**WAEC Further Mathematics Questions 2018 and OBJ and Theory Answers.**

Further Mathematics WAECQuestions 2018… In this article, I will be showing you past Further Mathematics objective and theory random repeatedquestionsfor free. You will also understand howWAEC FurtherMathematics questions are set and many other examination guides. Stay focus and read through.

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*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.

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**WAEC Further Mathematics Answers.**

1CBBDCAADCB

11DCBDCBCDCC

21ABDCDABBCB

31CDCADBCAAD

11DCBDCBCDCC

21ABDCDABBCB

31CDCADBCAAD

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

12)

P:F=4:1 =4x+1x=100

5x=100

x=100/5

x=20

pass=20*4=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)

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

P:F=4:1 =4x+1x=100

5x=100

x=100/5

x=20

pass=20*4=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

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

(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)=1/5*1/4*1/3

=1/60

5b)

pr(only age and fully gained admission)=4/5*3/4*1/3

=1/5

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

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)=1/5*1/4*1/3

=1/60

5b)

pr(only age and fully gained admission)=4/5*3/4*1/3

=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

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

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

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

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

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

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)

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)^2/2*10

=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

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

**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)^2/2*10

=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

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

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)

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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