Waec Gce 2016 Further Mathematics Obj And Theory Answers

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11-20: ACBABDBDAB
21-30: BDDBCBACCD
31-40: ACBDACDBAB
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1a) 8m + 2³m = 1/4
2²m + 2²m = 2-²
4³m = 2-²
Equate the base
6m = – 2
m = -2/4 = -1/3
OR
1a) 8^m +2^3m = 1/4
2^3m + 2^3m = 2^-2
4^3m = 2^-2
2^6m = 2^-2
Equate the bases
6m = -2
M = -2/6 = .....
1b) log15 base 4 = x
x = log15 base/log4 base 10 = 1.1761/0.6020
= 1.9536
therefore x approximately 1.954
OR
1(b) Log₄ 15 = x
x = Log₄ 15
= log15 ÷ log 4
= 1.18 ÷ 0.60
= 1.97
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2a)
(-2) = m(-2)² + n(-2)+2=0
4m-2n = – – – – – – – – (1)
F(1) = m(1)²+n(1)+2=3
m+n+2=3
m+n=3-2
m+n=1 – – – – – – – – – – – -(2)
X Equate (2) by 4
4m+4n=4 – – – – – – – – – – (3)
4m-2n= -2
-4m+4n=4/-6n = – 6
n=1
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4) Given y= 5/x² + 3
Y = 5(x² +3)-1
dy/dx = anxn-1
y = 5(x² + 3)-1
dy = 5(-1) (x2 +3)-1^1
dy/dx = 10x (x² + 3)-²
dy/dx = – 10/(x² +3)²
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5a)
^nP5 ÷ ^nC4 = 24
n!/(n-5)! ÷ n!/(n-4)!4! =24
n!/(n-5)! * (n-4)!4!/n! =24
(n-4)!4!/(n-5)!=24
(n-4)(n-5)!4!/(n-5)! = 24
n-4=1
n=4+1
n=5
b)PR= 5C3 (1/6)³ (5/6)^5-3
5!/(5-3)!3! (1/6)³ (5/6)²
5!/2!3! (1/6)³ (5/6)²
10 (1/216) (25/36)
=0.03215
Pr = 1-0.03215
=0.9678
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6) make a table containing
MARKS,TALLY,F&FX
UNDER F - 2,9,4,2,2,1
UNDER FX - 2,18,12,8,10,6
€F= 20
€FX = 56
6b) Mean = €fx / €f
= 56 / 20
= 2.8
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7a)
Given
h=15.4t-4.9t
Velocity V=dh/dt =15.4-9.8t
At maximum height V=0
Therefore 15.4-9.8t=0
9.8t=15.4
t=15.4/9.8
t=1.6secs
Time to reach maximum height is 1.6secs
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7b) maximum h=15.4t-9.4t^2
15.4(1.6)-4.9(1.6)^2
22.64-12.544
=12.096
Max height = 12.1m
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9a)X+6/(x+1)^2 = A/x+1 + B/(x+1)² + C/(x+1)³
X+6/ ~(x+1)³~ =A(x+1)² + B(x+1) + C / ~(x+1)³~
X+6 =A (x+1)² + B(x+1) + C
Let X+1=0 , X=-1
-1+6= A (-1+1)² + B(-1+1) +C
5= C
Therefore:- C=5
X+6= A(X2+2x+1) + Bx + B + C
X+6= Ax²+2Ax+A+Bx+B+C
comparing the coefficient of X²
A=0
Comparing the coefficient of X
1=2A+B
1=2(0)+B
B=1
x+6/(x+3)² = o/x+1 + 1/(x+1)² + 5(x+1)³
1/(x+1)² + 5/(x+1)³
OR
7b)
maximum h=15.4t-9.4t^2
15.4(1.6)-4.9(1.6)^2
22.64-12.544
=12.096
9a)X+6/(x+1)^2 = A/x+1 + B/(x+1)² + C/(x+1)³
X+6/ ~(x+1)³~ =A(x+1)² + B(x+1) + C / ~(x+1)³~
X+6 =A (x+1)² + B(x+1) + C
Let X+1=0 , X=-1
-1+6= A (-1+1)² + B(-1+1) +C
5= C
Therefore:- C=5
X+6= A(X2+2x+1) + Bx + B + C
X+6= Ax²+2Ax+A+Bx+B+C
comparing the coefficient of X²
A=0
Comparing the coefficient of X
1=2A+B
1=2(0)+B
B=1
x+6/(x+3)² = o/x+1 + 1/(x+1)² + 5(x+1)³
1/(x+1)² + 5/(x+1)³Max height = 12.1m
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10a)
3x^2+x-2 <= 0
3x^2+3x-2x-2 <= 0
3x(x+1) -2 (x+1) <= 0
(3x-2)(x+1) <= 0
3x-2 <= 0 or 8+1 <= 0
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14a) m=0.5kg
F-r=ma
0.5*10-R=0.5*2
5-r=1
R=5-1
R=4n
bi)Ra+Rb=80+M+100
Ra+Rb=180M
Rb=180+M-90
Rb=90+m ——-(i)
Taking moment about B
A.C.M=80*1.7*2*M
=136+2m
C.W.M=100*1.2+90+2*4
=120+216
=336
A.C.W=C.W.M
136+2M=336
2M=336-136
2M=200
M=200/2
=100N
-: M= 100N
ii)
Rb=90+M
=90+100
=190N
-: the reaction at B is 190N