| Date | May 2013 | Marks available | 1 | Reference code | 13M.1.sl.TZ1.15 |
| Level | SL only | Paper | 1 | Time zone | TZ1 |
| Command term | Write down | Question number | 15 | Adapted from | N/A |
Question
Consider the function \(f (x) = ax^3 − 3x + 5\), where \(a \ne 0\).
Find \(f ' (x) \).
Write down the value of \(f ′(0)\).
The function has a local maximum at x = −2.
Calculate the value of a.
Markscheme
\( f '(x) = 3ax^2 - 3\) (A1)(A1) (C2)
Note: Award a maximum of (A1)(A0) if any extra terms are seen.
−3 (A1)(ft) (C1)
Note: Follow through from their part (a).
\(f '(x) = 0\) (M1)
Note: This may be implied from line below.
\(3a(-2)^2 - 3 = 0\) (M1)
\((a =) \frac{1}{4}\) (A1)(ft) (C3)
Note: Follow through from their part (a).
Examiners report
Many candidates could find the derivative of the cubic function and find the value of the derivative at \(x = 0\). For part (c) many candidates calculated the value of the function rather than the derivative at \(x = - 2\).
Many candidates could find the derivative of the cubic function and find the value of the derivative at \(x = 0\).
Many candidates could find the derivative of the cubic function and find the value of the derivative at \(x = 0\). For part (c) many candidates calculated the value of the function rather than the derivative at \(x = - 2\). However only the best realized that the derivative is zero at the maximum and so calculated the value of \(a\).
