 4 = 3
1 0 1 −4
0 4
1
15
−5 −6 =
= 1.71
77
77
−9 5
i j k 4
1 4 4 = −6
0 5 6 5
3 3 0
−5 − 0 = −5
−6 3 −9
Answer
408
Q8
7(b)
7(a)
Question
−4 −2
1
53
−7 −5 =
= 7.21
√ 54 √ 54
2 5
0 4 −4
−5 − 2 = −7
0 −2 2
Craftear
A1
−2 x − 5 y + 5z = 25
3
M1 A1
M1
4
M1
M1 A1
Marks
−5(−5) = 25
i j k −2
5 0 2 = −5
0 1 1 5
Answer
409
7(c)
Craftear
B1 FT
55
−2
1
r = −135 + k −5
27
22
5
(B1 FT their common perpendicular)
8
M1 A1
A1
55
1
55
22
11
i − 5j + k
λ = OP = −135 =
27
27
27
27
22
4 − 5λ 0
7 + μ 1 = 0 −2λ + 2μ = −5
−2 + μ − 2λ 1
M1
4 − 5λ 5
4 − 5λ
−2
7 + μ 0 = 0 or 7 + μ = k −5
−2 + μ − 2λ 2
−2 + μ − 2λ
5
A1
M1 A1
5λ
4
4 − 5λ
OP = −5 , OQ = 2 + μ PQ = 7 + μ
2λ
−2 + μ
−2 + μ − 2λ
410
Q9
6(ii)
6(i)
Question
Craftear
4
A1 AEF
−x + 4 y + 5z = 0
M1 A1 Finds normal to plane.
M1 Uses point on plane.
2
k
−1
1 =t 4
−1 5
− (0) + 4 ( 0) + 5 ( 0) = 0
3
i j
n = 1 −1
5
M1 A1 Uses formula for shortest distance.
1 3
−1 ⋅ 2
0 −1
= 1 = 0.267
14
32 + 22 +12
Guidance
M1 A1 Finds direction of common perpendicular.
B1
Marks
i j k −9 3
JJJG JJJG
OC × AB = 1 −1 1 = −6 = t 2
1 2 7 3 −1
1
JJJG
AB = 2
7
Answer
411
Q10
3
Question
3
4
JJJG JJJG
OP = −1 , OQ = −2
7
1
λ = −3, µ = 1
8
JJJG
JJJG
A1 States OP and OQ .
M1 A1 Solves simultaneous equations.
A1 Deduces second equation. CWO.
−2 − λ 0
2 − λ − 6µ −6 = 0 ⇒ 6λ + 37 µ = 19
−7 + µ 1
˙
A1 Deduces one equation. CWO.
Craftear
Guidance
JJJG
M1 Uses that dot product of PQ with line directions is 0 .
JJJG
Or, alternatively, PQ is multiple of common perpendicular.
JJJG
M1 A1 Finds PQ .
Marks
−2λ − 6µ = 0
−2 − λ 1
2 − λ − 6 µ 1 = 0
−7 + µ 0
i j k
−2 − λ
1
Or 2 − λ − 6 µ = k 1 1 0 = k −1
−7 + µ
−6
0 −6 1
˙
6 + λ
4
−2 − λ
JJJG
JJJG
JJJG
OP = 2 + λ , OQ = 4 − 6 µ ⇒ PQ = 2 − λ − 6 µ
µ
7
−7 + µ
Answer
412
Q11
7(ii)
7(i)
Question
=
54
18
2
2
2
13 + 4 + 2
2
378
9
=−
( or 0.481rad )
=−
= 6 = 2.45
42
cos −1 −
− 90 = 27.6°
14
1 +1 + 0
2
1 −13
1 ⋅ 4
0 2
j k
2
−26 −13
2 4 5 = 8 ~ 4
2 6 1 4 2
i
2
5 +5 +2
2
1 −5
5 ⋅ 5
1 −2
i j k −5
2 4 5 = 5
1 1 0 −2
2
1
JJJG JJJG
AB = 4 , CD = 1
5
0
Answer
42
14
Craftear
Guidance
5
A1
M1 A1 FT Uses dot product of their normal with direction of line.
M1 A1 Finds normal to the plane.
5
M1 A1 Uses formula for shortest distance.
M1 A1 Finds direction of common perpendicular. Allow any nonzero scalar multiple.
B1 Find the directions of the lines.
Marks
413
Q12
10(i)(b)
10(i)(a)
Question
6
13
Craftear
A1 AEF
Equation of plane: −13 x + 5 y − z = 43
4
M1 Substitutes a point
M1 A1 Uses cross product to find normal to the plane
AEF
3
A1 AG
Using point on plane:
e.g. 5 ( 9 ) − 2 = 43
i
j k
6
30
6
−1 1 = −6i + j − k ~ −13i + 5 j − k
−
13
13 13
6
2 4
13
Normal to the plane:
⇒ µ = 1, λ =12, a = −
M1 Equates coordinates of points
λ a −6 − µ a
Point of intersection: 9 − λ = −5 + 2 µ
2 + λ 10 + 4 µ
Guidance
B1
Marks
λa
−6 − µ a
Point on l1 is 9 − λ and on l2 is −5 + 2µ
2+ λ
10 + 4µ
Answer
414
Q13
7(i)
Question
10(ii)
Answer
Craftear
A1
Use third equation to obtain a = 2 .
M1
Marks
5
A1
9 + 2λ = 7 + 2 µ
13 + 3λ = −2 − 3µ
⇒a =3
to obtain λ = −3 and µ = −2 .
Solve two equations
2
⇒ 26 ( a − 3) = 0
2
Guidance
A1
)
⇒ 15 18 + 13a 2 = ( 6 + 13a )
(
M1 Puts distance equal to 3 30
M1 A1 Takes dot product of the correct vectors
B1 Finds the magnitude of the cross product of the
direction vectors of the lines
3 30 36 + 26a 2 = 36 + 78a
−6 −6
−5 − 9 ⋅ −5a = 36 + 78a
10 − 2 a
−6
2
−5a = 36 + 26a
a
i
j k
a −1 1 = −6i − 5aj + ak
−a 2 4
415
Q14
8(i)
Question
7(iii)
7(ii)
M1 A1
−6 −12
1
Perpendicular distance =
6 ⋅ 0 = 10 = 3.16
160
−8 4
−x + 8 y − 4z = 3
i j k
−4 1 3 = −i + 8 j − 4k
0 1 2
−i + 2 j − 3k
2i +10 j + 6k
Answer
= 10 = 3.16
−6 −1 −2
6 × 2 = −10
−8 −3 −6
Craftear
Guidance
3
M1 A1 Uses point on plane to find cartesian equation, AEF.
M1 Finds normal to Π1 .
Marks
11
M1 A1 Find length PN M1, A1
M1, A1
Perpendicular distance from P to l2 is
= 10
M1 A1 Alt method: Find N (foot of perpendicular) in terms of
parameter and uses scalar product with n to find
3
parameter M1, A1 so PN = 0
−1
9 +1
3 × 3 + 0 ×1− 6 + 7
Cross product of direction of P to l2 with direction of l2
Use formula
Alt method: Finds equation of plane M1, A1: Finds foot of perpendicular from
P to plane M1 Hence length A1
Alt method: Find equation of plane 3x – z + 7 = 0 M1, A1
M1 A1
Normal to the plane is n = −12i + 4k
416
Q15
10(i)
Qu
uestion
8(iii)
8(ii)
2
1
JJJG JJJG 1 JJJG
AB = 3 , CD = 1.5 = AB
2
1 2
Craftear
5
Guiidance
1
B1 Or shows if parallel,, then m=3/2
Marks
A1 States vector equation of line.
M1 A1 Finds point common to both planes.
Point on both planes is, e.g. (1,1,1) .
r = i + j + k + λ ( 4i + 13j + 25k )
M1 A1 Finds direction of line of intersection.
i j k
3 1 −1 = 4i +13j + 25k
−1 8 −4
Answer
A1 Accept 1.26 rad.
⇒ θ = 72.5°
2
M1 Uses scalar product.
−1 3
8 ⋅ 1 = 9 11cos θ
−4 −1
417
10(iii)
10(ii)
2
2
= 2
=
( 3 − 2m ) 2 + ( 2m − 3 ) 2
−2 ( 3 − 2m ) + 0 + 0
o
o.
19
=
2
2
2
378
1 +4 +5
1 + 8 +10
12 + 22 + 22
⇒ θ = 12.2°
cosθ =
i j k 1
2 3 2 = −4 o.
o
−1 1 1 5
i
j k 2 1
2 3 2 = −4 ~ −2 o.
o
−2 −1 0 4 2
=
n
AC.n
i j k 3 − 2m
1
n = 2 3 2 = 0 ( so parallel to 0 )
−1
1 m 1 2m − 3
Craftear
−1
−4
2
1
−2
JJJG JJJG
JJJG
AB = 3 , CD = m and AC = −1 or AD = m −1 or BC= −4
−2
2
1
0
1
6
A1 CAO.
M1A1FT Uses formula for angle between two lines.
A1 Finds normal to ABD (AEF).
M1A1 Finds normal to plane ABC (AEF).
5
A1
M1 Uses formula for shortest distance.
M1A1 Finds common perpendicular using cross
product.
B1
418
Q16
12O(ii)
12O(i)
Question
2
uuur
AB = −2
2
2
uuur
CD = − 4
α −1
2
2
uuur
or CD = −4
2
i j k
1
Distance of D from AC =
4 −6 0 =
1 +1 +1
1 −1 −1
4
uuur
AD = −6
0
(AG)
Craftear
56
= 4.32 (or with CD)
3
= 2 2 ⇒ 2 2 2α 2 − 16α + 38 = 8
⇒ (α − 3)(α − 5 ) = 0 ⇒ α = 3 or 5
2
(α − 5 ) + ( α − 3 ) + 4
2 α − 5
−2 α − 3
−2 2
i j
k
5 − α α − 5
uuur uuur
AB × CD = 1 −1
1 = 3 − α ~ α − 3
2 −4 α −1 −2 2
2
uuur
AC = −2
−2
Answer
Total:
Total:
,
instead of AC
Guidance
3
M1A1 Use DP.AC = 0 to find
Find length
A1
(= -5/3)
B1 Alt method:
Let P be point on AC with parameter
7
A1
A1
M1
M1
M1A1 Substitutes their vectors into correct formula
M1A1
B1 May find
Marks
419
Q17
9(ii)
9(i)
Question
12O(iii)
k
Craftear
Total:
4
M1A1
5 × 2 − ( −3) + 7 × 1 = 20 ⇒ 5 x − y + 7 z = 20
Total:
M1A1
i j k 5
1 −2 −1 = −1
2 3 −1 7
4
A1
M1A1
Marks
7 × 1 + 2 − 3 × 1 = 6 ⇒ 7 x + y − 3z = 6
4
M1
Total:
M1A1
B1B1
Cartesian equation of Π1 is 7 x + y − 3z = const
i j k −7
1 −1 2 = −1
1 2 3 3
Answer
i j k 3
ABD: n 2 = 1 −1 1 = 2
2 −3 0 −1
cos θ = −32− 2+0
⇒ θ = 19.1° or 0.333 rads
14
j
−1
ABC: n1 = 1 −1 − = −1
1 −1 1 0
i
Guidance
420
Q18
6(i)
Question
9(iii)
Craftear
4
M1A1)
Find area using Area = ½ ab sin C or equivalent
A1
M1A1 OE
B1 2 correct required
(M1A1
1
3
212 + 32 + 182 = 13.9
86
2
2
JJJG
AC = −3i + 3 j + 3k
Marks
Alt method: Use scalar product to find angle
Area of triangle ABC =
JJJG JJJG
AB × BC = 21i + 3j + 18k (*)
JJJG
BC = −4i − 2j + 5k
Answer
13
⇒ θ = 78.7° or 1.37 rad.
59 75
JJJG
AB = i + 5 j − 2k
⇒ cos θ =
cos θ =
75
1 . −1
−3 7
49 + 1 + 9 25 +1 + 49
Guidance
Total:
3
A1
M1M1
421
6(iii)
6(ii)
Through ( 2, –1 , 1) Hence 7 +
+ 6 = const.
+ 6 = 19
From (*) Cartesian equation is 7 +
2
A1
M1
3
(M1A1)
A1 Area triangle = sin C × |AC|
M1A1 Alt method: Find angle at C
Craftear
Alt method: Use equation of BC to find D (foot of perpendicular)
in terms of parameter and scalar product to find parameter ,
λ= 8/15. Find length
1
= 4.15
430
5
JJJG JJJG
AB × BC
212 + 32 + 182
d=
=
JJJG
BC
4 2 + 2 2 + 52
422
423
i j k 17 1
AB × AC = 2 3 −8 = 34 ~ 2
−3 4 −5 17 1
So x + 2 y + z = const ⇒ const = 2 − 2 + 3 = 3 ( using a point ) ⇒
x + 2y + z = 3
1 3
3
3
9 + 1 + 4 1 + 4 +1 cosθ = 2 . −1 ⇒ cos θ =
=
14 6
84
1 2
⇒θ = 70.9° or 1.24 radians
i j k 5
Direction of line of intersection is 1 2 1 = 1
3 −1 2 −7
1 13
13 4
Finds point common to both planes is ( −1, 0 ,4 ) or , ,0 or ( 0, , )
5 5
7 7
−1 5
Equation of line of intersection is eg r = 0 + t 1 .
4 −7
M1 A1
M1
A1
[4]
M1 M1
A1
[3]
M1A1
M1
A1
[4]
Craftear
Q19 8
Q20
424
Qu
u
Soluti
11 (o) (i) Direction perpendicular to AB and CD:
i
j
k λ − 2
n= 0
1
1 = 4
4 3 − λ 1 −4
5
DB = −4 . Hence
−6
2
M1A1
5 λ − 2
4
− 4 .
−6 −4
or equivalent
= 3
( λ − 2 )2 + 1 6 + 1 6
(
⇒ ( 5λ − 2 ) = 9 λ 2 − 4λ + 36
)
M1A1
M1M1
A1
[7]
2
⇒ … ⇒ λ + λ − 20 = 0 (AG)
( λ + 5)( λ − 4 ) = 0 ⇒ λ = −5, 4
B1
i
j k −10
2
λ = 4 ⇒ a = 4 ⇒ Normal to ABD = 4 −1 1 = −29
−3
−1 3 7 11
M1A1
i
j k 44
2
λ = −5 ⇒ a = −5 ⇒ Normal to ABD = 4 8 1 = −29
−1 12 7 56
−3
A1
cos θ =
−440 + 841 + 616
2
2
2
10 + 29 + 11
⇒ θ = 66.1°
2
2
44 + 29 + 56
2
=
1017
1062 5913
M1A1
A1
[7]
Craftear
(ii)
Marks
Q21
11 E
JJJG JJJG
JJJG
M1A1
A1
Solves: s = –1 , t = –1 , k = 1
p = 3i + j + 2k and q = i + 2 j − 3k (Both required)
B1B1
−22
2
3
r = 1 + λ −19 + µ −1
2
5
5
Craftear
B1
i j k −22
3 −4 − = −19
2 −1 5 5
ALT METHOD:
Find PQ in terms of s,t M1
Calculate scalar products direction vectors for l1 and l2 M1,A1,A1
Solve simultaneous equations M1 A1
p and q A1
M1A1
M1A1
t − 3s + 2k = 4
−3t + 4s − k = −2
−t + 2s + 5k = 4
Obtains three equations. E.g. from OP + PQ = OQ
i j k 2
1 −3 −1 = −1
3 −4 −2 5
[3]
[7]
425
M1
A1
B1
i
j k 90 3
Normal to Π is 22 19 −5 = −120 = −4
2 −1 5 −60 −2
Cartesian equation of Π is 3x – 4y – 2z = 1 (Since P lies in Π )
−1
4
Vector equation of l3 is r = −1 + λ 3 (OE) (Since z = 0 .)
0
0
Craftear
B1
i j k −4 4
Direction of l3 is 3 −4 − = −3 ~ 3
0 0 1 0 0
[4]
426
427
110 (i)
(ii) (a)
5
3+λ
8+λ
p = 2 − 2λ , q = 3 + 2µ ⇒ QP = − 1 − 2λ − 2µ
3
17 + 3µ
− 14 − 3µ
M1A1
3+λ
1
− 2 . − 1 − 2λ − 2µ = 0 ⇒ 5λ + 4µ = − 5
0 17 + 3µ
M1A1
3+λ
0
2 . − 1 − 2λ − 2µ = 0 ⇒ − 4λ − 13µ = 53
− 3 17 + 3µ
A1
Solving :
λ = 3, µ = – 5
M1A1
Whence:
11
5
p = − 4 , q = − 7 .
3
1
A1
(8)
− 3
3
AP = − 6 , AQ = − 9
− 2
0
i j k
12
3 − 6 0 = 6 (CAO) ;
−45
−3 − 9 − 2
(b)
−3
uuur
1 1
AB = 1 ⇒ Volume = ×
3 2
−17
B1
Area =
1
2205
2
(=23.5)
− 3 12
1 . 6
−17 −45
= 735 (= 122.5)
2205
6
2205
M1A1
A1
M1A1
(6)
Total: 14
Alternative: for marks 3,4,5,6 and 7 in part (i)
i j
1 −2
0
k
6
0 = 3
2
2 −3
6k − λ = 3
3k + 2λ + 2 = −1
2k − 3µ = 17
Solving; k = 1 , λ = 3 and µ = – 5 (If k missing, or assumed to be 1, deduct 1
mark.)
(M1A1)
(A1)
(M1A1)
Craftear
Q22
Q23
3 + 2λ 1
λ . 2 = 3
2 − 2λ 1
3 + 2λ + 2λ + 2 − 2λ = 3 ⇒ λ = −1
P is (1, –1, 4)
(Accept position vector.)
1
n = 2
1
i
M1
A1
A1
(3)
B1
j
k
−5
w= 1 2 1 = 4
−3
2 1 −2
i
j k 10 5
Direction of PQ is − 5 4 − 3 = 2 ~ 1 (OE)
1 2 1 − 14 − 7
1
5
Equation of PQ is r = −1 + µ 1
4
−7
[N.B. For methods not using w at most three B1 marks.]
M1A1
(3)
M1A1
dM1A1
(4)
Total
10
Craftear
8
428
Q24
429
11 E
8 4
−1 2 0 = 4 ~ 2
− 1 0 4 2 1
E.
j k
1 4
0 . 2
0 1
4 2 + 2 2 + 12
=
4
(AG)
21
3 4i + 2 j + k 1
= (4i + 2 j + k )
21
21 7
1 − 3
Line AP: r = 0 + t 2
0 1
p=
M1A1
M1A1
[4]
B1
M1A1
0
1
1
For Q 1 − 3t = 0 ⇒ t = ⇒ q = 2
3
3
1
M1A1
[5]
E.
− 1
0
1
AB = 2 , BQ = − 4
3
0
1
B1
i
j
2
−1 2 0 = 1
0 − 4 1 4
cos
k
4 2
2 . 1
−1 1 4
21. 21
= cos −1
M1A1
8+ 2+ 4
14
2
= cos −1 = cos −1
(AG)
21
21
3
M1A1
[5]
Total
14
Craftear
i
Q25
430
6
5
AB = 4 CD = − 2
− 8
− 4
B1
i
j
− 16 2
Common perpendicular is 3 2 − 4 = − 8 ~ 1
5 − 2 − 4 − 16 2
1
AD = − 4 ⇒ shortest distance =
9
i
1 2
− 4 ⋅ 1
9 2
4 +1 + 4
=
16
or 5.33
3
− 2
Normal to Π1 is 1 − 4 9 = 31
3 2 − 4 14
i
j
k
M1A1
[5]
k
34
Normal to Π2 is 1 − 4 9 = 49
5 − 2 − 4 18
cos θ =
j
M1A1
M1A1
k
− 2 × 34 + 31× 49 +14 × 18
2
2 + 312 + 14 2 34 2 + 49 2 + 182
⇒ θ = 36.7° (CAO)
A1
M1A1
A1
[6]
Q26
OR
(i)
r = 4i – 2j + 2k + λ(i + 7j + k) + µ(3i + j – k) ⇒ A is in Π1.
(ii)
i j k −8 2
1 7 1 = 4 ~ − 1
3 1 − 1 − 20 5
(iii)
L is (12, –6, 6)
2x – y + 5z = 24 + 6 + 30 = 60
n = t (2i – j + 5k) ⇒ 4t + t + 25t = 60 ⇒ t = 2
n = 4i – 2j + 10k
(iv)
3
(8i – 4j + 4k) = 10i – 5j + 5k (AG)
4
M is (10, –5, 5) ⇒ NM = 6i – 3j – 5k
(6i – 3j – 5k) × (2i – j + 5k) = – i + 2j
20(−i + 2j 20
Perpendicular distance is
=
= 8.16
30
6
(Mark various alternative methods in a similar manner.)
m = 4i – 2j + 2k +
B1
[1]
M1A1
[2]
B1
B1
M1A1
A1
[5]
B1
B1
M1A1
M1A1
[6]
Craftear
11
Q27
431
i
(i)
k
− 3 1
1 − 2 − 3 = 3 ~ − 1
− 3 1
−2 1
3
M1A1
BA = 6i + 4j − 6k
Shortest distance
B1
6 1
4 . −1
−6 1
12 +12 +
= 43 ( = 2.31)
M1A1
(5)
Alternative
− 6 − 2λ − µ 1
− 4 + λ + 2µ . − 2 = 0 ⇒ −13λ − 14µ = 16
6 + 3λ + 3µ − 3
(M1)
− 6 − 2λ − µ − 2
− 4 + λ + 2µ . 1 = 0 ⇒ 14λ +113µ = −26
6 + 3λ + 3µ 3
(A1)
⇒ λ = − 529 , µ = 389
Shortest distance =
i
(ii)
j
(M1A1)
4
3
.
(A1)
(= 2.31
k
− 4 4
1 −1 1 = − 5 ~ 5
− 2 1 3 − 1 1
M1A1
Cartesian equation of Π: 4x + 5y + z = −12 − 5 + 2 = −15
(iii)
Distance of A from Π:
6 4
4 . 5
− 6 1
M1A1
4 2 + 52 + 12
= 38
42
M1A1
(4)
( = 5.86)
A1
(3)
[12]
Craftear
10
j
Q28
k 3
1 = − 1
1 − 6
M1A1
Obtains cartesian equation
3× 2 + 1× (−1) + (−2) × (−6) = 17
⇒ 3x − y − 6z = 17
M1
A1
Obtains area of triangle
ABC.
1 2
1
3 + 12 + 6 2 =
46 (=3.39)
2
2
M1 A1
Obtains length of
perpendicular
from D to triangle ABC.
9 − 6 − 12 − 17
M1A1
1
× Base
3
area × Height.
i
2
1
j
0
−3
3 2 + 12 + 6 2
=
26
46
1 1
26
13
×
46 ×
=
3 2
3
46
Uses
Either
or triple scalar product
method.
1 3
1 26 13
Or e.g. 5 ⋅ − 1 =
=
6
6
3
4 − 6
Obtains normal to ABD.
i
1
2
j
5
0
k 5
4 = 7
1 − 10
Uses scalar product
2
2
3 +1 + 6
to find angle between
normals
and hence angle between
Π1
and Π2.
⇒ cos θ =
2
4
M1A1
6
M1A1
3 5
5 + 7 + 10 cos θ = − 1 ⋅ 7
− 6 − 10
M1
68
A1
2
46 174
2
2
⇒ θ = 40.5º
Craftear
11 O Vector perpendicular to Π1
432
[14]
433
Q29
Finds PQ.
Finds direction of
common perpendicular.
− 3 + µ − 3λ
PQ = − 6 µ − 2λ
12 − 2 µ + λ
i
j
M1A1
k
− 10
2
3 2 − 1 = 5 ~ − 1
4
1 − 6 − 2 − 20
M1
(Or uses scalar product
of PQ with the
direction vector of each
line.)
Obtains two correct
equations. e.g.
Solves.
Finds p and q.
Finds common
perpendicular.
− 3 + µ − 3λ = 12µ + 4λ
− 24µ − 8λ = −12 + 2µ − λ
A1
A1
µ = 1 , λ = −2
M1A1
4 + 3λ − 2
1+ µ 2
p = 7 + 2λ = 3 q = 7 − 6 µ = 1
−1− λ 1
11 − 2 µ 9
i
AB × PQ = − 1
2
j
0
k 4
4 = 12
− 1 4 1
6
2
3
–1
PA 4 QA 6 PB 4 QB 6 − 2
–10
10
2
Uses triple scalar product
to find shortest distance.
6 4
4 ⋅ 12
− 2 1
16 +144 + 1
Alternative for last 4
marks:
24 + 48 − 2
161
=
70
161
A1
M1A1
Finds PA (or QA or PB
or QB)
=
8
= 5.52
Plane through e.g. P in direction PA :
4 12 29 0 . Award M1A1.
Then use of distance of point from line
70
. Award M1A1.
formula to get
161
Craftear
11 E
B1
M1A1
A1
6
[14]
Q30
434
8
Finds normal to Π1 .
i
j
k
1
0
1 = i + 3j − k
M1A1
1 −1 − 2
Finds Cartesian
equation.
Equation of Π1 :
Finds angle
between normals,
using scalar
product.
cosθ =
Finds direction of
line of intersection,
using vector
product.
i
Finds point
common to both
planes.
States vector
equation.
3 12
2 − 3 −1
11 6
2
=
⇒ θ = 75.7° or 1.32 rad.
66
j
A1
3
M1
A1
2
k
1 3 − 1 = 2i − 3 j − 7 k
2 −1 1
M1A1
Point on both planes is e.g. (6,2,0)
M1A1
r = 6i + 2j + t (2i − 3j − 7k )
(OE)
A1
5
[10]
8
Finds normal
AB = i + 2j + 2k, AC = −2i + j − 2k
and cartesian equation
AB × AC = −6i − 2j + 5k
M1A1
6x + 2 y − 5z = constant = 24 + 10− 30 = 4
M1A1
Finds where OD meets
plane
Obtains angles between
planes
Equation of OD: r = t (6i + 3j + 6k )
1
⇒ 36t + 6t − 30t = 12t = 4 ⇒ t =
3
E is the point (2,1,2).
Craftear
Q31
(4)
B1
M1A1
A1
(4)
Using (6i + 2j − 5k ) . 2i + j + 2k
⇒12 + 2 −10 = 36 + 4 + 25 4 + 1 + 4 sinθ
M1A1
⇒ θ = 9.5o (0.166 rad).
A1
(3)
[11]
435
11
OR
Obtains direction of
common perpendicular.
Uses result for shortest
distance between lines.
Solves equation.
i
j
k
n = 4 −1
1
0
m −1
1
= –mi + 4(1– m)j + 4k
1 − m
0 4 − 4m
− 4 4
m 2 + 16(1 − 2m + m 2 ) + 16
M1A1
=3
M1A1
A1
⇒ … ⇒ 19m 2 − 40m + 4 = 0
⇒ (19m − 2)(m − 2) = 0
M1
A1
⇒ m = 2, since m is an integer. (AG)
Finds relevant vectors.
1
0
− 1
CA = 0 and CD = 1 or AD = 1
− 4
1
5
B1
Use of cross-product.
i j k
− 4
1
1
0 1 1 =
1
17
17
1 0 −4
−1
M1
Obtains shortest
distance.
1
18
4 2 +12 + 12 =
17
17
A1
Finds 2nd vector in BCD
(CD may already have
been found.)
Finds normal vector to
BCD. (Normal to ACD
already found.)
Finds angle between
planes = angle between
normal vectors.
(= 1.03)
BC = 3i – j +5k
i
j
cos θ =
1
3
B1
k
n = 3 − 1 5 = –6i – 3j +3k ~ 2i + j – k
0
7
Craftear
Q32
M1
1
(4i − j + k) ⋅ (2i + j − k)
16 + 1 + 1 4 + 1 + 1
=
6
18 6
1
∴ Angle between planes = cos–1
(AG)
3
=
1
3
M1
A1
4
[14]
436
OR
OR
Alternatives for middle
part:
Or (a)
Vector from D to any
point on AC
1+ t
−1
− 5 − 4t
(B1)
Uses orthogonality to
obtain t.
1+ t 1
21
− 1 . 0 = 0 ⇒ t = −
17
− 5 − 4t − 4
(M1)
1
18
4 2 +12 + 12 =
17
17
(A1)
Finds magnitude of
perpendicular.
Or (b)
Finds length of AD (or
CD)
Finds projection of AD
(or CD) onto AC.
Finds perpendicular by
Pythagoras.
(= 1.03)
AD = 27
− 1 1
1 . 0
5 − 4
4 2 +12
27 −
(3)
(B1)
=
21
17
441
18
=
17
17
(M1)
(= 1.03)
(A1)
(3)
Craftear
11
437
Finds vector normal
to Π.
i
Cartesian equation
of Π.
Substitutes general
point of l2.
Finds value of
parameter.
Finds p.v. of
intersection.
Finds distance along
l from known point
to foot of
perpendicular from
given point to l.
F.t. on nonhypotenuse side
(must be real).
Writes a set of three
equations in
three unknowns for
the intersection of
l with Π.
Solves the set of
equations.
Finds p.v. of
intersection.
2 = 2i + 8 j + 7 k
1
M1A1
−2
8 2
3 + 8t 2
6 + 5t ⋅ 8 = 138 or 5 ⋅ 8 = 0
− 8 7
12 − 8t 7
A1
Independent of t ⇒ parallel, or ⇒ parallel.
A1
4
Π: 2x + 8y + 7z = 21
Sub. x = 5 + 2s , y = −4 − s , z = 7 + s
⇒ s = −2
and line meets Π at point with p.v. i – 2j + 5k
B1
M1
A1
A1
4
Take (9,11,2) as A, (3,6,12) as B and let C be foot of
perpendicular from A to l.
B1
AB = 62 + 52 +102 = 161
BC =
Finds distance from
point to known
point on l.
k
n= 1 −2
3
Dot product of this
with general point
on l1.
Deduces result.
j
1
2
2
2
(6 5 + 8 )(6i + 5j−10k)(8i + 5j−8k)
AC = 161−153 = 8or 2 2
=
(= 2.83)
153
= 153
153
M1
A1
A1
Alternatively:
5 + 2 s = 2 + λ + 3µ
(B1)
− 4 − s = 3 − 2λ + µ
7 + s = −1 + 2λ − 2 µ
⇒ s = −2 , ⇒ λ = 2 , µ = −1
and line meets Π at point with p.v. i – 2j + 5k
(M1A1)
(A1)
4
[12]
Craftear
9
438
Q33
9
Finds vector BA .
B1
BA = 6i + 5j – 10j
i
1
j
k
6 5 − 10
82 + 52 + 8 8 5 − 8
M1A1
1224
= 8 or 2 2 ( = 2.83)
153
=
A1
(4)
Alternatively:
(A)
Take (9,11,2) as A, (3,6,12) as B and let C be
foot of perpendicular from A to l.
Finds distance from point to
known point on l.
(B1)
F. on non-hypotenuse si
F.t.
(must be real).
AB =
BC =
2
2
6 + 5 + 10 = 161
1
2
2
2
(6i + 5j − 10k)(8i + 5j − 8k )
8 +5 +8
153
=
= 153
153
AC = 161−153 = 8 or 2 2
(M1)
(A1)
(A1 )
(4)
( = 2.83)
Craftear
Finds distance along l from
known point to foot of
perpendicular from given point
to l.
2
(B)
8t − 6
AC = 5t − 5
10 − 8t
Finds vector . AC
Uses AC perpendicular to l to
find t.
8t − 6 8
5t − 5 . 5 = 0 ⇒ t = 1
10 − 8t − 8
Finds length AC.
AC =
22 + 02 + 22 = 8
(B1)
(M1A1)
(A1)
[12]
(4)
Q34
4
Finds vector
product.
Finds area of
triangle.
Finds length of
perpendicular.
1
AB = 2
3
1
AC = 1
2
B1
i j k 1
AB × AC = 1 2 3 = 1
1 1 2 − 1
M1A1
1
1 1
Area of triangle ABC = 1 =
3
2 2
− 1
M1A1
1 2
1
3
1 + 2 2 + 32 d =
3⇒d =
2
2
14
A1
3
3
[6]
439
Q35
Finds normal vector to
plane.
x = 2+λ +µ
6
1 − 2 − 1 = 1 or y = −3 − 2λ + 2 µ
z = 1− λ − 2 µ
1 2 − 2 4
Uses known point to
find constant term.
Plane equation is 6x + y + 4z = constant.
Substitute (2, –3,1) ⇒ 12 – 3 + 4 = 13.
Or eliminate λ and µ.
⇒ Π1 : 6x + y + 4z = 13
Angle between normals
is equal to angle
between planes.
Solve plane equation
simultaneously.
(Note: may find
direction from vector
product and use with one
point.)
i
j
cosθ =
k
(6i + j + 4k).(3i − 2 j − 3k)
2
2
6 +1 + 4
2
2
2
3 +2 +3
2
=
M1A1
M1
4
53 22
A1
4
M1A1
A1
3
⇒ θ = 83.3º or 1.45 rad.
6x + y + 4z = 13 and 3x – 2y – 3z = 4
Obtains e.g. y + 2z = 1 and 3x + z = 6
Or two of (0,–11,6) , (11/6,0,1/2) , (2,1,0)
x 2 −1
y = 1 + t − 6 (OE)
z 0 3
Alternatively:
Direction of line from vector product.
Finds a point on line.
States equation of line.
M1
A1A1
A1
(M1A1)
(A1)
(A1)
4
[11]
Craftear
9
440
Craftear
PROOF BY INDUCTION
441
Q1
3
1
Prove by mathematical induction that 2 4n + 31 n - 2 is divisible by 15 for all positive integers n.
[6]
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442
Q2
3
6
(a) Prove by mathematical induction that, for all positive integers n,
n
/`5r + r j = n (n + 1) (2n + 1) .
r =1
4
2
1
2
2
2
[6]
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............................................................................................................................................................
443
7
(b) Use the result given in part (a) together with the List of formulae (MF19) to find
n
/ r in terms
r =1
of n, fully factorising your answer.
4
[3]
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444
Q3
3
5
The sequence of real numbers a1 , a2 , a3 , … is such that a1 = 1 and
a n +1 = ea n +
3
1
.
an o
(a) Prove by mathematical induction that ln an H 3 n -1 ln 2 for all integers n H 2 .
[6]
1
[You may use the fact that ln bx + l 2 ln x for x 2 0 .]
x
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(b) Show that ln an +1 - ln an 2 3 n -1 ln 4 for n H 2 .
[2]
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445
Q4
2
It is given that y = xe ax , where a is a constant.
Prove by mathematical induction that, for all positive integers n,
dny
= `a n x + na n-1j e ax .
dx n
[6]
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446
Q5
5
Prove by mathematical induction that, for every positive integer n,
d 2n-1
(x sin x) = (-1) n - 1 `x cos x + (2n - 1) sin xj .
dx 2n-1
[7]
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447
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448
Q6
2
4
Prove by mathematical induction that 7 2n - 1 is divisible by 12 for every positive integer n.
[5]
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449
Q7
2
The sequence u1, u 2, u3, f is such that u1 = 1 and un + 1 = 2un +1 for n H 1.
(a) Prove by induction that un = 2 n - 1 for all positive integers n.
[5]
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(b) Deduce that u2 n is divisible by un for n H 1.
[2]
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450
Q8
2
It is given that y = ln ax + 1, where a is a positive constant. Prove by mathematical induction that,
for every positive integer n,
n
dn y
n−1 n − 1!a
=
−1
.
[6]
dxn
ax + 1n
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451
Q9
8
(i) Prove by mathematical induction that, for z ≠ 1 and all positive integers n,
1 + z + z2 + à + zn−1 =
zn − 1
.
z−1
[5]
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452
Q10
1
3
Prove by mathematical induction that 33n − 1 is divisible by 13 for every positive integer n.
[5]
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453
Q11
2
It is given that f n = 23n + 8n−1 . By simplifying f k + f k + 1, or otherwise, prove by mathematical
induction that f n is divisible by 9 for every positive integer n.
[6]
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454
Q12
9
18
For the sequence u1 , u2 , u3 , à, it is given that u1 = 8 and
ur+1 =
5ur − 3
4
for all r .
(i) Prove by mathematical induction that
n
un = 4 54 + 3,
for all positive integers n.
[5]
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455
(ii) Deduce the set of values of x for which the infinite series
u1 − 3x + u2 − 3x2 + à + ur − 3xr + à
is convergent.
[2]
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Ð ln un − 3 = N 2 ln a + N ln b.
N
n=1
[3]
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(iii) Use the result given in part (i) to find surds a and b such that
456
Q13
3
4
The sequence of positive numbers u1 , u2 , u3 , à is such that u1 < 3 and, for n ≥ 1,
un+1 =
4un + 9
.
un + 4
(i) By considering 3 − un+1 , or otherwise, prove by mathematical induction that un < 3 for all positive
integers n.
[5]
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457
(ii) Show that un+1 > un for n ≥ 1.
[3]
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458
Q14
6
dr u
dr y
and
are denoted by
dxr
dxr
y r and u r respectively. Prove by mathematical induction that, for all positive integers n,
Q
P@ A
@ A
@ A
@ A
@ A
n
n
n
n
n
n
1
2
r
n
x
y =e
u+
u +
u +à+
u +à+
u .
[8]
0
1
2
r
n
@ A @
A @
A
k
k
k+1
[You may use without proof the result
+
=
.]
r
r−1
r
It is given that y = ex u, where u is a function of x. The rth derivatives
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9231/12/O/N/18
Craftear
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459
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© UCLES 2018
9231/12/O/N/18
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Craftear
................................................................................................................................................................
460
Q15
2
4
Prove, by mathematical induction, that 5n + 3 is divisible by 4 for all non-negative integers n.
[5]
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© UCLES 2017
9231/11/M/J/17
Craftear
................................................................................................................................................................
461
Q16
Ð
@
n
3
Prove, by mathematical induction, that
r=1
r ln
A
@
A
r+1
n + 1n
= ln
for all positive integers n. [6]
r
n!
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© UCLES 2017
9231/13/M/J/17
Craftear
................................................................................................................................................................
462
Q17
3
4
(i) Show that
dn n
dn+1 n+1
n
x
ln
x
=
n x + n + 1x ln x .
n+1
dx
dx
[2]
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
........................................................................................................................................................
(ii) Prove by mathematical induction that, for all positive integers n,
@
A
1
1
dn n
x ln x = n! ln x + 1 + + à +
.
dxn
2
n
[5]
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© UCLES 2017
9231/11/O/N/17
Craftear
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Q18
Prove by mathematical induction that, for all positive integers n, 10n + 3 × 4n+2 + 5 is divisible by 9.
[6]
Craftear
3
463
© UCLES 2016
9231/11/M/J/16
[Turn over
Q19
It is given that a diagonal of a polygon is a line joining two non-adjacent vertices. Prove, by
mathematical induction, that an n-sided polygon has 12 n n − 3 diagonals, where n ≥ 3.
[6]
Craftear
2
464
© UCLES 2016
9231/13/M/J/16
[Turn over
465
Q20
4
@
Using factorials, show that
A @ A @
A
n
n
n+1
+
=
.
r−1
r
r
[2]
Hence prove by mathematical induction that
@ A
@ A
@ A
@ A
n n
n n−1
n n−r r
n n
a + xn =
a +
a x+à+
a x +à+
x
0
1
r
n
for every positive integer n.
Craftear
[4]
© UCLES 2016
9231/11/O/N/16
[Turn over
Q21
3
466
4an
5
+
for every positive integer n.
5
an
Prove by mathematical induction that an > 5 for every positive integer n.
[5]
The sequence a1 , a2 , a3 , à is such that a1 > 5 and an+1 =
[2]
Craftear
Prove also that an > an+1 for every positive integer n.
© UCLES 2015
9231/11/M/J/15
[Turn over
467
Q22
3
r=1
Ð 2r − 1 .
∞
State the value of
Ð 2r − 1 = 2n + 1 .
n
Prove by mathematical induction that, for all positive integers n,
2
n
[6]
[1]
2
Craftear
r=1
1
1
© UCLES 2015
9231/13/M/J/15
[Turn over
Q23
3
468
Given that a is a constant, prove by mathematical induction that, for every positive integer n,
6
Craftear
dn
xeax = nan−1 eax + an xeax .
dxn
© UCLES 2015
9231/11/O/N/15
[Turn over
Q24
3
469
Prove by mathematical induction that, for all non-negative integers n,
112n + 25n + 22
[6]
Craftear
is divisible by 24.
© UCLES 2014
9231/11/M/J/14
[Turn over
Q25
It is given that φ n = 5n 4n + 1 − 1, for n = 1, 2, 3, à . Prove, by mathematical induction, that φ n
is divisible by 8, for every positive integer n.
[7]
Craftear
3
470
© UCLES 2014
9231/13/M/J/14
[Turn over
Q26
3
471
It is given that ur = r × r! for r = 1, 2, 3, . Let Sn = u1 + u2 + u3 + + un . Write down the values
of
2! − S1 ,
3! − S2 ,
4! − S3 ,
5! − S4 .
2
[1]
Prove, by mathematical induction, a formula for Sn , for all positive integers n.
[4]
Craftear
Conjecture a formula for Sn .
© UCLES 2014
9231/11/O/N/14
[Turn over
472
Q27
Prove by mathematical induction that 52n − 1 is divisible by 8 for every positive integer n.
[5]
Craftear
2
© UCLES 2013
9231/11/M/J/13
[Turn over
Q28
3
473
Prove by mathematical induction that, for every positive integer n,
7
Craftear
dn x
e sin x = ï2n ex sin x + 14 n0.
dxn
© UCLES 2013
9231/13/M/J/13
[Turn over
474
Q29
5
It is given that y = 1 + x2 ln 1 + x. Find
d3 y
.
dx3
[3]
Prove by mathematical induction that, for every integer n ≥ 3,
5
Craftear
dn y
n−1 2 n − 3!
.
n = −1
dx
1 + xn−2
© UCLES 2013
9231/11/O/N/13
[Turn over
Q30
9
475
Prove by mathematical induction that, for every positive integer n,
cos 1 + i sin 1n = cos n1 + i sin n1.
5
Craftear
Express sin5 1 in the form p sin 51 + q sin 31 + r sin 1, where p, q and r are rational numbers to be
determined.
[6]
© UCLES 2013
9231/13/O/N/13
[Turn over
Q31
2
476
Prove, by mathematical induction, that, for integers n ≥ 2,
[5]
Craftear
4n > 2n + 3n .
© UCLES 2012
9231/11/M/J/12
[Turn over
477
Q32
5
Let In denote ã
0
xn e−2x dx. Show that In = 12 nIn−1 , for n ≥ 1.
n!
.
2n+1
[6]
Craftear
Prove by mathematical induction that, for all positive integers n, In =
[2]
© UCLES 2012
9231/12/O/N/12
[Turn over
Q33
1
2
3
N
+
+
+ ... +
. Prove by mathematical induction that, for all positive
(N + 1)!
2! 3! 4!
integers N ,
1
SN = 1 −
.
[5]
(N + 1)!
Let SN =
Craftear
3
478
© UCLES 2012
9231/13/O/N/12
[Turn over
Q34
3ur − 2
for all r. Prove by
4
n
3
mathematical induction that un = 4 4 − 2, for all positive integers n.
[5]
For the sequence u1 , u2 , u3 , . . . , it is given that u1 = 1 and ur+1 =
Craftear
2
479
© UCLES 2012
9231/13/M/J/12
[Turn over
PROOF BY INDUCTION
MS
Craftear
480
1
Question
Q1
Craftear
A1
Hence, by induction, true for every positive integer n.
6
A1
M1 A1 Separates 2 4 k + 31k − 2 or considers difference.
is divisible by 15 because 15 × 24 k + 30 × 31k is divisible by 15.
Then 24k + 4 + 31k +1 − 2 = (15 +1)24 k + (30 +1)31k − 2
B1 States inductive hypothesis.
Assume that 2 4 k + 31k − 2 is divisible by 15 for some positive integer k.
Guidance
B1 Checks base case.
Marks
24 + 31 − 2 = 45 is divisible by 15
Answer
481
Q2
3(b)
3(a)
Question
4
r =1
2
)
2
4
5r 4 + r 2 = 1 k 2 (k +1)2 ( 2k +1)
2
(
k
2
M1 Take out the factor of 𝑘 + 1 OR expands the
summation expression and the target expression
for 𝑘 + 1 and collects like terms for both.
n
r 4 + 16 n(n +1)(2n +1) = 12 n 2 (n +1) 2 (2n +1)
4
1
30
2
r =1
r = n(n +1)(2n + 1)(3n + 3n − 1)
n
r =1
[5] r 4 = 16 n(n +1)(2n +1)(3n(n + 1) − 1)
r =1
n
5
Craftear
3
A1
4
r the subject and takes out all linear
2
r .
CAO
factors and the remaining term is of correct form.
M1 Makes
M1 Uses correct formula for
6
A1 States conclusion. Implication must be clearly
expressed.
2
So H k +1 is true. By induction, H n is true for all positive integers n.
2
M1 Considers sum to k +1.
A1 Factorises or having expanded, checks explicitly.
At least one intermediate step seen following the
award of M1 before reaching the answer.
3
( 2k + k + 10(k + 1) + 2 )
Guidance
B1 States inductive hypothesis [for some k] including
the algebraic form. If says for ALL k, then B0.
B1 Checks base case.
Marks
1
(k + 1)2 (2k 3 + 11k 2 + 20k + 12) = 12 (k + 1)2 (k + 2) 2 (2k + 3)
2
1
( k + 1) 2
2
r =1
1
( 5r + ) = 2 k (k +1) ( 2k +1) + 5(k + 1) + (k + 1)
k +1
r2
Assume that
1
5 × 14 + 12 = (2)2 ( 2 + 1) [ = 6] so H1 is true.
2
Answer
482
Q4
Q3
2
Question
3(b)
3(a)
Question
ln 2.
3k ln 2
Marks
(
)
dk y
= a k x + ka k −1 e ax .
dx k
)
( ) (
Craftear
So true when n = k + 1. By induction, true for all positive integers n.
(
d k +1 y
= a a k x + ka k −1 e ax + e ax a k = a k +1 + ( k + 1) a k e ax
k +1
dx
Assume that
dy
= axeax + eax = (ax +1)eax so true when n = 1.
dx
Guidance
6
A1 States conclusion.
M1 A1 Differentiates kth derivative.
B1 States inductive hypothesis.
M1 A1 Differentiates once using the product rule.
2
A1 AG
> 2 × 3n −1 ln 2 = 3n −1 ln 4
6
A1 States conclusion, all previous marks must be gained.
M1 A1 Applies inductive hypothesis and writes in required
form.
M1 Applies result given in part (a).
B1 States inductive hypothesis.
M1 Applies results given in parts (a) and (b).
)
Guidance
B1 Checks base case. Must be n =2 and exact.
Marks
1
3ln an + − ln an > 2ln an
an
Answer
So true when n = k + 1. By induction, true for all integers n ≥ 2.
≥
1
ln ak +1 = 3ln ak + > 3ln ak
ak
Assume that ln ak ≥ 3
k −1
a2 = 23 leading to ln a2 = 3ln 2 so true when n = 2.
Answer
483
Q6
Q5
2
Question
5
Question
d 2k
x sin x ) = (−1) k −1 (− x sin x + 2k cos x)
2k (
dx
A1
A1 Depends on previous M1 and A1.
7 2 k + 2 − 1 is divisible by 12.
So if is true for n = k (could be written earlier) it is also true for n = k +1 .
Hence, by induction, true for every positive integer n.
Craftear
M1 Separates 7 2 k − 1.
Then 7 2 k +2 − 1 = 49 ⋅ 7 2 k −1 = 48 ⋅ 7 2 k + 7 2 k −1
Or (7 2 k + 2 − 1) − (7 2 k −1) = 48 ⋅ 7 2 k
5
B1 States inductive hypothesis.
Assume that 7 2k − 1 is divisible by 12 for some positive integer k.
Guidance
B1 Checks base case.
Marks
7
A1 States conclusion.
M1 A1 Differentiates again.
M1 A1 Differentiates once. Must have correct LHS for A1.
7 2 − 1 = 48 is divisible by 12.
Answer
So, it is also true for n = k + 1 . Hence, by induction, true for all
positive integers.
d 2 k +1
x sin x ) = ( −1) k −1 (− x cos x − sin x − 2k sin x)
2 k +1 (
dx
= (−1)k (x cos x + (2k +1)sin x)
Then
B1 States inductive hypothesis.
Assume true for n = k , so
d 2 k −1
k −1
x sin x ) = ( −1) ( x cos x + (2k −1)sin x )
2 k −1 (
dx
Guidance
B1 Checks base case using product rule.
Marks
d
( x sin x ) = x cos x + sin x = (−1)0 ( x cos x + (2(1) − 1)sin x)
dx
Answer
484
Q8
Q7
2
Question
2(b)
2(a)
Question
(
)(
)
)
By induction, true for every positive integer n .
so true for n = k +1 .
k
d k +1 y
k!a k +1
k −1 (k −1)! a
k
=
−
ka
(−1)
=
(
−
1)
dx k +1
(ax + 1) k +1
(ax + 1) k +1
Then
Craftear
k
dk y
k −1 (k −1)! a
=
(
−
1)
k
(ax + 1) k for some positive integer k.
Assume that dx
so true for n = 1 .
dy
a
0!a1
0
=
= ( −1)
dx ax + 1
( ax + 1)1
Answer
n
n
u2 n 2 2 n − 1 2 − 1 2 + 1
= n
=
= 2n + 1
n
un
2 −1
2 −1
(
Guidance
6
A1 States conclusion.
M1 A1 Differentiates k th derivative.
B1 States inductive hypothesis.
M1 A1 Proves base case.
Marks
So, it is also true for n = k + 1 . Hence, by induction, true for all positive integers.
2
M1 A1
5
A1
M1 A1
B1
Assume that it is true for n = k , so uk = 2k − 1 .
Then uk +1 = 2(2k − 1) + 1 = 2k +1 − 1
B1
Marks
u1 = 1 = 21 −1
Answer
485
Q10
Q9
1
Question
8(i)
Question
k
k
k +1
z − 1 k z − 1 + z ( z − 1) z − 1
+z =
=
,
z −1
z −1
z −1
k
zk −1
z −1
Guidance
A1
A1 States conclusion.
is divisible by 13
H k ⇒ H k +1 By induction, 33n − 1 is divisible by 13 for every positive
integer n .
Craftear
M1 Separates 33k − 1
Then 33k + 3 − 1 = 3333k − 1 = 26 ⋅ 33k + 33k − 1
5
B1 States inductive hypothesis.
Assume that 33k − 1 is divisible by 13 for some positive integer k
Guidance
B1 Checks base case.
Marks
5
A1 States conclusion.
M1 A1 Combines fractions.
B1 States inductive hypothesis.
B1 Shows base case.
Marks
33 − 1 = 26 is divisible by 13
Answer
H k → H k +1 Hence, by induction, true for all positive integers.
so true when n = k +1
1 + z + … + z k −1 + z k =
Then
Assume that 1 + z + …+ z k −1 =
z1 − 1
z −1
So true when n =1 .
1=
Answer
486
Q12
Q11
9(i)
Question
2
Question
n
+ 3,
Craftear
So Pk ⇒ Pk+1. Therefore, by induction, Pn is true for all positive integers.
k +1
A1 States conclusion.
A1
M1 Proves inductive step.
k
1 5
uk +1 = 5 4 + 3 − 3 = correct step
4 4
5
= 4
4
B1 States inductive hypothesis.
Proves base case.
B1 States proposition.
Assume Pk is true for some k . Then
5
Pn : un = 4 + 3
4
5
Let n = 1 then 4 + 3 = 8 ⇒ P1 true.
4
Marks
Guidance
A1 = 23.23k + 8.8k–1 M1
= 8 f(k) A1
So if f (k) is divisible by 9, so is f (k +1), (and f (1) is divisible by 9), f ( n )
is divisible by 9 for every integer n ⩾1
Answer
A1 Alt method: f (k + 1) = 23k + 3 + 8k
= 9 23k + 8k −1 OE so f ( k +1) is divisible by 9.
6
A1 Correct split of powers
23.23k + 8.8k–1 + 23k + 8k–1 OE
)
M1 Uses expansion of f ( k + 1) .
f ( k + 1) + f ( k ) = 23k +3 + 8k + 23k + 8k −1 =
(
B1 Makes general statement.
Assume that f ( k ) is divisible by 9.
Guidance
B1 Checks base case.
Marks
We have that f (1) = 9 is divisible by 9.
Answer
487
9(iii)
9(ii)
n
N
ln(un − 3) =
N
∑
5 n
ln 4
n =1
4
N
(
)
1
5
N ( N + 1) ln + N ln 4
2
4
n =1
N
1
2
Use ∑n = N ( N +1) .
4
5x
4
<1⇒− < x <
4
5
5
n
= N 2 ln
Craftear
5
5
+ N ln 2 5 ⇒ a =
, b = 2 5 oe
2
2
Alt method: Writes series as an AP M1, uses summation formula M1 Correct
answer A1
=
∑ ∑
5
= ln n + ln 4
4 n =1
n =1
n =1
∑
N
So series is convergent for −1 <
n5
5x
5x
( un − 3) x = 4 x = 4 so r =
4
4
4
n
10
A1
M1
= N 2 ln
= Nln4 +
∑
n
(
ln(
)
5
) M1
4
∑n
5
5
+ N ln 2 5 ⇒ a =
, b=2 5
2
2
2
N ( N + 1)
4
5
= Nln4+ ln
= ln 4N ∏
4
n
M1
5
1
N
5
ln(un − 3) = ln ∏ 4
4
n=1
N
M1 Alt method:
A1
M1
A1
488
Q13
3(ii)
3(i)
Question
4uk + 9 −uk + 3
=
> 0 ⇒ uk +1 < 3
uk + 4
uk + 4
4un + 9
−u 2 + 9
− un = n
un + 4
un + 4
So un < 3 ⇒ un+1 − un > 0 .
un +1 − un =
Hence, by induction, un < 3 for all n.1 .
3 − uk +1 = 3 −
Then
Craftear
3
B1 Uses un < 3
M1 A1 Considers un+1 − un .
5
B1 States conclusion.
M1 A1
B1 States inductive hypothesis.
Assume that uk < 3
Guidance
B1 States base case.
Marks
u1 < 3 ( given )
Answer
489
Q14
6
Question
k
k r
k k
k 1
= ex u + u( ) + " + u( ) + " + u( ) +
1
r
k
0
k 1 k 2
k r +1
k k +1
e x u( ) + u( ) + " + u( ) + " + u( )
1
r
k
0
k +1)
Guidance
Craftear
A1 States conclusion.
So H k implies H k +1 so, by induction, H n is true for all n .1 .
8
B1 Shows reasoning for first and last term
correctly
M1A1 Shows application of
k k k + 1
+
=
.
r r −1 r
M1 Differentiates using product rule
B1 States inductive hypothesis.
M1A1 Shows base case using product rule
Marks
k +1
k +1 r
k + 1 ( k +1)
= ex
u + …
u + …
u
r
k + 1
0
k
k k r
k
= e x u + … +
u + … u ( k +1)
k
r r − 1
0
y(
Then
k
k 1
k r
k k
k
H k : y( ) = ex u + u( ) + " + u( ) + " + u( )
1
r
k
0
Assume that
1
1 1
1
1
y ( ) = e x u ( ) + ue x = e x u + u ( ) ⇒ H1 is true
1
0
Answer
490
Q15
2
Question
Craftear
P0 is true and Pk ⇒Pk+1, hence Pn is true for all non-negative integers n.
(or shows that 5k +1 + 3 = 5.5k + 5.3 − 4.3 = 5 5k + 3 − 4.3 )
)
Total:
5
A1
A1
= 20α −12 = 4 ( 5α − 3)
(
M1 Alt method:
Use f(k+1) – f(k)
5k +1 + 3 = 5 ( 4α − 3) + 3
M1 A1
B1 or e.g. 5k +3 = 4α for 2nd B1
Assume that Pk is true for some non-negative integer k.
Guidance
B1 Some explanation of what Pk being true means
Marks
Let Pn be the proposition that 5n+3 is divisible by 4
50 +3 =4 ⇒ P0 is true (allow P1)
Answer
491
Q17
Q16
3(i)
Question
3
Question
21
1 × ln2 = ln2 = ln ⇒ (H1 is true )
1!
( k +1)k
r +1
k +2
+ ( k + 1) ln
rln
=
ln
k!
r
k +1
r =1
(
)
(
dn n
x + ( n +1) x n lnx
n
dx
)
d n +1 n+1
d n n +1 1
x . + ( n + 1) x n lnx =
x
ln
x
=
n +1
n
x
dx
dx
Answer
Craftear
2
M1A1
Marks
AG
Guidance
6
A1
Thus Hk ⇒ Hk+1 and hence by PMI Hn is true for all positive integers.
Total:
A1
= ln
k
k +1
k + 1) ( k + 2 )
(
= ln
( k + 1)!( k + 1)k
k +1
k + 2)
(
= ln
( k + 1)!
B1
M1
∑
k +1
B1
B1
Marks
( k + 1)k ( k + 2 )k +1
k +1
k !( k +1)
Hence
∑
k
[ k +1]k
r +1
rln
Assume, for some positive integer k, that
= ln
k!
r
r =1
When n =1
Answer
Guidance
492
3(ii)
(
)
)
Craftear
5
A1
A1
1
1
= ( k +1)!lnx + 1+ + … +
⇒ Hk+1 is true
k +1
2
is true; hence, by PMI,
A1
M1
B1 Statement of Hk seen
1
1
= k !+ [ k +1] k !lnx + 1 + + … +
2
k
(
Check H1 is true and Hk is true ⇒ H
Hn is true for all positive integers n.
(
)
dk
1
1
x k lnx = k !lnx + 1 + + … +
k
2
k
dx
d k +1 k +1
dk
x
x
ln
=
x k + [ k + 1] x k lnx
k +1
k
dx
dx
Assume Hk is true ⇒
493
494
Q18
3
For n =1 10 +192 + 5 = 207= 9 × 23 ⇒ H1 is true.
B1
Assume Hk is true for some positive integer k ⇒ 10n + 3.4n +2 + 5 = 9α
Let f(n) = 10 n + 3.4n + 2 + 5
Hence f ( n +1) − f ( n ) =10n (10 −1) + 3.4n+2 ( 4 − 1)
B1
(
= 9 10 n + 4 n + 2
= 9β
M1
)
A1
Hence f ( n + 1) ( = 9 ( β + α ) ) ⇒ Hk+1 is true
A1
H1 is true and Hk ⇒ Hk+1 , hence by PMI Hn is true for all positive integers n.
(
)
N.B. Or can show f ( n +1) = 9 10α − 2.4 n +2 − 5 for M1A1A1. (3rd ,4th&5th
A1
[6]
marks)
Q19
1
n ( n − 3) = 0
2
A triangle has no diagonals ⇒ H3 is true.
1
Assume Hk is true: A k-gon has k ( k − 3) diagonals for some integer ≥ 3
2
Adding an extra vertex, a further (k – 1) diagonals can be drawn.
1
k 2 − 3k + 2k − 2 ( k + 1)( k − 2 )
k ( k − 3) + k − 1 =
=
2
2
2
1
= ( k + 1)( k +1− 3) (So Hk ⇒ Hk+1)
2
⇒Hn is true for all integers n 3 .
With n = 3 ,
M1
A1
B1
M1
A1
A1
[6]
Craftear
2
Q20
4
1
1
k
1
k
r
k
k
k k k + 1
k − r +1 r
x is:
Multiplying by ( a + x ) , the coefficient of a
+ =
r −1 r r
Hence Hn is true for all positive integers.
⇒ Hk+1 is true.
k
Craftear
( a + x ) = a k + a k −1 x +…+ a k −r x r +…+ x k
k
0
Assume Hk is true,i.e.
1
( a + x ) = a + x = a + x ⇒ H1 is true.
1
0
n n
n!
n!
n!
1
1
+
=
+
+ =
r −1 r ( r − 1)!( n − r + 1) ! r !( n − r ) ! ( r − 1) !( n − r ) ! n − r + 1 r
r + n − r +1
( n + 1)! = n +1
n!
=
=
( r − 1)!( n − r )! r ( n − r +1) r !( n − r + 1)! r
A1
M1
B1
B1
A1
M1
[4]
[2]
495
Q21
496
3
a1 > 5 (given) ⇒ H1 is true.
Assume Hk is true for some positive integer k, i.e. ak = 5 + δ, where δ > 0.
2
2
4a + 25 − 25ak (4ak − 5)(ak − 5)
4a + 25
ak +1 − 5 = k
−5= k
=
> 0 , ⇒ ak +1 > 5
5ak
5ak
5ak
Or
5
4
4
δ δ2
ak +1 = (5 + δ ) +
, = 4 + δ + (1 − +
− ...) for 0 < δ < 5
5
5+δ
5
5 25
3
= 5 + δ + 0(δ 2 ) [ ak+1 > 5, ( δ [ 5 is trivial).
5
Hk ⇒ Hk +1 and H1 is true, hence by mathematical induction, the result is true for all===
n ∈ Z + =(N.B. The minimum requirement is ‘true for all positive integers’.)
5
1
a k +1 − a k =
− ak
ak 5
1
5
< 1 and ak > 1 ⇒ ak +1 − ak < 0 ⇒ ak +1 < ak
5
ak
B1
B1
M1A1
(M1)
(A1)
A1
(5)
M1
A1
(2)
Total: 7
Q22
Hk : ∑
k
1
r = 1 (2r ) 2 − 1
=
k
is true for some integer k.
2k +1
1
1
1
= =
⇒ H1 is true.
2 − 1 3 2 ×1 + 1
1
k
k
1
2k 2 + 3k + 1
+
=
+
=
2k + 1 (2k + 2) 2 − 1 2k + 1 (2k + 1)(2k + 3) (2k + 1)(2k + 3)
(2k + 1)(k + 1)
k +1
=
=
(2k + 1)(2k + 3) 2[k + 1] + 1
∴ H k ⇒ H k +1
∴ (By Principle of Mathematical Induction) Hn is true for all positive integers n. (This mark requires
all previous marks.)
2
1
1
=
2
r = 1 (2r) −1
2
∑∞
B1
B1
M1A1
A1
A1
(6)
B1
(1)
Total
7
Q23
3
n =1 in formula gives a 0e ax + axe ax = e ax + axe ax
d
xe ax = e ax × 1 + x.aeax = eax + axe ax ⇒ H1 is true oe
dx
( )
Assume Hk is true, i.e.
d k +1
xe ax
dx k +1
dk
xe ax = ka k −1e ax + a k xe ax .
k
dx
( )
( ) = ka e + a e + a xe
k ax
Craftear
3
k ax
k +1
ax
= (k +1)a k eax + a k +1xeax
⇒ H k +1 is true, hence by PMI Hn is true for all positive integers n.
B1
B1
B1
M1
A1
A1
[6]
Total
6
Q24
3
497
Hk: f(k) = 112k + 25k + 22 = 24λ
B1
f(0) = 1 + 1 + 22 = 24 = 1 × 24 ⇒ H is true.
B1
f(k + 1) – f(k) = 112k + 2 + 25k + 1 + 22 – (112k + 25k + 22)
= 112k(121 – 1) + 25k(25 – 1)
= 112k × 24 × 5 + 25k × 24 = 24µ
M1
A1
A1
Alternatively:
f(k + 1) = 112k + 2 + 25k + 1 + 22
= 121.112k + 25.25k + 22 = (120 + 1)112k + (24 + 1)25k + 22 (OE)
= 120.112k + 24.25k + 24λ = 24µ
(M1)
(A1)
(A1)
⇒ f(k + 1) = 24µ + 24λ = 24(µ + λ) ⇒ Hk + 1 is true.
Hence by PMI Hn is true for all non-negative integers. (Must see non-negative integers.)
CSO: Final mark requires all previous marks.
A1
[6]
3
φ (1) = 5 × 5 – 1 = 24 which is divisible by 8 ⇒ H1 is true.
B1
Assume Pk is true for some positive integer k ⇒ φ (k) = 8l
φ (k + 1) – φ (k) = 5k + 1(4k + 5) – 1 – 5k(4k + 1) + 1
= 5k(20k + 25 – 4k – 1)
= 5k(16k + 24) = 8m
∴φ (k + 1) = 8(l + m)
Hence, by PMI, true for all positive integers n. (CWO – all previous marks required.)
B1
M1
A1
A1
A1
A1
[7]
Alternatively
φ (k + 1) = 5k + 1(4k + 5) – 1
= 5. (4k.5k) + 25.5k – 1
= 5(8l – 5k + 1) + 25.5k – 1
= 40l + 20.5k + 4
= 40l + 24.5k – 4.5k+ 4
= 40l + 24.5k – 4(5k– 1)
= 40l + 24.5k – 4(8l – 4k.5k)
= 8l + 24.5k + 16k.5k
= 8m
(M1A1)
(A1)
(A1)
Q26
3
2! – S1 = 1 , 3! – S2 = 1 , 4! – S3 = 1 , 5! – S4 = 1 (Two correct B1, all four correct B2)
Sn = (n +1)! – 1
2! – 1 = 2 – 1 = 1 ⇒ H1 is true.
Hk: Sk = (k +1)! – 1
(k +1)! – 1 + (k + 1) × (k + 1)!
= (k + 1)!(1 + k +1) – 1
= ([k +1] + 1)! – 1 Hence Hk ⇒ Hk+1
So result holds for all positive integers (by PMI).
Craftear
Q25
B2,1,0
(2)
B1
(1)
B1
B1
M1
A1
(4)
[7]
Q27
2
498
Proves base case.
Pn: 5 2 n − 1 is divisible by 8.
5 2 −1 = 24 = 3× 8 ⇒ P1 is true
B1
B1
States inductive hypothesis.
Assume Pk is true: 5 2 k − 1 = 8λ for some k.
5 2 k + 2 − 1 = 25.5 2 k − 1 = 24.5 2 k + 5 2 k − 1
= 3 × 8.5 2 k + 8λ
M1
Proves inductive step.
∴ Pk ⇒ Pk+1
A1
States conclusion.
(Since P1 is true and Pk ⇒ Pk+1)∴ Pn is for every
positive integer n (by PMI).
A1
5
[5]
Q28
Differentiates once.
Rearranges.
Shows true for n = 1 .
States inductive
hypothesis.
(May be seen by
implication.)
(
)
M1
A1
( )
B1
( 2 ) sin x + 14 kπ e + e cos x + 14 kπ
M1
=
( 2 ) e 12 sin x + 14 kπ + 12 cos x + 14 kπ
A1
=
( 2 ) e sin x + 14 (k +1)π ⇒ H
A1
Differentiates.
d k +1
=
dx k +1
Rearranges.
Shows H k ⇒ H k +1
and states
conclusion.
B1
d x
e sin x = e x sin x + e x cos x
dx
1
1
cos x
= 2e x
sin x +
2
2
1
= 2e x sin x + π ⇒ H1 true .
4
k
k
1
d
H k : k (e x sin x) = 2 e x sin x + kπ
4
dx
k +1
k
x
x
x
k +1
x
k +1
true.
Craftear
3
[7]
7
∴ ,by PMI, true for all positive integers n. (CWO)
Q29
5 (i)
Differentiates once,
twice
and three times.
(ii)
y′ = 2(1+ x )ln (1+ x ) + (1+ x )
y′′ = 2ln (1+ x ) + 3
2
y′′′ =
1+ x
(Allow B1 if constant term in previous line incorrect.)
d 3 y (− 1) .2.0!
2
=
=
⇒ H 3 is true.
3
1+ x
1+ x
dx
k −1.
d k y (− 1) .2.(k − 3)!
Hk : k =
for some k .
dx
(1 + x )k − 2
B1
B1
B1
3
2
Proves base case.
States inductive
hypothesis.
Differentiates
Proves inductive
step and
states conclusion.
d k +1 y
k −1
− ( k −1)
= (−1) .2(k − 3)!(−1)(k − 2 )(1 + x )
dx k +1
(− 1)k .2.(k − 2)! ⇒ H is true
=
k +1
(1 + x )k −1
Hence by PMI Hn is true for all integers [=3
B1
B1
M1
A1
A1
5
[8]
499
Q30
9
States inductive
hypothesis
Proves H k ⇒ H k +1
Hk: (cosθ + i sinθ ) = cos kθ + i sin kθ for some k
k
(cosθ + isinθ )k +1 = (cosθ + isin θ )(cos kθ + isin kθ )
= (cosθ coskθ − sinθ sin kθ ) + i(sinθ cos k θ + cosθ sin kθ )
= cos [k + 1]θ + 1sin[k +1]θ ∴Hk ⇒ H k+1
States conclusion
H1 is trivially true, so true for all positive integers
Uses de M’s Thm. to
find 2isin nθ
1
= (cos nθ + i sin nθ ) − (cos nθ − i sin nθ )
zn
= 2i sin nθ
zn −
B1
M1
A1
A1
A1
(5)
B1
5
and groups
= z 5 − 5 z 3 + 10 z −
Applies above result
and obtains sin5
B1
10 5
1
+
− 5
3
z z
z
1
1
1
= z 5 − 5 − 5 z 3 − 3 + 10 z −
z
z
z
M1
A1
32i sin5 θ = 2i sin5θ −10i sin3θ + 20i sinθ
M1
sin 5 θ =
A1
(
1
sin 5 θ − 5 sin 3θ + 10 sin θ
16
)
(6)
[11]
Q31
2
Craftear
Uses binominal
expansion
1
5
5
z − = (2i sin θ ) = 32i sin θ
z
(States proposition.)
(Pn: 4n > 2n + 3n)
Proves base case.
Let n = 2, 16 > 4 + 9 ⇒ P2 is true.
B1
States inductive
hypothesis.
Proves inductive step.
Assume Pk is true ⇒ 4k > 2k + 3k
B1
4 k +1 = 4.4 k > 4(2 k + 3 k ) = 4.2 k + 4.3 k
> 2.2 k + 3.3 k = 2 k +1 + 3 k +1
M1
A1
∴ Pk ⇒ Pk+1
States conclusion.
Hence result true, by PMI, for all integers n ≥ 2.
A1
(CWO)
5
[5]
500
Q32
5
Integrates by parts to
obtain reduction formula.
∞
∞
∫ xe
n −2 x
0
-2x
∞
− e −2 x
n −1 e
dx = x n
+
nx
dx
0
2 0
2
n
= I n −1
(AG)
2
∫
(States proposition.)
Pn : I n =
n!
2 n +1
Proves base case.
n=1
I0 =
Shows Pk ⇒ Pk +1 .
States conclusion.
∞
1
1
e -2x dx = − e -2x =
0
2
0 2
1 1 1 1!
I1 = × = = 2
∴ P1 true.
2 2 4 2
∫
∞
k!
for some integer k.
2 k +1
(k + 1)!
k + 1 k!
∴ I k +1 =
× k +1 = k +2
2
2
2
∴ Pk ⇒ Pk +1
Pk : I k =
Hence by PMI Pn is true for all positive integers n.
M1
A1
2
B1
B1
B1
M1A1
A1
[8]
6
3
1
( N + 1)!
Proves base case. S = 1 = 1 = 1 − 1 ⇒ H1 is true.
1
2! 2
2!
States inductive
1
Hk : Assume S k = 1 −
is true.
hypothesis.
(k + 1)!
Proves inductive ⇒ S = 1 − 1 + k + 1 = (k + 2)!−(k + 2) + (k + 1)
k +1
step.
(k + 1)! (k + 2)!
(k + 2)!
1
∴ Hk ⇒ Hk+1.
⇒ S k +1 = 1 −
(k + 2)!
(By PMI Hn is) true for all positive integers N.
States conclusion. ∴
Proposition.
Craftear
Q33
HN : S N = 1 −
B1
B1
M1
A1
A1
5
[5]
Q34
2
n
(States proposition.)
Proves base case.
States Inductive
hypothesis.
Proves inductive step.
3
(Pn : un = 4 − 2)
4
3
Let n = 1 4 ×
– 2 = 3 –2 = 1 ⇒ P1 true.
4
Assume Pk is true for some k.
3 k
34 − 2 − 2
k
4
3 3
6+2
= 4. . −
u k +1 =
4
4 4
4
B1
B1
M1
k +1
States conclusion.
3
= 4. − 2 ∴ Pk ⇒ Pk +1
4
∴ By PMI Pn is true ∀ positive integers.
A1
A1
5
[5]
0
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