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1. Solve the equation
xx 734 21
= [4 marks]
2. The function f is defined forx 1 by
f(x) =x
x1
+
(a) Show that f(x) increases asx increases. [3 marks]
(b) State the range of f. [1 marks]
3. The function f and g are defined as follows :
f :x ,1,22 + xxxg :x + xx ,4
(a) Find the value ofx such that gf(x) = 7. [2 marks]
(b) Find an expression for )(f1x
. [3 marks]
4. Find
(a) + .316
2 dxx
x[2 marks]
(b) .4dxxe
x
[3 marks]
5. A spherical balloon is being inflated in such a way that its volume is increasing at
a constant rate of 150 -13scm . At time tseconds, the radius of the balloon is r
cm.
(a) Find
dt
drwhen r= 50. [4 marks]
(b) Find the rate of increase of the surface area of the balloon when its radius is
50 cm. [2 marks]
[The formulae for the volume and surface area of a sphere are3
3
4rV = ,
.4 2rA = ]
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6. Evaluate dxx
x
ln
4
2 by using
(a) the substitutiony = lnx, [4 marks]
(b) the trapezium rule with five ordinates. [3 marks]
Giving each answer correct to three decimal places.
7. (a) When the polynomial f(x) = 54 234 ++++ bxaxxx is divided by 12 x , aremainder of 2x + 3 is obtained. Find the values ofa and b. [4 marks]
(b) Given that the roots of the equation 022 =++ qpxx are and.
Find in terms of and, the roots of the equation
0222222
=++ qxqxpxq . [5marks]
8. The complex number 31 i+ is denoted by u. Show that u is a root of the
equation ,0422 =+ zz and state the other root of this equation.
Express u in the form r(cos +i sin ), where r> 0 and .
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CONFIDENTIAL*
10. A straight line of gradient mpasses through the point (2, 1) and cuts thex-axis
andy-axis atA andB respectively. A pointPlies onAB such thatAP:PB = 1 : 2.
(a) Show that as m varies, the locus ofPis a curve with equation043 = yxxy
[6 marks]
(b) Find the perpendicular distance from the point (2, 1) to the tangent of the
curve 043 = yxxy at the origin. [5 marks]
11. Given matrix M =
12
311
102
k
(a) Find the value ofkif M is a singular matrix. [2 marks]
(b) Ifk= 1, find the values ofp, q and rsuch that 0IMMM 23 =+++ rqp where I is a 3 3 identity matrix. Deduce the inverse of matrix M, .M -1
Find the determinant of M and hence write down the adjoin of matrix M.
[11 marks]
12. Find the turning point of the curvex
xy
ln= and determine whether this is a
maximum or minimum point. [6 marks]
Sketch the graph ofx
xy
ln= . [4 marks]
Hence, show that xe ex for all positive values ofx. [5 marks]
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No Working/AnswerPartial
marks
Total
marks
1 xx 734 2
1
=
Let n = ,21
x 2734 nn =
0347 2 =+ nn 0)1)(37( =+ nn
7
3=n or n = 1 (N.A. 0>n )
7
32
1
=x
49
9=x
B1
M1
A1
A1
4
2 1,
1)( += x
xxxf
(a) 21
1)('x
xf = > 0 for x > 1
f(x) increases asx increases.
(b) Whenx = 1, f(x) = 2Since f(x) increases asx increases, f(x) 2
M1, M1
A1
B1
3
1
3 f :x ,1,22 + xxx
g :x + xx ,4
(a) gf(x) = 7g( )22 xx + = 7,x 1
( )22 xx + + 4 = 7 0322 =+ xx
0)3)(1( =+ xx x = 1 (sincex 1)
(b) Let f(x) =yxxy 22 +=
Completing the square, 1)1(2
yx +=+
yx +=+ 11 (sincex 1)
yx ++= 11
M1
A1
M1
2
3
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)(f1x
= x++ 11 ,x 3
A1
A1
4(a) ++=+ cxdxx
x)31ln(
31
6 22
(b) dxxe
x4
Let u =x andxe
dx
dv 4=
1=dx
du
xev 4
4
1=
dxxex4
= dxexe xx 44 41
4
1
= cexexx + 44
16
1
4
1
M1, A1
M1, A1
A1
2
3
5 (a) Let Vbe the volume of the spherical balloon and rbe the
radius. Then
3
3
4rV =
)3(3
4 2rdr
dV = = 24 r
dt
dr
dr
dV
dt
dV=
150 = 24 rdtdr
24
150
rdt
dr
= or 22
75
r
When r= 50, 2)50(4
150
=
dt
dr
015.0= =
200
3or 0.00477
B1
M1
M1
A1
4
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5 (b) LetAbe the surface area of the spherical balloon. Then
24 rA =
rdr
dA8=
dt
dr
dr
dA
dt
dA
=
r8=
015.0
When r= 50, =dt
dA)50(8
015.0 6=
the surface area is increasing at a rate of 6 12 scm
M1
A1
2
6(a)dx
x
x
ln
4
2
Lety = lnx
dy = dxx
1
whenx = 2, y = ln 2
whenx = 4, y = ln 4
dxx
x
ln
4
2 dyy4ln
2ln=
4ln
2ln
2
2
=
y
])2(ln)4[(ln2
122 =
721.0
B1
M1
M1
A1
4
6(b)Let ,
ln
x
xu = 5 ordinates = 4 strips each ofwidth = 0.5
nx 2 2.5 3 3.5 4
nu 0.3466 0.3665 0.3662 0.3579 0.3466
By using trapezium rule,
dxx
x
ln
4
2
=
+++
+3579.03662.03665.0
2
3466.03466.05.0
719.0
B1
M1
A1
3
7(a) f(x) = 54 234 ++++ bxaxxx 4
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f(1) = 541 ++++ ba2 (1) + 3 = 10 + a + b
5=+ba .f(1) = 5)1(41 +++ ba
2 (1) + 3 = 2 + ab
ab +=1
Substitute into,51 =++ aa62 =a3=a
From, 2)3(1 =+=b
3=a , 2=b
M1
M1
M1
A1
7(b) 022 =++ qpxx
Roots are andp=+ 2q=
)( +=p =2q
02 22222 =++ qxqxpxq
Substitute )( +=p and =2q
02)(22 =+++ xxx
0]2)[(22 =++ xx
0)( 222 =++ xx0))(( = xx
,
=x
=x
The roots are
and
.
B1
M1
A1
M1
A1
5
8 0422 =+ zz
2
1642 =z
31 iz =
31 i+ is a root (u) [Shown]
M1
10
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31 i is the other root.
++=
3sin
3cos31
iu =
+
3sin
3cos2
i
iu 32312 += = i322+
arg3
2)3(tan)(
12 == u
41242 =+=u
8623 ==u
arg =)( 3u ,83 =u
A1
B1
B1
M1
A1
A1
M1
A1
A19(a)
Sum of first n terms, =
+=n
r
r
n rS1
1 )23(
==
+=n
r
n
r
rr
11
123
++
= )1(2
12
13
)13(nn
n
)1(2
13nn
n
++
=
M1A1,M1
A1
4
9(b) Let common difference be d, d = 10.
Given 00010=nS .
[ ] 00010)1(22
=+ dnan
n
na00020
10)1(2 =+
)1(1000020
2 = nn
a
)1(500010
= nn
a
dnaTn )1( +=
)1(10)1(500010
+= nnn
m
B1
M1
7
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)1(500010
+= nn
[Shown]
Given .500
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CONFIDENTIAL*
11(a)
M =
12
311
102
k
M is a singular matrix
0
12
311
102
=
k
02
111
1
310
12
312 =
++
kk
0)2(1)61(2 =++ kk= 16
M1
A1
2
11(b)
=2M
121
311
102
121
311
102
=
603
176
123
=3M
121
311
102
603
176
123
=
91212
16518
843
0IMMM 23 =+++ rqp
91212
16518
843
+
603
176
123
p +
121
311
102
q +
100
010
001
r =0
024 =+ p 0212 =+ q 069 =+++ rqp
2=p 6=q 15=r
0IMMM 23 =+++ rqp1M
, =+++ -12 MIMM rqp I1M = (
1
rI)MM
2 qp
B1
B1
M1
M1
A2
M1
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= (15
1I)6M2M
2 +
15
1= I)6M2(M 2 +
151=
+
100
010
001
6
121
311
102
2
603
176
123
15
1=
241
714
127
=
15
2
15
4
15
115
7
15
1
15
415
1
15
2
15
7
(ii)21
1110
12
312M
+
=
)12()61(2 += 15=
MAdj(M)M =
1MMAdj(M) =
=
241
714
127
M1
A1
B1
B1
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12
x
xy
ln=
Using2v
dx
dvu
dx
duv
dx
dy
= with u = lnx and v =x
2
ln1
)(
x
xx
x
dx
dy
=
2
ln1
x
x=
0=dx
dy 0
ln12
=
x
x
1 lnx = 0
lnx = 1 x =e
whenx =eee
ey
1ln==
The turning point is .1,
ee
4
2
2
2
2)ln1(1
x
xxx
x
dxyd
=
4
ln22
x
xxxx +=
4
ln23
x
xxx += or 3
ln23
x
x+
x =e =2
2
dx
yd0
13