Neco 2020/2021 Further Mathematics obj & Theory Question and Answers Now Available

 


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Neco 2020 Further mathematics Question and Answers Now Posted


F/MATHS OBJ

01-10: EAAEDDBADB

11-20: DBEDDCBBAA

21-30: CEABCCCBDC

31-40: BBBEAEAAAD

41-50: BCAAAADACB

 

COMPLETED

 

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Theory
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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
====================================

(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
==========

 

(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

 

======================

 

(10a)

2x² — 5x — 3 = 0

Following general quadratic formula 

@ + ẞ = — b/a 

@ẞ= c/a 

x² — 5/2x — 3/2 = 0

— 5/2 = —(@ + ẞ) : (@ + ẞ) = 5/2 

— 3/2 = @ẞ

Find 1/@ + 1/ẞ

:. (@ + ẞ)/@ẞ = 5/2 ÷ (—3/2)

5/2 × (— 2/3) = — 5/3

Thus , 1/@ + 1/ẞ = — 5/3

 

@² + ẞ² = (@ + ẞ)² — [email protected]

= (5/2)² — 2(—3/2) 

= 25/4 + 3

= (25 + 12)/4 : 37/4

Hence, @² + ẞ² = 37/4

 

(10b)

Since they have equal roots 

D = 0

b² = 4ac

(q + 2)² = 4(q)²

q² + 4q + 4 = 4q²

3q² — 4q — 4 = 0

3q² — 6q + 2q — 4 = 0

3q(q — 2) + 2(q — 2) = 0

 

thus,

q = — 2/3 and 2

 

======================

 

(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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(10a)

View image above

 

(10b)

Since they have equal roots 

D = 0

b² = 4ac

(q + 2)² = 4(q)²

q² + 4q + 4 = 4q²

3q² — 4q — 4 = 0

3q² — 6q + 2q — 4 = 0

3q(q — 2) + 2(q — 2) = 0

 

thus,

q = — 2/3 and 2

 

 

(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)

Coming…….

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