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

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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

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(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

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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)

One Response

1. Doris Esi September 28, 2021
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