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# 2020 NECO FURTHER MATHEMATICS OBJ AND ESSAY QUESTIONS AND ANSWERS FURTHER MATHS OBJ:
11-20: DBEDDCBBAA
21-30: CEABCCCBDC
31-40: ACBCBEDCAB

(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

From equ(1) and when r=7

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

(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

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

(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…….

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

(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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