June 2016 Paper 2 Q2
2 The function \(\quad \mathrm{f}(x) = x^2 - 4x + 1 \quad\) has domain \(\ -1 \leqslant x \leqslant 5\)
Here is the graph of \(\quad y = \mathrm{f}(x)\)
(a) Write down the equation of the line of symmetry of the graph. [1 mark]
(b) Use the graph to work out the solutions of \(\quad x^2 - 4x + 1 = 5\)
Give your answers to 1 decimal place. [2 marks]
(c) Write down the range of \(\mathrm{f}(x)\) for domain \(\ -1 \leqslant x \leqslant 5\) [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(x = 2\) | B1 |
Additional guidance
| \(2 = x\) | B1 |
| \(y = 2\) | B0 |
| 2 | B0 |
| Answer | Mark | Comments |
|---|---|---|
| \(-0.8\) and 4.8 with no other answers | B2 | Both correct in either order B1 One correct and one incorrect or missing |
Additional guidance
| The word ‘and’ is not needed If their answer has both \(-0.8\) and 4.8 with no other solutions award B2 eg 4.8, \(-0.8\) with no other solutions | B2 |
| Answer | Mark | Comments |
|---|---|---|
| \(-3 \leqslant \mathrm{f}(x) \leqslant 6\) or \(6 \geqslant \mathrm{f}(x) \geqslant -3\) or [\(-3\), 6] | B2 | oe B1 \(\mathrm{f}(x) \geqslant -3\) or \(\mathrm{f}(x) \leqslant 6\) on their own or embedded within an interval for \(\mathrm{f}(x)\) or only \(-3\) and 6 chosen |
Additional guidance
| Allow as two inequalities \(\mathrm{f}(x) \geqslant -3 \quad \mathrm{f}(x) \leqslant 6\) | B2 |
| \(-3\) to 6 inclusive | B2 |
| \(\mathrm{f}(x)\) may be replaced by f or \(y\) or \(x^2 - 4x + 1\) for B2 and B1 | |
| B1 may be seen with an incorrect inequality eg1 \(-3 \lt \mathrm{f}(x) \leqslant 6\) eg2 \(-3 \leqslant \mathrm{f}(x) \lt 6\) eg3 \(-3 \leqslant \mathrm{f}(x) \leqslant 5\) | B1 B1 B1 |
| For B1 ignore incorrect notation if only \(-3\) and 6 chosen eg1 \(-3 \leqslant x \leqslant 6\) eg2 \(-3 \lt x \leqslant 6\) eg3 \(-3\) to 6 eg4 \(-3\) to \(6 = 9\) (working out a statistical range) eg5 \(-3 \geqslant \mathrm{f}(x) \geqslant 6\) eg6 \(-3\), 6 | B1 B1 B1 B1 B1 B1 |
| {\(-3, -2, -1, 0, 1, 2, 3, 4, 5, 6\)} | B0 |