WAEC Further Mathematics Questions 2018 Confirmed OBJ And Theory Answers

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WAEC Further Mathematics Questions 2018 Confirmed OBJ And Theory Answers

WAEC records mass failure in Nov/Dec WAsSCE

Further Mathematics WAEC Questions 2018… In this article, I will be showing you past Further Mathematics objective and theory random repeated questions for free. You will also understand how WAEC Further Mathematics questions are set and many other examination guides. Stay focus and read through.

WAEC Further Mathematics Questions

The West African Examinations Council (WAEC) is an examination board that conducts the West African Senior School Certificate Examination, for University and Jamb entry examination in West Africa countries. … In a year, over three million candidates registered for the exams coordinated by WAEC.

The West African Examination Council (WAEC) Further Mathematics Senior School Certificate Examination (SSCE) paper will take place on Wednesday, 9th April, 2018.

The 2018 WAEC Further Mathematics exam will comprise of Papers 2 & 1 Essay and Objective which will commence from 8.30am and end by 14.30am. That means the examination will last for three hours (3hrs) only.

In this post, we will be posting out samples of the WAEC Further Mathematics questions for candidates that will participate in the examination for practice purposes.

WAEC Further Mathematics Answers.

WARNING: If you use this answers, you will surely fail! Click Here For The Updated Answers

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21ABDCDABBCB
31CDCADBCAAD

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12)
P:F=4:1 =4x+1x=100
5x=100
x=100/5
x=20
pass=204=80%
fail= 20
1=20%
p(pass)=80/100=0.8
p(fail)=20/100=0.2
n=7
12ai)
P(at least 3passed)
P=0.8
Q=0.2
P(x=r)=n(rP^rq^n-r
P(x>/3)=1-P(x<2) P(x<2)=P(x=0)+P(x=1)+P(x=2) P(x=0)=7dgree (0.8)degree (0.2)^7 P(x=0)=0.0000128 P(x=1)=^7( (0.8)^1 (0.2)^6 =0.0003584 P(x<2)=7^C2 (0.8)^2 (0.2)^5 =0.0043008 P(X<2)=0.0000128+0.0003584+0.004300 =0.004672 P(x>3)=1-0.004672
=0.995321
=0.10(2d.p)
12aii)
P(between 3 and 6 failed)
P=0.2
q=0.8
P(36)
P(x=3) + P (x=4)+p(x=5)+P(x=6)
p(x=3) 7^C3 (0.2)^3 (0.8)^4
=0.114688
p(x=4)=7^C4 (0.2)^4 (0.8)^3
0.028672
P(x=5)=7^C5 (0.2)^5 (0.8)^2
=0.0043008
P(x=6)=7^C6 (0.2)^6 (0.8)^1
=0.0003584
p(36)
=0.114688+0.028672+0.0043008
+0.0003584
=0.1480192

=0.15(2d.p)

4)
(x^2+5x+1)sqroot(2x^3+mx^2+nx+11)=(2x-5)
remainder:30x+16
(x^2+5x+1)(2x-5)
=2x^3+10x^2+2x-5x^2-25x-5
=2x^3+10x^2-5x^2-25x-5
=2x^3+5x^2-23x+30x+16-5
=2x^3+5x^2+7x+11

Therefore m=5, n=7

5a)
pr(age)=4/5
pr(fully)=3/4
pr(must)=2/3
pr(age not admitted)=1-4/5
=1/5
pr(fully not admitted)=1-3/4
=1/4
pr(must not admitted)=1-2/3
=1/3
Therefore pr(none admitted)=1/51/41/3
=1/60
5b)
pr(only age and fully gained admission)=4/53/41/3

=1/5

12a)
tabulate
Marks| 1-10, 11-20, 21-30, 31-40, 41-50,51-60, 61-70, 71-80, 81-90, 91-100
F| 3, 17, 41, 85, 97, 115, 101, 64, 21, 6
C.B| 0.5-105, 10.5-205, 20.5-305, 30.5-405,40.5-505, 50.5-605, 60.5-705, 70.5-805,80.5-905, 90.5-1005

C.F| 0+3=3, 3+17=20, 20+41=61, 61+85=146,146+77=243, 243+115=358, 358+101=459,459+64=523, 523+21=544, 544+6=550

11a)
Given:
f(x)={(4x-x^2)dx
f(x)=2x^2 – x^3/3 + K
f(3)=2(3)^2 – (3)^2/3 + K =21
18 – 9 + K=11
9+K=21
K=21-9
K=12
Therefore
f(x)= -x^3 + 2x^2 + 12
11b)
i) Tn=a+(n-1)d
T2=a+(2-1)d
T2=a+d
T4=a+3d
T8=a+7d
GP
Tn=ar^n-1
T1=ar^1-1
T2=ar^2-1=ar
T3=ar^2
a+d=a …..equation (1)
a+3d=ar …..equation (2)
a+7d=ar^2 …..equation (3)
T3+T5=20
a+2d+a+4d=20
2a+6d=20
a+3d=10 …..equation (4)
…..equation (2)/…..equation (1)
ar/a=a+3d/a+d
r=a+3d/a+d
…..equation (3)/…..equation (2)
ar^2/ar=a+7d/a+3d
r=a+7d/a+3d
but r=r
a+3d/a+d=a+7d/a+3d
(a+3d)^2=(a+d)(a+7d)
a^2+6ad+ad^2
a^2+7ad+ad+7d^2
a^2+8ad+7d^2
a^2+6ad+9d^2=a^2
+8ad+7d^2
6ad+9d^2=8ad+7d^2
6ad-8ad=7d^2-9d^2
-2ad=2d^2
ad=dd

a=d

(9a)
1/1-cos tita + 1/1+cos tita
=1+cos tita + 1-cos tita//(1-cos tita) (1+cos tita)
= 2/1+cos tita – cos tita – cos^2 tita
= 2/1-cos^2 tita
Recall that :
Cos^2 tita + sin^2 tita = 1
.:. Cos^2 tita = 1-sin^2 tita
.:. 1/1-cos^2 tita + 1/1+cos tita
= 2/1-(1-sin^2 tita)
(9b)
At stationary points,
dy/dx=0.
y=x^0(x-3)
Let u=x^2,v=x-3.
du/dx=2x dv/dx=1.
dy/dx= Udv/dx + Vdu/dx
dy/dx=x^2(1)+(x-3)(2x)
.:. dy/dx=x^2+2x^2-6x
dy/dx=3x^2-6x
At stationary point,
dy/dx=0..
.:.3x^2-6x=0

Equation of line=> 3x^2-6x=0

14ai)
SKETCH THE DIAGRAM
14aii)
Using lami’s theory
T1/sin60=T2/sin30
48N/sin60=T2/sin30
48N/0.8660=T2/0.5
0.5(48)/0.8660=T2(0.8660)/0.8660
T2=24/0.8660
T2=27.7N
14b)
Using the equation of motion
H=U^2/2g
H=(20)^2/210
=20
20/20
H=20m
Timetaken to reach the maximum height
S=Ut+1/2at^2
20=0+1/2(100)t^2
20/5=5t^2/5
t^2=4
t=sqroot4

t=2S

10a)
i) (x^2-1) (x+2)=0
(x-1) (x+1) (x+2)
x=1, or -1 or -2
ii) 2x-3/(x-1)(x+1)(+2)
=A/x-1+B/x+1+C/x+2
2x-3=A(x+1)(x+2)+B(x-1)(x+2)
+C(x-1)(x+1)
let x+1=0,x=-1
2(-1)-3=B(-1-1)(-1+2)
-5/2=-2B/-2 B=5/2
let x-1 =0 x=1
2(1)-3=A(1+1)(1+2)
-1=CA, A=-1/6
Let x+2=0 x=-2
2(-2)-3=C(-2-1)(-2+1)
-7=3C, C=-7/3
10b)
X1 Y2
(3, 1)
r=sqr(x2-x1)^2+(y2-y1)^2
r=sqr(3+3)^2+(1-1)^2
r=sqr6^2+0=sqr36=6
the equatuon of a circle
(x-a)^2+(y-b)^2=r^2
(x-(-3))^2+(y-1)^2=6^2
(x+3)^2+(y-1)^2=36
x^2+6x*9+y^2-2y+1=36
x^2+y^2+6x-2y+9+1-36=0

x^2+y^2+6x-2y-26=0

1a)
g(x)=y
y=x+6
x=y-6
g^- f(x-6)
=4-5(x-6)/2=4-5x+30/2
=34-5x/2
1b)
coodinate=(x1+x2/2 ,y1+y2/2)
=(7-2/2,7-5/2)=(5/2,2/2)
=(5/2,1)

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