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Stanly18

Further maths
fmaths
1BDBCAADCAC
11DCDABBBBCA
21CDCBABDAAB
31DABABCBACA

(1)
(Log(base 2)m)^2-log(base 2)m^5=10
(log(base 2)m)(log(base tita)m)-3log
(base 2)m=10
(log(base 2)m)(log(base 2)m-3)=10
(log(base 2)m)9log(base 2)m-3)=10
log(base 2)m-3=10
log(base 2)m-3=10/log(base 2)m
log(base 2)m-log(base 2)m^3=10log
(base 2)m^-1
log(base 2)m-3log(base 2)m=-10log
(base 2)m
log(base 2)m/(3^-1)=log(base 2)m^-10
m/3^-1=-10
m=-10*3^-1
m=-10*1/3
m=-10/3
(2a)
given m*n =((m^2)-(n^2))/2mn
-3*2=(-3)^2-(2)^2
=(9-4)/-12
=5/-12
=-5/2
(2b)
to show that whether * is associative
=(m*n)*p=m*(n+p)
=(m+n)*p=((m^2)-((n^2))/2mn)*p
=((m^2)-((n^2)^2-p^2)/(((m^2)-
((n^2))/2mn)p
=(((m^2)-((n^2)^2)/4(m^2)(n^2))-
p^2)/2p((m^2)-(n^2))/2mn)
=(m^2)-((n^2)^2)/4(m^2)(n^2)
(p^2)/4(m^2)(n^2))* 2mn/(2p(m^2)-
(n^2))
=((m^2)-(n^2))^2-(2mnp)^2)/(2mn(2p)
((m^2)-(n^2))
=((m^2)-(n^2))^2-(2mnp)^2))/4pmn
(m^2)-(n^2))
also, m*(n+p)=m*((n^2)-(p^2))/2np)
=m^2-((n^2)/2np)^2)/2m((n^2)-
(p^2))^2/2m((n^2)-(p^2)/2np))
=(m^2)-((n^2)-(p^2))^2)/4np))/(m)
((n^2)-(p^2))/np
=(4(m^20np-((n^2)-(p^2))^2)/4np)/m
((n^2)-(p^2))/np
=(4(m^20np-((n^2)-(p^2))^2)/4np)/m
((n^2)-(p^2))/np
=(4(m^20np-((n^2)-(p^2))^2)/4np)*
(np)/m((n^2)-(p^2))
hence * is not associative
(6)
No of man =5,
No of woman =3,no of committee =3
total no of people =5+3=8.
A no of ways of foring the
committee=8C base3,
but nC base r=n!/(n-r)!r!,
8C base3=8!/(8-3)!3!,
=8!/5!3!= 8*7*6*5/5!*3*2=56ways.
6b.
prob ( of at least one woman is on the
committee)
= 5C2 + 3C1/8C5+8C3
= 5!/2!(5 - 2)! + 3!/11(3 -1)!= 5!/2!3! +
3!/1!2! = 120/2X6 + 6/1X2
8!/5!(8-5)! + 8! /3!(8-5)! = 8!/5!3! +
8!/3!3! = 40320/120X6 + 40320/6X6
= 120/12 + 6/2=10 + 3 = 13
40320/720 - 40320/36 = 56 + 1120 =
1176
(11a)
f(x)=x-3/2x-1,
g(x)=x-1/x-1,
g of =g(f(x),
g of =x-3/2x-1-1/x+3/2x-1+1,
g of =x-3-(2x-1)/2x-1/
x-3+(2x-1)/2x-1,
x-3-2x+1/x-3+2x-1,
-x-2/3x-4
=x+2/4-3x.
(11b) y=9x-x^3,
y:x(9-x^2),
y:x(3^2-x^2),
y :x(3-x)(3+x),
for the roots of y =0,
y:x(3-x)(3+x)=0,
:x=0,3 or -3,
y=9x-x^2,
dy/dx=9-3x^2,
at turnin point,
dy/dx=0,
9-3x^2=9,
x^2=3,
x=root3,
x=1.732,
For maximum and minimum values,
when x=1.732,
y=9(1.732)-(1.732)^3,
=15.588-5.196,
=10.392 minimum
(9a)
y=(x-3)(x^2+5),
let u =x-3,
du/dx=1,
for product rule,
dy/dx=Udy/dx+Vdu/dx,
dy/dx=(x-3)(2x)+(x^2+5)(1),
=2x^2-6x+x^2+5=3x^2-6x+5.
(9b)
IF (X+ 1)^2 is a factor of f(x)=x^3 +
ax^2+bx + 3
then (x+1)^2 = 0 is a solution of f(x)
ie x+1=0
or x = -1 (twice)
also (x+1)^2 = x^2=2x+1
when x=-1
f(x)=f(-1)=(-1)^3+a(-1)^2+b(-1)+3
= -1+a-b+3
=a-b+2=0
=a-b=-2
9b continues
also, f(0)= 0^3+a(0^2)+b(0)+3 (not
equal) to 0
F(1)= 1^3+a(1^2)+b(1)+3
=1+a+b+3=0
=a+b=-4
solving (1) &2(2)
a-b=-2
a+b=-4
2a=6
a=-6/2=-3
from(1)
-3-b=-2
-b=-2+3
-b-1
b=-1
a=-3,b=-1
(ii)
therefore f(x)
(3.)
3x^2+5x^2+1=0
Solving a=3, b=5, c=1
x= -b(+_)sqr root (b2-4ac/2a)
=( -5+_sqr root 5 ^2 - 4(3)(1))/(2x3)
= -5 +_ sqr root 25 - 12/6
= -5 +_ sqroot 13/6
= -5 + sqroot 13/6
= -5 + sqroot 13/6 or -5 - sqroot 13/6
27 (alpha sqr root3 + betasqr
root3)=27[(alpha + beta (alpha + beta)]
alpha + beta = -5/6 + sqroot13/6 +
-5/6 + sqroot13/6
= -10/6
alphabeta= (-5/6 + sqroot 13/6)(-5/6 -
sqroot 13/6)
= 25/36 - 13/36 = 25 - 13/36 = 12/36
= 1/3
then, 27 (alphasqroot3 + betasqroot3)
=27 [(alphasqroot3)=
27[(-10/6)sqroot3(1/3)(-10/6)
= 27 (-1000/216 + 10/6)
=27 (-1000+360/216)
=27 (-640/216) = -17280/216
=-80.
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