Date | November 2010 | Marks available | 3 | Reference code | 10N.2.sl.TZ0.7 |
Level | SL only | Paper | 2 | Time zone | TZ0 |
Command term | Write down | Question number | 7 | Adapted from | N/A |
Question
Let \(f'(x) = - 24{x^3} + 9{x^2} + 3x + 1\) .
There are two points of inflexion on the graph of f . Write down the x-coordinates of these points.
Let \(g(x) = f''(x)\) . Explain why the graph of g has no points of inflexion.
Markscheme
valid approach R1
e.g. \(f''(x) = 0\) , the max and min of \(f'\) gives the points of inflexion on f
\( - 0.114{\text{, }}0.364\) (accept (\( - 0.114{\text{, }}0.811\)) and (\(0.364{\text{, }}2.13)\)) A1A1 N1N1
[3 marks]
METHOD 1
graph of g is a quadratic function R1 N1
a quadratic function does not have any points of inflexion R1 N1
METHOD 2
graph of g is concave down over entire domain R1 N1
therefore no change in concavity R1 N1
METHOD 3
\(g''(x) = - 144\) R1 N1
therefore no points of inflexion as \(g''(x) \ne 0\) R1 N1
[2 marks]
Examiners report
There were mixed results in part (a). Students were required to understand the relationships between a function and its derivative and often obtained the correct solutions with incorrect or missing reasoning.
In part (b), the question was worth two marks and candidates were required to make two valid points in their explanation. There were many approaches to take here and candidates often confused their reasoning or just kept writing hoping that somewhere along the way they would say something correct to pick up the points. Many confused \(f'\) and \(g'\) .