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NECO 2016 FURTHER MATHS ANSWERS

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FURTHER MATHS OBJ:
1DCEBDBABAC
11ABDDDCCDAC
21CDBACAEBCB
31CEAEEDEBBD
41CEBCDDCDDA


(1a)
A=(2 -1)(1 3)
|A|=(2*3--1)
|A|=6+1
|A|=7
since the determinant of A is not 0 the A is not non singular matrix
To find A^-1
A^-1=(3 1)(-1 2)
then 1/7(3 1)(-1 2)
A^-1=(3/7 1/7)(-1/7 2/7)


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(2a)
3root2-3root5/5root2+2roo5
conjugate=5root2-2root5
=(3root2-3root5)(5root2-2root5)/(5root2+2root5)(5root2-2root5)
=(15*2-6root10-15root10-6*5)/(25*2-10root10+10root10-4*5)
=(30-31root10-30)/(50-20)
=-31root10/30

(2b)
cosx=adj/hyp=12/13
then tanx=opp/adj=5/12
siny=3/5
tany=3/4
tan(x+y)=tanx+tany/1-tanxtany
=(5/12+3/4)/(1-(5/12)(3/4)
=(14/12)/(1-15/48)
=36/125+18/125+16/125
=70/125
=14/25


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(3a)
Pr(A passed)=4/5,pr(A failed)=1/5
Pr(B passed)=3/5,Pr(B failed)=2/5
Pr(C passed)-2/5,Pr(C failed)=3/5
Pr(atleast one passed)
=4/5*2/5*3/5+3/5*1/5*3/5+2/5*1/5*2
=24/125+9/125+4/125
=47/125

(3b)
Pr(atleast one failed)
=4/5*3/5*3/5+3/5*2/5*1/5+4/5*2/5*2
=101/125
4)
(3+2y)^5 with (a+b)^5 shows
a=3,b=2y
Using pascals triangle the coefficient of (a+b)^5 are 1,5,10,10,5 and 1
(3+2y)^5
=(3)^5+5(3)^4(2y)+10(3)^3(2y)^2+10(3)^2(2y)^4+(2y)^5
=243+810y+1080y^2+720y^3+240y^4+32y^5
(3.04)^5=(3+0.04)^5
=3^5+5(3)^4(0.04)+10(3)^2(0.04)^2+10(3)(0.04)^4+(0.04)^5
=243+5*81(0.04)+10*27(0.0016)+10*9(0.00
=243+16.2+0.432+0.00576+0.000039+0.000
=259.63783


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(10a)
T4=ar^4=162---(1)
T8=ar^7=4374---(2)
Divide (2) by (1)
4374/162=ar^7/ar^4
2187/81=r^3
27=r^3
r=3root27
r=3
Recall(1)
T5=ar^4=162
162=a(3)^4
a=162/81=2

(10ai)
a=1st,T2=ar,T3=ar^2
a=2,T2=2*3=6,T3=2*3^2=18
therefore the three terms are 2,6 and 18

(10aii)
Sn=a(r^n-1)/(r-1)
=2(3^10-1)/(3-1)
Sn=2(3^10-1)/2
=59049-1
=59048

(10b)
a=6,c=60,Sn=330
Sn=n/2(a+c)
330=n/2(6+60)
300*2=n(66)
660=n(60)
n=660/66
=60
L=a+(n-1)d
60=6+(10-1)d
60=6+9d
60-6=9d
54=9d
d=54/9
d=6

(10c)
T2=ar=6---(1)
T4=ar^3=54---(2)
divide 2 by 1
54/6=ar^3/ar
9=r^2
r=root9
r=3
Recall(1)
ar=6
a(3)=6
a=6/3
a=2
Tn=ar^n-1
Tn=2(3)^n-1


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(11a)
5x^2-2x+3=0
Divide both sides by 5
x^2-2/5x+3/5=0--eq1
Compare eq1 above with
x^2-(alpha+Beta)x+alphabeta=o
alpha+beta=2/5
alphabeta=3/5

(11ai)
alpha^3+beta^3=(alpha+beta)(alpha+beta)^2-3alphabeta
=2/5(2/5)^2-3(3/5)
=2/5(4/25-9/5)
=2/5(-41/25)
=-82/125

(11aii)
Sum of alpha +1/beta and beta+1/alpha
=(alphabeta+1)/beta +(alpha beta + 1)/alpha
=alphabeta^2+alpha^2beta+beta/alphabeta
=alphabeta(alpha+beta)+(alpha+beta)/alphabeta
product root
(alpha +1/beta)(beta+1/alpha)
=alpha(beta+1/alpha)=1/beta(beta+1/alpha)
=alphabeta+1+1+1q/alphabeta
sum=alphabeta(alpha+beta)+(alpha+beta)/alphabeta
=3/5(2/5)+(2/5)
=6/25+2/5
=(6+10)/25
=16/25
Product alphabeta+2+1/alphabeta
=3/5+2+1/(3/5)
=3/5+2/1+5/3
=64/15
Recall
x^2-(alpha+beta)+alphabeta=0
x^2-16/25x+64/15=0

(11bi)
f(x)=x^3+2x-kx-6
x-2=0
x=+2
f(2)=2^3+2(2)^2-k(2)-6=0
8+8-2k-6=0
16-6-2k0
10-2k=0
2k=10
k=10/2
k=5

(11bii)
(x^2+4x+3)
(x-2)rootx^3+2x^2-5x-6
=x^3-2x^2
=4x^2-5x
=4x^2-8x
=3x-6
=3x-6
=0
=>x^2+4x+3
=x^2+3x+x+3
=x(x+3)+1(x+3)
=(x+1)(x+3)
the remaining factors of f(x) are (x+1) and (x+3)

(11biii)
Zero of f(x)
x-2=0,then x=2
x+1=0, then x=-1
x+3=0, then x=-3
Zeros of f(x) are 2,-1 and -3


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(12a)
w=weight of the plank
moment=force*perpendicular distance
clockwise moment=Anticlockwise
Make A the reference point
Pw=2.5m,
AM=(2.5-0.8)m
=1.7m g=10m/s
AB=3.2m,PA=0.8m
Convert 30kg to weight
=mg=30*10=300N
therefore 300*0.8=w*1.7
w=240/1.7=141.18N
Distance of its C.G from P=2.5m

(12b)
For R1=A, R2=B
R1=R2=2R1
Let Q be the reference point
300x*141.18*2.5=2R1(4.8+1)
300x+352.95=11.6R1
300x-11.6R1=-352.95---(1)
Let P in the reference point
2R1(0.8+4)=141.15*2.5
9.6R1=352.95
R1=352.95/9.6
=36.77N


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(15a)
Rank correlation=1-6ED^2/n(n^2-1)
Rank correlation =1-(6*50)/8(64-1)
=1-348/504
=1-0.691
=0.309

(15b )
Pr (at most 2 )= Pr ( 0 )+ Pr ( 1 )+ Pr ( 2)
n =1000, Pr = 0. 05 Pr ( 2 )= 0 . 95
Pr (x )= e ^- landa (landa ^ x )
Pr (0 . 05)< 0 . 1 and landa = 1 . 000 *0 . 05= 50
e ^- 50(50^0 ) / o!+ e ^- 50( 50^1 )/ 1!+ e^ -50( 50^2 ) / 2 !
1 / e ^50+ 50/ e ^50+ 250 . 2 e ^50
= 1/ e ^50[1 +50+ 25/ 2]


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COMPLETED
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