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VERIFIED NECO JUNE/JULY 2015
EXAM DAY: FRIDAY
SUBJECT: FURTHER MATH
Further maths obj
1BBEDEDABCA
11DCBBABDABD
21AECCBEACBA
31DAACBDEEBD
41CCECBDCADA
Theory
Further maths obj
1
P=15kg(15N)
Efx=8cos50degrees-10cos60degrees-150cos90de
=8(0.642 -10(0.5000)-150(0)
=5.142-5
=-0.1424N
Efy=8sin50degrees
+10sin60degrees-150sin90degrees
=8(0.7660)+10(0.8660)-150(1)
=6.128+8.660-150
=-135.242N
i)Resultant of force R=Sqroot
(Efx^2+Efy^2)
=sqroot(-0.1424)^2+(-135.242)^2
=sqroot( 0.020+18282.285)
=sqroot( 18282.305)
=135.21N
(1ii)
let titat be the direction odf the
resultant force
tan tita= Efx/Ef
tan tita=-135.212/-0.1424
=949.5225.
tita=tan^-1(949.5225)
=89.94 degree
(1iii) acceleration
from P-R=F
i.e 150-135.21=14.74
ma=14.79
a=14.79/m=14.79/150g
=0.986ms^-1
2
a)From S=ut-1/2at^2
29=3.89*15-1/2*a*15^2
29=58.09-0.5*a*225
29-58-58.05=-0.5*a*225
-29.05=-112.5a
a=29.05/112.5
a=0.26m/s^2
b)Total distance=21m+8m=29m
Total time=10+5=15s
from S=(v+u/2)t
v=0,s=29m,t=15s
29=(0+u/2)15
29=(u/2)15
2*29=15u
58=15u
=>u=58/15
therefore initial velocity u=3.87m/s
(6i)
from 3x-y+11=0
-y=-3x-11
y=3x+11(comparing to y=mx+c)
gradient m=3,
from equation of straight line
y-y1=m(x-x1)
(x1,y1)=(1,-5) or x1=1, y1=-5
y-(-5)=3(x-1)
y+5=3(x-1)
y=3x-3-5
y=3x-8
or y-3x+8=0 or 3x-y-8=0
(6ii)
from y=3x+4 and y=2x-1
let y1=3x+4 nand y2=2x-1
therefore m1=3, m2=2
let tita be the angle between the
two lines.
tan titat=m1-m2/(1+m1m2)=(3-2)/
(1+3(2))
tan tita= 1/1+6 =1/7
tita= tan^-1(1/7)
tita= 8.13 degree
4
i)seven male surgeons can not be
formed from a team consisting of
four men
Also three female surgeons cannot
be formed from two women
Hence the numbers of way is
i)0 ways
ii)0 ways
5
i)for -2<_x<_2
ie x=-2,-1,0,1,2
for x=-2,f(x)=(-2)^2+2=4+2=6
for x=-1,f(x)=(0)^2+2=1+2=3
for x=0,f(x)=(0)^2+2=0+2=2
for x=1,f(x)=1^2+2=1+2=3
for x=2,f(x)=2^2+2=4+2=6
Hence the co-domain are 2,3,6
ii) f is onto
Reason since f(-2)=f(2)
f(-1)=f(1)
therefore f is onto
iii)f(x)=x^2+2=y
=>x^2+2=y
x^2=y-2
x=sqroot(y-2)
f^-1(x)=sqroot(x-2)
f^-1(11)=sqroot(11-2)
=sqroot(9)
=3 0r -3
(7)
a)Given a*b=a+b+2ab where a,bER
To show wether * is commutative
a*b=a+b+2ab
b*a=b+a+2ba=a+b+2ab=a*b
Hence, the operation is
commutative
b)a*e=a+e+2ae=a=e*a
=>a+e+2ae=a
=>e+2ae=a-a
=>e(1+2a)=0
=>e=0/1+2a
=>e=0
the identity element e=0
7c)a*a^-1
=>a+a^-1+2aa^-1=0
a=a_1(1+2a)=0
a^1(1+2a)=0-a
a^-1(1+2a)=-a
a^-1=-a/1+2a
the inverse element a^-1=-a/1+2a
Any element E/ -a/1+2a has no
inverse
9
Given the curve f(x)=x^3-6^2-15x-1
dy/dx=3x^2-12x-15
of the turning point dy/dx=0
=>3x^2-12x-15=0
or x^2-4x-5=0
x^2-5x+x-5=0
x(x-5)+1(x-5)=0
(x-5)(x+1)=0
x=5 or x=-1
d^2y/dx^2=6x-12
i)The gradient of x=1=dy/dx x=1
=3(1^2)-12(1)-15
=3(1)-12-15
=3-12-15
=-24
9ii)when x=5,d^2y/dx^2 x=5
=6(5)-12
=30-12
=18
when x=-1,d^2y/dx^2 x=-1
=6(-1)-12
=-6-12
=-18
since d^2y/dx^2>1 when x=5, the y
is minimum
since d^2y/dx^2<1 when x=-1,then
y is maximum at x=-1
when x=5,f(x)=5^3-6(5^2)-15(5)-1
=125-150-75-1
=the maximum point is (5,-101)
when x=-1,f(x)=
(-1)^3-6(-1)^2-15(-1)-1
=-1-6+15-1
=7
Hence the maximum point is (-1,7)
10
Given thatalpha,beta are the roots of
2x^2-x+4
where a=2,b=-1,c=4
(alpha+beta)=-b/a, =-1/2
(alpha*beta)=c/a, =4/2=2
i)alpha^2+beta^2=(alpha
+beta)^2-2alpha*beta
=(-1/2)^2-2(2)
=1/4-4/1=1-16/4
=-15/4
ii)alpha^3+beta^3=(alpha
+beta)^3-39alpha*beta)
=(-1/2)^3-3(2)(-1/2)
=-1/8-6(-1/2)=-1/8+3
=-1+24/8
=23/8
iii)alpha/beta+beta/alpha
=(alpha^2+beta^2)/(alpha*beta)
=(-15/4)/2
=-15/8
10b)Given (2a+3)x^2-6x+4-a
the expression is a perfect square of
b^2=4ac
ie(-6)^2=4(2a+3)(4-a)
36=4(8a-2a^2+12-3a)
9=8a-2a^2+5a+12
-2a^2+5a=12-9=0
-2a^2+5a+3=0
2a^2-5a-3=0
2a^2-6a+a-3=0
2a(a-3)+1(a-3)=0
(a-3)(2a+1)=0
a-3=0 or 2a+1=0
a=3 or a=-1/2
10c)Given x-2y=2----eq1
5x^2+3y^2=3---eq2
from(1)x=2+2y---eq3
substitute for x in eq(2)
5x^2+3y^2=3
5(2+2y)^2+3y^2=3
5(4+8y+4y^2)=3y^2=3
20+40y+120y^2+3y^2=3
23y^2+40y+20-3=0
23y^2+23y+17y+17=0
23y(y+1)+17(y+1)=0
(23y+17)(y+1)=0
23y+17=0 or y+1=0
y=-17/23 or y=-1
From x=2+2y
when y=-17/23,x=2+2(-17/23)
=2-34/23
=46-34/23=12/23
when y=-1,x=2+2(-1)
=2-2=0
the solutions are x,y (12/23,-17/23)
or (0,-1)
(12ai)
please note we used n as
intersection here
p=-2i+2j, q=2i+4j and r=4i+5j to
show that P,Q and R are collinear.
we show that (PnQnQ)=0.
PnQnR=
matrices
row 1 -2,2,1|
row2: 2,4,1
row3:4,5,1
=-2(4-5)-2(2-4)+1(10-16)
=-2(-1)-2(-2)+1(-6)
2+4-6=0
hence P,Q and R are collinear.
(12aii)
note that this arrow(=>) will be on
top of PQ
|=>PQ|=|-2(2)+2(4)|=|-4+8|
=sqr root((-4)^2+8^2=sqr root
(16+64)= sqr root (80)
=sqr root(16*5)= 4root5
|=> QR|= |2(4)+4(5)|=|8+20|
=sqr root((8^2)+(20^2)
=sqr root(64+400)
=sqr root (464)= sqr root(16*29)
=4root 29
hence|=> PQ| br /> (12bi)
scalar product of r and s= r.s
< r= (49/32)i+3j=(49i+99j )/33, =>
r= 49i+ 99j
s= 495i-124j
r.s=(49i+99j).(495i-124j)
=(149*495)-
(99*124)=24235-12276=11979
12bii)let the angle between r and s
be tita
since r.s=|rs|cos tita
therefore cos tita=r.s/|rs|
|rs|=|(49i+99j)(495i-125j)|
=|24255i-12276j|
=sqroot(24255^2)+(-12276^2)
=sqroot(739005201)
=27184.6501
therefore cos
tita=11979/27184.6501
=0.4407
therefore tita= cos^-1(0.4407)
=63.85degrees
1BBEDEDABCA
11DCBBABDABD
21AECCBEACBA
31DAACBDEEBD
41CCECBDCADA
Theory
Further maths obj
1
P=15kg(15N)
Efx=8cos50degrees-10cos60degrees-150cos90de
=8(0.642 -10(0.5000)-150(0)
=5.142-5
=-0.1424N
Efy=8sin50degrees
+10sin60degrees-150sin90degrees
=8(0.7660)+10(0.8660)-150(1)
=6.128+8.660-150
=-135.242N
i)Resultant of force R=Sqroot
(Efx^2+Efy^2)
=sqroot(-0.1424)^2+(-135.242)^2
=sqroot( 0.020+18282.285)
=sqroot( 18282.305)
=135.21N
(1ii)
let titat be the direction odf the
resultant force
tan tita= Efx/Ef
tan tita=-135.212/-0.1424
=949.5225.
tita=tan^-1(949.5225)
=89.94 degree
(1iii) acceleration
from P-R=F
i.e 150-135.21=14.74
ma=14.79
a=14.79/m=14.79/150g
=0.986ms^-1
2
a)From S=ut-1/2at^2
29=3.89*15-1/2*a*15^2
29=58.09-0.5*a*225
29-58-58.05=-0.5*a*225
-29.05=-112.5a
a=29.05/112.5
a=0.26m/s^2
b)Total distance=21m+8m=29m
Total time=10+5=15s
from S=(v+u/2)t
v=0,s=29m,t=15s
29=(0+u/2)15
29=(u/2)15
2*29=15u
58=15u
=>u=58/15
therefore initial velocity u=3.87m/s
(6i)
from 3x-y+11=0
-y=-3x-11
y=3x+11(comparing to y=mx+c)
gradient m=3,
from equation of straight line
y-y1=m(x-x1)
(x1,y1)=(1,-5) or x1=1, y1=-5
y-(-5)=3(x-1)
y+5=3(x-1)
y=3x-3-5
y=3x-8
or y-3x+8=0 or 3x-y-8=0
(6ii)
from y=3x+4 and y=2x-1
let y1=3x+4 nand y2=2x-1
therefore m1=3, m2=2
let tita be the angle between the
two lines.
tan titat=m1-m2/(1+m1m2)=(3-2)/
(1+3(2))
tan tita= 1/1+6 =1/7
tita= tan^-1(1/7)
tita= 8.13 degree
4
i)seven male surgeons can not be
formed from a team consisting of
four men
Also three female surgeons cannot
be formed from two women
Hence the numbers of way is
i)0 ways
ii)0 ways
5
i)for -2<_x<_2
ie x=-2,-1,0,1,2
for x=-2,f(x)=(-2)^2+2=4+2=6
for x=-1,f(x)=(0)^2+2=1+2=3
for x=0,f(x)=(0)^2+2=0+2=2
for x=1,f(x)=1^2+2=1+2=3
for x=2,f(x)=2^2+2=4+2=6
Hence the co-domain are 2,3,6
ii) f is onto
Reason since f(-2)=f(2)
f(-1)=f(1)
therefore f is onto
iii)f(x)=x^2+2=y
=>x^2+2=y
x^2=y-2
x=sqroot(y-2)
f^-1(x)=sqroot(x-2)
f^-1(11)=sqroot(11-2)
=sqroot(9)
=3 0r -3
(7)
a)Given a*b=a+b+2ab where a,bER
To show wether * is commutative
a*b=a+b+2ab
b*a=b+a+2ba=a+b+2ab=a*b
Hence, the operation is
commutative
b)a*e=a+e+2ae=a=e*a
=>a+e+2ae=a
=>e+2ae=a-a
=>e(1+2a)=0
=>e=0/1+2a
=>e=0
the identity element e=0
7c)a*a^-1
=>a+a^-1+2aa^-1=0
a=a_1(1+2a)=0
a^1(1+2a)=0-a
a^-1(1+2a)=-a
a^-1=-a/1+2a
the inverse element a^-1=-a/1+2a
Any element E/ -a/1+2a has no
inverse
9
Given the curve f(x)=x^3-6^2-15x-1
dy/dx=3x^2-12x-15
of the turning point dy/dx=0
=>3x^2-12x-15=0
or x^2-4x-5=0
x^2-5x+x-5=0
x(x-5)+1(x-5)=0
(x-5)(x+1)=0
x=5 or x=-1
d^2y/dx^2=6x-12
i)The gradient of x=1=dy/dx x=1
=3(1^2)-12(1)-15
=3(1)-12-15
=3-12-15
=-24
9ii)when x=5,d^2y/dx^2 x=5
=6(5)-12
=30-12
=18
when x=-1,d^2y/dx^2 x=-1
=6(-1)-12
=-6-12
=-18
since d^2y/dx^2>1 when x=5, the y
is minimum
since d^2y/dx^2<1 when x=-1,then
y is maximum at x=-1
when x=5,f(x)=5^3-6(5^2)-15(5)-1
=125-150-75-1
=the maximum point is (5,-101)
when x=-1,f(x)=
(-1)^3-6(-1)^2-15(-1)-1
=-1-6+15-1
=7
Hence the maximum point is (-1,7)
10
Given thatalpha,beta are the roots of
2x^2-x+4
where a=2,b=-1,c=4
(alpha+beta)=-b/a, =-1/2
(alpha*beta)=c/a, =4/2=2
i)alpha^2+beta^2=(alpha
+beta)^2-2alpha*beta
=(-1/2)^2-2(2)
=1/4-4/1=1-16/4
=-15/4
ii)alpha^3+beta^3=(alpha
+beta)^3-39alpha*beta)
=(-1/2)^3-3(2)(-1/2)
=-1/8-6(-1/2)=-1/8+3
=-1+24/8
=23/8
iii)alpha/beta+beta/alpha
=(alpha^2+beta^2)/(alpha*beta)
=(-15/4)/2
=-15/8
10b)Given (2a+3)x^2-6x+4-a
the expression is a perfect square of
b^2=4ac
ie(-6)^2=4(2a+3)(4-a)
36=4(8a-2a^2+12-3a)
9=8a-2a^2+5a+12
-2a^2+5a=12-9=0
-2a^2+5a+3=0
2a^2-5a-3=0
2a^2-6a+a-3=0
2a(a-3)+1(a-3)=0
(a-3)(2a+1)=0
a-3=0 or 2a+1=0
a=3 or a=-1/2
10c)Given x-2y=2----eq1
5x^2+3y^2=3---eq2
from(1)x=2+2y---eq3
substitute for x in eq(2)
5x^2+3y^2=3
5(2+2y)^2+3y^2=3
5(4+8y+4y^2)=3y^2=3
20+40y+120y^2+3y^2=3
23y^2+40y+20-3=0
23y^2+23y+17y+17=0
23y(y+1)+17(y+1)=0
(23y+17)(y+1)=0
23y+17=0 or y+1=0
y=-17/23 or y=-1
From x=2+2y
when y=-17/23,x=2+2(-17/23)
=2-34/23
=46-34/23=12/23
when y=-1,x=2+2(-1)
=2-2=0
the solutions are x,y (12/23,-17/23)
or (0,-1)
(12ai)
please note we used n as
intersection here
p=-2i+2j, q=2i+4j and r=4i+5j to
show that P,Q and R are collinear.
we show that (PnQnQ)=0.
PnQnR=
matrices
row 1 -2,2,1|
row2: 2,4,1
row3:4,5,1
=-2(4-5)-2(2-4)+1(10-16)
=-2(-1)-2(-2)+1(-6)
2+4-6=0
hence P,Q and R are collinear.
(12aii)
note that this arrow(=>) will be on
top of PQ
|=>PQ|=|-2(2)+2(4)|=|-4+8|
=sqr root((-4)^2+8^2=sqr root
(16+64)= sqr root (80)
=sqr root(16*5)= 4root5
|=> QR|= |2(4)+4(5)|=|8+20|
=sqr root((8^2)+(20^2)
=sqr root(64+400)
=sqr root (464)= sqr root(16*29)
=4root 29
hence|=> PQ| br /> (12bi)
scalar product of r and s= r.s
< r= (49/32)i+3j=(49i+99j )/33, =>
r= 49i+ 99j
s= 495i-124j
r.s=(49i+99j).(495i-124j)
=(149*495)-
(99*124)=24235-12276=11979
12bii)let the angle between r and s
be tita
since r.s=|rs|cos tita
therefore cos tita=r.s/|rs|
|rs|=|(49i+99j)(495i-125j)|
=|24255i-12276j|
=sqroot(24255^2)+(-12276^2)
=sqroot(739005201)
=27184.6501
therefore cos
tita=11979/27184.6501
=0.4407
therefore tita= cos^-1(0.4407)
=63.85degrees
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