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END OF YEAR MARKING GUIDE
The paper was set in accordance to the UNEB setting system covering the
sections of VECTORS, TRIGONOMETRY, ANALYSIS, GEOMETRY and ALGEBRA.
1.
o
o
Solve the equation 2 cos 2  3 sin   1 ; 0  x  360 .


2 1  2 sin 2   3 sin   1 ,
4 sin 2   3 sin   1  0
4 sin 2   4 sin   sin   1  0 , 4 sin   1sin   1  0
1
sin    ,   94.48o , 345.52 o , sin   1 ,   90 o
4
Comment:
Popular question as the students did well to use the double angles but they were
denied marks for using calculators to solve the quadratic equation.
2
2.
Differentiate from first principles: y  3x  cos 2 x .
y  y  3x  x   cos 2x  x 
2

y  3x  x   cos 2x  x   3x 2  cos 2 x
2

y  3x 2  6 xx  3 x    2 sin 2 x  x sin x  3x 2
2
y
dy
 6 x  3x  2 sin 2 x  x  ,
 6 x  2 sin 2 x
x
dx
Comment:
The students did not recall the concepts of differentiating from first principles.
Poorly done amongst the students.
3x  2
3.
Express in partial fractions:
2 x  12 3  x 
3x  2
A
B
C
Let



2
2
2 x  1 3  x  2 x  1 2 x  1 3  x 
3x  2  A2 x  13  x   B3  x   C 2 x  1
22
7
11
3x  2
22
7
11
A
, B , C



2
2
25
5
25
2 x  1 3  x 252 x  1 52 x  1 253  x
Comment:
Partial fractions was a popular question but the concept of LINEAR REPEATED
FACTOR was confused with a QUADRATIC FACTOR.
2
4.
a 2  b 2 sin  A  B 

Prove that in any triangle ABC ,
sin  A  B 
c2
From L.H.S

a 2  b 2 4 R 2 sin 2 A  sin 2 B

c2
4 R 2 sin 2 C
sin A  sin B sin A  sin B 

sin 2  A  B 
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
sin  A  B  sin C
2 sin

AB
2

cos A 2 B 2 cos A 2 B sin
sin 2  A  B 
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AB
2

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END OF YEAR MARKING GUIDE

Comment:
2 sin
AB
2

cos A 2 B 2 cos A 2 B sin
sin 2  A  B 
A B
2


sin  A  B sin  A  B 
sin  A  B 
as R.H.S

2
sin  A  B 
sin  A  B 
Poorly attempted. Application of the sine rule. The same question appeared in
UNEB Maths paper 1 2011 in section B.
5.
A right circular cone is to be made such that its slant height is  metres. Show
2 4
that the maximum volume of the cone is
.
9 3
Solution:  2  r 2  h 2
 r 2   2  h2


Volume V  r 2 h
so V  h( 2  h 2 )
3
3
dV  2
 (  3h 2 ) but for max
dh 3
dV


0
Volume,
So, ( 2  3h 2 )  0 h 
dh
3
3
 Maximum Volume

2 
2 4
=
V  ( 2  )
.
3
3
3
9 3
Comment:
Application of maximum and minimum was poorly done. Need to revise the topic.
6.
x3  8
The gradient of a curve at a point x, y  is
. Find the equation of the curve
x2
given that it passes through the point 2, 4 .
Comment:
7.
8
dy x 3  8 dy
 x  2 y   x  8 x 2 dx
,

2
dx
dx
x
x
2
x
8
y
  c , x  2, y  4 , so 4  2  4  c , thusb c   2
2
x
x2 8
Equation of the curve is y 
 2
2
x
Analysis required integration but after splitting the terms. Most of the students
failed to interpret the question and begun differentiating, sketching the curve e.t.c
indicating that most were living on cram work.
The gradient of the side PQ of the rectangle PQRS is 3 4 . The coordinates of
the opposite corners Q , S are respectively 6, 3 and  5, 1 . Find the equation
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of PR .
y3 3
 ,  4 y  3x  6
x6 4
__
y 1
4
  ,  3 y   4 x  17 since they are perpendicular.
Find the eqn of RS 
x5
3
__
Find the eqn of PQ 
Solving the two simultaneously, we get the coordinates of P as P 2,  3
Let M be the midpoint of QS , thus the coordinates are M  12 , 2 .
__
__
3 2
M is also the midpoint of PR , gradient of PR 
2
 2  12
__
y3
 Equation of PR :
 2 , giving y  2 x  1
x2
Comment:
Coordinate geometry was poorly done despite the question appearing so many
times. The students seem not to revise the work given to them.
x 1 y 2 3  z


8.
Find the angle between the line
and the plane
2
3
4
4x  3y  2z  1  0 .
__
d  2i  3 j  4k , is the direction vector of the line.
n  4i  3 j  2k , vector normal to the plane.


cos 90 o   
 2  4
   
 3 .  3
 4  2
   

898

25
29
90
o

   30.45o    59.55 o
4  9  16 . 16  9  4
29. 29
Comment:
The question was popular but the students refused to indicate vector symbols and
they also failed to read the direction vector of the line correctly.
9a)
Solve the equation: cos 3x  sin 2 x  sin 3x  cos 2 x ; 0  x  
cos 3x  cos 2 x  sin 3x  sin 2 x
2 cos
5x
x
5x
x
cos  2 sin
cos  0
2
2
2
2
x
5x
5x 
2 cos  cos
 sin
0
2
2
2 
2 cos
x
x
 0,
 90 o , x  
2
2
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5x
5x 

 sin
 cos
0
2
2 

Comment:
b)
5x
 45 o , 225 o , 405 o
2
5 x  5 9
  9
 ,
,
, x ,
,
2
4 4 4
10 2 10
The question had appeared in their MOT but like earlier mentioned they do not
revise what is given to them. Those who managed failed to convert the angles to
radians.
tan
5x
 1,
2
Express 10 sin x cos x  12 cos 2 x in the form R sin 2 x    , hence or otherwise solve
10 sin x cos x  12 cos 2 x  7  0 in the range 0 o  x  360 o .
10 sin x cos x  12 cos 2 x = 5 sin 2 x  12 cos 2 x
Let 5 sin 2 x  12 cos 2 x  R sin 2 x cos   R cos 2 x sin 
12
 5  R cos  , 12  R sin  , thus tan  
  67.38o
5
R  5 2  12 2  13
5 sin 2 x  12 cos 2 x  13 sin( 2 x  67.38o ) as required.
10 sin x cos x  12 cos 2 x  7  0 , 13 sin( 2 x  67.38o )   7
2 x  67.38o  212.59 o , 327.41o , 2 x  145.21o , 260.03o
Comment:
10a)
Thus, x  72.61o , 130.02 o
Trigonometry was not popular despite the question being set again.
2
to solve the equation x 4  x 3  16 x 2  2 x  4  0 .
x
x 4  x 3  16 x 2  2 x  4  0 Divide thru by x 2
Use the substitution y  x 
2
4
4 
2
 2 0
x 2  2   x    16  0
x x
x
x

4
y2  4  x2  2
y 2  4  y  16  0
x
y 2  y  12  0 ,
 y  3 y  4  0 , y   4, y  3
x 2  x  16 
 4  24
3  17
And x 2  3x  2  0 , x 
2
2
The question was popular as these were taught in s5 first term. The students who
used calculators were denied the marks. Popular question, a MUST do to all.
Thus x 2  4 x  2  0 , x 
Comment:
b)
x  2y
y  2z
2x  z


, x yz2
3
4
5
x  2y
y  2z
2x  z


  then x   3  2 y , y  4  2 z , z  5  2 x
Let
3
4
5
So x   3  24  2z  , x   11  4 z
Solve the equations:
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z  5  2 11  4z  , z  3 and y   2
From x  y  z  2 , we have   2  3  2 thus   1
x  1, y   2, z  3
Comment:
Popular question, a MUST do to all.
11a)
The sum of two numbers is 24. Find the two numbers if the sum of their squares
is to be minimum.
Let the numbers be x and y .
x  y  24 ,
Let the sum be S such that S  x 2  y 2
S  x 2  24  x   2 x 2  48 x  576
dS
 0 , i.e 4 x  48  0, x  12 Thus the numbers are x  12, y  12 .
For max or min,
dx
Comment:
Analysis still on application of maximum and minimum. Poorly done yet so simple
and straight forward.
2
b)
A point P on the curve is given parametrically by x  3  cos  and y  2  sec  .
Find the:
i)
equation of the normal to the curve at the point  
When  
ii)

3
5

5

and y  2  sec  4 , thus the point is P , 4  .
3
3 2
3
2

dx
dy
 sin  and
 sec  tan  thus the gradient function is given by
d
d
dy
1
1
 dy
 sec  tan  

4
when   ,
2
dx
sin  cos 
3 dx
1
Gradient of the normal is  , so the equation of the normal is
4
y4
1

To get 8 y  2 x  37
x  52
4

, x  3  cos


Cartesian equation of the curve.
cos   3  x , and from y  2 
1
1
, so cos  
cos 
y2
1
to get 3 y  2 x  xy  7
y2
Popular question, a MUST do to all. Well done by most of the students.
Thus 3  x 
Comment:
12a)
Solve the equation 2 tan   sin 2 sec   1  sec  in the range 0 o    2 .
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2 sin  2 sin  cos  cos   1


cos 
cos 
cos 
1
 5
,  ,
2
6 6
Most students cancelled the trigonometric function and missed the marks. Those
who managed to get some angles did not convert them to radians.
1  cos 2 sin   1  0
Comment:
b)
Prove that tan  A  B  


Either cos    1,    or sin  
tan A  tan B
, hence or otherwise solve the equation
1  tan A tan B
tan   45o  6 tan  , where  180 o    180 o .
sin  A  B 
From the LHS, tan  A  B  
cos A  B 
sin A cos B  cos A sin B

, divide through by cos Acos B
cos A cos B  sin A sin B
tan A  tan B

as R.H.S
1  tan A tan B


For A   , B  45o , for tan   45o  6 tan  , we get
tan   tan 45
tan   1
 6 tan  ,
 6 tan 
1  tan  tan 45
1  tan 
Comment:
13a)
6 tan 2   5 tan   1  0 , 6 tan 2   3 tan   2 tan   1  0
1
tan    ,    18.43o , 161.57 o
3 tan   12 tan   1  0
3
1
tan    ,    26.57 o , 153.43o
2
Popular question, a MUST do to all. Well done.
 2
1 
 2
1 
 
 
 
 
The vector equation of two lines are r  1    1  and r   2     2  where t is a
0
 2
t 
1 
 
 
 
 
constant. If the two lines intersect find:
(i)
t and the position vector of the point of intersection.
2   
2   




r  1    and r   2  2   , if they do intersect then r  r
 2 
t   




2    2   


 
Thus 1      2  2   implies that 2    2   , so   
 2   t   


 
Also 1    2  2  , for    , then 1    2  2  so,    1  
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Using 2  t   , then  2  t   1 , gives t   1
1 
 
Position vector for point of intersection is given by r   0  or r  2i  2k .
  2
 
(ii)
the angle between the two lines giving your answer to the nearest degree.
Let d1  i  j  2k and d 2  i  2j  k be the directional vectors.
i  j  2k  . i  2 j  k   5
Thus the angle is given by cos  
12  12  2 2 . 12  2 2  12 6
cos  
b)
5
,   33.56 o  34 o to the nearest degree.
6
The position vector of points P and Q are 2i  3 j  4k and 3i  7 j  12k
respectively. Determine;
i)
the size of PQ.
ii)
The Cartesian equation of PQ.
  2  3   1 
     
 PQ = (12  (4) 2  8 2 = 9
i)
PQ =  3     7     4 
  4   12   8 
     
 2 
 1 
 
 
ii)
Vector equation PQ is rPQ    3      4 
 4 
 8 
 
 
 x  2   
  

x2 y3 z4


(  )
  y     3  4   Cartesian equation is;
1
4
8
 z   4  8 
  

Comment:
Popular question, a MUST do to all. Well done question.
14a)
2
A polynomial function Px has a factor of x  3x  4 and leaves a remainder of 5
when divided by x  2 . Determine the remainder when the polynomial is divided


by x 2  3x  4 x  2 .
Px   x  4x  1x  2Qx   ax 2  bx  c
P 2  4a  2b  c  5
P 1  a  b  c  0
P4  16a  4b  c  0
5
5
 10
5
5
10
,c
Solve to get a  , b 
, thus R x   x 2  x 
6
2
3
6
2
3
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Comment:
b)
Solve for x and y :
x log 4 128  y log 8 2  6
log 2 x  13 log 2 y 3  2 log 4 6
x log 2 128 y log 2 2
7x y
  6 …..(i)
we get;

6
2 3
log 2 4
log 2 8
Comment:
15a)
Popular question, but the majority failed to interpret the question. Remainder
theorem.
xy  6 …….(ii)

log 2 xy  log 2 6
7 6 y
So
y 2  18 y  63  0

.  6
2 y 3
( y  21)( y  3)  0
y  3, x  2
y  21, x  72
Popular question but the students did not use or apply the rules of logarithms.
The curve y  ax 2  bx  c has a maximum point at 2, 18 and passes
through the point 0, 10 . Find the values of a, b, c .
x  0, y  10 so, c  10
dy
 2ax  b , x  2, 4a  b  0
dx
Comment:
b)
x  2, y  18 so, 4a  2b  c  18 ,  4a  2b  8
Poorly done yet it was so straight forward.
b  8, a   2
Find the equations of the tangent and normal to the curve

a 
 .
x 2  3 y 2  2a 2 at the point  a,
3


3
dy
dy
x
a 
2x  6 y
 0 , so,
, thus, at  a,
 grad of T is 

dx
3
dx
3y
3

a
y 3
3

Equation is
, 3 y   3x  2 3a
xa
3
Grad of normal is 3 , thus equation of normal is
y
Comment:
a
3
 3 , equation is 3 y  3 3 x  2 3a .
xa
Popular question, but many failed to differentiate a constant to get zero.
© Prepared by Ssekyewa Edward Maths Dept 2011
Email: [email protected]
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