i'll just type what i can remember from memory
these are my answers
1i) p = 3/7
2i) use expansion of e^x and sin (2x) up to x²
2ii) n = 9/4
3i) u_n = S_n - S_n-1
3ii) u_n+1 = u_n + 4
4i) implicit differentiation
4ii) solve dy/dx = 0
5i) y = ½(x-2)³ - 6
6i) use gc to find beta and gamma
6ii) forgot
6iii) 9/4
6iv) required set of k = (-1 , 3)
7) 10 ln (2) minutes
temperature tends to 20°C
8i) z1 = 2(cos (pi/3) + i sin (pi/3))
z2 = (√2)(cos (-3pi/4) + i sin (-3pi/4))
8ii) (z1/z2)* = (√2)(cos (11pi/12) + i sin (11pi/12))
8iv) (0,0) and (2,0)
9i) x = cube root [(200+200k)/3]
9ii) y/x = (100/x²)/x = 100/x³ = 300/(200+200k) = 3/(2+2k)
9iii) when k = 0, y/x = 3/2
when k = 1, y/x = 3/4
therefore, 3/4 <= y/x < 3/2
9iv) square ends means y/x = 1
k = ½
10i) direction vector of line l is a multiple of the normal to
plane p
or you can do vector product of these 2 vectors and get a zero
vector which implies that these 2 vectors are parallel
10ii) (8.5 , 2 , 1.5)
10iii) B has coordinates (19, -19, -30)
10iv) Area = ½|(-63, 567, -399)| = 348 (to nearest whole
number)
11i) find dy/dx, then find equation of tangent
11ii) A = (2p, 2p) , B = (2/p, -2/p)
length of OA = (√8)p , length of OB = (√8)/p
area of triangle OAB = ½(base)(height) = ½[(√8)p][(√8)/p] = 4
therefore it is independent of p
11iii) x²/4 - y²/4 = 1 (hyperbola)
x-intercepts: (-2, 0) and (2, 0)
asymptotes: y = +/- x
there are also answers on this website:
http://www.exampaper.com.sg/blog/wp-content/uploads/gce-a-level-2010-h2-maths-9740-paper-1-solutions.pdf
