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WAEC 2019 Further Mathematics Theory and Objective Answers /Expo

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WAEC 2019 Further Mathematics Theory and Objective Answers /Expo by : 1:06 am On April 9, 2018

WAEC 2019 Further Mathematics exam will be holding on 9th May, 2019. As we have been doing for other exams, we are here to help you with the Further Mathematics Theory and Objective Answers. Most importantly, you should subscribe to Earboard.com Waec Runz so that you get correctly solved answers to the Further Mathematics questions.

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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


(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


(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


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


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


(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

 


(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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