Examination Style Paper
1.
The point Plies on the curve with equation
y  3x  ln 2 x 
2
The x-coordinate of P is 0.5.
Find an equation of the tangent to the curve at the point P.
Total: 6 marks
2.
Express
3x 2  2 x
2
 2
 x  1 3x  2  x  1
as a single fraction in its simplest form
Total: 7 marks
3.
g( x)  5sin x  3cos x
Given that g(x) = R sin( x   ), where R  0 and 0    90o
(a) find the value of R and the value of .
(4)
(b) (i) Write down the maximum value of g(x).
(1)
(ii) Find, to the nearest degree, the smallest positive value of x for which the
maximum value occurs.
(2)
Total: 7 marks
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Figure 1
4.
y
(3, 1)
M
A  4 kg 
–6
 
B  m kg 
x
6
Figure 1 shows the graph of y = f(x) 6  x  6 . The point M (3,1) is the minimum
turning point of the graph.
Sketch on separate diagrams the graphs of :
(a) y = f(x) - 1
(2)
(b) y = |f(x + 3)|
(3)
(c) y = f(|x|)
(3)
Show on each graph the coordinates of any minimum turning points.
Total: 8 marks
5.
f ( x)  x5  x 2  20
(a) Show that the equation f(x) = 0 can be written as
20
x  3 1 2
x
(3)
The equation f(x) = 0 has a root  in the interval (1, 2).
(b) Use the iterative formula
20
xn 1  3 1  2
xn
with x0  2 , to find, to 2 decimal places, the values of x1 , x2 , x3 and x4 .
(c) By choosing a suitable interval, prove that = 1.88 correct to 2 decimal
places.
(4)
(3)
Total: 10 marks
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6.
The functions f and g are defined by
f : x  ln(3  2 x)
x  , x  1.5
g : x  e2 x  1
x
(a) Find gf(-1)
(b) Find f
1
(4)
 x
(c) Find the exact value of x for which f
1
(3)
 x  g  x .
(5)
Total: 12 marks
7.
p
, 0  x  2 and p>0 and q > 0,
q
find cosec2x in terms of p and q.
(i) Given that sin x 
(ii) Solve for 0  x   , giving your answers as multiples of ,
2 cot 2 2 x  3cosec2 x  0
(4)
(8)
Total: 12 marks
8.
(a) Differentiate with respect to x
(i) y  x3e 2 x 1
(4)
(ii) y 
2
sin( x  1)
2x
(4)
(b) Given that x  3 tan  2 y  1 , find
dy
in terms of x.
dx
(5)
Total: 13 marks
Total for paper: 75 marks
THE END
© Edexcel Limited 2008
3
Worked Solutions
1.
2
x  0.5  y  ln1  32  
9
4
B1
dy
1 
 2  ln 2 x  3 x    3
dx
x 
x  0.5 
dy
3

 2  0    2  3  15
dx
2

Differentiate using the
chain rule, then let x=0.5
to obtain the gradient of
the tangent.
M1A1
Use y  y1  m  x  x1 
A1
M1A1
Equation of tangent is
y  94  15  x  0.5  or y  15 x  5.25
2.
x  3x  2 
3x 2  2 x
x


 x  1 3 x  2   x  1  3x  2  x  1
 x  1   x  1 x  1
2

Factorise numerator and
cancel (3x - 2) terms
M1A1
Use difference of two squares
B1
3x 2  2 x
2
x
2
 2


 x  1 3x  2  x  1 x  1  x  1 x  1

x  x  1  2
 x  1 x  1

x2  x  2
 x  1 x  1

 x  1  x  2  x  2

 x  1  x  1 x  1
Put both fractions over a
common denominator
and add.
Factorise and cancel
(x - 1) terms.
M1
A1
M1A1
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3.
5sin x  3cos x  R sin x cos   R cos x sin 
5 = Rcos
Compare
coefficients of sinx
and cosx
3 = Rsin
3

Aa
Use sin(x - a) formula
So R  52  32  34  5.83
M1A1
3
tan     30.96...  31o
5
M1A1
5
(b)
y  sin  has a maximum value of 1 when   90o
(i)
Maximum value of g(x) =
(ii)
When x    900 and so x  90    121o
34
Compare with a
simple sin
curve as g(x) is
simply Rsin
B1
M1A1
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4.
(a)
y
Move graph down
1 unit
x
–6
6
(3,0)
B1 (Shape
including min)
B1 (For (3, 0)
(b)
y
Move 3 units to the
left.
Reflect negative y
portion in x-axis
(0,1)
B1 (Move to left)
B1 (For modulus)
B1 for (0, 1)
x
–6
6
(c)
y
Reflect the positive
x portion in the yaxis
(-3,1)
(3,1)
B1 (For RHS with
min at (3, 1))
B1 (For LHS being
a reflection)
B1 for (-3, 1)
x
–6
6
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5.
(a)
x5  x 2  20  0
x5  x 2  20
20
x3  1  2
x
20
x  3 1 2
x

(b) x0  2 ,
x1  1.817...
Divide by x 2 and
then take the cube
root.
M1
M1
A1cso
3 1
20 3
 6
4
x2  1.918...
x3  1.860...
Use the given
iterative formula,
starting with 2.
M1
A1
A1
A1
x4  1.892...
(c)
So
6.
(a)
f(1.885) = 0.24….. > 0
f(1.875) = - 0.34…. < 0
Change of sign so there is a root in the interval
 = 1.88 to 2 decimal places.
f(-1) = ln5
g  ln 5   e2 ln 5  1
2
 eln 5  1  25  1 = 26
(b)
y = ln( 3 - 2x)
e y  3  2x
2x  3  e y
3  ey
3  ex
1
x
 f  x 
2
2
Choose an interval so that
all values in the interval
round to 1.88 to 2 decimal
places.
Check for change of sign.
First find f(-1) then substitute
this answer into g. Remember
that lnx and e x are inverse
functions
M1
A1
A1
B1
M1
M1A1
Put y = f(x) and rearrange to
make x the subject.
Remember inverse function
properties and also to rewrite
the final answer as f 1  x  .
M1
Form a suitable equation
M1
M1
A1
(c)
So
3  ex
 e2 x  1
2
2e 2 x  e x  2  3  0
 2e  1 e  1 = 0
x
x
ex 
1
2
 x  ln  12    ln 2
[ e x  1 has no solutions ]
Rearrange and treat the
equation as a quadratic in e x .
Remember that e x must
always be >0.
M1A1
M1A1
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7.
q2  p2
p2
p
(i) sin x  so cos x  1  2 
q
q
q
cosec2 x 
So
1
1

sin 2 x 2sin x cos x
1
cosec2 x 
2
(ii)

q 2  p2
p

q
q

q2
2 p q2  p2

2 cosec2 2 x  1  3cosec2 x  0
2cosec2 2 x  3cosec2 x  2  0
 2cosec2 x  1 cosec2 x  2   0
cosec2 x 
1
or  2
2
Use sin 2 x  cos 2 x  1 to
find an expression for cosx
M1A1
Use the definition of cosec
and the formula for sin2
M1
Substitute expression for sinx
and cosx
A1
Use cosec2  1  cot 2  and
rearrange to form a quadratic.
M1
A1
Factorize and solve
M1A1
1
Write as sinq = … equations
has no solution
and remember to find 2nd
2
and 3rd solutions.
B1M1M1
1
   7 11
cosec2 x  2  sin 2 x    2 x     ,
,
2
 6 6 6
7 11
x
,
Divide by 2
A1
12 12
cosec2 x 
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8.

Or
(ii)

d 2 x 1
e
 2e 2 x 1
dx
(a) (i)
By chain rule
dy
 x3 2e 2 x 1  3 x 2 e 2 x 1
dx
[  x 2 e2 x 1  2 x  3 ]
Use product rule


2
2
dy 2 x cos( x  1)  2 x  sin x  1  2

dx
4x2



]
2 x 2 cos x 2  1  sin x 2  1
Or
[ 
(b)
dx
 3sec2  2 y  1  2  6sec2  2 y  1
dy
2 x2
sec2   1  tan 2  and tan 2  2 y  1 
 x2 
dx
 6 1  
dy
9 

dy
3
And

dx 2 9  x 2
So

B1
Using the quotient
rule
M1A1A1
M1A1A1A1
Use chain rule to differentiate
with respect to y
x2
9
Use sec2   1  tan 2 
Remember
dy 1

dx dx
dy
M1A1
M1
M1A1

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