Advanced Higher Maths 2011-2012 : Discussion and Help

Has anyone written up an answer doc? I am happy to make one, which we can all edit...

My answers: (can't be bothered with latex, sorry :P)

1a) (3-2x-x^2)/(x^2+1)^2
b) (1-sin2x)exp(tanx)
2. r=1/2, n=9
3. z=-3, z=-1+/-2i
4. -5376
5. 3x+7y-4z=5
6. 4+4x+3x^2+(5/3)x^3+...
7a) looks like a tick :P
b) y=1 for x>=-2, y=-1 for x<-2
8. 2sqrt3 + 4pi/3
9. k=-1
10. 3412
11a)1/sqrt(1-x^2)
b) x - sqrt(1-x^2)*sin-1x + c
12. -0.139 m^3 s^(-1)
13. dy/dx=(t^2-3)/(t+2)
d^2y/dx^2=(t+3)(t+1)/(t+2)^3
Min at t=sqrt3, max at t=-sqrt3
2 points of inflexion at t=-1 and t=-3 (and since the y-values are different they are separate points)
14a)This was disgusting. I'm using f for lambda here, for no reason other than that it's distinctive and l looks like 1.
x=(f-7)/(4(f+2))
y=(3f+9)/(4(f+2))
z=3/(2(f+2))
b) 0z=3 - impossible. System is inconsistent, no solutions.
c) x=22.75, y=-6.75, z=-15. Small change in coefficient = disproportionately large change in solution, so system is ill-conditioned.
15a) 1/(9(x-1))-1/(9(x+2))-1/(3(x+2)^2)
b) y=(1/9)(x-1)(ln|(x-1)/(x+2)|+3/(x+2))+C(x-1)
16a)n=1 obviously true, assume true for n=k>=1, then I had a really short n=k+1 step - used the inductive hypothesis, then polar multiplication, and that was it really.
b)the big fraction is equal to 0+i, so the real part is 0.

These aren't guaranteed or anything, but they seem alright.

Don't know how to start a poll but if can please do so at the top of this thread, was wondering how people found this in comparison to other years past papers?

...

What is the exact answer for Q12. I got -6pi/125 is that right?

I agree with all!

Exact value for 12 is -111pi/2500

Do you want to make a google doc with these answers or shall i?

You didn't use the combination function.

I got -111pi/2500 = to around -0.1395

My answers: (can't be bothered with latex, sorry :P)

1a) (3-2x-x^2)/(x^2+1)^2
b) (1-sin2x)exp(tanx)
2. r=1/2, n=9
3. z=-3, z=-1+/-2i
4. -5376
5. 3x+7y-4z=5
6. 4+4x+3x^2+(5/3)x^3+...
7a) looks like a tick :P
b) y=1 for x>=-2, y=-1 for x<-2
8. 2sqrt3 + 4pi/3
9. k=-1
10. 3412
11a)1/sqrt(1-x^2)
b) x - sqrt(1-x^2)*sin-1x + c
12. -0.139 m^3 s^(-1)
13. dy/dx=(t^2-3)/(t+2)
d^2y/dx^2=(t+3)(t+1)/(t+2)^3
Min at t=sqrt3, max at t=-sqrt3
2 points of inflexion at t=-1 and t=-3 (and since the y-values are different they are separate points)
14a)This was disgusting. I'm using f for lambda here, for no reason other than that it's distinctive and l looks like 1.
x=(f-7)/(4(f+2))
y=(3f+9)/(4(f+2))
z=3/(2(f+2))
b) 0z=3 - impossible. System is inconsistent, no solutions.
c) x=22.75, y=-6.75, z=-15. Small change in coefficient = disproportionately large change in solution, so system is ill-conditioned.
15a) 1/(9(x-1))-1/(9(x+2))-1/(3(x+2)^2)
b) y=(1/9)(x-1)(ln|(x-1)/(x+2)|+3/(x+2))+C(x-1)
16a)n=1 obviously true, assume true for n=k>=1, then I had a really short n=k+1 step - used the inductive hypothesis, then polar multiplication, and that was it really.
b)the big fraction is equal to 0+i, so the real part is 0.

These aren't guaranteed or anything, but they seem alright.

Ops I forgot that Omg. Really scared that I won't get my B ;(

I didn't even see a link between the 2 parts of the penultimate question.

Did you get r =3?

For the last question, I originally tried multiplying by the complex conjugate. Eventually, though, I just changed the angles in the numerator to (pi/36) and changed the power to 22-I checked wolfram alpha and it says it's the same thing. I then cancelled out the denominator which left (cos(pi/36)+isin(pi/36))^18
Then, using de moivre's (cos on its own for quickness) cos(18pi/36) =cos pi/2=0?

Would that be correct?

Did you not use de moivre's before multiplying by the conjugate?