Here's an unofficial mark scheme type thing that I've made. Some answers may not be in their correct places, and some
could be wrong. I also probably left out a few questions, so let me
know which ones. I'm pretty sure most of them are correct though, since I was able to check most of them.
1. 0.4986 or something similar.
2. You should end up with a quadratic and then get two solutions that are greater IN MAGNITUDE (modulus) than 1, and so cannot be solutions since the sine of angle cannot have a greater magnitude than 1.
I got x=22 and x=143.
I got phi=0, 180, 124.85 (I think).
3. You had to use the sine rule to get the two values of the angle, 138 and 42 I think.
Acute angle would therefore be 17.
So area=33.33.
4. nth term is a+(n-1)d, sub a and d in.
It's your typical proof, just with numbers in, like Jan 2013 if I recall correctly.
I think I got a=-7 and d=3.
5. r=-1/3, a=216.
Sum to infinity clearly 162.
Show that r satisfies the equation. Modulus of r less than 1, then pick the right solution.
I don't know whether I've got the order of this question right.
6. Integral is something like 3x^1/2 - (18x^7/3)/7 + c.
50/3. But everyone got different answers
.
7. Basic log proof.
1/2 and 2/3 I think.
8. (3,-1)
Just work out 1/2 the diameter PQ to find radius r = sqrt29.
Got an angle of 66.8 I think.
6 for length of ST.
9. Use 1/2r^2(theta-sintheta) to find red segment. Find white segment by doing area of whole circle - area of red segment. Should get 1.04 or something to 2.1 something for ratio, so white is 2 times red (angle was 2.6 radians).
Hope that's helpful, and again, any mistakes or questions left out please let me know.
P.S. I'll give explanations to any of my answers if you don't understand how I got them.