Mathematics · May / June 2023 · Paper 41 · Question 4
YMA01
As printed in the paper
The question as text
4. Relative to a fixed origin O, • the point A has position vector 4i + 8j + k • the point B has position vector 5i + 6j + 3k • the point P has position vector 2i – 2j + k The straight line l passes through A and B.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
4(a)[2]
2
4
2
4
2
2
4
2
3!
4
B1: Obtains an expansion with a constant term of 8
3
1
1
−
2
M1: Attempts the binomial expansion of
−
2 x
to get the third and/or fourth term with an acceptable
4
structure.
3
5
3
5
7
−
×−
7
−
×−
×−
9
−
2
−
3
2
2
1
1
2
2
2
1
1
Look for
2
or
2
±
2 x
±
2 x
2
4
3!
4
1
with or without the brackets on the
±
2 x
A1: Correct simplified or unsimplified expansion .
Expression at the top of the page is acceptable
A1: Two correct and simplified terms of
8
+
24
x
+
60
x
2
+
140
x
3
A1: All correct and simplified
8
+
24
x
+
60
x
2
+
140
x
3
Direct expansion in (c)
1
1
1
−
1
1
1
1
1 1
1
x
2
2
2
x
....
B1: First two terms which may be unsimplified
−
=
+
−
+
4
2
4
2 4
2
1 1
1
−
3
−
2
2 2
1
1
M1: Correct form for the third term
2
−
2 x
2
4
1
1
1
A1: Correct simplified expansion
−
x
−
x
2
+
...
2
2
4
2(a)
(
)
x
=
2
4
−
8
y
+
y
2
=
13
y
2
−
8
y
−
9
=
0
y
=
...
y =
9
2
x
→
2 ln 2
x
d
y
d
y
−
4
xy
→−
4
x
−
4
y
OR
y
2
→
2
y
d
x
d
x
d
y
d
y
2 ln 2
x
−
4
x
−
4
y
+
2
y
=
0
d
x
d
x
d
y
d
y
(
)
2
y
−
4
x
=
4
y
−
2 ln 2
x
=
...
d
x
d
x
d
y
4
y
−
2 ln 2
x
or d
y
2 ln 2
x
−
4
y
d
y
ln2e
x
ln 2
−
4
y
=
=
or
=
d
x
2
y
−
4
x
d
x
4
x
−
2
y
d
x
4
x
−
2
y
d
y
4 9
( )
−
2 ln 2
2
(
)
2, 9
→
=
d
x
2 9
( )
4 2
( )
−
36
−
4ln 2
(
)
y
−
''9''
=
''
''
x
−
2
10
36
−
4ln 2
y
=
0
0
−
''9''
=
''
''
(
x
−
2
)
x
=
...
10
4ln 2
+
9
−
8ln 2 18
−
x
=
o.e. e.g.
x
=
2ln 2 18
−
−
4ln 2
+
36
.
(a)
M1: Substitutes x = 2 into the equation for C, forms a quadratic equation in y and then solves to obtain at least
one value for y
A1:
y =
9
only
d
y
(b) It is acceptable to use
y
'
in this question
d
x
B1: Correct differentiation of 2x . Allow also
2
x
=
e
x
ln2
→
e
x
ln2
ln 2
d
y
d
y
M1: Differentiates
−
4
xy
→±
Ax
±
By
OR differentiates
y
2
→
k y
d
x
d
x
A1: Fully correct differentiation. Allow versions such as 2 ln 2d
x
4 d
4 d
2 d
=
0
x
−
x y
−
y x
+
y y
M1: Attempts to make d
y
x the subject with the 2 terms in d
y
x coming from differentiating
y and 4xy
2
−
d
d
A1: Any correct expression for d
y
2
y
−
2
x
−
1
ln 2
x . Note that you can isw after a correct answer.
is correct
d
y
−
2
x
M1: Uses x = 2 and their y value from part (a) to find the gradient at P and attempts to form the equation of the
tangent at P. Condone poor attempts at differentiation but d
y
x must have both x and y terms.
d
If the form y = mx + c is used they must proceed as far as c = ...
dM1: Substitutes y = 0 into their tangent equation and rearranges to find x. Dependent upon previous M
It is possible for a method to combine this with the previous M.
Look for methods like the following, which score M1, dM1
where 'm' is the value of their d
y
−
''9''
=
''
m
''
(
x
−
2
)
x
=
...
x at x = 2
d
or ''9''
where 'm' is the value of their d
y
x =
''
m
''
x
=
...
x at x = 2
2
−
d
A1: Correct expression in the required form. Must come from a correct d
y
x
d
Do not be concerned about the order of the terms on the numerator and denominator
3(a)
8
x
−
5
A
B
8
x
5
A
(
4
x
3
)
B
(
2
x
1
)
A
...
B
...
+
−
=
−
+
−
=
=
(
)(
)
2
x
−
1 4
x
−
3
2
x
−
1
4
x
−
3
A
=
1 or
B
=
2
8
x
−
5
1
2
+
(
2
1 4
)(
3
)
2
1
4
3
x
−
x
−
x
−
x
−
(b)
1
2
1
1
(
)
+
d
x
=
ln 2
x
−+
1
ln 4
x
−
3
+
c
2
x
−
1
4
x
−
3
2
2
Follow through their A and B
(c)
3
k
1
1
1
1
ln 2
(
x
1
)
ln 4
(
x
3
)
ln 6
(
k
1 12
)(
k
3
)
ln 2
(
k
1
)(
4
k
3
)
−
+
−
=
−
−
−
−
−
2
2
2
2
k
(
)(
)
1 ln
6
k
−
1 12
k
−
3
=
(
)(
)
2
2
k
−
1 4
k
−
3
(
)(
)
(
)(
)
1
6
k
−
1 12
k
−
3
1
6
k
−
1 12
k
−
3
ln
=
ln 20
=
20
(
)(
)
(
)(
)
2
2
k
−
1 4
k
−
3
2
2
k
−
1 4
k
−
3
(
)(
)
(
)(
)
6
k
−
1 12
k
−
3
=
20 2
k
−
1 4
k
−
3
88
k
2
−
170
k
+
57
=
0
88
k
2
−
170
k
+
57
=
0
k
=
...
3
k =
2
(a)
M1: ''Correct'' P.F form (condoning slips) with strategy to find at least one constant or partial fraction.
(
)
(
)
Look for an attempt at the correct form (condoning slips)
5 8
−
x
=
A
4
x
−
3
+
B
2
x
−
1
followed by
values for A or B
A1: One correct fraction or constant
A1: Correct partial fractions which may be awarded for sight of correct PF in part (b).
Award if the correct answer is just written down with little or no working.
Award if the correct partial fractions are seen in (b)
(b)
α
α
M1:
For
d
x
=
β
ln 2
x
−
1
OR
d
x
=
β
ln 4
x
−
3
2
x
−
1
4
x
−
3
(
)
Allow for example brackets instead of moduli, e.g.
ln 2
x
−
1
ln 2
x
−
1
α
Condone the omission of moduli/brackets, for example
1 x
d
=
β
ln 2
x
−
1
for this mark only
2
x
−
A
A
B
B
A1ft: For
d
x
=
ln 2
x
−
1
OR
d
x
=
ln 4
x
−
3
Follow through their A and B
2
x
1
2
4
x
3
4
−
−
ln 2
(
1
)
ln 2
1
Allow for example brackets instead of moduli, e.g .
x
−
x
−
2
x
−
1
4
x
−
3
2
4
A
B
A
B
(
)
(
)
or
d
x
+
d
x
=
ln 2
x
−
1
+
ln 4
x
−
3
2
x
−
1
4
x
−
3
2
4
Follow through their numerical A and B. There is no requirement for the + c
ISW after a correct answer. This can be awarded (but not implied) from part (c)
1 ln 2
(
)(
)(
)
(
)(
) (
)
Note that
x
−
1
4
x
−
3
+
c
and
ln
2
x
−
1 4
x
−
3
+
c
are also correct
2
(c)
M1: Attempts the use of the correct limits and uses correct log work to combine terms to obtain a single ln term.
It is dependent upon having an answer to part (b) involving two ln terms.
Condone slips/ bracketing errors .
1
1
✓
1
1
✓
1
1
1
E.g
(
)
(
)
(
)
(
)
(
)
(
)
(
)
ln 6
k
−
1
+
ln 12
k
−
3
−
ln 2
k
−
1
+
ln 4
k
−
3
=
ln 6
k
−
1
+
ln 12
k
−
3
−
ln 2
k
−
1
+
2
2
2
2
2
2
2
(
)(
)(
)
1 ln
6
k
−
1 12
k
−
3
4
k
−
3
=
(
)
2
2
k
−
1
dM1: Eliminates ln’s, multiplies up and collects terms to form a polynomial equation in k.
It is dependent upon the previous M
A1: Correct 3TQ. The = 0 may be implied by subsequent work
ddM1: Solves 3TQ. Dependent upon both previous M's.
Can be solved via a graphical calculator (you may need to check the answers)
3
A1: Correct value and no others. Condone
x =
2
Solution with limited working in (c)
Case I
(
)(
)
6
k
−
1 12
k
−
3
1
1
3
ln
=
ln 20
x
or
k
=
SC 11000
2
(
2
k
1 4
)(
k
3
)
2
2
−
−
Case II
(
)(
)
6
k
−
1 12
k
−
3
3
=
20
x
or
k
=
SC 11100
(
)(
)
2
k
−
1 4
k
−
3
2
Attempts direction vector by subtracting(
k and (
)
)
4(a)
5
+
6
+
3
4
+
8
j k either way
+
i
j
i
around.
(
)
(
)
E.g.
r
=
4
i
+
8
j
+
k
+
λ
i
−
2
j
+
2
k ,
r
=
5
i
+
6
j
+
3
k
+
μ
i
−
2
j
+
2
k
(b)
4
+
λ
2
2
+
λ
E.g
PC
=
8
−
2
λ
−−
2
=
10
−
2
λ
1
+
2
λ
1
2
λ
(
)
Uses
PC
.
i
−
2
j
+
2
k
=
0
E.g.
1 2
(
)
2 10
(
2
)
2 2
0
...
+
λ
−
−
λ
+ ×
λ
=
λ
=
(
)
(
)
(
)
E.g I
λ
=
2
=
4
+
2
+
8
−
4
+
1
+
4
k
c
i
j
(
)
(
)
(
)
E.g II
μ
=
1
=
5 1
+
+
6
−
2
+
3
+
2
k
c
i
j
(
6, 4, 5
)
(c)
(
)
OP
′=
p
+
2
PC
=
2
i
−
2
j
+
k
+
2 4
i
+
6
j
+
4
k
(
)
10, 10, 9
(d)
PP
′=
2
PC
2 4
2
6
2
4
2
...
=
+
+
=
4 17
General rule in this question: If no method is shown look for two ''correct'' components for their vector
(a) There are many correct versions so please check the candidates answer carefully
M1: Attempts the direction of l by subtracting(
k and (
)
)
5
+
6
+
3
4
+
8
j k either way around.
+
i
j
i
If no method is shown look for two correct components of (
)
±
1
i
±
2
j
±
2
k or a multiple of this
A1: Any correct equation including the lhs of ‘r =’
4
1
4
1
4
i
=
8
+
λ
−
2
l
=
8
+
λ
−
2
=
8
+
λ
−
2j
Allow in the form
r
but
and is
r
A0
1
2
1
2
1
2k
(b)
M1: Attempts the general vector from P to l forming PC
where C is a general point on l.
4
+
λ
2
Look for an attempt to subtract their
'' 8
−
2
λ
''
and
−
2
either way around
1 2
+
λ
1
(
)
dM1: Attempts
PC
.their
i
−
2
j
+
2
k
=
0
and solves for λ.
It is dependent upon the previous M and the scalar product of this vector must be attempted with their
gradient for the line l to produce and solve a linear equation for λ.
ddM1: Uses their λ to find C. Scored for substituting their correctly found λ into their equation for l
A1: Correct coordinates or vector.
6i
Only penalise an error such as
4j
once, the first time that it is made. Hence award if a mark has already been
5 k
M1: Correct strategy for OP′
Can be implied by two correct coordinates using their P and C
For example look for
OP
′=
OP
+ ×
2
PC
using the coordinates of C found in part (b).
A1: Correct coordinates or vector
(d)
M1: Correct method for PP′
using their values for P and P' or P and C or P' and C . E.g
PP
′=
2
There are many ways to do this but it must be a complete method, not a distance squared.
A1: CAO
Alternative part (b) using shortest distance
M1: Attempts the general vector from P to l.
4
+
λ
2
Look for an attempt to subtract their
'' 8
−
2
λ
''
and
−
2
either way around
1 2
+
λ
1
dM1: Finds an expression for d or d2 , differentiates and sets = 0 to find λ.
As it is a quadratic expression an equivalent method would be via completing the square .
2
d
D
2
2
(
)
2
(
)
2
(
)
2
FYI :
D =
2
+
λ
+
10
−
2
λ
+
2
λ
=
9
λ
−
36
λ
+
104
=
18
λ
−
36
=
0
λ
=
2
d
λ
ddM1: Uses their λ to find C.
A1: Correct coordinates or vector
Alternative part (b) using Pythagoras' Theorem
M1: Attempts the general vector from P to l.
4
2
+
λ
Look for an attempt to subtract their
'' 8
2
''
and
2
either way around
−
λ
−
1 2
1
+
λ
2
2
2
2
2
2
dM1: Uses
PC
+
CB
=
PB
or
PC
+
CA
=
PA
to set up and solve an equation in λ.
2
2
2
E.g Using
PC
CB
PB
the equations are
+
=
(
2
)
2
(
10
2
)
2
(
2
)
2
(
)
2
(
2
2
)
2
(
2
)
2
77
+
λ
+
−
λ
+
λ
+
λ
−
+
−
λ
+
λ
−
=
2
( )
18
λ
−
54
λ
+
36
=
0
λ
=
2 1
ddM1: Uses their λ to find C.
A1: Correct coordinates or vector
(
)
P′
10,10, 9
l
(
)
C
6, 4, 5
(
)
B
5, 6, 3
P
(
2,
2,1
)
−
4(b)[4]
2
x
→
2 ln 2
x
d
y
d
y
−
4
xy
→−
4
x
−
4
y
OR
y
2
→
2
y
d
x
d
x
d
y
d
y
2 ln 2
x
−
4
x
−
4
y
+
2
y
=
0
d
x
d
x
d
y
d
y
(
)
2
y
−
4
x
=
4
y
−
2 ln 2
x
=
...
d
x
d
x
d
y
4
y
−
2 ln 2
x
or d
y
2 ln 2
x
−
4
y
d
y
ln2e
x
ln 2
−
4
y
=
=
or
=
d
x
2
y
−
4
x
d
x
4
x
−
2
y
d
x
4
x
−
2
y
4(c)[2]
d
y
4 9
( )
−
2 ln 2
2
(
)
2, 9
→
=
d
x
2 9
( )
4 2
( )
−
36
−
4ln 2
(
)
y
−
''9''
=
''
''
x
−
2
10
36
−
4ln 2
y
=
0
0
−
''9''
=
''
''
(
x
−
2
)
x
=
...
10
4ln 2
+
9
−
8ln 2 18
−
x
=
o.e. e.g.
x
=
2ln 2 18
−
−
4ln 2
+
36
.
(a)
M1: Substitutes x = 2 into the equation for C, forms a quadratic equation in y and then solves to obtain at least
one value for y
A1:
y =
9
only
d
y
(b) It is acceptable to use
y
'
in this question
d
x
B1: Correct differentiation of 2x . Allow also
2
x
=
e
x
ln2
→
e
x
ln2
ln 2
d
y
d
y
M1: Differentiates
−
4
xy
→±
Ax
±
By
OR differentiates
y
2
→
k y
d
x
d
x
A1: Fully correct differentiation. Allow versions such as 2 ln 2d
x
4 d
4 d
2 d
=
0
x
−
x y
−
y x
+
y y
M1: Attempts to make d
y
x the subject with the 2 terms in d
y
x coming from differentiating
y and 4xy
2
−
d
d
A1: Any correct expression for d
y
2
y
−
2
x
−
1
ln 2
x . Note that you can isw after a correct answer.
is correct
d
y
−
2
x
M1: Uses x = 2 and their y value from part (a) to find the gradient at P and attempts to form the equation of the
tangent at P. Condone poor attempts at differentiation but d
y
x must have both x and y terms.
d
If the form y = mx + c is used they must proceed as far as c = ...
dM1: Substitutes y = 0 into their tangent equation and rearranges to find x. Dependent upon previous M
It is possible for a method to combine this with the previous M.
Look for methods like the following, which score M1, dM1
where 'm' is the value of their d
y
−
''9''
=
''
m
''
(
x
−
2
)
x
=
...
x at x = 2
d
or ''9''
where 'm' is the value of their d
y
x =
''
m
''
x
=
...
x at x = 2
2
−
d
A1: Correct expression in the required form. Must come from a correct d
y
x
d
Do not be concerned about the order of the terms on the numerator and denominator
3(a)
8
x
−
5
A
B
8
x
5
A
(
4
x
3
)
B
(
2
x
1
)
A
...
B
...
+
−
=
−
+
−
=
=
(
)(
)
2
x
−
1 4
x
−
3
2
x
−
1
4
x
−
3
A
=
1 or
B
=
2
8
x
−
5
1
2
+
(
2
1 4
)(
3
)
2
1
4
3
x
−
x
−
x
−
x
−
(b)
1
2
1
1
(
)
+
d
x
=
ln 2
x
−+
1
ln 4
x
−
3
+
c
2
x
−
1
4
x
−
3
2
2
Follow through their A and B
(c)
3
k
1
1
1
1
ln 2
(
x
1
)
ln 4
(
x
3
)
ln 6
(
k
1 12
)(
k
3
)
ln 2
(
k
1
)(
4
k
3
)
−
+
−
=
−
−
−
−
−
2
2
2
2
k
(
)(
)
1 ln
6
k
−
1 12
k
−
3
=
(
)(
)
2
2
k
−
1 4
k
−
3
(
)(
)
(
)(
)
1
6
k
−
1 12
k
−
3
1
6
k
−
1 12
k
−
3
ln
=
ln 20
=
20
(
)(
)
(
)(
)
2
2
k
−
1 4
k
−
3
2
2
k
−
1 4
k
−
3
(
)(
)
(
)(
)
6
k
−
1 12
k
−
3
=
20 2
k
−
1 4
k
−
3
88
k
2
−
170
k
+
57
=
0
88
k
2
−
170
k
+
57
=
0
k
=
...
3
k =
2
(a)
M1: ''Correct'' P.F form (condoning slips) with strategy to find at least one constant or partial fraction.
(
)
(
)
Look for an attempt at the correct form (condoning slips)
5 8
−
x
=
A
4
x
−
3
+
B
2
x
−
1
followed by
values for A or B
A1: One correct fraction or constant
A1: Correct partial fractions which may be awarded for sight of correct PF in part (b).
Award if the correct answer is just written down with little or no working.
Award if the correct partial fractions are seen in (b)
(b)
α
α
M1:
For
d
x
=
β
ln 2
x
−
1
OR
d
x
=
β
ln 4
x
−
3
2
x
−
1
4
x
−
3
(
)
Allow for example brackets instead of moduli, e.g.
ln 2
x
−
1
ln 2
x
−
1
α
Condone the omission of moduli/brackets, for example
1 x
d
=
β
ln 2
x
−
1
for this mark only
2
x
−
A
A
B
B
A1ft: For
d
x
=
ln 2
x
−
1
OR
d
x
=
ln 4
x
−
3
Follow through their A and B
2
x
1
2
4
x
3
4
−
−
ln 2
(
1
)
ln 2
1
Allow for example brackets instead of moduli, e.g .
x
−
x
−
2
x
−
1
4
x
−
3
2
4
A
B
A
B
(
)
(
)
or
d
x
+
d
x
=
ln 2
x
−
1
+
ln 4
x
−
3
2
x
−
1
4
x
−
3
2
4
Follow through their numerical A and B. There is no requirement for the + c
ISW after a correct answer. This can be awarded (but not implied) from part (c)
1 ln 2
(
)(
)(
)
(
)(
) (
)
Note that
x
−
1
4
x
−
3
+
c
and
ln
2
x
−
1 4
x
−
3
+
c
are also correct
2
(c)
M1: Attempts the use of the correct limits and uses correct log work to combine terms to obtain a single ln term.
It is dependent upon having an answer to part (b) involving two ln terms.
Condone slips/ bracketing errors .
1
1
✓
1
1
✓
1
1
1
E.g
(
)
(
)
(
)
(
)
(
)
(
)
(
)
ln 6
k
−
1
+
ln 12
k
−
3
−
ln 2
k
−
1
+
ln 4
k
−
3
=
ln 6
k
−
1
+
ln 12
k
−
3
−
ln 2
k
−
1
+
2
2
2
2
2
2
2
(
)(
)(
)
1 ln
6
k
−
1 12
k
−
3
4
k
−
3
=
(
)
2
2
k
−
1
dM1: Eliminates ln’s, multiplies up and collects terms to form a polynomial equation in k.
It is dependent upon the previous M
A1: Correct 3TQ. The = 0 may be implied by subsequent work
ddM1: Solves 3TQ. Dependent upon both previous M's.
Can be solved via a graphical calculator (you may need to check the answers)
3
A1: Correct value and no others. Condone
x =
2
Solution with limited working in (c)
Case I
(
)(
)
6
k
−
1 12
k
−
3
1
1
3
ln
=
ln 20
x
or
k
=
SC 11000
2
(
2
k
1 4
)(
k
3
)
2
2
−
−
Case II
(
)(
)
6
k
−
1 12
k
−
3
3
=
20
x
or
k
=
SC 11100
(
)(
)
2
k
−
1 4
k
−
3
2
Attempts direction vector by subtracting(
k and (
)
)
4(a)
5
+
6
+
3
4
+
8
j k either way
+
i
j
i
around.
(
)
(
)
E.g.
r
=
4
i
+
8
j
+
k
+
λ
i
−
2
j
+
2
k ,
r
=
5
i
+
6
j
+
3
k
+
μ
i
−
2
j
+
2
k
(b)
4
+
λ
2
2
+
λ
E.g
PC
=
8
−
2
λ
−−
2
=
10
−
2
λ
1
+
2
λ
1
2
λ
(
)
Uses
PC
.
i
−
2
j
+
2
k
=
0
E.g.
1 2
(
)
2 10
(
2
)
2 2
0
...
+
λ
−
−
λ
+ ×
λ
=
λ
=
(
)
(
)
(
)
E.g I
λ
=
2
=
4
+
2
+
8
−
4
+
1
+
4
k
c
i
j
(
)
(
)
(
)
E.g II
μ
=
1
=
5 1
+
+
6
−
2
+
3
+
2
k
c
i
j
(
6, 4, 5
)
(c)
(
)
OP
′=
p
+
2
PC
=
2
i
−
2
j
+
k
+
2 4
i
+
6
j
+
4
k
(
)
10, 10, 9
(d)
PP
′=
2
PC
2 4
2
6
2
4
2
...
=
+
+
=
4 17
General rule in this question: If no method is shown look for two ''correct'' components for their vector
(a) There are many correct versions so please check the candidates answer carefully
M1: Attempts the direction of l by subtracting(
k and (
)
)
5
+
6
+
3
4
+
8
j k either way around.
+
i
j
i
If no method is shown look for two correct components of (
)
±
1
i
±
2
j
±
2
k or a multiple of this
A1: Any correct equation including the lhs of ‘r =’
4
1
4
1
4
i
=
8
+
λ
−
2
l
=
8
+
λ
−
2
=
8
+
λ
−
2j
Allow in the form
r
but
and is
r
A0
1
2
1
2
1
2k
(b)
M1: Attempts the general vector from P to l forming PC
where C is a general point on l.
4
+
λ
2
Look for an attempt to subtract their
'' 8
−
2
λ
''
and
−
2
either way around
1 2
+
λ
1
(
)
dM1: Attempts
PC
.their
i
−
2
j
+
2
k
=
0
and solves for λ.
It is dependent upon the previous M and the scalar product of this vector must be attempted with their
gradient for the line l to produce and solve a linear equation for λ.
ddM1: Uses their λ to find C. Scored for substituting their correctly found λ into their equation for l
A1: Correct coordinates or vector.
6i
Only penalise an error such as
4j
once, the first time that it is made. Hence award if a mark has already been
5 k
M1: Correct strategy for OP′
Can be implied by two correct coordinates using their P and C
For example look for
OP
′=
OP
+ ×
2
PC
using the coordinates of C found in part (b).
A1: Correct coordinates or vector
(d)
M1: Correct method for PP′
using their values for P and P' or P and C or P' and C . E.g
PP
′=
2
There are many ways to do this but it must be a complete method, not a distance squared.
A1: CAO
Alternative part (b) using shortest distance
M1: Attempts the general vector from P to l.
4
+
λ
2
Look for an attempt to subtract their
'' 8
−
2
λ
''
and
−
2
either way around
1 2
+
λ
1
dM1: Finds an expression for d or d2 , differentiates and sets = 0 to find λ.
As it is a quadratic expression an equivalent method would be via completing the square .
2
d
D
2
2
(
)
2
(
)
2
(
)
2
FYI :
D =
2
+
λ
+
10
−
2
λ
+
2
λ
=
9
λ
−
36
λ
+
104
=
18
λ
−
36
=
0
λ
=
2
d
λ
ddM1: Uses their λ to find C.
A1: Correct coordinates or vector
Alternative part (b) using Pythagoras' Theorem
M1: Attempts the general vector from P to l.
4
2
+
λ
Look for an attempt to subtract their
'' 8
2
''
and
2
either way around
−
λ
−
1 2
1
+
λ
2
2
2
2
2
2
dM1: Uses
PC
+
CB
=
PB
or
PC
+
CA
=
PA
to set up and solve an equation in λ.
2
2
2
E.g Using
PC
CB
PB
the equations are
+
=
(
2
)
2
(
10
2
)
2
(
2
)
2
(
)
2
(
2
2
)
2
(
2
)
2
77
+
λ
+
−
λ
+
λ
+
λ
−
+
−
λ
+
λ
−
=
2
( )
18
λ
−
54
λ
+
36
=
0
λ
=
2 1
ddM1: Uses their λ to find C.
A1: Correct coordinates or vector
(
)
P′
10,10, 9
l
(
)
C
6, 4, 5
(
)
B
5, 6, 3
P
(
2,
2,1
)
−
4(d)[2]
2
4
2
4
2
2
4
2
3!
4
B1: Obtains an expansion with a constant term of 8
3
1
1
−
2
M1: Attempts the binomial expansion of
−
2 x
to get the third and/or fourth term with an acceptable
4
structure.
3
5
3
5
7
−
×−
7
−
×−
×−
9
−
2
−
3
2
2
1
1
2
2
2
1
1
Look for
2
or
2
±
2 x
±
2 x
2
4
3!
4
1
with or without the brackets on the
±
2 x
A1: Correct simplified or unsimplified expansion .
Expression at the top of the page is acceptable
A1: Two correct and simplified terms of
8
+
24
x
+
60
x
2
+
140
x
3
A1: All correct and simplified
8
+
24
x
+
60
x
2
+
140
x
3
Direct expansion in (c)
1
1
1
−
1
1
1
1
1 1
1
x
2
2
2
x
....
B1: First two terms which may be unsimplified
−
=
+
−
+
4
2
4
2 4
2
1 1
1
−
3
−
2
2 2
1
1
M1: Correct form for the third term
2
−
2 x
2
4
1
1
1
A1: Correct simplified expansion
−
x
−
x
2
+
...
2
2
4
2(a)
(
)
x
=
2
4
−
8
y
+
y
2
=
13
y
2
−
8
y
−
9
=
0
y
=
...
y =
9
2
x
→
2 ln 2
x
d
y
d
y
−
4
xy
→−
4
x
−
4
y
OR
y
2
→
2
y
d
x
d
x
d
y
d
y
2 ln 2
x
−
4
x
−
4
y
+
2
y
=
0
d
x
d
x
d
y
d
y
(
)
2
y
−
4
x
=
4
y
−
2 ln 2
x
=
...
d
x
d
x
d
y
4
y
−
2 ln 2
x
or d
y
2 ln 2
x
−
4
y
d
y
ln2e
x
ln 2
−
4
y
=
=
or
=
d
x
2
y
−
4
x
d
x
4
x
−
2
y
d
x
4
x
−
2
y
d
y
4 9
( )
−
2 ln 2
2
(
)
2, 9
→
=
d
x
2 9
( )
4 2
( )
−
36
−
4ln 2
(
)
y
−
''9''
=
''
''
x
−
2
10
36
−
4ln 2
y
=
0
0
−
''9''
=
''
''
(
x
−
2
)
x
=
...
10
4ln 2
+
9
−
8ln 2 18
−
x
=
o.e. e.g.
x
=
2ln 2 18
−
−
4ln 2
+
36
.
(a)
M1: Substitutes x = 2 into the equation for C, forms a quadratic equation in y and then solves to obtain at least
one value for y
A1:
y =
9
only
d
y
(b) It is acceptable to use
y
'
in this question
d
x
B1: Correct differentiation of 2x . Allow also
2
x
=
e
x
ln2
→
e
x
ln2
ln 2
d
y
d
y
M1: Differentiates
−
4
xy
→±
Ax
±
By
OR differentiates
y
2
→
k y
d
x
d
x
A1: Fully correct differentiation. Allow versions such as 2 ln 2d
x
4 d
4 d
2 d
=
0
x
−
x y
−
y x
+
y y
M1: Attempts to make d
y
x the subject with the 2 terms in d
y
x coming from differentiating
y and 4xy
2
−
d
d
A1: Any correct expression for d
y
2
y
−
2
x
−
1
ln 2
x . Note that you can isw after a correct answer.
is correct
d
y
−
2
x
M1: Uses x = 2 and their y value from part (a) to find the gradient at P and attempts to form the equation of the
tangent at P. Condone poor attempts at differentiation but d
y
x must have both x and y terms.
d
If the form y = mx + c is used they must proceed as far as c = ...
dM1: Substitutes y = 0 into their tangent equation and rearranges to find x. Dependent upon previous M
It is possible for a method to combine this with the previous M.
Look for methods like the following, which score M1, dM1
where 'm' is the value of their d
y
−
''9''
=
''
m
''
(
x
−
2
)
x
=
...
x at x = 2
d
or ''9''
where 'm' is the value of their d
y
x =
''
m
''
x
=
...
x at x = 2
2
−
d
A1: Correct expression in the required form. Must come from a correct d
y
x
d
Do not be concerned about the order of the terms on the numerator and denominator
3(a)
8
x
−
5
A
B
8
x
5
A
(
4
x
3
)
B
(
2
x
1
)
A
...
B
...
+
−
=
−
+
−
=
=
(
)(
)
2
x
−
1 4
x
−
3
2
x
−
1
4
x
−
3
A
=
1 or
B
=
2
8
x
−
5
1
2
+
(
2
1 4
)(
3
)
2
1
4
3
x
−
x
−
x
−
x
−
(b)
1
2
1
1
(
)
+
d
x
=
ln 2
x
−+
1
ln 4
x
−
3
+
c
2
x
−
1
4
x
−
3
2
2
Follow through their A and B
(c)
3
k
1
1
1
1
ln 2
(
x
1
)
ln 4
(
x
3
)
ln 6
(
k
1 12
)(
k
3
)
ln 2
(
k
1
)(
4
k
3
)
−
+
−
=
−
−
−
−
−
2
2
2
2
k
(
)(
)
1 ln
6
k
−
1 12
k
−
3
=
(
)(
)
2
2
k
−
1 4
k
−
3
(
)(
)
(
)(
)
1
6
k
−
1 12
k
−
3
1
6
k
−
1 12
k
−
3
ln
=
ln 20
=
20
(
)(
)
(
)(
)
2
2
k
−
1 4
k
−
3
2
2
k
−
1 4
k
−
3
(
)(
)
(
)(
)
6
k
−
1 12
k
−
3
=
20 2
k
−
1 4
k
−
3
88
k
2
−
170
k
+
57
=
0
88
k
2
−
170
k
+
57
=
0
k
=
...
3
k =
2
(a)
M1: ''Correct'' P.F form (condoning slips) with strategy to find at least one constant or partial fraction.
(
)
(
)
Look for an attempt at the correct form (condoning slips)
5 8
−
x
=
A
4
x
−
3
+
B
2
x
−
1
followed by
values for A or B
A1: One correct fraction or constant
A1: Correct partial fractions which may be awarded for sight of correct PF in part (b).
Award if the correct answer is just written down with little or no working.
Award if the correct partial fractions are seen in (b)
(b)
α
α
M1:
For
d
x
=
β
ln 2
x
−
1
OR
d
x
=
β
ln 4
x
−
3
2
x
−
1
4
x
−
3
(
)
Allow for example brackets instead of moduli, e.g.
ln 2
x
−
1
ln 2
x
−
1
α
Condone the omission of moduli/brackets, for example
1 x
d
=
β
ln 2
x
−
1
for this mark only
2
x
−
A
A
B
B
A1ft: For
d
x
=
ln 2
x
−
1
OR
d
x
=
ln 4
x
−
3
Follow through their A and B
2
x
1
2
4
x
3
4
−
−
ln 2
(
1
)
ln 2
1
Allow for example brackets instead of moduli, e.g .
x
−
x
−
2
x
−
1
4
x
−
3
2
4
A
B
A
B
(
)
(
)
or
d
x
+
d
x
=
ln 2
x
−
1
+
ln 4
x
−
3
2
x
−
1
4
x
−
3
2
4
Follow through their numerical A and B. There is no requirement for the + c
ISW after a correct answer. This can be awarded (but not implied) from part (c)
1 ln 2
(
)(
)(
)
(
)(
) (
)
Note that
x
−
1
4
x
−
3
+
c
and
ln
2
x
−
1 4
x
−
3
+
c
are also correct
2
(c)
M1: Attempts the use of the correct limits and uses correct log work to combine terms to obtain a single ln term.
It is dependent upon having an answer to part (b) involving two ln terms.
Condone slips/ bracketing errors .
1
1
✓
1
1
✓
1
1
1
E.g
(
)
(
)
(
)
(
)
(
)
(
)
(
)
ln 6
k
−
1
+
ln 12
k
−
3
−
ln 2
k
−
1
+
ln 4
k
−
3
=
ln 6
k
−
1
+
ln 12
k
−
3
−
ln 2
k
−
1
+
2
2
2
2
2
2
2
(
)(
)(
)
1 ln
6
k
−
1 12
k
−
3
4
k
−
3
=
(
)
2
2
k
−
1
dM1: Eliminates ln’s, multiplies up and collects terms to form a polynomial equation in k.
It is dependent upon the previous M
A1: Correct 3TQ. The = 0 may be implied by subsequent work
ddM1: Solves 3TQ. Dependent upon both previous M's.
Can be solved via a graphical calculator (you may need to check the answers)
3
A1: Correct value and no others. Condone
x =
2
Solution with limited working in (c)
Case I
(
)(
)
6
k
−
1 12
k
−
3
1
1
3
ln
=
ln 20
x
or
k
=
SC 11000
2
(
2
k
1 4
)(
k
3
)
2
2
−
−
Case II
(
)(
)
6
k
−
1 12
k
−
3
3
=
20
x
or
k
=
SC 11100
(
)(
)
2
k
−
1 4
k
−
3
2
Attempts direction vector by subtracting(
k and (
)
)
4(a)
5
+
6
+
3
4
+
8
j k either way
+
i
j
i
around.
(
)
(
)
E.g.
r
=
4
i
+
8
j
+
k
+
λ
i
−
2
j
+
2
k ,
r
=
5
i
+
6
j
+
3
k
+
μ
i
−
2
j
+
2
k
(b)
4
+
λ
2
2
+
λ
E.g
PC
=
8
−
2
λ
−−
2
=
10
−
2
λ
1
+
2
λ
1
2
λ
(
)
Uses
PC
.
i
−
2
j
+
2
k
=
0
E.g.
1 2
(
)
2 10
(
2
)
2 2
0
...
+
λ
−
−
λ
+ ×
λ
=
λ
=
(
)
(
)
(
)
E.g I
λ
=
2
=
4
+
2
+
8
−
4
+
1
+
4
k
c
i
j
(
)
(
)
(
)
E.g II
μ
=
1
=
5 1
+
+
6
−
2
+
3
+
2
k
c
i
j
(
6, 4, 5
)
(c)
(
)
OP
′=
p
+
2
PC
=
2
i
−
2
j
+
k
+
2 4
i
+
6
j
+
4
k
(
)
10, 10, 9
(d)
PP
′=
2
PC
2 4
2
6
2
4
2
...
=
+
+
=
4 17
General rule in this question: If no method is shown look for two ''correct'' components for their vector
(a) There are many correct versions so please check the candidates answer carefully
M1: Attempts the direction of l by subtracting(
k and (
)
)
5
+
6
+
3
4
+
8
j k either way around.
+
i
j
i
If no method is shown look for two correct components of (
)
±
1
i
±
2
j
±
2
k or a multiple of this
A1: Any correct equation including the lhs of ‘r =’
4
1
4
1
4
i
=
8
+
λ
−
2
l
=
8
+
λ
−
2
=
8
+
λ
−
2j
Allow in the form
r
but
and is
r
A0
1
2
1
2
1
2k
(b)
M1: Attempts the general vector from P to l forming PC
where C is a general point on l.
4
+
λ
2
Look for an attempt to subtract their
'' 8
−
2
λ
''
and
−
2
either way around
1 2
+
λ
1
(
)
dM1: Attempts
PC
.their
i
−
2
j
+
2
k
=
0
and solves for λ.
It is dependent upon the previous M and the scalar product of this vector must be attempted with their
gradient for the line l to produce and solve a linear equation for λ.
ddM1: Uses their λ to find C. Scored for substituting their correctly found λ into their equation for l
A1: Correct coordinates or vector.
6i
Only penalise an error such as
4j
once, the first time that it is made. Hence award if a mark has already been
5 k
M1: Correct strategy for OP′
Can be implied by two correct coordinates using their P and C
For example look for
OP
′=
OP
+ ×
2
PC
using the coordinates of C found in part (b).
A1: Correct coordinates or vector
(d)
M1: Correct method for PP′
using their values for P and P' or P and C or P' and C . E.g
PP
′=
2
There are many ways to do this but it must be a complete method, not a distance squared.
A1: CAO
Alternative part (b) using shortest distance
M1: Attempts the general vector from P to l.
4
+
λ
2
Look for an attempt to subtract their
'' 8
−
2
λ
''
and
−
2
either way around
1 2
+
λ
1
dM1: Finds an expression for d or d2 , differentiates and sets = 0 to find λ.
As it is a quadratic expression an equivalent method would be via completing the square .
2
d
D
2
2
(
)
2
(
)
2
(
)
2
FYI :
D =
2
+
λ
+
10
−
2
λ
+
2
λ
=
9
λ
−
36
λ
+
104
=
18
λ
−
36
=
0
λ
=
2
d
λ
ddM1: Uses their λ to find C.
A1: Correct coordinates or vector
Alternative part (b) using Pythagoras' Theorem
M1: Attempts the general vector from P to l.
4
2
+
λ
Look for an attempt to subtract their
'' 8
2
''
and
2
either way around
−
λ
−
1 2
1
+
λ
2
2
2
2
2
2
dM1: Uses
PC
+
CB
=
PB
or
PC
+
CA
=
PA
to set up and solve an equation in λ.
2
2
2
E.g Using
PC
CB
PB
the equations are
+
=
(
2
)
2
(
10
2
)
2
(
2
)
2
(
)
2
(
2
2
)
2
(
2
)
2
77
+
λ
+
−
λ
+
λ
+
λ
−
+
−
λ
+
λ
−
=
2
( )
18
λ
−
54
λ
+
36
=
0
λ
=
2 1
ddM1: Uses their λ to find C.
A1: Correct coordinates or vector
(
)
P′
10,10, 9
l
(
)
C
6, 4, 5
(
)
B
5, 6, 3
P
(
2,
2,1
)
−
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