I personally wouldn't but I guess (?) you could say it's just a bunch of ones? Not something i'd personally agree with but I can see where someone could come from with that.
Can someone verify my answer for 5 part (iv) of the 2015 paper? I got g(a,b) = s(b,a,p(g(a,m(b)))) as my function which I think satisfies the condition g(a,b) = a+b but I'm not 100% sure
Its one of the circle theorems that angles within a circle made from lines connecting one end of a chord to a point on the circle to the other end of the chord are equal, i.e. x=y
Note this only holds if they are both in the same segment of the circle.
You're welcome. Another good site that I have found for everyone (incase you already haven't seen it) is this site: http://www.drfrostmaths.com/resources/page.php?id=7, it has powerpoints going through questions and also explains how to approach some of the more challenging questions, I've found it pretty useful.
Can someone verify my answer for 5 part (iv) of the 2015 paper? I got g(a,b) = s(b,a,p(g(a,m(b)))) as my function which I think satisfies the condition g(a,b) = a+b but I'm not 100% sure
Your function would return a+b if b was greater than (or equal to) 0, but part (iv) states it wants a function for non-positive values of b. So you basically need to swap some things. The mark scheme has as the first answer g(a, b) = s(p(b), m(g(a, p(b))), a), which is what you'd get if you swapped the second and third parameters, replaced p with m and vice versa, and the p(b) instead of just b comes from the fact that you'd get an off-by-one-error otherwise.
Can someone explain 1D paper 2015? Thanks for any help
First work out the integrals in their general form for both f(x) and g(x). Then substitute in the values of A and test option a. Because the numerator on either fraction should be A^6 you can compare the denominators of both fractions to see which result is true.
First work out the integrals in their general form for both f(x) and g(x). Then substitute in the values of A and test option a. Because the numerator on either fraction should be A^6 you can compare the denominators of both fractions to see which result is true.
The issue I'm having is with integrating the first integral sorry, up until practicing for the MAT I hadn't encountered integrals with what appear to be two variables (x and t) unless im just being stupid (very possible haha). Could you just explain the integration for f(x)? Thanks
The issue I'm having is with integrating the first integral sorry, up until practicing for the MAT I hadn't encountered integrals with what appear to be two variables (x and t) unless im just being stupid (very possible haha). Could you just explain the integration for f(x)? Thanks
Yeah I know how you feel, I only encountered them until very recently..
Expanding the bracket gives you x^2*t^2, because the integral is with respect to t (dt) you can move the x^2 out of the bracket, then just treat the integral as normal but with t's instead.. the x^2 can be treated as a constant. Integrating gives you [t^3/3] between 1 and 0, which evaluates to 1/3, but the constant needs to be multiplied with it as-well, which gives you x^3/3 for f(x).
In terms of the fomulas that we need to remember are there any that I have missed?
- Sum of an arithmetic series - Sum of a geometric series (& sum to infinity) - Trapezium rule - Binomial expansion for +ive exponents - Cosine rule + other geometry formulae - Trig identities in C2 - Laws of logs
I havent seen a solution yet where you had to find out the coordinates on a circle/parabola and you use simultaneous equations but subbing in x = .....
In terms of the fomulas that we need to remember are there any that I have missed?
- Sum of an arithmetic series - Sum of a geometric series (& sum to infinity) - Trapezium rule - Binomial expansion for +ive exponents - Cosine rule + other geometry formulae - Trig identities in C2