Waec 2017 Further Mathematics Answers – May/June Expo - GurusGist

NIGERIA ANSWERS
=================
F/maths Obj
1CBBDCAADCB
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+0.0043008
=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)

12b)
n=5
P(sin) =1/6-p
p(x=3)=5^C3 (1/6)^3 (5/6)^2
=10*1/216*25/36
=250/216*36
=41.6
==================

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

(5)
USING THE BINOMIAL PROBABILITY FUNCTION
n(rp rq n-r
n=10
p=8%(0.08)
q=92%(0.94)
Prob(aprotimate – under >)
=10/2(0.08) 3(0.92) 7 + 10/4(0.08) 4(0.92) 6+ 10/5(0.08) 5(0.92) 5 + 10/6(0.08) 6(0.92) 4+10/7(0.08) 7(0.92) 3+10/8(0.08) 8(0.92) 2+10/9(0.08) 9(0.92) 1+10/10(0.08) 10(0.92) 0
=0.0343+0.052+0.0005+0.00003+0.00000195+…….Aproxtimately 0.0400
================

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)
==================