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WAEC 2016/2017 Further Mathematics Objectives & Theory Answers Posted Here [13/05/2016]


CURRENT EXAMINATION:

WAEC Further Mathematics Questions & Answers ;
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FURTHER MATHS OBJ:
1-10 ABACCDCCCD
11-20 CDDDDABDBC
21-30 BDADABACBC
31-40 CCBBABABDD
(2)
(5,2)(-4,k)(2,1)
(y3-y2)/(x3-x2)=(y2-y1)/(x2-x1)
(1-k)/2-(-4)=(k-2)/(-4-5)
(1-k)/(2+4)=(k-2)/-9
(1-k)/6=(k-2)/-9
-9(1-k)=6(k-2)
-9+9k=6k-12
9k-6k=-12+9
3k=-3
k=-1

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




(3a)
If f(x+2)=6x^2+5x-8)
To find f(5)
Therefore f(x+2)=f(5)
where x+2=5
x=5-2
x=3
therefore f(5)=6(3)^2+5(3)-8
=6(9)+15-8
=54+7
=61

(3b)
(7root2+3root3)/(4root2-2roo3)*(4root2+2root3)/(4root2+2root3)
(24*2+14root6+12root6+6*3)/(16*2+8root6-8root6-4*3)
(48+26root6+18)/(32-12)
=(66+26root6)/20
=66/20+(26root6/20)
=33/10+(13root6/10)
=3.3+1.3root6

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


(4)
(x^2+5x+1)sqroot(2x^3+mx^2+nx+11)=(2x-5)
remainder:30x+16
(x^2+5x+1)(2x-5)
=2x^3+10x^2+2x-5x^2-25x-5
=2x^3+10x^2-5x^2-25x-5
=2x^3+5x^2-23x+30x+16-5
=2x^3+5x^2+7x+11
Therefore m=5, n=7

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



(5a)
pr(age)=4/5
pr(fully)=3/4
pr(must)=2/3
pr(age not admitted)=1-4/5
=1/5
pr(fully not admitted)=1-3/4
=1/4
pr(must not admitted)=1-2/3
=1/3
Therefore pr(none admitted)=1/5*1/4*1/3
=1/60

(5b)
pr(only age and fully gained admission)=4/5*3/4*1/3
=1/5

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



(6a)
(1+4+k+k+4+11)/5=k+1
(20+2k)/5=k+1
20+2k=5(k+1)
20+2k=5k+5
20-5=5k-2k
15=3k
k=15/3
k=5

(6b)
1,4,5,9and 11
TABULATE
x:1,4,5,9,11
x-x^-:-4,-1,0,4,6
(x-x^-)^2:16,1,0,16,36
E(x-x^-)^2=69
Hence=E(x-x^-)^2/n
=69/5
=13.8

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



(7)
m1=3
u1=8m/s
m2=?
u2=5m/s
v=6m/s
m1u1+m1u2=(m1+m2)v
3*8+m2*5=(3+m2)6
24+5m2=18+6m2
24-18=6m2-5m2
m2=6

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


(10a)
(1+x)^7
7Co(1)^7(x)^0+7C1(1)^6(x)+7C2(1)^5(x)^2+7C3(1)^4(x)^3+7C4(1)^3(x)^4+7C5(1)^2(x)^5+7C6(1)(x)^6+7C7(1)^0(x)^7
=1+7x+21x^2+35x^3+35x^4+21x^5+7x^6+x^7

(10b)
35 21 7
a=35
d=T2-T1
=21-35
d=-14


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

(11a)
Kp2=72
K!/(k-2)!=72
K(k-1)(K-2)!/(K-2)!=72
K^2-K=72
K^2-K-72=0
K^2-9k+8k-72=0
K(K-9)+8(k-9)=0
(K+8)(K-9)=0
k=-8,K=9
We consider positive value of K=9

(11b)
The equation 2cos^2tita-5costita=3
Let cos tita=x
2x^2-5x=3
using quadratic formular
a=2,b=-5,c=-3
5+_root(25+24)/4
=5+_root(49)/4
=(5+_7)/4
=(5+7)4=3 or (5-7)/4=-2/4=-1/2
since x cos tita
cos tita=-0.5
tita=cos^-1(-0.5)
tita=120degrees
  

==============================================================
(13ai)
M=2
P=5
C=3
total=10
If the books of the same subject are to stand together
No of arrangements=2!*5!*3!*3!
=2*120*6*6
=8640arrangements

(13aii)
Only the physics textbook must stand together
No of arrangements=5!*6!
=120*720
=86400arrangements

(13b)
P=13/20
q=1-13/20=7/20
pr(atleast 3 speak E)=1-Pr(2 speak E)
=(1-8C1p^1q^7+8C2p^2q^6)
=1-(8*(13/20)*(7/20)^7+28(13/20)^2*(7/20)^6
=1-(0.003346+0.0217467)
=1-0.0251
=0.9749
=0.975(3s.f)

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

(15a)
at max height
V=0m/s
g=10m/s^2
V^2=u^2-2gh
0^2=30^2-2*10H
0=900-20H
20H=900
H=900/20
H=45m

(15b)
time taken to get to max height
V=u-gt
0=30-10t
10t=30
t=30/10
t=3secs
Time taken to return=2t
=2*3=6secs

(15c)
H=40m
H=ut-1/2gt^2=40
30t-1/210t^2=40
30t-5t^2=40
5t^2-30t+40=0
t^2-2t-4t+8=0
(t^2-2t)-(4t+8)=0
t(t-2)-4(t-2)=0
(t-2)(t-4)=0
t-2=0 or t-4=0
t=2secs or t=4secs


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