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Reply 60
Original post by The tesseract
That's not too bad if you just learn some tricks on how to integrate. If not, do something like +1-1 it works lol. Something like t^2/(1+t^2). You add 1 then subtract 1 on the numerator so you have the 1+t^2 bits cancel and you're left with a constant + a standard arctan integral. easy

Are you also allowed to use a double-substitution for Weierstrass questions?
Reply 61
Screenshot 2023-06-22 10.30.45.pngAnyone know how to do this?
Reply 62
Original post by Lucysc
hey if anyone has done the 2022 paper, does anyone get how you know to reject y=8x+24 for question 7? it says correct reason for rejection is not needed for the mark but im confused as to how you can tell., both going through x = -3. Other solution has to be from a tangent meeting with the extension of the 8 - x^2 graph

You get the 2 lines, both pass through x = -3, which means the 'extra' tangent cant also be on the part of the 8 - x^2 graph on the sketch. So it must be a tangent to the extension of the 8 - x^2 graph below the x - axis. This line would have a steeper gradient (you can check easily by sketching), and is the line you reject so reject y = 8x + 24 - solution with steeper gradient
Original post by moheat13
Screenshot 2023-06-22 10.30.45.pngAnyone know how to do this?

First and foremost, plug in your values of y and dy/dx to find what d^2y/dx^2 is when x = 0.

Next, differentiate it again and do essentially the same thing to find d^3y/dx^3 when x = 0.

Then it’s just plugging the y and derivatives into the maclaurin series, as x = 0.
Reply 64
Original post by moheat13
Screenshot 2023-06-22 10.30.45.pngAnyone know how to do this?


thats a taylor/maclaurin series question: solve for 2nd derivative at x = 0 by plugging in values. Then implicitly differentiate whole expression to get an expression with 3rd derivative, and sub in all your values to get a value for 3rd derivative at x = 0. Then series solution is y = y + dy/dx x + d2y/dx2 x^2/2 + d3y/dx3 x^3/6
Bumping this I guess
Reply 66
That was a decent paper, heres all the answers I could remember and the inequality graph (if this is against the rules please lmk :wink:)
Reply 67
Original post by habeebm
That was a decent paper, heres all the answers I could remember and the inequality graph (if this is against the rules please lmk :wink:)

Are the intersections right here? I remember getting a surd for one intersection
(edited 10 months ago)
Reply 68
Original post by habeebm
That was a decent paper, heres all the answers I could remember and the inequality graph (if this is against the rules please lmk :wink:)

The first range for the mod graph question was x < (- 7 - root57) /2
(edited 10 months ago)
I got coordinates of R as 7/13cos , -21/52 sin
Reply 70
I got 4costheta , -3 sintheta. For R which makes the exact same ellipse and therefore the same eccentricity
(edited 10 months ago)
Does anyone remember how many marks for that limits part of the question?
Reply 72
Original post by Maximus_2005
Does anyone remember how many marks for that limits part of the question?

3 marks
Reply 73
I honestly think I got everything right up until the vectors question. Feeling a big fat 0/15 for that one. Might have gotten the last bit on the limits question wrong though, I got K as 3 but thinking about it now I'm not sure that one was right.
Reply 74
Original post by 1014
The first range for the mod graph question was x < (- 7 - root57) /2
Also the values of the coefficients of cos theta and sin theta in the ellipse question are wrong because the eccentricity of that ellipse isn't roo7/4


I got root(7)/4 for e.
Reply 75
Original post by habeebm
That was a decent paper, heres all the answers I could remember and the inequality graph (if this is against the rules please lmk :wink:)


Got the same for the ellipse, for the eccentricity proof did you say that the coordinates of R are scalars of the original ellipse (scalar = 7/25)?
(edited 10 months ago)
Guys what was your final answer to the integration question? Does anyone remember? T formula one
Reply 77
Original post by Maximus_2005
Guys what was your final answer to the integration question? Does anyone remember? T formula one

I got:
(1/root5)arctan((3tan(x/2) + 1)/root5)
Or something like that (can't fully remember).
You had to complete the square for the denominator and then do u = (t+2/3)^2 to have it ready for the arctan integration.
15 for the last part of the last question? I was very unsure for that one
(edited 10 months ago)
Original post by moheat13
I got:
(1/root5)arctan((3tan(x/2) + 1)/root5)
Or something like that (can't fully remember).
You had to complete the square for the denominator and then do u = (t+2/3)^2 to have it ready for the arctan integration.


Thank god! I got the same answer

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