P425/1
Pure math
Paper 1
3 hours
GREENHILL ACADEMY
MOCK EXAMINATIONS 2016
UGANDA ADVANCED CERTIFICATE OF EDUCATION
INSTRUCTIONS TO CANDIDATES:
- Attempt all the eight questions in section A and any five questions from section B.
- Clearly show all the necessary working
- Begin each answer on a fresh sheet of paper
- Silent, simple non-programmable scientific calculators may be used.
SECTION A (40 MARKS)
1.
2.
3.
4.
5.
6.
7.
Find the fourth root of 8  6i
The 5th term of an arithmetic progression is 12 and the sum of the first five terms is 100.
Determine the first term and the common difference
2 x
Solve the inequality 0.6  3.6 to four significant figures3
Find the angle subtended by the plane 2 x  3 y  7 z  4  0 and a line
x  4 5  y z 1


.
3
5
2
Express cos x  3 sin x in the form Rcosx    , hence, solve the equation
cos x  3 sin x  0 , for 0 o  x  360 o .
10 Inx
dx
Evaluate: 
1
x
A is a point 1, 3 and B is the point 4, 6 . P is the variable point which moves such
__ 2
8.
__ 2
that AP  PB  34 . Show that the locus of P describes the circle, hence state the
centre and the radius of the circle
dy
Solve the differential equation x1  y   y  0 given that y  1 when x  e .
dx
SECTION B (60MARKS)
9a)
Use a substitution t  tan

to solve the equation 3 cos   5 sin    1 , for
2
0 o  x  360 o .
b)
Show that for any triangle ABC, sin A  sin B  sin C  4 cos
A
B
C
cos cos .
2
2
2
10a)
Given that z  x  iy is a complex number, find the equation of the locus
b)
Solve the equation z 3  1  0
11a) Find:
dx
 4  5 cos x
4
using t  tan
1
 1
x
2
dx .
b)
Evaluate:
12a)
Prove by induction that  2r  1 
1
x
z 1 i
 1.
z  2  3i
n
r 1
2


n
4n 2  1 .
3
1  ax
 1  4 x  12 x 2  cx 3  . . . . Find the values of the constants a, b, c .
1  bx
b)
Given that
13a)
Given vectors a  3i  2j  k and b   2i  j  4k . show that 𝒂 and 𝒃 are
perpendicular.
b)
Two planes have equation x  2 y  2 z  2 and 3𝑥 − 3𝑦 + 6𝑧 = 3. Find
i)
The acute angle between the two planes
ii)
14a)
The line of intersection in vector form of the two planes.
Find the equation of a circle through the points A0, 1 B2, 3 and C0, 5 .
b)
Given that x  1  2 sin  and y   3  2 cos  .
i)
Show that x and y represents a circle. Hence its state the centre and radius
ii)
Find the equation of the common chord of intersection of the circles in (b) (i)
above and a circle 𝑥 2 + 𝑦 2 − 2𝑥 + 6𝑦 = 0
15.
Sketch the curve
16a)
Find the particular solution of the equation x
b)
3  2x
x(3  x)
dy
 y  2 given y(2)  4.
dx
The rate at which mangoes fall off a tree is directly proportional to the number of
mangoes remaining on the tree. a certain tree initially had 100 fruits, 2hours later the
fruits had reduced to 80mangoes
1
(i) Find the numbers of mangoes which had remained after 3 2 hours
(ii) calculate the time it takes for just 1 mango to remain
END
S 6 MARKING GUIDE MOCK TERM 2 2016
z  8  6i
1.
z  8 2  6 2  10
Let
6
8
z  10(cos 0.205  i sin 0.205 )
  tan 1 ( )  0.205
z
1
4
1
 10 4 (cos 0.205  i sin 0.205 )
1
1
4
1
 10 4 (cos( 1 .0.205 )  i sin( 1 0.205 )  10 4 (0.99  0.16i )
4
4
5th  12
2.
s5  100
a  4d  12............1
5
(2a  4d )  100
2
a  2d  20.........2
a  28
and  d  4
(0.6)−2𝑥 < 3.6
3.
ln 0.6 2 x  ln 3.6
(2 x)( 0.51082)  1.28093
x  1.254
4.
x  4 5  y z 1



3
5
2
x  4  3 ,
let
y  5  5
z  1  2
 2(4  3 )  3(5  5 )  7(1  2 )  4  0

34
5
x  24.4, y  29, z  14.6
cos x  3 sin x  R cos( x   )
5.
cos x  3 sin x  R cos  cos x  Rs 3in sin x
R cos   1, R sin   3  tan   3andR 2  5
   60o , R  5
cos x  3 sin x  5 cos( x  60o )
5 cos( x  60o )  0
( x  60o )  cos 1 (0)
x  60o  900 ,2700  x  300 ,2100
6
1
ln xdx
1 x
let ..t  ln x
10

1
dx
x
dx  xdt
dt 
10

1
1
ln xdx 
x
ln10

0
1
t.xdt
x
ln10
t 2 
 
 2 0
7.
(ln 10) 2
 2.651(3dp)
2
letP ( x, y )
( x1) ( y 3) ( x4) ( y 6) 34
2
2
2
2
x 2  2 x  1  y 2  6 y  9  x 2  8 x  9  y 2  12 y  36  34
2 x 2  2 y 2  10 x  14 y  12  0
x 2  y 2  5x  7 y  6  0
 2 f  5,  2 g  7
f  2.5,  g  3.5
r  (2.5) 2  (3.5) 2  6  3.54
Circle of center (2.5,3.5) and radius 3.54
8.
dy
 y0
dx
dy
x(1  y )
 y
dx
(1  y )
1
dy  dx
y
x
1
1
1
1
(1  )dy  dx   (1  )dy   dx y
y
x
y
x
y  ln y  ln x  C
x(1  y )
y  1, x  e1
1 0  1 C  C  0
9a)
y  ln y  ln x  yx  e y
3 cos   5 sin   1
1 t2   t2 
  5
  1
3
2 
2 
1

t
1

t

 

2
2
3  3t  5t  (1  t 2 )
4  7t 2
t
tan 

2
7
2

2
2
or tan  
2
7
7
 37.10 0r

 142.9 0
2
2
0
   74.1 ,285.80
b)
L.H .S  sin A  sin B  sin A
A
A
BC
B C
cos  2 sin(
) cos(
)
2
2
2
2
A
A
180o  A
B C
 2 sin cos  2 sin(
) cos(
)
2
2
2
2
A
A
A
B C
 2 sin cos  2 sin( 900  ) cos(
)
2
2
2
2
A
A
A
BC
 2 sin cos  2 cos cos(
)
2
2
2
2
A A
BC 
 2 cos sin  cos(
)
2
2
2 
 2 sin
A  180o  ( B  C )
B C 
 2 cos sin
 cos(
)
2
2
2 
 2 cos
A
BC
B C 
cos(
)  cos(
)

2
2
2 
A
B
C
2 cos cos 

2
2
2
A
B
C
 2 cos 2 cos cos  R.H .S
2
2
2
z 1 i
 1, let z  x  yi
z  2  3i
 2 cos
10(a)
x  yi  1  i
1
x  yi  2  3i
x  12   y  12
x  22   y  32
x  12   y  12
1

x  22   y  32
x 2  2x  1  y 2  2 y  1  x 2  4x  4  y 2  6 y  9
8 y  6 x  11  0
f ( z)  z 3  1
b)
f (1)  (1) 3  1  0
 Z  1..root  Z  1 factor
Z 2  Z 1
Z  1 z3  1
Z3  Z2
 Z 2 1
Z2 Z
Z 1
Z 1
forZ  Z  1
2
Z
11a)
 (1)  1  4(1)(1)
1
3
 Z  i
2
2
2
dx
 4  5 cos x
x
1
x
2dt
1 t2
t  tan ,  dt  sec 2 dx  dx 

cos
x

2
2
2
1 t2
1 t2


2dt


2
dx
2dt
2dt
2dt
1 t





2
2
2
2
 4  5 cos x  
(3  t )(3  t )
1 t 
4  4t  5  5t
9t
 4  5(1  t 2 ) 
2
A
B
let 


 2  A(3  t )  B (3  t ),
(3  t )(3  t ) 3  t (3  t )
1
1
 A B
3
3
2dt
1 dt
1 dt
1
1
 (3  t )(3  t )  3  3  t  3  3  t  3 ln( 3  t )  3 ln( 3  t )  ln A
1
1
x
 ln A(9  t 2 )  ln A(9  tan 2 )
3
3
2
b)
4
1
 1
1
x
dx,
let ..1  x  p  x  ( p  1) 2  dx  2( p  1)dp
4
3
3
1
1
1
3
1 1  x dx  2 p .2( p  1)dp  22 (1  p )dp  2 p  ln p2  23  ln 3  (2  ln 2)  2(1  ln 2  ln
12a)
∑𝑛𝑟=1(2𝑟 − 1)2 =
1
3
𝑛(4𝑛2 − 1)
1
12  33  52  ....  (2n  1) 2  n(4n 2  1)
3
for..n  1  LHS  1,  RHS  1  hold . for..n  1
1
for..n  2  LHS  1  9  10,  RHS  (2)(15)  10  hold . for..n  2
3
assume.it .holds. for.n  k
1
12  33  52  ....  (2k  1) 2  k (4k 2  1)
3
it .holds. for.n  k  1
1
12  33  52  ....  (2k  1) 2  (2(k  1)  1) 2  k (4k 2  1)  (2k  1)) 2
3
1
1
 1
RHS  k (4k 2  1)  (2k  1) 2  (2k  1)  k (2k  1)  (2k  1)  (2k  1) 2k 2  k  6k  3
3
3
 3
1
1
 RHS  (2k  1) 2k 2  5k  3  (2k  1)( k  1)( 2k  3)
3
3
for., n  2,  k  1  2, k  1



1
12  33  (3)( 2)(5)  10,  holds. for.n  2
3
for.n  3  k  1  3, k  2
1
12  33  52  (5)(3)(7)  35, holds. for.n  3
3
it holds for n=1, n=2,n=3,Therefore it holds for all positive values of
n
b)
1+𝑎𝑥
1+𝑏𝑥
= 1 − 4𝑥 + 12𝑥 2 + 𝑐𝑥 3

(`1  ax)(1  bx) 1
(1)( 2)
(1)( 2)( 3)


 1  ax 1  (1)(bx) 
(bx) 2 
(bx) 3  . . .
2!
3!


2 2
3 3
 (1  ax)(1  bx  b x  b x . . .)
 (1  bx  b 2 x 2  b 3 x 3  ax  abx 2  ab 2 x 3 )
 1  (a  b) x  (b  ab) x 2  (ab 2  b 3 ) x 3
2
 a  b  4  b  ab  12.and .  ab 2  b 3  C
2
a  b  4  b  (b  4)b  12
2
b  b  4b  12  b  3  a  1  C  (1)(9)  (27)  36
a  3i  2 j  k
b  2i  j  4k
a.b  0
2
13a)
2
3   2
 2 .  1   0
  
2 4 
628  0
b)
00
𝒂 = 3𝑖 + 2𝑗 + 𝑘, 𝒃 = −2𝑖 − 𝑗 + 4𝑘 .
1 
n1  2 
 2
3 
n2   3
6 
n1 .n2  n1 n2 cos 
1  3 
2 . 3  12  2 2  (2) 2 32  (32 )  6 2 cos 
  
 2 6 
3  6  12  9 54 cos 
 15  3 54 cos 
 15
 132.9 0
3 54
int er sec tion
  cos 1
let ...z  
 x  2 y  4   2  3 x  3 y  6  3
 x  2 y  2  4 .....................(1)
 3 x  3 y  3  6 ...................( 2)
3eqn(1)  2eqn(2)
3 x  6 y  6  12
6 x  6 y  6  12
9 x  12
x
4
3
4
1
 y   2
3
3
4

4


 
3


3
x
 
0
  1
 

1 
 y     2   r       2 
z  3
1 

3
  
 

0 


 


 
 2 y  2  4 
14.(a
)
A(0,1) , B( 2,3) and C(0,5) generally, x 2  y 2  2 fx  2 gy  d  0
0 2  12  2 f (0)  2 g (1)  d  0  2 g  d  1..........................(1)
2 2  32  2 f (2)  2 g (3)  d  0  4 f  6 g  d  13...............(2)
0 2  52  2 f (0)  2 g (5)  d  0  10 g  d  25.....................(3)
 f  0,  g  3  d  5
x 2  y 2  2(0) x  2(3) y  5  0  x 2  y 2  3 y  5  0
Is the equation of the circle
(b)
𝑥 = 1 + √2 𝑠𝑖𝑛 𝜃 and 𝑦 = −3 + √2 𝑐𝑜𝑠 𝜃.
(i)
x 1
y5
.......and ......... cos  
2
2
2
2
sin   cos   1
sin  
( x  1) 2 ( y  5) 2

2
2
2
 ( x  1) 2  ( y  5) 2  4
this is the eqn of the circle of centre (1,-5) with radius 2
(ii)
 ( x  1) 2  ( y  5) 2  4
x 2  2 x  1  y 2  10 y  25  4  x 2  y 2  2 x  10 y  22...........(1)
x 2  y 2  2 x  6 y  0..................( 2)
eqn..(1)  eqn(2)
11
2
Is the eqn of the common chord
 4 y  22  y  
15
Intercepts: For y  0, x   1 , thus  1, 0 .
15
Turning points:



dy 3 3x  x 2  3x  3 2 x  3

0
2
dx
3x  x 2

9 x  3x  6 x  9 x  6 x  9  0 ,
x  1x  3  0
2
2
x   3, x  1 and y 
L
Sign of
dy
dx

Very
popular
but low
scores,
need to
do better.
x  3
3x 2  6 x  9  0
1
y3
3
R
L


x 1

1

1, 3 min
  3,  max
3

As y    , x  0, x  3 so, vertical Asymptotes: x  0, x  3
As x   , y  0 . y  0 is the horizontal asymptote.
Region table:
x
3x  3
3 x
y
x  1
1  x  0
0 x3
x3
















16(a)
dy
dy
dy
1
y2 x
 2 y 
 dx
dx
dx
2 y x
dy
1
 2  y   x dx   ln( 2  y)  ln x  ln A
1
 Ax,
2 y
1
1
y (2)  4. 
 2A  A  
2
4
For
1
1
 x
2 y
4
(b)
Let N= number of mangoes present at any time t
dN
dN
N
 kN
dt
dt
dN
dN
 kdt  
   kdt  ln N  kt  C  N  e kt C
N
N
 kt
N  Ae ...where.. A  e C
x
at t  0, N = 100
100  A
N  100e kt .,
 80 
at t  1, N = 80  80  100e  k  k   ln 
  0.2231
 100 
N  100e 0.2231t
(i ) N  ?, t  3.5
N  100e ( 0.2231)( 3.5)
N  46mangoes
(ii )t  ?, N  1
1  100e 0.2231t
1
)  t  20 min
100
20minutes will elapse for 1 mango to remain
 0.2251t  ln(