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WAEC 2018 FURTHER MATHS THEORY /ESSAY QUESTION & ANSWER IS OUT .

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1)
X-3|2 2|+4|5 2 |+3|5 2|= -24
| -4 6-x| |2 6-x| |2 -4|
X-3(12-2x+8)+4(30-5x-4)+3(-20-4) = -24
X-3(20-2x)+4(26-5x)+3(-24)= -24
20x - 2x² - 60 + 6x + 104 - 20x - 72 = -24
-2x² + 6x - 60+104-72+24 = 0
-2x² + 6x - 4 = 0

Divide through by -2
X² - 3X + 2 = 0
X² - 2X - X + 2= 0
X(X-2)-1(X-2)=0
(X - 1)(X - 2)=0
X=1 OR X=2

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2)
Log 3x - 3logx³+ 2 = 0
Log 3x - 3log3³/log3x + 2 = 0
Log3x - 3/log3x + 2 = 0
P - 3/p + 2 = 0 where P = log3x
P² - 3 + 2p = 0
P² + 2p - 3 = 0
(p + 3) ( p - 1) = 0
P = -3 OR p = 1
But log3x = p
when p = -3
Log3x = -3
X = 3-³ = 1/27
And when P = 1
Log3x = 1
X = 3¹ = 3
And X = 1/27 
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3a)
U=x-2 hence x=u+2
therefore 
x^2+5/(x-2)^4
=(u+2)^3+5/(u+2-2)^2
=(u+2)^3+5/u^4

3b)
(u+2)^3+5/u^4
=(u+2)^3/u^4+5/u^4

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No 4
Sn = n/2[2a+(n-1)d]
S12 = 12/6[2a + 11d]
S12= 6[2a + 11d] = 168
2a + 11d = 28-------(1)
Also 
T3 = a + 2d = 7---------(2)
Multiplying 1 by 2
2a + 4d = 14
Eqn 1 minus Eqn 3 gives 
7d = 14
d = 2
Putting this in eqn 2
a + 2(2) = 7
a + (4) = 7
a = 7 - 4
a = 3
Common difference = 2
First term = 3
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5)
tabulate
X,y,d,d^2
A 8,6,2,4
B 5,3,2,4
C 1,4,-3,9
D 7,8,-1,1
E 2,5,-3,9
F 6,7,-1,1
G 3,1,2,4
H 4,2,2,4
Ed^2 =4+4+9+1+9+1+4+4
=36
r=1-6£d^2/n(n^2-1)
=1-6*36/8(8^2-1)=1-216/504
=1-0.43
=0.57

(6)
It follow that:
P(n) = 1/3 and p(k)¹ = 1 - 1/3 = 2/3
P(T) = 1/5 and P(T)¹ = 1- 1/5 = 4/5
Hence,
probability that only one if the them will be solve the questions will be:
= (1/3 × 4/5) + (1/5 × 2/3)
= 4/15 + 2/15
= 6/15
= 2/5
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(7)
m = 3i - 2j ; n = 2i + 3j ; p = i + 6j
Therefore:
4(3i - 2j) +2(2i +3j) -3(-i + 6j)
12i - 8j + 4i + 6j + 3i - 18j
12i + 4i + 3i - 8j + 6j - 18j
19i = 20j
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(8a)
m1 = 20kg
u1 = 8ms-1
m2 = 30kg
u2 = 50ms-1
(a) in same direction
m1u1 + m2u2 = (m1+m2)v
20 × 80 + 30 × 50 = (20+30)v
1600+1500 = 50v
3100/50 = 50/50
V = 62ms-1
(b)
In opposite direction
m1u1 - m2u2 = (m1 + u2)V
20×30 - 30×50 = (20+30)V
1600 - 1500 = 50V
100/50 = 50V/50
V = 2m/s

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