| Date | May 2011 | Marks available | 2 | Reference code | 11M.2.hl.TZ1.2 |
| Level | HL only | Paper | 2 | Time zone | TZ1 |
| Command term | Show that | Question number | 2 | Adapted from | N/A |
Question
Consider the function \(f(x) = {x^3} - 3{x^2} - 9x + 10\) , \(x \in \mathbb{R}\).
Find the equation of the straight line passing through the maximum and minimum points of the graph \(y = f (x)\) .
Show that the point of inflexion of the graph \(y = f (x)\) lies on this straight line.
Markscheme
\(f'(x) = 3{x^2} - 6x - 9\) (\(= 0\)) (M1)
\(\left( {x + 1} \right)\left( {x - 3} \right) = 0\)
\(x = - 1\); \(x = 3\)
(max) (−1, 15); (min) (3, −17) A1A1
Note: The coordinates need not be explicitly stated but the values need to be seen.
\(y = - 8x + 7\) A1 N2
[4 marks]
\(f''(x) = 6x - 6 = 0 \Rightarrow \) inflexion (1, −1) A1
which lies on \(y = - 8x + 7\) R1AG
[2 marks]
Examiners report
There were a significant number of completely correct answers to this question. Many candidates demonstrated a good understanding of basic differential calculus in the context of coordinate geometry whilst other used technology to find the turning points.
There were a significant number of completely correct answers to this question. Many candidates demonstrated a good understanding of basic differential calculus in the context of coordinate geometry whilst other used technology to find the turning points. There were many correct demonstrations of the “show that” in (b).
