NECO-FUTHER-MATHS-ANSWERS

F/Maths-Obj.
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Completed.

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Theory- Section A Answer All.
1,2,3,4,5,6,7,8
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(1)
∫dy/dx, dx=∫2x-3dx
Y=x²-3x+c
at the minimum point y=3 and x=0
3=0²-3(0)+c
3=c
Substituting C=3 at equation (1)
Y=x²-3x+3
The equation of the curve is y=x²-3x+3
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(2i)
F(x) = x³ – 6x² + 9x
FD/dx (fx) = 3x² – 12x + 9
Using standard deviation

(2ii) Gradient of f(x) at point A (2,2)
d/dx f(x) = 3x² – 12x + 9

At point A , x=2
= 3(2)² – 12(2) + 9
= 3(4) – 12(2) + 9
= 12 -24 + 9
= -3

(2iii)
Equation of Tangent at point A

y-y¹= m ( x-x¹)
but m= -3
at point A, y¹= 2¹ x¹= 2
y-2=-3(x-2)
y-2 =-3x+6
y=-3x +6 + 2=> y= 8-3x
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(3)
M=α+β, N=α β

2(x+2)²+λ(x+1)=0
2(x²+4x+4)+λ(x+1)=0
2x²+8x+8+λx+λ=0

2x²+8x+λx+8+λ=0
2x²+(8+λ)x+8+λ=0

a=2, b=8+λ, c=8+λ

(i) M-N= -(8+λ)
LHS RHS
M=α+β= -b/a= -(8+λ)/2, N=c/9=8+λ/2

From LHS
M-N = – (8+λ)/2 – (8+λ)/2
= – (8+λ)(1/2+1/2)
= – (8+λ)(1)
M-N = – (8+λ)
:. LHS = RHS

(ii)
λ=? M=2
M = – (8+λ)/2
2/1= – (8+λ)/2
4 = – (8+λ)
-4 = 8+λ
λ = -4-8
λ = -12
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(4)
(1-√3)²(x+y√3) = 2√3 -2
(1-√3)(1-√3)(x+y√3) = 2√3 -2
(1- √3 – √3 + √9)(x+y√3) = 2√3 – 2
(1-2√3 + 3)(x+y√3) = 2√3 – 2
(1-2√3 + 3)(x+y√3) = 2√3 – 2
(1+3 – 2√3)(x+y√3)= 2√3 – 2
(4-2√3)(x+y√3) = 2√3 – 2
x+y√3 = 2√3 – 2/4 – 2√3
x+y√3 = 2(√3 – 1)/2(2- √3)
x+y√3 = √3 -1/2- √3
x+y√3 = √3 – 1/2 – √3 * 2+√3/2+√3
x+y√3 = (√3 -1)(2+√3)/(2- √3)(2+ √3)
x+y√3 = 2√3 + √9 – 2- √3/4+2√3 – 2√3 – √9
x+y√3 = 2√3 + 3 – 2 – √3/4+2√3 – 2√3 – 3
x+y√3 = 1+ 2√3 – √3/4-3+2√3 – 2√3
x+y√3 = 1+ √3/1
x+y√3 = 1+ √3
x=1 and y=1
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(5)
Mass,m=150g, g=9.8m/s²
When the lift moves with a constant velocity acceleration
a=o
(i) Reaction,R=w=mg
R=mg
=150×9.8
=1470N

(ii) When the lift moves up word with acceleration 4.5m/s²
F=ma=R-mg
: . R=ma+mg
R=m(a+g)
R=150(4.5+9.8)
=150×14.3
=2145N
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(6i)
Given; p = 5i +2 , q=4i-7j
|q-3p|= |4i-7j + 15i – 6j|
=|19i-13j|
=√19²+(-13)²
=√361+169
=√530
=23.02

(6ii)
p+q =(-5i+2j)+(4i-7i)
=-i-5j
=(-5i+2j)(4i-7j)/
√-5i²+2j²*4i²-7j²

=20-14/√25+4*16+49
=6/√94=6/9.695=0.6189
θ=cos-¹(0.6189)
θ=51.76
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(7)
Tabulate.

At the end of the Year.
1|2|3|4|5|6

Maintenance cost (₦)
25,000|32,000|42,000|60,000|85,000|120,000|

Capital cost (₦)
250,000|330,000|440,000|550,000|630,000|770,000|

Total cost (₦)
275,000|362,000|482,000|610,000|715,000|890,000|

Average annual cost (₦)
275,000|181,000|160,667|152,500|143,000|148,333|

Initial value =₦950,000
Capital cost=initial value- Resale price
Total cost = capital cost + Running cost
Average annual cost= total cost /no of years.
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(8i)
Any woman and man can be included means there is no restriction hence :
Required combination
=6c²*8c³
=6!/4!,2! * 8!/5!,3!
=6*5/2*1 * 8*7*6/3*2*1
=15*56
=840 ways

(8ii)
For a particular man to be in the committee, thens
Ways of selecting will be
= 1*8-1C3-1
= 1*7c²
=7c²

Hence,
Total selection =6c²*7c²
=15*7!/5!,2!
=15*7*6/2
=15*21
=315ways
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Theory- Section B Part I Answer two(2).
9,10.
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(9a)
By formula area of the sector is given by :
A = 1/2rθ , where θ is in radians
Therefore,
θr²/2 = 147

θr² = 294
θ = 294/r² ——-(1)
Also,
Perimeter of the sector will be 56cm
So,
P = θr + 2r
Therefore,
θr + 2r = 56
θr = 56-2r
θ = 56-2r/r

(9b)
Set the RHS of both equations equal since the LHS are equal
:. 294/r² = 56-2r/r
:. 294/r = 56-2r
294 = 56r – 2r²
= 2r² – 56r + 294 = 0
= r² – 28r + 147 = 0
= (r-7)(r-21) = 0
Either:
r-7 = 0
r=7
OR
r-21 = 0
r=21
And,
When r=21
θ=294/21² = 0.7rad
From equ(1) and when r=7
θ=294/7² =6rad
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(10a)
Given
2x²-5x-3=0
Therefore
∝+β=5/2 and ∝β =3/2
(i)from
1/∝+1/β=∝+β/∝β=5/2/-3/2=5/2*2/3=-5/3

(ii)∝³+β³=(∝+β)[(∝+β)²-3*β]
=5/2[(5/2)²-3(-3/2)]
=5/2(25/4+9/2)
=5/2*43/4
=215/8
=26⅞

(10b)
Guren:
x² + (q+2)x+ q²= 0
For equal roots :
be – Hac =0
Me have;
(q+2)² – 4q²=0
q²+4q+4-4q²=0
3q²-4q-4=0
(3q+2) (q-2) =0

Either :
3q²+2=0
3q=-2
q=-⅔

OR
q-2=0
q=2

(10c)
Given:
P(2,5), Q(1,4) and R (3, 8)
Using Matrices
approaches, transform the vertices of the triagle into a square given Matrix as
below.
T = [2,5,1]
[1,4,1]
[3,8,1]
Hance, Area of the tringle will be:
A=±½|2,5,4|
|1,4,1|
|3,8,1|

A=±½[1|5,1|
|8,1|- 4|2,1|
|3,1|+|2,5|3,8|]

A=±½[(5-8)-4(2-3)+(16-15)]

A=±-(-3 + 4 + 1) = ½*2 = 1 squared unit
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Theory- Section B Part II Answer one(1).
13.
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(13ai)
Given: mass ,m =10kg
Force,F = 40N
Time, t = 0.5secs
Impulse, I = Ft = 40×0.5 = 20Ns

(13aii) Ft = m(v-u) where u= 0 (at rest)
20 = 10(v-0)
20 = 10v
V = 20/10 = 2m/s
Final speed = 2m/s

(13aiii)
Given: u=0 ; v=2m/s ; t=0.5secs

S= 1/2(u+v)t
S= 1/2(0+2)×0.5
S= 0.5 metres
Distance = 1/2 metre or 50cm

(13b)
Range R , = Time of flight × Horizontal component of speed

75 = T×35×cos38°
T = 75/35cos38° = 2.719secs

Vertical displacement= vertical component × Time of flight of speed
= Usinθ × T
= 35sin38 × 75/35cos38
= 75Tan38°
= 58.596 metres
~ 58.6 metres
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Theory- Section B Part III Answer one(1).
14
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(14a)
TABULATE

UNDER KADUNA
Ibadan: 900
Abeokuta: 1100
Benin: 1000
Supply: 200+x

UNDER ABUJA
Ibadan: 800
Abeokuta: 1000
Benin: 900
Supply: 1000+x

UNDER MINNA
Ibadan: 700
Abeokuta: 900
Benin: 800
Supply: 1000

UNDER DEMAND
Ibadan: 1200+x
Abeokuta: 1050
Benin: 250

(i) Total demand = Total Supply
2500+x = 2200+2x

Collect like term’s
2500-2200 = 2x – x
300 = x
:. x=300

(ii)
TABULATE

UNDER KADUNA
Ibadan: 900(500)
Abeokuta: 11000
Benin: 1000
Supply: 500

UNDER ABUJA
Ibadan: 800(1000)
Abeokuta: 1000(800)
Benin: 900
Supply: 1300

UNDER MINNA
Ibadan: 700
Abeokuta: 900(750)
Benin: 800(250)
Supply: 1000

UNDER DEMAND
Ibadan: 15000
Abeokuta: 1050
Benin: 250
Supply: 2800

Transportation cost = 900*500+800*1000+1000*300+900*750+800*250

= 450000+800000+300000+675000+200000
= 2425000

(14b)
Pr(Ayo pass) = 2/9, Pr(Ayo fail)= 1-Pr(Ayo pass)
= 1-2/9= 7/9

Pr (Doris pass) = 1/3 , Pr (Doris fail) = 1-Pr(Doris pass)
= 1-1/3= 2/3

Pr (Musa pass) = 3/4 , Pr (Musa fail) = 1-Pr(Musa pass)
= 1-3/4 = 1/4

(i) Pr (only one will pass)
P(AD¹M¹) or Or(A¹DM¹) or Or(A¹D¹M)
(2/9*2/3*1/4)+(7/9*1/3*1/4)+(7/8*2/3*3/4)
4/108+7/108+42/108 = 53/108