FURTHER MATHS (OBJ & Essay) Answers

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Further Maths Essay:
(1)
P=15kg(15N)
Efx=8cos50degrees-10cos60degrees-150cos90degrees
=8(0.6428)-10(0.5000)-
150(0)
=5.142-5
=-0.1424N
Efy=8sin50degrees
+10sin60degrees-150sin90degrees
=8(0.7660)+10(0.8660)-
150(1)
=6.128+8.660-150
=-135.242N
(i) Resultant of force
R=Sqroot(Efx^2+Efy^2)

=sqroot(-0.1424)^2
+(-135.242)^2
=sqroot(0.020+18282.285)
=sqroot(18282.305) =135.21N
(ii) Let titat be the
direction odf the resultant
force
tan tita= Efx/Ef
tan tita=-135.212/-0.1424
=949.5225. tita=tan^-1(949.5225)
=89.94 degree
(iii) Acceleration
from P-R=F
i.e 150-135.21=14.74
ma=14.79
a=14.79/m=14.79/150g
=0.986ms^-1

(2a)
From S=ut-1/2at^2
29=3.89*15-1/2*a*15^2
29=58.09-0.5*a*225
29-58-58.05=-0.5*a*225
-29.05=-112.5a a=29.05/112.5
a=0.26m/s^2
(b)
Total distance=21m, 8m=29m
Total time=10, 5=15s
from S=(v u/2)t
v=0,s=29m,t=15s
29=(0 u/2)15 29=(u/2)15
2*29=15u
58=15u
u=58/15
Therefore Initial Velocity
u=3.87m/s

(5i)
for -2<_x<_2
ie x=-2,-1,0,1,2
for x=-2,f(x)=(-2)^2+2=4+2=6
for x=-1,f(x)=(0)^2+2=1+2=3
for x=0,f(x)=(0)^2+2=0+2=2
for x=1,f(x)=1^2+2=1+2=3
for x=2,f(x)=2^2+2=4+2=6
Hence the co-domain are 2,3,6
ii) f is onto
Reason since f(-2)=f(2)
f(-1)=f(1)
therefore f is onto
iii)f(x)=x^2+2=y
=>x^2+2=y
x^2=y-2
x=sqroot(y-2)
f^-1(x)=sqroot(x-2)
f^-1(11)=sqroot(11-2)
=sqroot(9)
=3 0r -3
7

NOTE:
^ MEANS Raise to power
tita means tita sign
degree is a degree sign
root or sqroot means square root sign ie _/

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