NECO GCE 2024/2025 Further Maths / Further Mathematics – It true that there are several smart secrets to achieving academic success; but of them all, using relevant materials is by far the smartest, strongest and best way to succeed. Above all, you must study your books before entering the exam hall.
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
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 for
ce 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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