MAT Prep Thread - 2nd November 2016

I see makes sense now

Btw are you taking step aswell?

He's already at Cambridge.

He's already at Cambridge.

No. I hope we haven't confused you.

A turning point is a turning point if and only if f'(x)=0 (the gradient is 0 at that point) AND f''(x)=/=0 (the sign of the gradient changes).

That would seem to exclude f(x) = x^4 having a turning point at x=0.

My. You're right! I tried to make my explanation easy to understand but that seemed to remove the rigour...

A turning point is a point where the derivative changes sign. That should be right :P

if you look at part iii, the f(k)...=-f(2-k)... so it has been reflected in the x-axis
then it also changes from (1+t) to (1-t) essentially this is a reflection the line x=1 because for example if you sub t=0 you get the same function so that is the axis of symmetry
when you reflect or move a graph the area stays the same

Btw zacken just came across your step hacks and tricks thread loving it, thanks man

Can I go through the multiple choice answers for the 2003 paper please? Anyone done it?

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Hi, I got b, b, c, c, a, c, a, d, a, d.

We wouldn't need to know D or G right?

And how did you get a for I? I got b

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Explain your reasoning.

I simply did the sum from 1 to 20, because order isn't accounted and you can have 2 of the same.

Not sure if that's right? Yourself?

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Sorry I meant c) instead of a).

We have 20 choices for our first candy. For each choice of the first candy, we have 20 choices for our second candy. This takes order into account so 20(20)/2=200.

Would g(a,b) = s( b-1 , p(a) , p(f(a,m(b)) work for 2015 Q5 part IV?

The idea being that its one less than f(a,b) which = a + b + 1 as g(a,b) has one less iteration due to x = b-1 instead of x=b

Sorry I meant c) instead of a).

We have 20 choices for our first candy. For each choice of the first candy, we have 20 choices for our second candy. This takes order into account so 20(20)/2=200.