Neco 2016 FURTHER MATHEMATICS (Obj And Theory) Answers

F/MATH 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)
======================
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
======================
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*2/5
=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*2/5
=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)^3 + 5(3)(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)^2(0.04)^3 + 5(3)(0.04)
^4 + (0.04)^5
=243 + 5*81(0.04) + 10*27(0.0016) + 10*9(0.000064) +
15(0.0000026) + 0.0000001
=243 +16.2 + 0.432 + 0.00576 + 0.000039 + 0.0000001
=259.63783
======================
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
======================
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
======================.
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
======================
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+250.2e^50
=1/e^50[1+50+25/2]
======================
(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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