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Paper Reference(s)

6666/01
Edexcel GCE
Core Mathematics C4
Gold Level (Hard) G1
Time: 1 hour 30 minutes
Materials required for examination Items included with question papers
Mathematical Formulae (Green) Nil

Candidates may use any calculator allowed by the regulations of the Joint
Council for Qualifications. Calculators must not have the facility for symbolic
algebra manipulation, differentiation and integration, or have retrievable
mathematical formulas stored in them.

Instructions to Candidates
Write the name of the examining body (Edexcel), your centre number, candidate number, the
unit title (Core Mathematics C4), the paper reference (6666), your surname, initials and
signature.

Information for Candidates

A booklet ‘Mathematical Formulae and Statistical Tables’ is provided.
Full marks may be obtained for answers to ALL questions.
There are 8 questions in this question paper. The total mark for this paper is 75.

Advice to Candidates
You must ensure that your answers to parts of questions are clearly labelled.
You must show sufficient working to make your methods clear to the Examiner. Answers
without working may gain no credit.

Suggested grade boundaries for this paper:

A* A B C D E
66 59 48 43 37 29

Gold 1 This publication may only be reproduced in accordance with Edexcel Limited copyright policy.
©2007–2013 Edexcel Limited.
 PhysicsAndMathsTutor.com

π
⌠2
1. Use integration to find the exact value of  x sin 2 x dx .
⌡0
(6)
January 2011

2.

Figure 1

3
Figure 1 shows part of the curve y = . The region R is bounded by the curve, the x-axis,
√ (1 + 4 x)
and the lines x = 0 and x = 2, as shown shaded in Figure 1.

(a) Use integration to find the area of R.

(4)

The region R is rotated 360° about the x-axis.

(b) Use integration to find the exact value of the volume of the solid formed.
(5)
January 2009

3. Using the substitution u = 2 + √(2x + 1), or other suitable substitutions, find the exact
value of

4
1
∫ 0 2 + √ (2 x + 1)
dx

giving your answer in the form A + 2ln B, where A is an integer and B is a positive constant.
(8)
June 2013 (R)

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4. A curve C has parametric equations

π π
x = 2sin t, y = 1 – cos 2t, − ≤t≤
2 2

dy π
(a) Find at the point where t = .
dx 6
(4)
(b) Find a cartesian equation for C in the form

y = f(x), –k ≤ x ≤ k,

stating the value of the constant k.

(3)
(c) Write down the range of f(x).
(2)
June 2013

 1  1
   
5. The line l 1 has equation r =  0  + λ  1  .
 −1  0
   

 1  2
   
The line l 2 has equation r =  3  + µ  1 .
6  −1
   

(a) Show that l 1 and l 2 do not meet.

(4)

The point A is on l 1 where λ = 1, and the point B is on l 2 where µ = 2.

(b) Find the cosine of the acute angle between AB and l 1 .

(6)
June 2007

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6. (a) Use the substitution x = u2, u > 0, to show that

1 2
∫ x(2 x − 1)
dx =
∫ u (2u − 1)
du

(3)
(b) Hence show that

9
1 a
∫ 1 x(2 x − 1)
dx = 2ln  
b

where a and b are integers to be determined.

(7)
June 2013

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7.

Figure 2
Figure 2 shows a sketch of the curve C with parametric equations
π
x = 27sec3 t , y = 3tan t , 0≤t≤
3
π
(a) Find the gradient of the curve C at the point where t = .
6
(4)
(b) Show that the cartesian equation of C may be written in the form
2 1
y ( x 3 − 9) 2 ,
= a≤x≤b
stating values of a and b.
(3)

Figure 3
The finite region R which is bounded by the curve C, the x-axis and the line x = 125 is shown
shaded in Figure 3. This region is rotated through 2π radians about the x-axis to form a solid of
revolution.

(c) Use calculus to find the exact value of the volume of the solid of revolution.
(5)
June 2013 (R)

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8.

Figure 2

Figure 2 shows a cylindrical water tank. The diameter of a circular cross-section of the tank
is 6 m. Water is flowing into the tank at a constant rate of 0.48π m3 min−1. At time t minutes, the
depth of the water in the tank is h metres. There is a tap at a point T at the bottom of the tank.
When the tap is open, water leaves the tank at a rate of 0.6πh m3 min−1.

(a) Show that, t minutes after the tap has been opened,

dh
75 = (4 – 5h).
dt

(5)

When t = 0, h = 0.2

(b) Find the value of t when h = 0.5

(6)
June 2010

TOTAL FOR PAPER: 75 MARKS

END

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Question
Scheme Marks
Number

x cos 2 x cos 2 x
1. ∫ x sin 2 x dx =
−
2
+
∫ 2
dx M1 A1 A1

sin 2 x
= ... + M1
4
π
π
[ ... ]02 = M1 A1
4

[6]

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Question
Scheme Marks
Number

2 2
3 − 12
2 (a) Area(R) = ∫
0 (1 + 4 x)
dx
= ∫ 3(1 + 4 x)
0
dx

Integrating 3(1 + 4 x) − to give

1
2

2 M1
± k (1 + 4 x) 2 .
1
 3(1 + 4 x) 12 
= 
2 .4
1
  0 Correct integration.
A1
Ignore limits.
2
=  32 (1 + 4 x) 2 
1

 0

Substitutes limits of 2 and 0 into a

= ( 3
2 )
9 − ( 32 (1) ) changed function and subtracts the M1
correct way round.

= 9
2
− 3
2
= 3 (units) 2 3 A1
(4)
(Answer of 3 with no working scores
M0A0M0A0.)

 3 
2 2
Use of V = π ∫ y 2 dx .
(b)
Volume = π 
0 
∫
 (1 + 4 x) 

dx
Can be implied. Ignore limits and B1
dx .

2
9
= (π ) ∫ dx
0
1 + 4x

2 ± k ln 1 + 4 x M1
= (π )  94 ln 1 + 4x  0 9
4
ln 1 + 4x A1

Substitutes limits of 2 and 0

= (π ) ( 94 ln 9 ) − ( 94 ln1) and subtracts the correct way round.
dM1
Note that ln1 can be implied as equal to 0.

So Volume = 94 π ln 9 9
π ln 9 or 92 π ln 3 or 18
π ln 3 A1 oe
4 4
isw
Note the answer must be a one term exact Note
= that 94 π ln 9 + c (oe.) would (5)
value. Note, also you can ignore subsequent be awarded the final A0.
working here.
[9]

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Question
Scheme Marks
Number
4

∫
1
3. dx , u =
2 + (2 x + 1)
0 2 + (2 x + 1)
Either
du −
1
dx M1
du −
1
dx =± K (2 x + 1) 2
or = ± λ (u − 2)
= (2 x + 1) 2 or = u−2 dx du
dx du 1
du − dx
Either = (2 x + 1) 2 or = (u − 2) A1
dx du
 
∫ ∫
1 1
 =dx  (u − 2) du Correct substitution
 2 + (2 x + 1)  u (Ignore integral sign and du ). A1

∫
 2
= 1 −  du An attempt to divide each term by u. dM1
 u
± Au ± B ln u ddM1
= u − 2ln u
u − 2ln u A1 ft
Applies limits of 5 and 3 in u
{So [u − 2ln u ] } = ( 5 − 2ln 5) − ( 3 − 2ln 3)
5
3
or 4 and 0 in x in their integrated
function and subtracts the correct
M1
way round.
3 A1
= 2 + 2ln   3
5 2 + 2ln   cao
 5  cso
[8]

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Question
Scheme Marks
Number
π π
4. 2sin t ,
x= 1 − cos 2t
y= {=
2sin t} ,
2
−
2
t
2
dx dy
dx dy At least one of or
= 2cos t , = 2sin 2t or dt dt B1
dt dt
(a) correct.
dy
= 4sin t cos t dx dy
dt Both and are correct. B1
dt dt
dy
Applies their divided by their
dy 2sin 2t 
4cos t sin t  dt
So,
= =  = 2sin t 
dx 2cos t  2cos t  dx π
and substitutes t = into M1;
 2π  dt 6
2sin  
π dy  6 ; 1 dy
At =
t= , = their .
6 dx π  dx
2cos  
6 dy A1 cao
Correct value for of 1
dx cso
(4)
(b) y =1 − cos 2t =1 − (1 − 2sin 2 t ) M1
= 2sin 2 t
x
2
x2 x2
So, y = 2   or y = or y= or equivalent.
2 2 2
A1 cso
  x 
2
isw
y =−2 2 1 −   
  2  

Either k = 2 or − 2  x  2 B1
(3)
(c) Range: 0  f ( x)  2 or 0  y  2 or
B1 B1
0f 2
(2)
[9]

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Question
Scheme Marks
Number

5. (a) If l 1 and l 2 intersect then:

1  1  1 2
       
 0  + λ1= 3 + µ 1 
 −1 0 6  −1
       

i : 1 + λ = 1 + 2µ (1)
Any two of j : Writes down any two of these
λ= 3 + µ (2) M1
equations correctly.
k: −1 = 6 − µ (3)

Solves two of the above

=
(1) & (2) yields λ 6,
= µ 3
equations to find …
=
(1) & (3) yields λ 14,
= µ 7 either one of λ or µ correct A1
=
(2) & (3) yields λ 10,
= µ 7
both λ and µ correct A1

checking eqn (3), -1 ≠ 3 Complete method of putting their

values of λ and µ into a third
Either checking eqn (2), 14 ≠ 10 B1
equation to
checking eqn (1), 11 ≠ 15
show a contradiction.

or for example:
this type of explanation is also
checking eqn (3), LHS = -1 , RHS = 3 allowed for B1 .
⇒ Lines l 1 and l 2 do not intersect
[4]

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Question
Scheme Marks
Number

Only one of either

2
  
OA =  1  or
2 5  −1
       
5. (b) λ =1 ⇒ OA = 1 & µ=2 ⇒ OB = 5 5 B1
   4   
 −1   OB =  5  or A(2,1, − 1)
 4
 
or B(5,5,4) .
(can be implied)

5  2  3
        
AB = OB − OA =  5  −  1  =  4  or
 4   −1  5 
Finding the difference

      between their OB and
M1
 −3 
   OA .
BA =  −4  (can be implied)
 −5 
 

AB = 3 i + 4 j + 5k , d 1 = i + j + 0k & θ is M1
angle

 Applies dot product

AB • d 1 3 + 4 + 0 formula between d 1 M1
cos θ =  = ±   
AB . d 1  50 . 2  and their ± AB.
Correct expression. A1

cos θ = 7
10
7
10
or 0.7 or 7
100 A1 cao
but not 7
50 2
[6]

10
marks

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Question
Scheme Marks
Number
dx du 1 − 12 du 1
6. (a) {x =u 2
⇒}
du
=2u or = x
dx 2
or =
dx 2 x
B1

 
∫ ∫
1 1
 dx  = 2
2u du M1
 x(2 x − 1)  u (2u − 1)

∫
2 A1 *
= du
u (2u − 1) cso
(3)
(b) 2 A B
≡ + ⇒ 2 ≡ A(2u − 1) + Bu
u (2u − 1) u (2u − 1)
u= 0 ⇒ 2= − A ⇒ A= −2
M1 A1
u = 2 ⇒ 2= 2B ⇒ B =4
1 1

Integrates
M N
+ , M ≠ 0 , N ≠ 0 to
∫ ∫
2 −2 4 u (2u − 1)
So = du + du M1
u (2u − 1) u (2u − 1)
obtain any one of ± λ ln u or
± µ ln(2u − 1)
− 2ln u + 2ln(2u − 1) At least one term correctly
= A1 ft
followed through
−2ln u + 2ln(2u − 1) . A1 cao
So, [ −2ln u + 2ln(2u − 1)]1
3

Applies limits of 3 and 1

in u or 9 and 1 in x in
= ( −2ln 3 + 2ln(2(3) − 1) ) − ( −2ln1 + 2ln(2(1) − 1) ) their integrated function M1
and subtracts the correct
way round.
− 2ln 3 + 2ln 5 − (0)
=
5 5 A1 cso
= 2ln   2ln  
3 3 cao
(7)
[10]

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Question
Scheme Marks
Number
π
7.=x 27sec
= 3
t , y 3tan t , 0t 
3
dx
At least one of or
dt
B1
dy
dx dy correct.
(a) = 81sec 2 t sec t tan t , = 3sec 2 t dt
dt dt
dx dy
Both and are
dt dt B1
correct.
dy
dy 3sec 2 t  1 cos t cos 2 t  Applies their
= =  = =  dt
3
dx 81sec t tan t 27sec t tan t 27 tan t 27sin t  M1;
dx
divided by their
π dy 3sec 2 ( π6 ) 4  3 1 dt
At t = , = = =  = 
6 dx 81sec ( 6 ) tan ( 6 ) 72  54 18 
3 π π 4 A1 cao
72 cso
(4)
2 2
 y
2
  x   x 
{1 + tan=t sec 2 t } ⇒ 1 +  =
3
(b) 2
 3  =  M1
3   27    
 27 
 
2 1

⇒1+ =
y2 x 3
9 9
2
⇒ 9 + y2 = x 3 ⇒ y= ( 2
x3 − 9 ) 2
* A1 * cso
=a 27
= and b 216 or 27  x  216 =a 27
= and b 216 B1
(3)
2
 1

∫ (
For π  x − 9  )
2
3 2
2
 2 125 1
 125
 
∫ ( ) ∫ (x )   B1
2
(c) π
V=  x3 − 9 2
 dx or π 3
− 9 dx
 
∫(x )
27   27 2
or π 3
−9
5
Either ± Ax 3 ± Bx or
125
3 53
M1
3 5 
= {π }  x 3 − 9 x  x oe
5  27 5
3 53
x − 9 x oe A1
5
 3 5
 3 5

= {π }   (125) 3 − 9(125)  −  (27) 3 − 9(27)  
 dM1
 5  5 
= {π } ( (1875 − 1125) − (145.8 − 243) )
4236 π 4236 π
= or 847.2π or 847.2π A1
5 5
(5)
[12]

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Questio
n Scheme Marks
Number

dV
8. (a) = 0.48π − 0.6π h M1 A1
dt
dV dh
V= 9π h ⇒ = 9π B1
dt dt
dh
π
9= 0.48π − 0.6π h M1
dt
dh
Leading to 75 = 4 − 5h  cso A1 (5)
dt

75
(b)
∫4 − 5h
dh = ∫ 1dt separating variables M1

−15ln ( 4 − 5h ) =t ( +C ) M1 A1
−15ln ( 4 − 5h ) =
t +C
When t = 0 , h = 0.2
−15ln 3 = C M1
t =15ln 3 − 15ln ( 4 − 5h )
When h = 0.5
 3 
t=15ln 3 − 15ln1.5 =15ln  =15ln 2 awrt 10.4 M1 A1 (6)
 1.5 

[11]

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Question 1

This proved a good starting question and full marks were common. A few candidates
differentiated the expression, using the product rule. However, the great majority realised that
integration by parts was necessary and such errors as were seen usually arose from integrating
sin 2x and cos 2x incorrectly. Both errors of sign and multiplying (rather than dividing) by 2
were not uncommon. Most knew how to use the limits and complete the question. In some cases
π
the numerically correct answer, , was obtained after incorrect working. In these cases, the final
4
accuracy mark was not awarded.

Question 2

Q2 was generally well answered with many successful attempts seen in both parts. There were
few very poor or non-attempts at this question.
1
In part (a), a significant minority of candidates tried to integrate 3(1 + 4 x) . Many candidates,
2

1
however, correctly realised that they needed to integrate 3(1 + 4 x) − . The majority of these
2

candidates were able to complete the integration correctly or at least achieve an integrated
1
expression of the form k (1 + 4 x) . Few candidates applied incorrect limits to their integrated
2

expression. A noticeable number of candidates, however, incorrectly assumed a subtraction of

zero when substituting for x = 0 and so lost the final two marks for this part. A minority of
candidates attempted to integrate the expression in part (a) by using a substitution. Of these
candidates, most were successful.

∫
In part (b), the vast majority of candidates attempted to apply the formula π y 2 dx , but a few of
2
them were not successful in simplifying y . The majority of candidates were able to integrate
9 9
to give ln 1 + 4 x . The most common error at this stage was for candidates to omit
1 + 4x 4
dividing by 4. Again, more candidates were successful in this part in substituting the limits
correctly to arrive at the exact answer of 94 π ln 9. Few candidates gave a decimal answer with no
exact term seen and lost the final mark.

Question 3

This question was generally well answered with about 57% of candidates gaining at least 7 of the
8 marks available and about 46% of candidates gaining all 8 marks. About 10% of candidates,
however, did not score on this question.
Most candidates followed the advice given in the question and used the substitution
1
du −
u = 2 + √(2x + 1). Most candidates differentiated this correctly to give either = (2 x + 1) 2 or
dx
1
dx du −
= u – 2 although a few differentiated incorrectly to give = 12 (2 x + 1) 2 . The majority
du dx
(u − 2)
were then able to apply the substitution and reach an integral of the form k ⌠  du. Whilst
⌡ u

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most candidates reaching this stage then correctly divided through by u and integrated term by
term to reach an expression of the form k(ln u – u), a few resorted to integration by parts or
partial fractions and were generally less successful. Most candidates applied changed limits of 5
and 3 correctly to their integrated function in u and a few converted back to a function in x and
used limits of 4 and 0 correctly. Disappointingly a number of candidates did not express their
answer in the form A + 2 ln B and gave answers such as 2 – 2 ln 53 or 2 + ln 259 .

Question 4
This question discriminated well between candidates of all abilities, with about 59% of
candidates gaining at least 6 of the 9 marks available and about 17% of candidates gaining all 9
marks. Part (a) was found to be accessible to most, part (b) less well answered and part (c) often
either not attempted or incomplete.

In part (a), the majority of candidates were able to apply the process of parametric differentiation
π dy
followed by substitution of t = into their . Occasional sign errors were seen in the
6 dx
differentiation of both x and y and the 2 was not always treated correctly when differentiating
dy
y= 1 − cos 2t resulting
= in λ sin 2t ( λ ≠ 2 ) . Other common mistakes included obtaining
dt
dy
t ± 2sin 2t or 2sin t for . A small number of candidates believed rewriting y as 2sin 2 t
dt
π
made the differentiation of y easier. Most candidates showed the substitution of t = into their
6
dy dy
but many wrote down the numerical answer. Some candidates achieved = 1 erroneously
dx dx
dx dy dx dy
by either dividing by or from writing =− 2 cos t , − 2sin 2t. Few candidates
=
dt dt dt dt
dy
formed the Cartesian equation (required in part (b)) and used this to correctly find .
dx

In part (b), sign errors, bracketing errors, manipulation errors, an inability to remember the
2
x x2
double angle formulae for cos 2t , and expanding   to give were all common mistakes.
2 4
1
The candidates who used cos 2t ≡ 1 − 2sin 2 t were more successful in achieving y = x 2 when
2
2 2 2
compared to those who used either cos 2t ≡ 2 cos t − 1 or cos 2t ≡ cos t − sin t. A small
minority of candidates made t the subject in one of their parametric equations and found a correct
 x 
alternative Cartesian form of y= 1 − cos  2sin −1    . Those candidates who attempted to find k
  2 
1
usually found the correct answer of k = 2 , although some incorrectly stated that k = or
2
k = λπ .

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Part (c) was not attempted by some and those that did often that gave partial or incorrect
solutions such as f ( x)  0 , f ( x)  0 , 0 < y < 2, 0  x  2, f ( x)  2, − 2  y  2, etc.
A number of candidates also used incorrect notation for range.

Question 6

This question discriminated well between candidates of all abilities, with about 44% of
candidates gaining at least 5 of the 10 marks available and about 28% of candidates gaining all 10
marks. Part (a) was found to be accessible, but most of the marks in part (b) depended on
2
candidates identifying the correct strategy for integrating with respect to u.
u (2u − 1)

Part (a) required candidates to ‘show that’, so it was expected that solutions would show clear
dx du 1 − 12
steps to the printed answer given. Most candidates found either = 2u or = x , although
du dx 2
du
a few replaced dx with instead of with 2u du. In some cases dx was replaced erroneously
2u
by du or omitted throughout. The function of x was usually converted to a function of u
correctly. There were some incorrect algebraic moves seen in attempting to reach the final answer
especially in cases containing an error with dx . The final answer was sometimes written with the
integral sign or du missing – thus not fully showing the answer required.

In part (b), only a minority realised that partial fractions were needed, often after attempting
integration in several different ways. Once expressed in partial fractions, correct integration

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5
usually followed leading to a correct answer of 2 ln   . Some candidates did make algebraic
3
−4
slips when forming their partial fractions or integrated incorrectly to give − 4 ln(2u − 1) .
(2u − 1)

2 2 2  1 1
Some candidates rewrote incorrectly as + or 2  2 −  . Other
u (2u − 1) u (2u − 1)  2u u
2 1 1 
candidates wrote × and integrated this to give ( 2 ln u )  ln(2u − 1)  or tried
u (2u − 1) 2 
unsuccessfully to use a method of integration by parts. There were consequently many other
incorrect versions of the integrated function which often, but not always, included a ‘ln’ term as
suggested by the answer given in the question. Most candidates applied changed limits of 3 and 1
to their integrated function in u. Some candidates, however, converted back to a function in x and
used limits of 9 and 1. Erroneous limits included 0 and 3 for u, and 1 and 81 for x.

Question 7

This question was challenging and discriminated well between candidates of all abilities, with
about 40% of candidates gaining at least 9 marks of the 12 marks available and about 8% gaining
all 12 marks. A significant number of candidates scored full marks in part (c).

In part (a), the majority of candidates were able to apply the process of parametric differentiation
π dy
followed by substitution of t = into their . Some candidates could not differentiate 27 sec3 t
6 dx
π dy
correctly whilst others did not achieve 181 after substituting t = into a correct .
6 dx
In part (b), many candidates were able to use a correct trigonometric identity in order to eliminate
t and achieve an equation in x and y. Whilst most candidates used 1 + tan2 t ≡ sec2 t, some
sin t
candidates were successful with using tan t ≡ and sin2 t + cos2 t ≡ 1. Inevitably there was
cos t
some fudging to arrive at the correct answer and some candidates wrote down and applied an
incorrect identity. Some candidates did not attempt to find the values of a and b and those that
did often gave incorrect values such as a = 0, b = 27 or b as infinity.
⌠
In part (c), the majority of candidates were able to apply volume formula π  y 2 dx on
⌡
2 1
⌠
y = ( x 3 − 9) 2 . A number of candidates, however, used incorrect formulae such as 2π  y 2 dx or
⌡
⌠ 2 ⌠
 y dx or even  y dx .
⌡ ⌡
2
Most candidates where confident with integrating x − 9 although some applied incorrect
3

x-limits of 0 and 27 or 0 and 125 to their integrated function. A significant minority did not use
the hint in part (b) and attempted to find the volume by parametric integration. Whilst a number

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⌠
of them were able to write down 729π  tan 3 t sec 3 t dt , it was rare to see this integrated
⌡
correctly.

Question 8

Many found part (a) difficult and it was quite common to see candidates leave a blank space here
and proceed to solve part (b), often correctly. A satisfactory proof requires summarising the
dV
information given in the question in an equation, such as= 0.48π − 0.6π h , but many could
dt
dh
not do this or began with the incorrect = 0.48π − 0.6π h . Some also found difficulty in
dt
obtaining a correct expression for the volume of water in the tank and there was some confusion
as to which was the variable in expressions for the volume. Sometimes expressions of the form
V = π r 2 h were differentiated with respect to r, which in this question is a constant. If they started
appropriately, nearly all candidates could use the chain rule correctly to complete the proof.

Part (b) was often well done and many fully correct solutions were seen. As noted in the
introduction above, some poor algebra was seen in rearranging the equation but, if that was done
correctly, candidates were nearly always able to demonstrate a complete method of solution
although, as expected, slips were made in the sign and the constants when integrating. Very few
candidates completed the question using definite integration. Most used a constant of integration
(arbitrary constant) and showed that they knew how to evaluate it and use it to complete the
question.

Statistics for C4 Practice Paper G1

Mean score for students achieving grade:

Max Modal Mean
Qu ALL A* A B C D E U
score score %
1 6 76 4.55 5.92 5.34 4.47 3.21 2.22 1.41 0.64
2 9 70 6.27 7.98 6.00 4.33 3.11 2.49 0.91
3 8 69 5.54 7.67 6.55 4.78 3.02 2.21 1.16 0.34
4 9 9 63 5.63 8.28 7.06 5.89 4.72 3.59 2.44 1.17
5 10 60 5.95 8.06 6.01 4.55 3.23 2.10 1.10
6 10 10 54 5.43 9.51 7.34 5.44 3.86 2.63 1.58 0.65
7 12 55 6.63 10.35 7.67 5.03 3.19 2.83 1.78 0.36
8 11 40 4.36 10.25 6.65 3.33 1.38 0.48 0.18 0.06
75 59 44.36 56.65 40.95 28.26 20.30 13.14 5.23

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