2017 /2018 WAEC WASSCE FURTHER MATHS OBJ ESSAY THEORY QUESTION AND ANSWER EXPO / RUNS FREE <<on: March 24, 2017, 09:36:36am
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Ghana members Essay Further Maths
Where ever you see
^ it means raise to power
/ it means division
* it means multiplication
X it means normal x
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(1a) 8m+2³m (2³)m+2³m=¼
2³m+2³m=¼
2(2²m)=¼
2³m=⅛
2³m=8-¹
2³m=2-³
Since bases are the same, powers are same
3m=-3
m=-3/3
m=-1
(1b)log15 base 4=x
X=log15 base/log4 base 10=1.1761/0.6020=1.9536
Therefore x approximately=1.954
(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
(4) ?given y=5/x²+3 ?y=5(X²+3)-1? day/dx=anxn-1 ?y=5
(X²+3)-1?dy=5(-1) (x2+3)-1
n1 ?day/dX=10x(x²+3)-2
dy/dx=-10/(x²+3)²
(5a) nnp5÷nc4=24
n!/(n-5)!÷n!/(n-45)1=24
n!(n-5)!*(n+4)=24
(n-4)!4!/(n-5)1=24
(n-4)(n-5)!4!/(n-5)+!=24
n-4=1
n=4+1
n=5
(5b) 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
(6a) 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=€fd/€f
=56/20
=2.8
(7a) given h=15.4t-4.9t²
Velocity v=CH/dt=15.4-9.8t
At maximum height, v=0
15.4-9.8t=0
9.8t=15.4
t=15.4/9.8
t=1.6secs
Time to reach maximum height=1.6secs
(7b) maximum height=15.4t-4.9t²
=15.4(1.6)-49(1.6)²
=24.64-12.544
=12.096
Maximum height=12.1m
(9a) x+6/(x+1)²=A/x+1+B/(x+1)²+(/(x+1)³
X+6/(x+1)³ cancel the (3+1)³at the middle =A(x+1)²+B(x+1)+c/(x+1)³ also cancel (x+1)³at the middle
X+6=A(x+1)²+B(x+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 x
1= 2A+B
1=2(0)+b
B=1
Good luck
Excellent Results. ERGN
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