NECO FURTHER MATHEMATICS QUESTIONS AND ANSWERS 2017

NECO FURTHER MATHEMATICS QUESTIONS AND ANSWERS 2017

NECO FURTHER MATHEMATICS QUESTIONS AND ANSWERS 2017 – Past questions and answers, 100% real answers, neco expo, neco exam runz, complete further maths expo, exam paper leaked on exam day – Get more update and answers on all exams today.

NECO FURTHER MATHEMATICS QUESTIONS AND ANSWERS 2017

Finally here comes the Further maths objectives Answers.

First and foremost  – 1BBEDEDABCA
Afterward  -11DCBBABDABD
Above all -21AECCBEACBA
Finally  -31DAACBDEEBD

First and Foremost

1
P=15kg(15N)
Efx=8cos50degrees-10cos60degrees-150cos90degrees
=8(0.6428)-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.

NECO FURTHER MATHEMATICS QUESTIONS AND ANSWERS 2017

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)
Finally =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)
Finally =89.94 degree.

NECO FURTHER MATHEMATICS QUESTIONS AND ANSWERS 2017

(1iii) acceleration
from P-R=F
i.e 150-135.21=14.74
ma=14.79
a=14.79/m=14.79/150g
Finally =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
Above all  -a=0.26m/s^2.

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NECO FURTHER MATHEMATICS QUESTIONS AND ANSWERS 2017

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
Above all the initial velocity u=3.87m/s.

NECO FURTHER MATHEMATICS QUESTIONS AND ANSWERS 2017

(6i)
from 3x-y+11=0
-y=-3x-11
y=3x+11(comparing to y=mx+c)
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
finally =y=3x-8
or y-3x+8=0 or 3x-y-8=0.

NECO FURTHER MATHEMATICS QUESTIONS AND ANSWERS 2017

(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)
Above all =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.

NECO FURTHER MATHEMATICS QUESTIONS AND ANSWERS 2017

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,

Above all 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.

NECO FURTHER MATHEMATICS QUESTIONS AND ANSWERS 2017

(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.

Furthermore.
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.

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NECO FURTHER MATHEMATICS QUESTIONS AND ANSWERS 2017

9

Furthermore.

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.

Furthermore

=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
Above all 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 =

NECO FURTHER MATHEMATICS QUESTIONS AND ANSWERS 2017

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)t (13) diagram frictional force Fr=mg sin tita =5.6* 9.8* 0.5878 =32.26N

Reaction R= mgcos tita =5.6 * 9.8 cos^2 36 56* 9.8 * 0.8090 =44.40N..

(a) Above all the force parallel to the plane Fr==32.26N.

(b) Finally from Fr= UR =0.45* 44.40 =19.98N.

6a ) Finally – greatest height (H )= ( U ^2 sin ^2 tita )/ 2 g =956 ^ 2 sin ^2 ( 36)) / 2 (10) =( 3136 * 0. 5878 ^2 )/ 20 = 1083 .516 /20 =54 .196 =54 .18 6b ) horizontal range( R) = ( U ^ 2 sin2 tita ) /g =( 56^ 2 sin 2( 36) )/ 10 = (3136 * sin 72)/ 10 =( 3136* 0 .9511 )/ 10 =2982 .6496 /10 =298 . 26495 =298 . 26m 6c ) time of flight =2 sin tita / g =( 2 *56 sin 36) / 10 =112 * 0 .5878 =65 .8336 /10= 6 .5834 — – — — – — — – — — – — — – — — – — — 15a) capital demand=N12,000* N120 =N5,040,000 total purchasing cost =N140*12,000 =N1,680,000 holding cost amount =20% =20/100 * N5,040,000 =N1,008,000 Economic order= N1,680,000 + N1,008,000 =N2688000 ii) total value cost per n annum =N268800/12,000 =N224 15b) mean x=74 s.d=15 i) Pr(x>85)=x- xbar/ sigma
=85-74 /15
=11/15 =0.73
=0.2673

ii)
Finally Pr(x<62)=(62-74)/15 =-12/15=-0.8 =0.2881 iii) Pr(62 =0.5-.2673 + 0.5- 0.2881 =0.4446

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