MULTIPLE CHOICE SECTION
1)E 2) C 3)D 4)A 5)E 6)A 7)D 8)D 9)D 10)C 11)D 12)B 13)C 14)D 15)E 16)C 17)B 18)E 19)B 20)B 21)D 22)C
ANALYSIS SECTION
1)a)i) 0.1678
ii) 0.2936
iii) 0.214
b)i) 0.2951
ii) 0.2849
c)i) Negative quasi-parabolic curve over [2,6] (somewhat skewed to the right), maximum turning point at (4.31,0.38)
(Best to draw a line along the x-axis to represent the '0 elsewhere')
ii) 0.1211
iii) 4.1333 hours
2)a)i) -7/a
ii) sqrt(a) [positive root as a>1]
b)i) 7
ii) 1/e
c)i) A=(7/2a)(a^2-1)
ii) (7a)/2-7/(2a)=7 => a=1+sqrt(2)
iii) Given integral is less than area found in c)i) = 7, => given integral < 7, => a<e
3)a) 191 minutes; t~3.19 hours = 191.67 days = 192 minutes after rounding, BUT at 192 minutes he is DEAD, therefore 191 minutes is correct
b) 3.6 hrs > 3.19 hrs
c) Show that T = 2[(sqrt(9+x^2)/5)+(9-x)/13]
d) dT/dx=2[x/(5sqrt(9+x^2) - 2/13] = 0 at x=1.25, sign diagram to justify minimum
e) 2.49 hrs < 3.19 hrs
f) A at (0,16), C at (1,24)
g) z=16/d + 8, no need for domain ['equation', not 'function']
h) Six days: z(0)=16, z increases by 8 every day => z(1)=24, z(2)=32, z(3)=40, z(4)=48, z(5)=54>50 => Need to take antidote on six days (Initially I read this incorrectly - five days will probably not be accepted, unless perhaps you have a strong argument that conveys your logic [-1])
4)a)i) f'(pi/2)=1
ii) m[normal] = -1 => y-1=-(x-pi/2) => y=-x + (pi/2 + 1)
b) {-pi/2, pi/2, 3pi/2, 5pi/2} ([-1] again for me =P)
c) a=1-pi/2
d)i) 0.5[cos(x/2)+sec^2(x/2)]
ii) x=2pi
e) Graph of h(x), asymptotes at x=-pi, x=pi, x=3pi
y-intercept at (0,2), stationary point of inflexion at (2pi,2)
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