Mark Scheme (Final) January 2009
GCE
GCE Core Mathematics C4 (6666/01)
Edexcel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holborn, London WC1V 7BH
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General Marking Guidance
All candidates must receive the same treatment. Examiners must mark the first candidate in exactly the same way as they mark the last. Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do rather than penalised for omissions. Examiners should mark according to the mark scheme not according to their perception of where the grade boundaries may lie. There is no ceiling on achievement. All marks on the mark scheme should be used appropriately. All the marks on the mark scheme are designed to be awarded. Examiners should always award full marks if deserved, i.e. if the answer matches the mark scheme. Examiners should also be prepared to award zero marks if the candidates response is not worthy of credit according to the mark scheme. Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and exemplification may be limited. When examiners are in doubt regarding the application of the mark scheme to a candidates response, the team leader must be consulted. Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response.
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January 2009 6666 Core Mathematics C4 Mark Scheme
Question Number 1. (a) Scheme C: y 2 3 y = x3 + 8 Marks
dy = dx
2y
dy dy 3 = 3x 2 dx dx
Differentiates implicitly to include either dy dy dy M1 ky or 3 . (Ignore = .) dx dx dx
Correct equation. A1 A correct (condoning sign error) attempt to dy dy 3 . M1 combine or factorise their 2 y dx dx Can be implied.
3x 2 2y 3
dy ( 2 y 3) = 3 x 2 dx
dy 3x 2 = dx 2 y 3
A1 oe
[4]
(b)
y = 3 9 3(3) = x3 + 8
Substitutes y = 3 into C. M1
x3 = 8 x = 2
Only x = 2
A1
(2,3)
dy 3(4) dy = =4 dx 6 3 dx
dy = 4 from correct working. dx Also can be ft using their x value and y = 3 in the A1 dy 3x 2 = correct part (a) of dx 2 y 3
[3] 7 marks
1(b) final A1
2
Note if the candidate inserts their x value and y = 3
into
dy 3x dy = , then an answer of = their x 2 , may indicate a dx 2 y 3 dx correct follow through.
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Question Number 2. (a)
Scheme
Marks
Area(R) =
3 1 dx = 3(1 + 4 x) 2 dx (1 + 4 x) 0
Integrating 3(1 + 4 x) 3(1 + 4 x) = 1 2 .4
1
1 2
1 2
to give
1
0
2
k (1 + 4 x) 2 .
M1
Correct integration. A1 Ignore limits.
= 3 (1 + 4 x) 2 2 0
=
3 2
9 ( 3 (1) ) 2 = 3 (units) 2
Substitutes limits of 2 and 0 into a changed function and subtracts the M1 correct way round. 3 A1
[4]
9 2
3 2
(Answer of 3 with no working scores M0A0M0A0.)
(b)
3 dx Volume = (1 + 4 x) 0
Use of V = y 2 dx . Can be implied. Ignore limits and dx .
B1
= ( )
0
9 dx 1 + 4x k ln 1 + 4 x
9 4
= ( ) 9 ln 1 + 4x 0 4
M1 A1
ln 1 + 4x
= ( ) ( 9 ln 9 ) ( 9 ln1) 4 4
Substitutes limits of 2 and 0 dM1 and subtracts the correct way round.
9 4
So Volume =
9 4
ln 9
ln 9 or 9 ln 3 or 2
18 4
ln 3
A1 oe isw [5]
9 marks
Note the answer must be a one term exact value. Note, also you can ignore subsequent working here.
Note that ln1 can be implied as equal to 0.
Note that =
9 4
ln 9 + c (oe.) would be awarded the final A0.
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Question Number 3. (a)
Scheme
27 x 2 + 32 x + 16 A(3 x + 2)(1 x) + B (1 x) + C (3 x + 2) 2
Marks
Forming this identity M1 Substitutes either x = 2 or x = 1 3 into their identity or equates 3 terms or substitutes in values to M1 write down three simultaneous equations. Both B = 4 and C = 3 A1 (Note the A1 is dependent on both method marks in this part.)
x = 2 , 12 3 x = 1,
64 3
+ 16 = ( 5 ) B 3
20 3
= (5)B B = 4 3
27 + 32 + 16 = 25 C 75 = 25 C C = 3
Equate x2:
27 = 3 A + 9C 27 = 3 A + 27 0 = 3 A A=0
x = 0, 16 = 2 A + B + 4C 16 = 2 A + 4 + 12 0 = 2 A A = 0
Compares coefficients or substitutes in a third x-value or B1 uses simultaneous equations to show A = 0.
[4]
(b)
f ( x) =
4 3 + 2 (3 x + 2) (1 x) Moving powers to top on any one M1 of the two expressions
= 4(3 x + 2) 2 + 3(1 x) 1
2 = 4 2 (1 + 3 x ) + 3(1 x) 1 2
= 1(1 + 3 x ) + 3(1 x) 1 2
2
(2)(3) 3 x 2 = 1 1 + ( 2)( 32x ); + ( 2 ) + ... 2! (1)(2) + 3 1 + (1)( x); + ( x) 2 + ... 2!
= 1 3 x + 27 x 2 + ...} + 3 1 + x + x 2 + ...} 4
Either 1 ( 2)( 32x ) or 1 (1)( x) from either first or dM1; second expansions respectively Ignoring 1 and 3, any one A1 correct {..........} expansion. Both {..........} correct. A1
= 4 + 0 x ; + 39 x 2 4
4 + (0 x) ;
39 4
x2
A1; A1
[6]
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Question Number 3. (c)
Scheme
Marks
Actual = f (0.2) =
1.08 + 6.4 + 16 (6.76)(0.8) 23.48 2935 = = 4.341715976... = 5.408 676
Attempt to find the actual value of f(0.2) or seeing awrt 4.3 and believing it is candidates actual f(0.2). Candidates can also attempt to find M1 the actual value by using A B C + + 2 (3x + 2) (3x + 2) (1 x) with their A, B and C.
Or
Actual = f (0.2) = 4 3 + 2 (3(0.2) + 2) (1 0.2) 4 2935 = + 3.75 = 4.341715976... = 6.76 676
39 4
Estimate = f (0.2) = 4 +
(0.2) 2
= 4 + 0.39 = 4.39
%age error = 4.39 4.341715976... 4.341715976... 100
Attempt to find an estimate for f(0.2) using their answer to (b) M1
their estimate - actual 100 actual
M1
= 1.112095408... = 1.1 % (2sf )
1.1% A1 cao
[4] 14 marks
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Question Number 4. (a)
Scheme
Marks
d 1 = 2 i + j 4k , d 2 = q i + 2 j + 2 k
As 2 q d1 d 2 = 1 2 4 2
= (2 q) + (1 2) + (4 2)
Apply dot product calculation between two direction vectors, ie. M1 (2 q ) + (1 2) + (4 2)
d1 d 2 = 0 2q + 2 8 = 0
2q = 6 q = 3 AG
Sets d1 d 2 = 0 and solves to find q = 3
A1 cso [2]
(b)
Lines meet where: 11 2 5 q 2 + 1 = 11 + 2 17 4 p 2
i : 11 2 = 5 + q First two of j : 2 + = 11 + 2 k : 17 4 = p + 2 (1) + 2(2) gives: (2) gives: 15 = 17 + = 2 (1) (2) (3) Need to see equations (1) and (2). Condone one slip. M1 (Note that q = 3 .) Attempts to solve (1) and (2) to find one of either or dM1 Any one of = 5 or = 2 A1 Both = 5 and = 2 A1
2 + = 11 4 = 5
(3) 17 4(5)
= p + 2(2)
Attempt to substitute their and into their k component to give an ddM1 equation in p alone. p =1 A1 cso [6]
p = 17 20 + 4 p = 1
(c)
11 2 r = 2 + 5 1 17 4
5 3 or r = 11 2 2 1 2
Substitutes their value of or into M1 the correct line l1 or l2 .
1 7 or (1, 7, 3) 3
1 Intersect at r = 7 or 3
(1, 7, 3)
A1 [2]
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Question Number
Scheme
Marks
(d)
uuur Let OX = i + 7 j 3k be point of intersection
1 9 8 uuur uuur uuu r AX = OX OA = 7 3 = 4 3 13 16
uuur Finding vector AX by finding the uuur uuu r difference between OX and OA . Can M1 uuur be ft using candidates OX .
uuu uuu uuu r r r uuu r uuur OB = OA + AB = OA + 2 AX
9 uuu r OB = 3 + 13 8 2 4 16
9 3+ 13
uuur 2 their AX
dM1
7 uuu r uuu r Hence, OB = 11 or OB = 7 i + 11 j 19 k 19
7 11 or 7 i + 11 j 19 k 19 or ( 7, 11, 19 )
A1
[3] 13 marks
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Question Number
Scheme
Marks
5. (a)
Similar triangles
r 16 2h = r= h 24 3
2
Uses similar triangles, ratios or trigonometry to find either one of these M1 two expressions oe.
AG
1 1 2h 4 h3 V = r 2h = h = 3 3 3 27
Substitutes r =
2h 3
into the formula for the A1 volume of water V.
[2]
(b)
From the question,
dV =8 dt
dV =8 dt dV 12 h 2 4 h2 = or dh 27 9
B1
dV 12 h 2 4 h 2 = = dh 27 9
B1
dh dV dV 9 18 = = 8 = 2 dh dt dt 4 h h2
Candidates
dV dV ; M1; dt dh
12 h 2 9 18 or oe A1 8 or 8 2 4 h h2 27 1 18 or 8 144
When h = 12,
dh 18 1 = = dt 144 8
A1 oe isw
[5] 7 marks
Note the answer must be a one term exact value. Note, also you can ignore subsequent working after 18 . 144
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Question Number
Scheme
Marks
6. (a)
tan
x dx
NB : sec 2 A = 1+ tan 2 A gives tan 2 A = sec2 A 1 = sec 2 x 1 dx = tan x x ( + c )
The correct underlined identity. M1 oe
Correct integration A1 with/without + c
[2]
(b)
x ln x dx
1
3
u = ln x du = 1 dx x dv 3 x 2 v = 2 = dx = x =
1 2 x2
Use of integration by parts formula M1 in the correct direction. Correct expression. A1 An attempt to multiply through k , n , n 2 by 1 and an x xn attempt to ... integrate(process the result); M1 correct solution with/without + c A1 oe
[4]
1 1 1 ln x 2 . dx 2 2x 2x x 1 1 ln x + 2 2x 2
1 dx x3
1 1 1 ln x + 2 ( + c ) 2 2x 2 2x
Correct direction means that u = ln x .
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Question Number
Scheme
Marks
(c)
e3 x dx 1 + ex Differentiating to find any one of the B1 three underlined
Attempt to substitute for e 2 x = f (u ) , dx 1 = x and u = 1 + e x their du e M1* dx 1 or e3 x = f (u ) , their = and du u 1 u = 1 + ex .
du dx 1 dx 1 x = ex , = x , = u = 1 + e dx du e du u 1
e 2 x .e x dx = 1 + ex
(u 1) 2 .e x 1 . x du u e
or =
(u 1)3 1 . du u (u 1)
(u 1) 2 du u u 2 2u + 1 du u 1 du u
(u 1) 2 du u
A1
= u2+
An attempt to multiply out their numerator to give at least three terms and divide through each term by u dM1* Correct integration with/without +c A1 Substitutes u = 1 + e x back into their integrated expression with at least dM1* two terms.
u2 2u + ln u ( + c ) 2
(1 + e x ) 2 = 2(1 + e x ) + ln(1 + e x ) + c 2 = 1 + e x + 1 e 2 x 2 2e x + ln(1 + e x ) + c 2 2 = 1 + e x + 1 e 2 x 2 2e x + ln(1 + e x ) + c 2 2 = 1 e 2 x e x + ln(1 + e x ) 3 + c 2 2 = 1 e 2 x e x + ln(1 + e x ) + k 2
AG
1 2
e 2 x e x + ln(1 + e x ) + k
must use a + c and " 3 " combined. A1 cso 2
[7] 13 marks
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Question Number
Scheme
Marks
7. (a)
At A, x = 1 + 8 = 7 & y = (1) 2 = 1 A(7,1)
A(7,1)
B1
[1]
(b)
x = t 3 8t ,
y = t2 ,
dx dy = 3t 2 8 , = 2t dt dt dy 2t = 2 dx 3t 8 At A, m(T) = 2( 1) 2 2 2 = = = 2 3(1) 8 38 5 5 Their
dy dt
divided by their Correct
dx dt
M1 A1
dy dx
Substitutes for t to give any of the A1 four underlined oe:
Finding an equation of a tangent with their point and their tangent gradient or finds c and uses dM1 y = (their gradient) x + " c " .
T : y ( their 1) = mT ( x ( their 7 ) ) or 1 =
2 5
(7) + c
c =1
14 5
= 9 5
2 Hence T : y = 5 x 9 5
gives T : 2 x 5 y 9 = 0
AG
2x 5 y 9 = 0
A1
cso [5]
(c)
2(t 3 8t ) 5t 2 9 = 0
Substitution of both x = t 3 8t and M1 y = t 2 into T
2t 3 5t 2 16t 9 = 0 (t + 1) {(2t 2 7t 9) = 0} (t + 1) {(t + 1)(2t 9) = 0} A realisation that ( t + 1) is a factor. dM1
t=
9 2
{t = 1 (at A)}
2
t=
9 2
at B
A1
x = ( 9 ) 8( 9 ) = 2 2
2
729 8
36 =
441 8
= 55.125 or awrt 55.1
Candidate uses their value of t to ddM1 find either the x or y coordinate One of either x or y correct. A1 Both x and y correct. A1 awrt
y = ( 9 ) = 81 = 20.25 or awrt 20.3 2 4 Hence B ( 441 , 81 ) 8 4
[6]
12 marks
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Note: dM1 denotes a method mark which is dependent upon the award of the previous method mark. ddM1 denotes a method mark which is dependent upon the award of the previous two method marks. Oe or equivalent.
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January 2009 6666 Core Mathematics C4 Appendix Question 1
Question Number Aliter 1. (a) Way 2 Scheme Marks
C: y 2 3 y = x3 + 8 Differentiates implicitly to include either dx dx M1 kx 2 . (Ignore = .) dy dy Correct equation. A1 2 y 3 = 3x 2 1 dx 1 = dy dy dx
dx = dy
2 y 3 = 3x 2
dx dy
( )
dy dx
Applies
( )
dM1
dy 3x 2 = dx 2 y 3
3x 2 2y 3
A1 oe
[4]
Aliter 1. (a) Way 3
C: y 2 3 y = x3 + 8
gives x 3 = y 2 3 y 8 x = ( y 2 3 y 8)
1 3
2 dx 1 3 = ( y 2 3 y 8 ) ( 2 y 3) dy 3
Differentiates in the form
1 3
( f ( y ) ) ( f ( y ) ) .
2 3
M1
Correct differentiation. A1
dx 2y 3 = 2 3 dy 3 ( y 2 3 y 8) 3 ( y 2 3 y 8) dy = dx 2y 3
2 3 2 3
Applies
dy 1 = dx dx dy
( )
dM1
3 ( x3 ) dy dy 3x 2 = = dx 2 y 3 dx 2y 3
3 ( x3 )
2 3
2y 3
or
3x 2 2y 3
A1 oe
[4]
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Question 2
Question Number Aliter 2. (a) Way 2 Scheme
2
Marks
Area(R) =
3 (1 + 4 x)
dx = 3(1 + 4 x)
0
1 2
dx
{Using substitution u = 1 + 4 x du = 4 } dx {change limits: When x = 0 , u = 1 & when x = 2 , u = 9 } So, Area(R) =
3u
1
1 1 2 4
du
Integrating u
1 2
3 u1 2 = 4 ( 1 ) 2 0 = 3 u 2 2 1
1
to give k u 2 . M1
1
Correct integration. A1 Ignore limits.
3 2
9 ( 3 (1) ) 2
Substitutes limits of either ( u = 9 and u = 1) or in x, ( x = 2 and x = 0 ) into a changed M1 function and subtracts the correct way round . 3 A1
[4]
=
Aliter 2. (a) Way 3
9 2
3 2
= 3 (units) 2
2
Area(R) =
3 1 dx = 3(1 + 4 x) 2 dx (1 + 4 x) 0
{Using substitution u 2 = 1 + 4 x 2u du = 4 1 udu = dx } dx 2 {change limits: When x = 0 , u = 1 & when x = 2 , u = 3 }
So, Area(R) =
3 1 u 2
u du =
3
3 2
du
Integrating to give k u . M1 Correct integration. A1 Ignore limits.
3 = 2
u 1
= ( 3 (3) ) ( 3 (1) ) 2 2
Substitutes limits of either ( u = 3 and u = 1) or in x, ( x = 2 and x = 0 ) into a changed M1 function and subtracts the correct way round . 3 A1
[4]
9 2
3 2
= 3 (units) 2
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Question 3
Question Number Aliter 3. (a) Way 2 Scheme Marks
27 x 2 + 32 x + 16 A(3x + 2)(1 x) + B(1 x) + C (3 x + 2) 2 (1) x 2 terms : 27 = 3 A + 9C x terms : 32 = A B + 12C (2) constants: 16 = 2 A + B + 4C (3)
(2) + (3) gives 48 = 3 A + 16C (4) (1) + (4) gives 75 = 25C C = 3 (1) gives 27 = 3 A + 27 0 = 3 A A = 0 (2) gives 32 = B + 36 B = 36 32 = 4
Forming this identity M1
equates 3 terms. M1
Both B = 4 and C = 3 Decide to award B1 for A = 0
A1 B1
[4]
3. (a)
If the candidate assumes A = 0 and writes the identity 27 x 2 + 32 x + 16 B(1 x) + C (3x + 2) 2 and goes on to find B = 4 and C = 3 then the candidate is awarded M0M1A0B0. If the candidate has the incorrect identity 27 x 2 + 32 x + 16 A(3 x + 2) + B(1 x) + C (3 x + 2) 2 and goes on to find B = 4, C = 3 and A = 0 then the candidate is awarded M0M1A0B1. If the candidate has the incorrect identity 27 x 2 + 32 x + 16 A(3 x + 2) 2 (1 x) + B (1 x) + C (3 x + 2) 2 and goes on to find B = 4, C = 3 and A = 0 then the candidate is awarded M0M1A0B1.
3. (a)
3. (a)
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Question Number
Scheme
Marks
Aliter 3. (b) Way 2
f ( x) =
4 3 + (3 x + 2) 2 (1 x) Moving powers to top on any one M1 of the two expressions
= 4(3 x + 2) 2 + 3(1 x) 1
= 4(2 + 3 x) 2 + 3(1 x) 1
( 2)( 3) 4 = 4 (2) 2 + ( 2)(2) 3 (3 x); + (2) (3x) 2 + 2! (1)(2) ( x) 2 + ... + 3 1 + (1)( x); + 2!
=4
Either (2) 2 ( 2)(2) 3 (3x) or 1 (1)( x) from either first or dM1; second expansions respectively Ignoring 1 and 3, any one A1 correct {..........} expansion. Both {..........} correct. A1
1 4
3 27 4 x + 16 x 2 + ...} + 3 1 + x + x 2 + ...}
= 4 + 0 x ; + 39 x 2 4
4 + (0 x) ;
39 4
x2
A1; A1
[6]
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Question Number
Aliter 3. (c) Way 2
Scheme
Marks
Actual = f (0.2) =
1.08 + 6.4 + 16 (6.76)(0.8) 23.48 2935 = = 4.341715976... = 5.408 676
39 4
Attempt to find the M1 actual value of f(0.2)
Estimate = f (0.2) = 4 +
(0.2) 2
= 4 + 0.39 = 4.39
4.39 100 %age error = 100 4.341715976...
Attempt to find an estimate for f(0.2) using their answer to (b) M1
their estimate 100 100 actual
M1
= 100 101.1120954 = 1.112095408... = 1.1 % (2sf )
1.1% A1 cao
[4] 3. (c)
Note that:
%age error =
4.39 4.341715976... 4.39
100
Should be awarded the final marks of M0A0
= 1.0998638... = 1.1 % (2sf )
3. (c)
Also note that: 4.341715976... %age error = 100 100 4.39 = 1.0998638... = 1.1 % (2sf ) so be wary of 1.0998638 Should be awarded the final marks of M0A0
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Question 4
Question Number
4. (a)
Aliter 4. (b) Way 2
Scheme
Marks
2q + 2 8 is sufficient for M1. Only apply Way 2 if candidate does not find both and . Lines meet where: 11 2 5 q 2 + 1 = 11 + 2 17 4 p 2
i : 11 2 = 5 + q
First two of j : 2 + = 11 + 2 k : 17 4 = p + 2
(2) gives = 9 + 2 (1) gives 11 2(9 + 2 ) = 5 3
(1) (2) (3)
Need to see equations (2) and (2). Condone one slip. M1 (Note that q = 3 .)
Attempts to solve (1) and (2) to find one of either or dM1
11 18 4 = 5 3 gives: 11 18 + 5 = = 2 Any one of = 5 or = 2 A1
(3) gives 17 4(9 + 2 ) = p + 2
Candidate writes down a correct equation containing p and one of either A1 or which has already been found. Attempt to substitute their value for ( = 9 + 2 ) and into their ddM1 k component to give an equation in p alone.
p =1
(3) 17 4(9 + 2(2)) = p + 2(2)
17 20 = p 4 p = 1
A1 cso
[6]
4. (c)
If no working is shown then any two out of the three coordinates can imply the first M1 mark. 1 Intersect at r = 7 or 3
M1 1 7 or (1, 7, 3) 3
(1, 7, 3)
A1
[2]
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Question Number Aliter 4. (d) Way 2
Scheme
uuur Let OX = i + 7 j 3k be point of intersection
Marks
1 9 8 uuur uuur uuu r AX = OX OA = 7 3 = 4 3 13 16
uuu uuur uuu r r uuur uuur OB = OX + XB = OX + AX
Finding the difference between their uuur uuu r OX (can be implied) and OA . 1 9 M1 uuur AX = 7 3 3 13
1 8 uuu r OB = 7 + 4 3 16 7 uuu r uuu r Hence, OB = 11 or OB = 7 i + 11 j 19 k 19
uuur uuur their OX + their AX 7 11 or 7 i + 11 j 19 k 19 or ( 7, 11, 19 )
dM1
A1
[3]
Aliter 4. (d) Way 3
At A, = 1. At X, = 5. Hence at B, = 5 + (5 1) = 9 11 2 uuu r OB = 2 + 9 1 17 4 7 uuu r uuu r Hence, OB = 11 or OB = 7 i + 11 j 19 k 19
B = ( their X ) + ( their X their A ) B = 2 ( their X ) ( their A )
M1
Substitutes their value of into the line l1. dM1 7 11 or 7 i + 11 j 19 k 19 or ( 7, 11, 19 )
[3]
A1
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Question Number
Aliter 4. (d)
Scheme
uuu r OA = 9i + 3j + 13k
Marks
uuur and the point of intersection OX = i + 7 j 3k
Way 4
9 Minus 8 1 3 Plus 4 7 13 Minus 16 3
Finding the difference uuur between their OX (can be uuu r implied) and OA . 1 9 M1 uuur AX = 7 3 3 13
1 Minus 8 7 7 Plus 4 11 3 Minus 16 19
7 uuu r uuu r Hence, OB = 11 or OB = 7 i + 11 j 19 k 19
uuur uuur their OX + their AX 7 11 or 7 i + 11 j 19 k 19 or ( 7, 11, 19 )
dM1
A1
[3]
Aliter 4. (d)
uuu r and OB = ai + bj + ck uuur and the point of intersection OX = i + 7 j 3k As X is the midpoint of AB, then
uuu r OA = 9i + 3j + 13k
Way 5
(1, 7, 3) =
9 + a 3 + b 13 + c , , 2 2 2
Writing down any two of these equations correctly. M1
a = 2(1) 9 = 7 b = 2(7) 3 = 11 c = 2( 3) 13 = 19 An attempt to find at least two of a, b or c. dM1 7 11 or 7 i + 11 j 19 k 19 or ( 7, 11, 19 ) or a = 7, b = 11, c = 19
[3]
7 uuu r uuu r Hence, OB = 11 or OB = 7 i + 11 j 19 k 19
A1
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Question Number Aliter 4. (d) Way 6
Scheme
uuur Let OX = i + 7 j 3k be point of intersection
Marks
1 9 8 uuur uuur uuu r AX = OX OA = 7 3 = 4 3 13 16
uuur and AX = 64 + 16 + 256 = 336 = 4 21
Finding the difference uuur between their OX (can be uuu r implied) and OA . 1 9 uuur AX = 7 3 M1 3 13 uuur Note AX = 336 would imply M1.
1 11 2 10 + 2 uuur uuur uuu r BX = OX OB = 7 2 + = 5 3 17 4 20 + 4
uuur uuur Hence BX = AX = 336 gives
( 10 + 2 )
+ ( 5 ) + ( 20 + 4 ) = 336
2 2
Writes distance equation uuur 2 of BX = 336 where uuur uuur uuu r BX = OX OB and dM1 11 2 uuu r OB = 2 + 17 4
100 40 + 4 2 + 25 10 + 2 + 400 160 + 16 2 = 336 21 2 210 + 525 = 336 21 2 210 + 189 = 0
2 10 + 9 = 0 ( 1)( 9) = 0
11 2(9) uuu r At A, = 1 and at B = 9, so, OB = 2 + 9 17 4(9) 7 uuu r uuu r Hence, OB = 11 or OB = 7 i + 11 j 19 k 19 7 11 or 7 i + 11 j 19 k 19 or ( 7, 11, 19 )
[3]
A1
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Question 5
Question Number
5. (a)
Scheme
Marks
Similar shapes either
1 3
(16) 2 24 24
V = h
3
or
V h = 2 1 24 3 (16) 24 V h = 2 1 24 3 r (24)
3
1 3
r 2 (24) 24
V = h
Uses similar shapes to find either one of M1 these two expressions oe.
or
4 h3 h V = 2048 = 27 24
3
AG
Substitutes their equation to give the correct formula for the volume A1 of water V.
[2]
5. (a)
Candidates simply writing: V = 4 1 3 h 9 3 or 1 16 V = h3 3 24
2
would be awarded M0A0.
dV = 8 or V = 8t dt
1 1
(b)
From question,
1
dV = 8 V = 8t ( + c ) dt
1 1
B1
27V 3 27(8t ) 3 54t 3 2t 3 h= h= = = 3 4 4
27(8t ) 3 54t 3 2t 3 or or 3 4
B1
dh 2 1 2 = 3 t 3 dt 3
1 3
dh 2 = k t 3 ; M1; dt dh 2 3 1 2 = 3 t 3 dt 3
3
1
A1 oe
12 When h = 12, t = = 32 2 3 So when dh 1 2 3 1 3 2 3 h = 12, = = = 3 dt 8 32 1024
1 2 1
1 8
A1 oe
[5]
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Question 7
Question Number
7. (a)
Scheme
Marks
It is acceptable for a candidate to write x = 7, y = 1, to gain B1.
A(7,1)
B1
[1]
Aliter (c) Way 2
x = t 3 8t = t (t 2 8) = t ( y 8)
So, x 2 = t 2 ( y 8) 2 = y ( y 8) 2 2 x 5 y 9 = 0 2 x = 5 y + 9 4 x 2 = (5 y + 9) 2 Hence, 4 y ( y 8) 2 = (5 y + 9) 2 4 y ( y 2 16 y + 64) = 25 y 2 + 90 y + 81 4 y 3 64 y 2 + 256 y = 25 y 2 + 90 y + 81 4 y 3 89 y 2 + 166 y 81 = 0 Forming an equation in terms of y M1 only.
( y 1)( y 1)(4 y 81) = 0
( y 1) is a factor.
A realisation that
dM1
Correct factorisation A1 y=
81 4
= 20.25 (or awrt 20.3)
Correct y-coordinate (see below!)
x 2 = 81 ( 81 8) 2 4 4
x=
Hence B ( 441 , 81 ) 8 4
441 8
= 55.125 (or awrt 55.1)
Candidate uses their y-coordinate ddM1 to find their x-coordinate. Decide to award A1 here for A1 correct y-coordinate. Correct x-coordinate A1
[6]
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Question Number
Aliter 7. (c) Way 3
Scheme
Marks
t=
So x =
( y)
3
( y)
Forming an equation in terms of y M1 only.
2 x 5 y 9 = 0 yields 2
( y)
16
3
( y ) 5y 9 = 0 ( y) 9 = 0 ) } ) }
( y)
5 y 16
( (
y +1 y +1
81 4
){( ){(
3
2y 7 y 9 = 0
A realisation that
y + 1 is a factor. dM1
y +1 2 y 9 = 0
)(
Correct factorisation. A1
Correct y-coordinate (see below!)
y=
= 20.25 (or awrt 20.3)
81 4 81 4
x=
( ) 8( )
441 8
x=
= 55.125 (or awrt 55.1)
Hence B ( 441 , 81 ) 8 4
Candidate uses their y-coordinate ddM1 to find their x-coordinate. Decide to award A1 here for A1 correct y-coordinate. Correct x-coordinate A1
[6]
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