That's good. People say they find C4 easier than C3, would you say that is justified
I can see why they'd prefer it, neither are particularly different in difficulty imo, depends on what type of question you prefer + How you adjust to vectors
I can't remember the exact question but I'll try to explain how I did it. You are told that P is a point on both curves and that x=2 so you know they intersect therefore you equate them then substitute in x=2 to have one equation with a's and b's. You are then told the gradient of the curves are equal at P so differentiate both curves sub in x=2 and equate the differentials to get a second equation with a's and b's. Then solve the equations with a's and b's simultaneously to get a=6 b=5. I hope this helps?
In (6) when I re-substituted into Y I copied the equation wrong to get 8 instead of 18, giving b=0 and a=4. Can anybody tell me how many marks lost this will be.
In (6) when I re-substituted into Y I copied the equation wrong to get 8 instead of 18, giving b=0 and a=4. Can anybody tell me how many marks lost this will be.
Just below average I would say. 56 an A. I guess you're doing C4. How is that compared to C3? I'm doing it next year so is there trig stuff in there as well?
Yeah i'd agree, lower than usual but not by a lot. C4 is quite similar to C3; there's a lot of integration and differentiation again, the only difference being they involve sin/cos/tan etc. As long as you know the identities it isn't too difficult. There's a few bits on vectors as well. I enjoy C4 more than C3 and, besides vectors, I'd say it's easier as well. Good luck!
Just below average I would say. 56 an A. I guess you're doing C4. How is that compared to C3? I'm doing it next year so is there trig stuff in there as well?
I think it will definitely be lower than that. Perhaps 52 or 53. It was much harder than usual.