Mathematics Edexcel Advanced Subsidiary GCE
Core 3 (6665) January 2010
Link to past paper on Edexcel website:
www.edexcel.com
The above link takes you to Edexcel’s website. From there you click
QUALIFICATIONS, GCE from 2008, under the subject list click MATHEMATICS,
click QUESTION PAPERS, click JANUARY 2010
If you have any difficulties downloading the question paper call us for help.
These solutions are for your personal use only. DO NOT photocopy or pass
on to third parties. If you are a school or an organisation and would like to
purchase these solutions please contact Chatterton Tuition for further
details.
Question 1
To put this as a single fraction we need to have a common denominator.
Before we do this though see if either fractions simplify.
For the first fraction we can factorise the denominator.
( )
=
()
()()
=
()
We now have
()
-
www.chattertontuition.co.uk 0775 950 1629
Page 1
Mathematics Edexcel Advanced Subsidiary GCE
Core 3 (6665) January 2010
To get the common denominator multiply both denominators together
3(x -1)(3x + 1)
If we multiply the first fraction by
then we will have the denominator we
want
()
If we multiply the second fraction by
then we will have the denominator
()
we want
()
()()
()
– ()() =
()
()()
Tidying up the numerator
()()
=
()()
www.chattertontuition.co.uk 0775 950 1629
Page 2
Mathematics Edexcel Advanced Subsidiary GCE
Core 3 (6665) January 2010
Question 2
a) x3 + 2x2 – 3x – 11 = 0
we need to pick one of the x terms and make that the subject. The trick is
which one?
I can see that the equation we want has 3x + 11 as the numerator so if we add
both sides by 3x + 11 then we might get close
x3 + 2x2 = 3x + 11
factorise the LHS
x2(x + 2) = 3x + 11
divide both sides by (x + 2)
x2 =
square root both sides
x=
b) if you set up your calculator correctly then you can do this without having to
keep keying the whole lot in. Key in 0 = and then AC. Now key in the equation
using ans in place of the xn value.
Every time you press equal you will get the next x value
()
The first iteration gives x2 = 2.345
Pressing the = button again gives x3 = 2.037
Pressing the = button again gives x4 = 2.059
www.chattertontuition.co.uk 0775 950 1629
Page 3
Mathematics Edexcel Advanced Subsidiary GCE
Core 3 (6665) January 2010
c) first rearrange the equation so that we have the equation equal to 0
-x=0
to show that = 2.057 to 3 decimal places we must look at 2.0565 and 2.0575
and if we get a sign change then that shows that must be between those two
values (ie is 2.057 to 3 decimal places)
when x = 2.0565 we get
(.)
.
- 2.0565 = 0.000826
when x = 2.0575 we get
(.)
.
- 2.0575 = -0.000248
Therefore we do have a sign change so = 2.057
www.chattertontuition.co.uk 0775 950 1629
Page 4
Mathematics Edexcel Advanced Subsidiary GCE
Core 3 (6665) January 2010
Question 3
a) From the formulae booklet we know that
cos(x + ) = cosxcosα – sinxsinα
So set 5cosx – 3sinx = Rcosxcosα – Rsinxsinα
Comparing coefficients
5 = Rcosα
-3 = -Rsinα
Using the identity that sin2α + cos2α = 1
Square both equations and add them
52 + (-3)2 = R2cos2α + R2sin2α = R2(sin2α + cos2α) = R2
25 + 9 = R2
34 = R2
R = √34
Using the fact that
= tanα
Divide the second equation by the first
!
=
= -tanα
tanα =
inverse tan (arctan) both sides
α = 0.5404 c
so we have √34 cos(x + 0.5404)
www.chattertontuition.co.uk 0775 950 1629
Page 5
Mathematics Edexcel Advanced Subsidiary GCE
Core 3 (6665) January 2010
b) we can use part a) to say that
√34 cos(x + 0.5404) = 4
Divide both sides by √34
cos(x + 0.5404) = 0.6859943
inverse cos (cos-1) both sides
x + 0.5404 = 0.8148
before we make the final adjustment to get x we need to find where else cosy
= 0.6859943
cos gets repeated at by subtracting from 2π
2π – 0.8148 = 5.4684
x + 0.5404 =0.8148 or 5.4684
Now subtract 0.5404 from both sides
x = 0.27c or 4.93c
www.chattertontuition.co.uk 0775 950 1629
Page 6
Mathematics Edexcel Advanced Subsidiary GCE
Core 3 (6665) January 2010
Question 4
i) we need to use the quotient rule here
"#
"
=
$%& &%$
%'
%'
(
Let u = ln (x2 + 1)
")
= "
Let v = x
"(
=1
"
"#
"
=
'
,( )
' *+
ii) x = tany =
=
–
,( )
!#
#
we need to take care here as the x and y have been interchanged from their
normal positions
again we can use the quotient rule (but note how I have changed the x and y)
"
"#
=
$%& &%$
%- %(
Let u = siny
")
= cosy
"#
Let v = cosy
"(
= -siny
"#
"
"#
=
##!#!#
./ #
=
./ # #
./ #
=
./ #
= sec2y
www.chattertontuition.co.uk 0775 950 1629
Page 7
Mathematics Edexcel Advanced Subsidiary GCE
Core 3 (6665) January 2010
If we start with the basic identity cos2y + sin2y = 1 and divide throughout by
cos2y we get
./ #
./ #
+
#
./ #
=
./ #
This is the same as
1 + tan2y = sec2y
So we have
"
= sec2y = 1 + tan2y = 1 + x2
"#
So
"#
"
=
www.chattertontuition.co.uk 0775 950 1629
Page 8
Mathematics Edexcel Advanced Subsidiary GCE
Core 3 (6665) January 2010
Question 5
The graph should be symmetrical about the y axis. It crosses the x axis when x
= -1 and 1
www.chattertontuition.co.uk 0775 950 1629
Page 9
Mathematics Edexcel Advanced Subsidiary GCE
Core 3 (6665) January 2010
Question 6
i) there are two transformations happening here. The first is inside the bracket
so is just affecting the x values. This one is a reflection in the y axis. The other
change is outside of the brackets so affects the y values. This is a translation of
1 unit in the positive y direction
ii) there are two transformations happening here. The first is inside the
bracket so is just affecting the x values. When the x values are affected they
do the opposite to what you would think. We are adding 2 units so we move
the curve 2 units in the negative x direction. The other change is outside of the
brackets so affects the y values. This is a translation of 3 units in the positive y
direction
Overall we have vector translation of 0
1
www.chattertontuition.co.uk 0775 950 1629
Page 10
Mathematics Edexcel Advanced Subsidiary GCE
Core 3 (6665) January 2010
iii) Again we have two transformations here. This time they are both stretches.
The first is affecting the x values as it is inside the brackets. When the x values
are affected then the effect is the opposite to what you would expect. The
curve is squashed by a scale factor of ½ parallel to the x axis. The other stretch
is of scale factor 2 parallel to the y axis.
www.chattertontuition.co.uk 0775 950 1629
Page 11
Mathematics Edexcel Advanced Subsidiary GCE
Core 3 (6665) January 2010
Question 7
a) there are two ways to do this; we could either use the quotient rule with u
as 1 and v as cosx, or we can set sec x = cos-1x and then use the chain rule
Quotient Rule
$%& &%$
%'
%'
(
"#
=
let u = 1
")
=0
"
"
let v = cos x
"(
= - sin x
"
"#
"
=
( ./ )( )
=
!
=
!
=
x
x
./ = secx tanx
= secx tanx
./ Chain Rule
Let u = cos x
y = u-1
"#
= -u-2
")
")
= -sin x
"#
=
"#
=-u-2 x -sin x = (
"
"
"
"#
")
x
")
"
./ ) x (-sin x) =
!
=
./ www.chattertontuition.co.uk 0775 950 1629
./ Page 12
Mathematics Edexcel Advanced Subsidiary GCE
Core 3 (6665) January 2010
b) we need to use the product rule here
"#
"
=
(")
"
+
)"(
"
u = e2x
")
= 2e2x
"
v = sec 3x
"(
= 3 sec 3x tan 3x (from part a)
"
"#
"
= 2e sec 38 + 3 e2x sec 3x tan 3x = e2x sec 3x (2 + 3tan 3x)
www.chattertontuition.co.uk 0775 950 1629
Page 13
Mathematics Edexcel Advanced Subsidiary GCE
Core 3 (6665) January 2010
c) turning points are where the gradient is equal to 0
"#
set equal to 0 and solve
"
e2x sec 3x (2 + 3tan 3x) = 0
e2x will not equal 0
sec 3x = 0 also cannot equal 0
2 + 3tan 3x = 0 (this can be solved)
subtract 2 from both sides
3tan 3x = -2
divide both sides by 3
tan 3x =
take inverse tan (arctan) of both sides
3x = arctan ( ) = -0.588c
Before we adjust for the 3x we need to find where else this gets repeated
For the tan curve we just keep adding or subtracting π
3x = -0.5880c, 2.5536c
divide both sides by 3
x = -0.196c (any more would be out of the range)
When x = -0.196c, we can substitute into y = e2x sec 3x
y = 0.812
a = -0.196 and b = 0.812
(-0.196, 0.812)
www.chattertontuition.co.uk 0775 950 1629
Page 14
Mathematics Edexcel Advanced Subsidiary GCE
Core 3 (6665) January 2010
Question 8
Starting with the basic identity
sin2x + cos2x = 1
Divide each term by sin2x
./ +
=
We have
1 + cot2x = cosec2x
Replace cosec2 2x with 1 + cot2 2x
1 + cot2 2x – cot 2x = 1
We now have a quadratic involving cot 2x
Rearrange and set the quadratic equal to 0
cot2 2x – cot 2x = 0
Factorise out the cot 2x
cot 2x (cot 2x – 1) = 0
So either cot 2x = 0 or cot 2x – 1 = 0 in which case cot 2x = 1
cot 2x = 0
cot 2x is the same as
9: cot 2x = 0 means tan 2x = ∞
2x = 90⁰ or 270⁰
x = 45⁰ or 135⁰
cot 2x = 1
cot 2x is the same as
9: cot 2x = 1 means tan 2x = 1
2x = 45⁰ or 225⁰
x = 22.5⁰ or 112.5⁰
x = 22.5⁰, 45⁰, 112.5⁰, 135⁰
www.chattertontuition.co.uk 0775 950 1629
Page 15
Mathematics Edexcel Advanced Subsidiary GCE
Core 3 (6665) January 2010
Question 9
i) a) take e of both sides
3x – 7 = e5
Add 7 to both sides
3x = e5 + 7
Divide both sides by 3
< = x=
Question asked for exact answer so don’t put in your calculator, just leave it as
it is
b) take ln of both sides
ln(3xe 7x+2) = ln 15
split the log
ln (3x) + ln (e7x+2) = ln 15
bring the powers to the front
xln 3 + (7x + 2)ln e = ln 15
lne = 1
xln 3 + (7x + 2) = ln 15
xln 3 + 7x + 2 = ln 15
subtract 2 from both sides
xln3 + 7x = -2 + ln 15
factorise
x(ln 3 + 7) = -2 + ln 15
divide both sides by ln3 + 7
, x=
, Question asked for exact answer so don’t put in your calculator, just leave it as
it is
www.chattertontuition.co.uk 0775 950 1629
Page 16
Mathematics Edexcel Advanced Subsidiary GCE
Core 3 (6665) January 2010
ii) a) f(x) = e2x + 3
y = e2x + 3
to find the inverse function (f-1) swap the x and y around and then rearrange to
make y the subject
x = e2y + 3
subtract 3 from both sides
x – 3 = e2y
take ln of both sides
ln (x – 3) = 2y
divide both sides by 2
,()
y=
now switch y back to f-1(x)
,()
f-1(x) =
the domain of the inverse function is the same as the range of the original
function
the range of e2x is > 0
the range of e2x + 3 is > 3
domain of f-1(x) will be x > 3
b) when we have two functions fg(x) we work from the inside outwards
g(x) = ln (x – 1)
so fg(x) = f(ln(x – 1)) = e2ln(x – 1) + 3
the 2 can move to be a power of (x – 1)
fg(x) = ? @0–1 + 3
fg(x) = (x – 1)2 + 3
anything squared is always ≥ 0 but we also know that from g(x) that x > 1 so (x
– 1)2 > 0 and fg(x) > 0 + 3
range is y > 3
www.chattertontuition.co.uk 0775 950 1629
Page 17
Mathematics Edexcel Advanced Subsidiary GCE
Core 3 (6665) January 2010
If you found these solutions helpful and would like to see some more then visit
our website
www.chattertontuition.co.uk
Chatterton Tuition is a tuition agency local to the area of North Yorkshire.
However we also offer online tuition via Skype. Please call, email or visit our
website for more details.
It should be noted that Chatterton Tuition is responsible for these solutions.
The solutions have not been produced nor approved by Edexcel. In addition
these solutions may not necessarily constitute the only possible solutions.
www.chattertontuition.co.uk 0775 950 1629
Page 18