Waec 2017 further maths answers (expo)

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FURTHERMATHS OBJ:
1-10: CBADCBADBB
11-20: DCBDCCDBCC
21-30: ACDCAABBCB
31-40: CDCDDBCAAD

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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=712ai)
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(3<m>6)
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(3<m>6)
=0.114688+0.028672+0.0043008
+0.0003584
=0.1480192
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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

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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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7)
m1=3
u1=8m/s
m2=?
u2=5m/s
v=6m/s
m1u1+m1u2=(m1+m2)v
3*8+m2*5=(3+m2)6
24+5m2=18+6m2
24-18=6m2-5m2
m2=6
(7b)
m2u2-m1u1=V(m1+m2)
6*5-3*8=V(3+6)
30-24=9v
9v=6
v=6/9
v=0.67m/s

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(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

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(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

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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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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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(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


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13ai)
M=2
P=5
C=3
total=10
If the books of the same subject are to stand together
No of arrangements=2!*5!*3!*3!
=2*120*6*6
=8640arrangements
(13aii)
Only the physics textbook must stand together
No of arrangements=5!*6!
=120*720
=86400arrangements
(13b)
P=13/20
q=1-13/20=7/20
pr(atleast 3 speak E)=1-Pr(2 speak E)
=(1-8C1p^1q^7+8C2p^2q^6)
=1-(8*(13/20)*(7/20)^7+28(13/20)^2*(7/20)^6
=1-(0.003346+0.0217467)
=1-0.0251
=0.9749
=0.975(3s.f)  ==================================

2)
x+1/3x^2-x-2
3x^2-x-2/-6(-3 2)
3x^2-3x+2x-2
3x(x-1)+2(x-1)
(3x+2)(x-1)
x+1/(3x+2)(x-1)=A/3x+2+B/x-1
x+1/(3x+2)(x-1)=A(x-1)+B(3x+2)/(3x+2)(x-1)
x+1=A(x-1)+B(3x+2)
3x+2=0
x=-2/3
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8a)
60,56,70,63,50,72,65,60
mean=£x/n=60+56+70+63+50+72+65
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