Math HL P2 TZ2 - 24 hours passed

Any guess for new grade boundary for this years paper?
We have to see the difficulty of paper 3 but i really hope it is between 70~73%

See my earlier post:

are you talking about the overall grade including IA, paper 1,2,3,?

I heard although boundaries for each paper is around 75~80%, the overall boundary is always like 80%.

If the overal boundary is 75%, then I am not that hopeless

Wow. I actually did all of that!I just didn't put it in the calculator right. Method marks at least I suppose.

For Q13, last part, what were you supposed to put for the y1,y2,y3 values? could you get it exact ? I gave up and just gave the numbers to 3sf from the GDC. Someone told me that you couldn't get exact answers for this, is that true ?

And could someone help me remember what were the questions in this P2 section A?

There was:
- Venn-diagram probability
- Acceleration (differential equation)
- Vector / gradient (perpendicular)
- Vaccination probability
- Polynomial exercise with part a) as show that b^2>24c
- Table arrangements
- Integration by substitution
- Related rates of change
- Transformations and inequalities (ln(2x+1))

There were 10 questions, I can only remember 9. What am I missing?

The one with the company making machines and calling people every 10 minutes.

The one with the company making machines and calling people every 10 minutes.

the day before I was revisin,lookin at the question was like" ok, i know this,, next.."

No, wasn't that the first long question on section B?

it didnt say exact. so dont you dare worrying about it.

i got the ratio of 0.215. anyone similar??? cuz im not sure

The one with the company making machines and calling people every 10 minutes.

Yeah, I got 0.2**, i can't remember, but probably what you got. But the question said "you should simplify", or something along these lines. What the hell does that mean ?? simplify to what ? And it said "should", what kind of a vague statement was that....

"Show that (2/(x+5)) + (3/(x+4)) = (....../......)" Something in that line.. Wasn't it?

answers for Q 12 (well this is what i wrote)

1. because PQ = 50, x+10<50. therefore x<40

2. well... draw extra line from Q to a radius on the other circle so that the line is parallel to MN.

then cos ((x-10)/50)

3. (a) square root of (2500 - (x-10)^2) or (50^2) - (x-10)^2

(b) when x = 10

4. (a) (alpha) = 2pi - 2(theta)

theta = arccos((x-10)/50)
therefore, (alpha) = 2pi - 2(theta)

(b) (beta) = 2pi - 2(pi/2) - 2(pi/2 - theta) = 2 (theta)
= 2 arccos((x-10)/50)


5. since arc = r(theta) where theta is angle in radius,
perimeter = x(2pi - 2(theta)) + 10(2 arccos((x-10)/50)) + 2 (square root of (2500 - (x-10)^2) or (50^2) - (x-10)^2)


ii and iii... well. put it in your calculator, and do it!

well the thing is x<40

the graph obviously goes to 60 but the point is that what value do you have when x=40 i think...

nah... i got it wrong anyway. (putting 10-x instead of x-10) haha