| Date | May 2015 | Marks available | 2 | Reference code | 15M.1.sl.TZ2.14 |
| Level | SL only | Paper | 1 | Time zone | TZ2 |
| Command term | Find | Question number | 14 | Adapted from | N/A |
Question
The number of fish, \(N\), in a pond is decreasing according to the model
\[N(t) = a{b^{ - t}} + 40,\;\;\;t \geqslant 0\]
where \(a\) and \(b\) are positive constants, and \(t\) is the time in months since the number of fish in the pond was first counted.
At the beginning \(840\) fish were counted.
Find the value of \(a\).
After \(4\) months \(90\) fish were counted.
Find the value of \(b\).
The number of fish in the pond will not decrease below \(p\).
Write down the value of \(p\).
Markscheme
\(a{b^0} + 40 = 840\) (M1)
Note: Award (M1) for substituting \(t = 0\) and equating to \(840\).
\(a = 800\) (A1)(C2)
\(800{b^{ - 4}} + 40 = 90\) (M1)
Note: Award (M1) for correct substitution of their \(a\) (from part (a)) and \(4\) in the formula of the function and equating to \(90\).
\({b^4} = 16\;\;\;\)OR\(\;\;\;\frac{1}{{{b^4}}} = \frac{1}{{16}}\;\;\;\)OR\(\;\;\;b = \sqrt[4]{{16}}\;\;\;\)OR\(\;\;\;b = {16^{\frac{1}{4}}}\) (M1)
Notes: Award second (M1) for correctly rearranging their equation and eliminating the negative index (see above examples).
Accept \(\frac{{800}}{{50}}\) in place of \(16\).
OR
(M1)(M1)
Notes: Award (M1) for a decreasing exponential and a horizontal line that are both in the first quadrant, and (M1) for their graphs intersecting.
For graphs drawn in both first and second quadrants award at most (M1)(M0).
\(b = 2\) (A1)(ft) (C3)
Note: Follow through from their answer to part (a) only if \(a\) is positive.
\(40\) (A1) (C1)
