S6 REVISION FOR PURE MATHEMATICS
Allow me to discuss this dozen of questions with you.
1.
The nth term of an arithmetic progression (A.P) is
first n terms of the progression is
3n  1
. Show that the sum of the
6
n
3n  1 .
12
Review
 nth term of an AP, u n  l  a  n 1d
 sum of n terms of an AP, s n 
 nth term of a GP, u n  a r n 1


n
2a  n  1d   n a  l 
2
2

For u n 


3n  1
2
5
; First term, u1  a  ; second term, u 2 
6
6
6
So, common difference, d  u 2  u1 
Here, s n 
2.

a 1 r n
a rn  1

sum of n terms of a GP, s n 
1 r
r 1
a
sum to infinity, s  
1 r
5 2 1
 
6 6 2
n 1
1
n
3n  1
2   n  1   

2 3
2  12
Given that log2 3  p and log4 5  q , prove that log 45 2 
1
.
2 p  q 
Review of laws of logarithms;

loga b  loga c  loga bc ; loga b  loga c  loga

n log a b  log a b n ; loga b 

a loga b  b
b
c
logc b
1

logc a logb a
For this question,
Maths Dept @ KCB 2020 Prepared by James Lubega - HOD
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RHS 
1
1
1
1



2 p  q  2log2 3  log4 5

log 2 5  2 log2 3  log 2 5
2log2 3 

log 2 4 


3.
1
1

 log 45 2  LHS
log 2 9  log 2 5 log 2 45
e x  e3x
d2y

If e  tan 2 y , prove that
.
d x 2 2 1  e2x 2
x

Review: chain rule,
d  dy  dt
dy dy dt d 2 y
  
  ;
2
dt  dx  dx
dx dt dx d x
d u
 
dx  v 
Quotient rule;

v
du
dv
u
dx
dx
2
v
For e x  tan 2 y , differentiating both sides
e x  2 sec2 2 y




dy
dy
dy
 2 1  tan 2 2 y
 2 1  e2x
dx
dx
dx



e x  e3x
d 2 y 1  e 2 x e x  e x  2e 2 x
dy
ex



;
.
2
2
dx 2 1  e 2 x d x 2
2 1  e2x
2 1  e2x

4.




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 and c.
For turning/stationary points,
dy
 0.
dx
y  ax 2  bx  c
At 0 , 10, 10  a0  b0  c : c  10
At 2 , 18, 18  4a  2b  10 ; 2a  b  4 …(i)
Now
dy
 2ax  b
dx
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At 2 , 18,
dy
4a  b  0 …(ii)
dx
Now (ii) – (i); a   2 ; b  8
5.
Use the substitution t  tan x to show that
Recall, if t  tan x , then sin 2 x 
For t  tan x ,



4
0
1
1
dx  .
1  sin 2 x
2
1  t2
2t
2t
;
cos
2
x

; tan 2 x 
.
2
2
1 t
1 t
1  t2

dt
dt
 sec2 x  1  t 2  dx 
dx
1  t2
Also change the limits i.e.
Now


4
0
dx

1  sin 2 x

0

4
T
0
1
1
1
0
X
1
2t
1  t2

dt

1  t2
1
dt
 1  t 
2
0
1
 1 
1
1

    1 
2
2
1  t  0
15
6.
1
Find the coefficient of x in the expansion of  x 3  4  .
x 

17
Review: Expansion, a  bn  a n  nC1. a n 1b  nC2 . a n  2b 2  ...  b n
The expansion for a general term;
1  x n
 1  nx 
nn 1 2 nn  1n  2  3
x 
x  ...
2!
3!
Maths Dept @ KCB 2020 Prepared by James Lubega - HOD
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For the r  1th term, U r 1  nCr a n r b r ; r  0 ,1, 2, ...
1
Now for  x 3  4 

15
x  ;
 
U r  1  15C r . x 3
15  r
n  15 , a  x 3 , b 
 
. x 4
r
1
x4
 15C r . x 45  7 r
For the term in x 17 ; 17  45  7r  r  4
Therefore the coefficient required is 15C 4 
7.
Given that y  ln1  sin x  , deduce that
15!
 1365
15  4!4!
d2y
 ey  0 .
2
dx
First apply the laws of logarithms and then the chain rule.
y  ln1  sin x   e y  1  sin x
2
dy
d2y
 dy 
e
 cos x ; e y
 e y     sin x
2
dx
dx
 dx 
y
2


dy y d 2 y
 cos x 
 ey y    ey  1
Eliminating sin x and
,e
2
dx
dx
 e 
ey
2
d2y
y  1  sin


e
 e2y
d x2


x
   e y  1




2
y


d2y
y  1 e  1 
e

e
 ey  1  0
2
2y


dx
e


y


e2y
d2y
 e2y  2 e y  e2y  e y  0
2
dx
e2y
d2y
d2y
y

e

0

 ey  0
2
2
dx
dx
Maths Dept @ KCB 2020 Prepared by James Lubega - HOD
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8.
 3
 1
  1
Given that a    , b    and c    , find the values of t such that t a  b and
0
 1
  3
t b  c are
perpendicular .
 3  1  3t  1
 1   1   t  1 
 ; t b  c  t       

t a  b  t       
 0  1  1 
 1   3   t  3 
For perpendicular vectors t a  b and t b  c , t a  b  t b  c  0
 3t  1  t  1 
  
  0  3t  1t  1  1t  3  0
 1   t  3
So, 
9.
3t 2  t  4  0 ; 3t  4 t  1  0
4
t
, t  1
3
 3  3 
 1 
   
 
Show that the vectors  4  ,   1 and   2  are coplanar.
1  2 
 1 
   
 
Review, if vectors a , b and c are coplanar, then a . b c  0 or b . a c  0 or
c . a b  0 .
 3  3   9 
    

 4     1    3 
 1   2    15
    

 1   3   3   1   9 
        

Now   2  .  4     1    2  .   3   9  6   15  0
 1   1   2   1    15
        

 1   3   3 
     
Since   2  .  4     1  0 , the vectors are coplanar.
 1   1   2 
     
10.
Find
dx
x
x
.
Maths Dept @ KCB 2020 Prepared by James Lubega - HOD
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1
Let u 
dx
x
11.
x
du 1  2
x;
 x  dx  2u du
dx 2

Solve the equation
For

2 u du
2

du  2 lnu  1  C  2 ln 1  x  C
2
u 1
u
u

x 1
2x
3
 2.
2x
x 1
x 1
2x
x 1
3
 2 , let a 
:“Here notice a reciprocal”
2x
2x
x 1
 a
 a 
3

a  1  0 ; a  9 or a  1

a 3
a
2;
2
 2 a  3  0.

When a = 9,
x 1
1
9; x
2x
17
When a = 1,
x 1
 1 ; x  1
2x
Remember to test for such an equation with surds.
For x  
3
1
 3  1  1  RHS
, LHS  9 
17
9
For x   1 , LHS  1  3 1   2  RHS

12.
x 
1
17
The points A4 , 0, B0 , 3 and Px , y  are such that PA and PB are always
perpendicular. Show that the locus of P is a circle. Hence find the centre and
radius of the circle.
Grad PA =
y0
y3 y3
y


; Grad PB =
x4 x4
x0
x
Review: If two lines with gradients m1 and m2 are perpendicular, then m1  m2   1
Maths Dept @ KCB 2020 Prepared by James Lubega - HOD
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
y   y  3
. 
  1
 x 4  x 
For this case, 
y  y  3   x x  4  x 2  y 2  4 x  3 y  0 of the form
x 2  y 2  2 gx  2 fy  c  0 , hence a circle.
By comparison; 2 g   4, g   2 ; 2 f   3 , f  


3
and c  0
2
3
2
Centre is  g ,  f    2 , 
Radius, r  g 2  f 2  c 
 22
2
 3
     0  2.5 units
 2
NOTE:
(i)
(ii)
(iii)
(iv)
(v)
You are required to continue doing practice wherever you are. The
pandemic will soon come to pass.
Do practice all the content you were given right from s5.
Such series of discussions will continue, God willing.
I wish you God’s blessings in everything you do.
Stay home, maintain social distance and revise. Thank you.
Maths Dept @ KCB 2020 Prepared by James Lubega - HOD
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