c ii was dodgy in my opinion, and I didn't think there were any solutions, but after checking the answers it made sense.
g(x) is a log graph over a restricted domain, and which has undergone transformations, the important ones to recognise are the reflection in the x, and is being stretched by a factor of k from the x axis. Everything else is resolved algebraically. Have a look here (
https://www.desmos.com/calculator/sqtxc0yott) if you can't visualise it in your head. If T is a solution, and k is maximised, then T is going to occur where x=5. Solving g(5)=T for k gives you the answer in terms of T. The answers cleaned it up (but I don't think you have to).
I think my method for 4f might be a bit unorthodox, but here's how I did it.
On my CAS:
Entered
:=15+a\cdot e^{0.04t}\cdot sin(\frac{\pi t}{3})\])
graphed
|x<60\: and\: a=1\])
to see what it looks like.
The gradient of a sine graph is at its most when it is in the middle of its range, and the gradient of an exponential keeps increasing, so I decided that the maximum gradient occurs at t=60
then
)|t=60,a)\])
---- d/dt (h(t))|t=60 means the gradient of h(t) at t=60
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