MULTIPLE CHOICE
1)D 2)E 3)B 4)E 5)B 6)C 7)D 8)D 9)B 10)C 11)D 12)A 13)E 14)B 15)D 16)C 17)A 18)E 19)C 20)E 21)A 22)B
ANALYSIS SECTION
1)a) Maximum (1,1.5)
b)i) 9x^4-26x^2+1=0
b)ii) (0.2,0.5), (1.7,1.4)
c) Graph rising steeply to maximum at (1,1.5), tapering off to around (4,1)
Mark in the previously stated inflection points at (0.2,0.5), (1.7,1.4)
d)i) 2pi*integral[(18x^3)/(3x^2+1)^2 dx,0 to 1/sqrt(3)]
d)ii) V = 2pi * integral[1/u - 1/u^2 du,1 to 2]
d)iii) V = pi[2loge(2)-1] cubic units
2)a) a=4.5 m/s^2
b) v=12 m/s
c) a=(180-3v)/40 m/s^2
d) t=1.8 seconds
e)i) From top anticlockwise, the forces are T, 100, 80g, 390
e)ii) Tcos(theta)=195sqrt(3)-100
and Tsin(theta)=979
e)iii) tan(theta)=4.118 (3 d.p.)
e)iv) T=1007 N
3)a)i) y^2=sin^2(t/3)*cos^2(t/3), via sine double angle formula
a)ii) y=±sqrt[x^2(1-x^2)], by substituting the parametric equation for x into y
b) Graph: a figure 8 on its side, symmetrical about the axes; (see Neobeo's post#6 in this thread for visualisation...
with ±1/2 instead of ±1/4 on the y-axis)
x-intercepts (-1,0), (1,0), four stationary points in each quadrant at (±1/sqrt(2),±0.5)
c) 6pi seconds
d) sqrt(2)/3 m/s
e)i) (1/3)*integral[sqrt(cos^2(t/3)+cos^2(2t/3))dt,0 to 6pi]
e)ii) 6.1 metres
(Re. 'solution' in post#6:
)
4)a) [(x-10)^2]/9 + (y-5)^2 = 1
b) Graph of the ellipse, endpoints clockwise from y-maximum: (10,6),(13,5),(10,4),(7,5)
Centre (10,5), semi-major axis length=3, semi-minor axis length=1
c)i) t=6 months
c)ii) 500 foxes
d)i) dy/dx=[xy-10y]/[25x-5xy] by related rates
d)ii) Use implicit differentiation to find the derivative of the given expression, such that dy/dx=[xy-10y]/[25x-5xy] in d)i)
e) Minimum number of rabbits: 6590 (to the nearest ten rabbits)
and Maximum number of rabbits: 14430 (to the nearest ten rabbits)
5)a) By conversion to polar form, sqrt(3)/2+0.5i is a root of z^3=i
b) The three roots are z={-i, sqrt(3)/2+0.5i, -sqrt(3)/2+0.5i}
c) The intersection points between the two relations are (-sqrt(3)/2, 1.5), (0,0)
d) Circle of radius=1, centre (0,1), and linear y=-sqrt(3)x
e) Shade area to the right of the linear graph, contained within the circle
f) A~2.53 square units - hamtarofreak's working:
http://vcenotes.com/forum/index.php/topic,7511.msg93089.html#msg93089======
Final edit [2]: itute confirms all these answers
http://www.itute.com/wp-content/uploads/2008_vcaa_specialist_mathematics_exam_2_solutions.pdfExam 2 mark = 93% = SAC mark
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