Waec 2016 Further Mathematics Obj And Theory Answers – May/ June Expo

Note
/ means DIVISION
* means Multiplication
tita You Should Know that.
rais means Raise to power or
^ means Raise to power
sqr rut/square rut means Square root
—–(1) means Equation 1
•= •= •= •= •= •= •= •= •= •= •= •
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
•=+•=+•=+•=+•=+•=+•=+•=+•=+•=+•=+•
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
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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
•=+•=+•=+•=+•=+•=+•=+•=+•=+•=+•=+•
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
•=+•=+•=+•=+•=+•=+•=+•=+•=+•=+•=+•
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
7b)
m2u2-m1u1=V(m1+m2)
6*5-3*8=V(3+6)
30-24=9v
9v=6
v=6/9
v=0.67m/s
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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
•=+•=+•=+•=+•=+•=+•=+•=+•=+•=+•=+•
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
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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
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13a)
M=2=mathematics=2
P=5=physics=5
C=3=chemistry=3
i)
M.P.C=2! x 5! x3! =2*1*5*4*3*2*1*3*2*1
=2*120*6=1,440ways
ii)
p=5!
=5*4*3*2*1
=120ways
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12a)
tabulate
Marks| 1-10, 11-20, 21-30, 31-40, 41-50, 51-60, 61-70, 71-80, 81-90, 91-100
F| 3, 17, 41, 85, 97, 115, 101, 64, 21, 6
C.B| 0.5-105, 10.5-205, 20.5-305, 30.5-405,
40.5-505, 50.5-605, 60.5-705, 70.5-805,
80.5-905, 90.5-1005
C.F| 0+3=3, 3+17=20, 20+41=61, 61+85=146, 146+77=243, 243+115=358, 358+101=459, 459+64=523, 523+21=544, 544+6=550
12ii)
CLICK HERE FOR GRAPH
•=+•=+•=+•=+•=+•=+•=+•=+•=+•=+•=+•
Keep Refreshinq...
While We Solve The Quesion.
•= •= •= •= •= •= •= •= •= •= •= •

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