CONFIRM 2012 GCE FURTHER MATHEMATICS OBJECTIVE AND THEORY ANSWERS

FURTHER MATH OBJECTIVE LOADING…..

FURTHER MATH THEORY.

(7a)exactly 8 heads=9/241,
(7b)between 2 and 5 heads=(24+37+10+60)/241=131/241.

(7c)at most 1 head=(3+8)/241=11/241

Further maths (Rule 1)^ means raised topowereg (i) 3x^2 means3xraised to power 2(ii)10^-4 means 10raised to power -4.(Rule 2)5! means 5 factorial

PLS KNOW THAT COMMA(,) MEANS NEXT LINE.

DRAW UR TABLE FOR X, F and FX.

Under F, put 1,2,3,4,5,6,7,8,9.
Under FX, put 2,3,m,8,10 and so on….
under FX, put 2,6,3m,32,50 and so on…

DONT WRITE THE ABOVE NOTE.

(14a)X:1,2,3,4,5,6,7,8,9.
F:2,3,m,8,10,5,3,3,n.
Fx:2,6,3m,32,50,30,21,24,9n. Mean=Efx/Ef=4.725.
4.725=(165+3m+9n)/(34+m+n).
34+(m+n)=40.
m+n=40-34=6.
m=6-n.
165+3m+9n=40*4.725=189.
3m+9n=189-165.
3m+9n=24.
3(6-n)+9n=24.
18-3n+9n=24.
6n=24-18.
6n=6.
n=1.
but m+n=6.
m+1=6.
m=6-1.
m=5.

(14b) prob.that a candidate scored more than 5.
6marks–>5candidate.
7marks–>3candidates.
8marks–>3candidates.
9marks–>1candidate.
P(more than 5)=12/40=0.3

(17.) Mass of body=20kg.
fx1=30cos60.
fx2=40cos30.
fy1=30sin60.
fx=30cos60+40cos30.
fy2=40sin30,fy=30sin60+40sin30.
F=resultant.
fy=30(0.866)+40(0.5)=25.98+20=45.98.
fx=30(0.5)+40(0.866)=15+34.64=49.64.
using pythogras theorem.
f squared=(45.98)squared +(49.64)squared.
f=square root of 2114.1604+2464.1296.
f=square root of (4578.29).
f = 67.67N.

(17b)fx=49.64N.
f=67.67N.
fy=45.98N.
tan(tita)=fy/fx=45.98/49.64.
tan(tita)=0.9263.
tita=42.8 degree.
direction of resultant=90-42.8=47.2degrees

(17c)acceleration of the body.
Force=ma.
20*a=67.67N.
a=67.67/20=33.835m/s2

(1.) ( log 7^3/2-log5^3/2)/(log 7/5).
3/2log(7/5) / log (7/5) =3/2 ans

3a) r = root 2-2/root 2-1=root 2-2/root 2-1 * root 2+1/root2+1.
2-2root2+root2-2/2-1=-root2/1.
r=-root2.

(3b) r=-root2,a=root2-1.
Tnth=ar power(n-1).
T3rd=(root2-1)(-root2)power(3-1)=(root2-1)(-root2)power2.
T3rd=2(root2-1)

(6) y=3x-1/2x.
<>y+y=3(<>x+x)-1/2(<>x+x).
<>y=3(<>x+x)-1/2(<>x+x)-3x+1/2x.
<>y=3<>x-(2<>x-x+2x)/4x(<>x+x)<>y=<>x[3+2/x2].
<>y/<>x tends to zero,dy/dx=3+2/2×2

Content:

(2) (k+3)x2+(6-2k)x+k-1=0.
For real root.
b2-4ac>0.
b2=4ac.
let a=k+3.
b=6-2k and c=k-1.
(6-2k)squared=4(k+3)(k-1).
36-12k-12k+4k2=4k2+8k-12.
36+12=8k+24k.
32k=48.
k=48/32.
k=3/2=1.5.

10a)4 root7+3 root 2/10+2 root 14..
To rationalise=(4 root 7+3 root 2)/(10+2 root14)*(10-2 root 14)/10-2 root 14)..
=(40 root 7-8 root 7*14)+(30 root 2-6 root 28)/(100-20 root 14)+(20 root 14-4*14)..
=(40 root 7-8 root 7*7*2)+(30 root 2-6 root 4*7)/(100-56)..
=(40 root 7-12 root 7-8)*(7 root 2+30 root 2)/44..
=(28 root 7-26 root 2)/44,=(14 root 7-13 root 2)/22.. =1/22(14 root 7-13 root 2)..

10bii) (dy)/(dx)=(-2px^5-4qx^3)/(x^8).
Let u=-2px^5-4qx^3.
(du)/(dx)=-10px^4-12qx^2.
and v=x^8.
(dv)/dx)=8x^7.
x^2 (d^y)/(dx^2)+7x(dy)/(dx)+8y.
(d^y)/(dx^2)=x^8(10px^4-12qx^2)-(-2px^5-4qx^3)8x^7/(x^8)^2.
=(-10px^12-12qx^10+16px^12+32qx^10)/(x^16).
=(16px^12-10px^12+32qx^10-12q)/(x^16).
=(6px^12+20qx^10)/(x^16).
x^2(d^y)/(dx^2)+7x(dy)/(dx)+8y=o.
(x^2)(x^16)(6px^12+20qx^10)+(7x)/(x^8)(-2px^5-4qx^3)+(8)/(x^4)(px^2+q)..

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