# WAEC FURTHER MATHEMATICS questions and answers 2020

here is the questions and answers for waec further maths QUESTIONS and answers…. we will post the main 2020 questions when its out, and the answers to it. just stay with us

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)

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

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)=1/5*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

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

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)^2/2*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

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

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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