2017/2018 NECO Further Maths OBJ Essay Theory Questions And Answers
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NECO Further Maths / Further Mathematics OBJ
Further maths obj Answer (Past Questions & Answer).
Warning: These Objective answers are NOT for this year.
- Question 1 - 10 BBEDEDABCA
- Question 11 - 20 DCBBABDABD
- Question 21 - 30 AECCBEACBA
- Question 31 - 40 DAACBDEEBD
- Question 41 - 50CCECBDCADA
NECO Further Maths / Mathematics Theory Essay
Further maths Essay Answer (Past Questions & Answer).
Warning: These Theory/Essay answers are NOT for this year.
Answer 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
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 of 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
Question 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
Question (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
Question (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
Question 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
Question 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
Question 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
Question 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)t
Question 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) the force parallel to the plane Fr==32.26N
(b) from Fr= UR =0.45* 44.40 =19.98N 6
a) 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
Question 15
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
i) total value cost per n annum =N268800/12,000 =N224
15bi) mean x=74 s.d=15 i) Pr(x>85)=x- xbar/ sigma
=85-74 /15
=11/15 =0.73
=0.2673
ii)
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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