Maths HL P1 & 2 2017

Liordi wrote: »

I got 21 but I don't know if I did it properly.

Cheers, you both got 21 so I'll assume it's right got 23 for that, not bad considering I had no idea what to do for most of that question.

Anyone get 33 degrees for the angle in the circle question?

Hi Nanook5, I've been through Paper 1 completely at this stage and, based on my own calculations, I have the following:

Q1

(a) f(x) can be written as 2(x-(7/4))^2 - 129/8 so I believe you have made a small slip in your calculation - I verified my own answer here by expanding it again and I do end up with the original f(x) = 2x^2-7x-10 so I'm almost 100% this is correct

(b) Because of the error in the previous part, your answer for minimum point is different to mine but I'm sure you won't lose any marks for this. My answer is (7/4,-129/8)

(c)

(i) Our techniques match here but you have a small slip where you say -7 squared becomes -49. As it happens, the -4ac part well exceeds +49 so you still end up with a positive value, proving there must be 2 distinct real roots involved

(ii) When it comes to the formula here, your technique is spot on again but again a small slip in the b^2 part where you never squared it and just left it as -7, introducing a small error. My answers here are x1 = 7/4+sqrt(129/16) and 7/4-sqrt(129/16). I verified my answers here by plugging them into the original equation for f(x) and they work out fine.

Q2.

(a) Our answers match here, I got -8-8.sqrt(3)i which I verified by expanding z^4 to get the same answer.

(b) Our answers match here also, I just wrote t in its simplest form as -6 instead

Q3.

(a) You have a slip in your calculation here, the correct answer is (2/3)x-1. This can be easily shown to be correct by just differenciating f(x) using the "quick" method.

(b) Our answers match exactly here

Q4.

(a) You are practically right here, our techniques match but I end up with n = 12.4702, as you do. However, correct to the nearest day, it will be day 13 (rounding n upwards) before the % of the substance in the solution first falls below 0.01%.

(b) we get exactly the same answer here

Q5.

(a) We get exactly the same root values here, and these are definitely right as I have verified them by plugging them back into the original equation for f(x)

(b) Our techniques are very similar here and we both get exactly the same values for max and min

(c) You look not to have been sure how to do this part. Basically the question is saying that the graph of f(x) is being shifted upwards (or downwards) by the value of some constant value a, such that f(x) ends up having only 1 real root. If it has only 1 real root this means the graph of f(x) only cuts the x-axis once. So, if you can imagine it, it means that a must be some value such that the local maximum must now be below the x-axis (therefore the graph of f(x) in its locaity never cuts through the x-axis) or else a must be some value such that the local minimum must now be above the x-axis(again ensuring the graph of f(x) in its locality never cuts through the x-axis). If this is indeed the case, then it must be that a<-9 or a>(100/27). So again, if you can imagine it, for all values of a less than -9 the local maximum of the graph (which had originally a y-value of +9) is "pulled down" below the x-axis and never reaches it, giving just 1 real root. Similarly, for all values of a > (100/27) the local minimum of the resulting graph (which originally had a y-value of -100/27 is now "pulled up" above the x-axis and never reaches it, giving just 1 real root also.

So the range of values for a here are a<-9 and a>100/27