A remark on Q13c)ii).

It had a nice trap where you had to recognise that f'(x) and g'(x) are undefined at the points (1,0) and (-1,0) but these are points where the f(x) and g(x) are defined.

So technically speaking you can only say that since their derivatives are

then

Therefore you can only really claim that

This would suggest that simply substituting in say x = 1 to find c has some rigour issues as it doesn't quite fit that domain (even though it works out).

I think the intended solution was for students to recognise that since

then f(x) and g(x) share common points at x = -1 and x = 1.

From part (i), they have the same gradient behaviour in the domain -1 < x < 0 but have a common point at x = -1 on the "edge" of that domain. This means that

Similarly, they have the same gradient behaviour in the domain 0 < x < 1 but have a common point at x = 1 on the "edge" of that domain. This means that