NECO 2015 Further Maths Obj and Essay Questions and Answers

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	2043	Paper III: Objective – Further Mathematics	2hrs	10:00am – 12:00noon
Monday 29th June	2042	Paper II: Essay – Further Mathematics	2hrs 30mins	12:00noon – 2:30pm

FURTHER MATHEMATICS TYPE A:
FURTHER MATHS OBJ ANSWERS
:
1-10: CDABBACDAA
11-20: DABDBCDBBD
21-30: DADCACCDDC
FURTHER MATHS ESSAY ANSWERS
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Note : Always Indicate the paper type that you Used. Don’t Panic, You will be Marked Right.

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NECO FURTHER MATHEMATICS 2015 ANSWERS ESSAY
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Indicate the Unit of any Answer
===================(1)===================

(1a)
f(x)=2x^3 + bx -11x +C
if x+3 and x-2 are factors
therefore x=-3, x=2 are solutions
f(2)= 2(2^3)+b(2^2)-11(2)+C=0
2(8)+b(4)-11(2)+C=0
16+4b-22+C=0
=>4b+C=6 eqn(1)
f(-3)=2(-3^3)+ b(-3^2) -11(-3)+C=0
2(-277)+b(9)-11(-3)+C=0
-54+9b+333+C
=>9b+C=21
solving (1) and (2) simultanueosly,
4b+c=6 —–(1)
9b+c=21
(2)-(1)
put b=3 in (1)
4(3)+c= 6
12+c=6
c=6-12
=-6
b=3, c=-6

(1b)
f(x)=2x^3 + bx^2 -11x +c
f(x)=2x^3 + 3x^2 -11x -6

(1c)
f(-2)=2(-2^3)+ 3(-2^2) – 11(-2) -6
=2(-8)+ 3(4)-11(-2)-6
=-16+12+22-6
=12

===================(2)===================

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(5) From the diagram
final velocity(v) = u +at
=20+5(4)
=20+20
=40m/s
total distance = area of trapezium OABD +Area of a triagngle BCD
time between DC = t from
v=u+at
t=v – u/a
= 40-0/10
= 4s total distance
= 1/2(20+4)4 +1/2(4) * 40
= 1/2(60)4 + 1/2(4) *40
= 30*4+ 2*40 =120+80
= 200m

===================(6)===================

6a)
greatest height(H) = (U^2 sin^2tita)/2g
=956^2 sin ^2 (36)) / 2(10)
=(3136 * 0.5878 ^2)/20
=1083.516/20
=54.196
=54.18m (approximately)

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 (approximately)

6c) time of flight =2sin tita/ g
=(2*56 sin 36) / 10
=112* 0.5878
=65.8336/10
= 6.5834
= 6.58s (approximately)

===================(7)===================

===================(8)===================

===================(9)===================

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===================(12)===================

===================(13)===================

13) From the diagram

Frictional force Fr = mg sin tita
=5.6* 9.8* 0.5878
Fr =32.26N

Normal Reaction R= mg cos tita
=5.6 * 9.8 cos^2 3656* 9.8 * 0.8090
R = 44.40N

a)The force of Normal reaction (R=44.40N) is parallel to the plane of Frictional force (Fr = 32.26N).

b)From Fr = UR
=0.45* 44.40
=19.98N

===================(14)===================

===================(15)===================

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