Hey, I managed to get every question on the 2004 paper except for 7) a) iii). Was hoping someone could help me out with it, thank you.
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Also, for question 5) b) iv), the solutions i found assumed that the intersection lies on y=x. When I did the question, I just let f(x) = f^-1(x) and factorised to get the required equation. They just let x = f(x). We can do this? I can see the intuition but I don't know if it's correct mathematical reasoning.
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Thank you!
Hey there!
With the first question, note that the equation of the height of the tide at the harbour and the height of the tide at the wharf is effectively the same function, but shifted.
ie. the equation of the tide's height at the harbour, where b is the midpoint of the minimum and the maximum, and z is the height of the tide is
Note that when t=0 (ie. 1am, since tides occur 1 hour earlier), the tide reaches a maximum ie. α=0, since z=b+3 and we know that the range is 6.
Hence, the full equation will just be
Now, since the water level will always be decreasing, we just need to know when the water level falls below 2 metres above low tide ie. 1 metre below the midpoint, and we just need to make sure the ship leaves before that time. ie. we have that
Solving for t,
ie. the ship must reach the harbour prior to 4:48, meaning it must leave by 4:28 am.
The intuition behind all this is just having some equation of motion, given a certain height and then solving for the time it reaches that height, if that makes sense.
Hope this makes sense
For the second question, spnmox is pretty much spot on; for any function, its inverse relation is effectively just the same thing, but flipped in y=x. So essentially wherever it intersects y=x, its relation will also do the same. Nowhere else does an intersection occur for a function and its inverse; hence we can just use x=f(x) as opposed to f(x)=f
-1(x) which is generally harder to solve.