NECO 2021/2022 Further Mathematics Obj & Essay/Theory Questions and Answers is Out

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Further Mathematics Obj Theory

Further Mathematics Obj Theory questions

(No 2)
2.1. F(x) = x26x2 +9x
The derivative is f'(x) = dyan |: f'(x)= dy = (x2 - 6x2 + 9x)
= 3x2-122+ a 20 Gradient > F') So, f'(x) (122) = 3 (2) 2-12 (2) + 9
- - 3 2 (11) At A (2,2), the equation
= 12-24 +9
-TIT -THE
le Vintag 1 - log 2 Fope
DL-2
=> -3 = y 2 y-2 = -3(2-2)
y-2 = -3x+6 3x ty-2-
6 0 Hence, the required equation is 3x+y-8 = 0
=====================================

(No: 4).
+ (1-13) Ex+y53) = 283-2
3) (x + y 3) = 283 -2
n be 14 -estricho-r3) (1-13) (x + y 13) -
Ki-283+3) (ar yr3 = 283 - 2
(4-283) (x + y + 3) = 283 - 2 2(2-13 (x + y 13 = 2(13 x + 483 - 2 Cr3 - 1)
- (2-13 xx yra = √3 -1
2 - 13
ind
= √3-1 x
2 + 13
V3-) x
2
2-15 2+13 = 213 + 3-2-13
4+283-2/3 - 3
=
3
+
1
nry 3 = 1 + 3 Hence, by comparing L.H.S and R. H.S. x=1 and y=1

(No: 9)

(a
QUESTION 13 ) Given! mass, n = 10kg
force, F = CON
time, t=0.5 sees. Impulse, I = Ft = 40X0.5 = 2ONS
(5)
where uso at rest)
Ft = m(v-u) 20= 10 (v-o)
20 = LOV v= 2% = 2mls final speed= amb
Thwhile souten ton
1
)
Giveni uzo; v= 2mls, t= 0.5 secs.
sa - Cutut sa Coto) x 0.5
s=0.5 metres distance = £ metre or soon
e)
Sx = Uy custt A
L


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