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Past WAEC Further Mathematics Answers

1-10DCADBCCDCC
11-20CDBDDDDBAC
21-30BCDCAABBAB
31-40BCDDCBBBDB
41-50ACBACCBBCA

THEORY

(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 OR 3

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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)x-8,5,1,7,2,6,3,4
y-6,3,4,8,5,7,1,2
solution

Coefficient of Correlation = ∑xy / √(∑x²∑y²)

Where:

x = (X – Mean)

y = (Y – Mean)

Data table
X x x² Y y y²
8 3.5 12.25 6. 1.5. 2.25
5 0.5 0.25 3 -1.5. 2.25
1 -3.5 12.25 4 -0.5 0.25
7 2.5 6.25 8 3.5 12.25
2 -2.5 6.25 5 0.5 0.25
6 1.5 2.25 7 2.5 6.25
3 -1.5 2.25 1 -3.5 12.25
4-0.5 0.25 2 -2.5 6.25

Mean of the set of Data for X = ∑X / N

∑X = 8 + 5 + 1 + 7 + 2 + 6 + 3 + 4

∑X = 36
starlight
Mean of the set of Data for X = 36 / 8

Mean of the set of Data for X = 4.5

Mean of the set of Data for Y = ∑Y / N

∑Y = 6 + 3 + 4 + 8 + 5 + 7 + 1 + 2

∑Y = 36

Mean of the set of Data for Y = 36 / 8

Mean of the set of Data for Y = 4.5

From the Data Analysis we have:

∑xy = 5.25 + -0.75 + 1.75 + 8.75 + -1.25 + 3.75 + 5.25 + 1.25

∑xy = 24

∑x² = 12.25 + 0.25 + 12.25 + 6.25 + 6.25 + 2.25 + 2.25 + 0.25

∑x² = 42

∑y² = 2.25 + 2.25 + 0.25 + 12.25 + 0.25 + 6.25 + 12.25 + 6.25

∑y² = 42

starlight
Coefficient of Correlation = ∑xy / √(∑x²∑y²)

Coefficient of Correlation = 24 / √(42 x 42)

Coefficient of Correlation = 24 / √(1764)

Coefficient of Correlation = 24 / 42

Coefficient of Correlation = 0.5714

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(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

====================================

(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

============================
(8)
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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(9)
Draw the pie chart diagram

radius is the same hence
(y – 2)² + (X – 3)² = (y + 4)² + (X – 5)² = (y + 1)² + (X + 2)²
Taking the first pair
Y² – 4y + 4 + x² – 6x + 9 = y² + 8y + 16 + x² – 10x + 25
– 6x – 4y + 13 = -10x+8y+41
4x – 12y = 28
X = 3y = 7——–(1)

Taking the second pair
Y² + 3y + 16+x² – 10x+25 = y²+ 2y + 1 + x²+ 4x + 4
-10x + 3y + 41 = 4x + 2y + 5
14x – 6y = 36
7x – 3y = 18———(2)
Eqn 2 minus Eqn 1
6x = 11
X = 11/6
Put this into Eqn (1)
11/6 – 3y = 7
3y = 11/6 – 7
3y = 11 – 42/6
3y = -31/6
y = -31/18

(a) coordinates of centre is (11/6, -31/18)

(b) Radius r = √(y-2)²+(x-3)²
r = √-31/18 -2)² + 11/6 – 3)²
r= √13.85 + 1.36
r = 3.9

(c) (X – 11/6)² + (y + 31/18)² = 3.9²
(X – 11/16)² + (y + 31/18)² = 15.21

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