I think I put 2^(13/3), think the question was 5*4^(2/3)+3*16^(1/3), write 4 and 16 in powers of 2 and you'll get 2^(4/3)*8, then write 8 as a power of 2 and simplify to 2^(13/3) if I remember correctly.
Guys the answer to the last question was 27, but for some stupid reason, I left the answer as:
8 x (√27/12)^3
Now, I checked on my calculator, and it is equal to 27, but I'm not sure about whether my form is acceptable. My intermediate working was all right, so do you think that they will dock marks, and if so, how many?
Yeah I left mine as 18root27/12 which is also 27 so I'm hoping we will still get full marks as it is equivalent but hopefully if we do drop marks it will be only 1 mark
Indices: Q5 pt 2: 2^13/2 Because: it was 8(2^4/3) = 2^3 x 2^4/3 = 2^13/3. Quite sure its correct as lots of people around me got the same answer
Got 27 for last Question, although i left it as a surd.
Im also sure the radius was 0<r<=2 As when you put y=0 into the circle equation, you got x=-2 Given the centre of the mini circle was at the origin, and had to be within the larger circle, the furthest it could have been was 0 -> -2 giving it a radius of 2 or less? It was only 2 marks so I cant see how it would be more complex than that?
Hoping grade boundaries low (56-58), definitely much harder than past papers. Everyone was of the same opinion, few of further maths kids didnt finish it and theyre the 100UMS type.
I also got 2^13-3
For the last question I left my answer as 8*(27/12)^3/2, which does equal 27, just hope that the examiner picks up on this, i should get it correct, right?
also, I'm pretty sure the set of values for r was 0<r<3√5 -5 we know the radius of the circle is 3√5. We can calculate using: L=SQRT((x2-x1)^2 + (y2-y1)^2) plugging in values for the origin (0,0) and the centre of the circle (4,3) (can't remember if that's correct) then you get the distance 5. Therefore the maximum distance must me 3√5 -5, and since the radius can never be 0, you produce the set of values I mentioned before, right?
How many marks was the 4+12x-2x^2 question worth? And what did u guys get for the k inequality question? Plz reply, ur help will be greatly appreciated.
I think I put 2^(13/3), think the question was 5*4^(2/3)+3*16^(1/3), write 4 and 16 in powers of 2 and you'll get 2^(4/3)*8, then write 8 as a power of 2 and simplify to 2^(13/3) if I remember correctly.
Yes but at the time after the exam everyone was saying it was hard, happens every year
but you can see the change in structure between the last 5 years papers and this year paper, they must admit that this year paper was the hardest but still easier than c2