You know from the graph in the previous question that the point of intersection does not occur when sinx=0. Therefore you can divide by it as you know it can't be 0
What did you get for the 2N thing? I got -44 and 125/3 which I think are both wrong.
I got 24
It was something like n=N∑2N​un​=2750 iirc. The sigma notation is indicating the sum of all the terms from n=N to n=2N which is the same as saying S2N​−SN−1​ (just like the previous question where n=10∑20​un​≡S20​−S9​ (because you'd be excluding u10​ if you did S20​−S10​).
So finding S2N​ and simplifying:
S2N​=22N​[2×5+3(2N−1)]=N(6N+7)=6N2+7N
Finding SN−1​ and simplifying:
SN−1​=2N−1​[10+3(N−2)]
This gets you brackets to expand:
=2(N−1)(3N+4)​=23N2+N−4​=1.5N2+0.5N−2
very fiddly but the question depends on you being able to exploit the algebra
We can form a quadratic from this: 4.5N2−6.5N−2748=0
Then I used the quadratic formula, only bothering to calculate −b+b2−4ac​ in the numerator because it was clear that using ± would mean that I would get a negative solution as well as a positive one, which wouldn't make sense in the context of sequences.
For 9ii) I put (3/2pi - a) + 3/2pi instead of 3pi - a, how many marks would I lose?
That's literally exactly the same as 3pi-a... Did it not cross your mind to just add the 3/2 pi together? (For the record, you'll lose no marks but that's such a weird thing to write )
Could someone draw out the graph question drawing sin1/3 second last part on 9 . And draw out how they did the last question ? So confused would really appreciate it !!!! Xx
That's literally exactly the same as 3pi-a... Did it not cross your mind to just add the 3/2 pi together? (For the record, you'll lose no marks but that's such a weird thing to write )
Haha I did this too, the panic of exams does some very strange things...
That's literally exactly the same as 3pi-a... Did it not cross your mind to just add the 3/2 pi together? (For the record, you'll lose no marks but that's such a weird thing to write )
I know yea haha, I just had no time left thank you anyway
That might be necessary. I remeber completing it think what I have written doesnt look like 4 marks. But if you did that you would get the next answer right. When they say simplist form I am sure they mean without fractions. Everybody may have lost a mark if you are correct
That would be a godsend honestly, I got screwed over badly by the 2N question and could easily have lost 4 maybe even all 6 marks there. Apart from that though, thought it went very well, if boundary is 53 or something, then where do you reckon 100 UMS will be? 67/68? How'd you find it in general aswell?
Used my fp1 knowledge for that question :P. I thought it was a paper that required you to be careful, but otherwise was quite straight forward. I know i've lost all my marks on the last question cus of radians smh
I was considering this but when you are working to find solutions dont you do 360+ and 180- to get a set of solutions then whatever is combined with the x on the other side gets factored in to the solutions giving you the real set of solutions. The 1/3 may have only become relevant in the x intercept of the x axis for cos whatever it was. For example 2x=cos^-1(1/2), 2x=60, from here you would do 720-60 and 360+60.