Higher November 2021 Paper 1 Q28
28 Here is the graph of \(\quad y = \cos x \quad\) for \(\ 0° \leqslant x \leqslant 360°\)

In parts (a) and (b) the graph of \(\quad y = \cos x \quad\) is shown as a dashed line.
(a) On the grid below, draw the graph of \(\quad y = \cos(x - 90°) \quad\) for \(\ 0° \leqslant x \leqslant 360°\) [1 mark]

(b) On the grid below, draw the graph of \(\quad y = 1 + \cos x \quad\) for \(\ 0° \leqslant x \leqslant 360°\) [1 mark]

(c) Rita tries to draw the graph of \(\quad y = \cos(-x) \quad\) for \(\ 0° \leqslant x \leqslant 360°\)
Here is her graph.

Give a reason why Rita’s graph is incorrect. [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| Correct graph (translated 90° to the right) | B1 | mark intention |
Additional guidance
Condone the graph starting at (90, 1)
Ignore the curve outside the domain \(0 \leqslant x \leqslant 360\)
| Answer | Mark | Comments |
|---|---|---|
| Correct graph (translated 1 up) | B1 | mark intention |
Additional guidance
Ignore the curve outside the domain \(0 \leqslant x \leqslant 360\)
| Answer | Mark | Comments |
|---|---|---|
| Correct statement | B1 | eg this is \(y = -\cos x\) \(\cos 0 = 1\) it’s upside down it should be the same as \(\cos x\) |
Additional guidance
| It has been reflected in the \(x\)-axis instead of the \(y\)-axis | B1 |
| It should have been reflected in the \(y\)-axis | B1 |
| It starts at \(-1\) (instead of 1) | B1 |
| 180 is above the \(x\)-axis | B1 |
| Correct curve drawn | B1 |
| \(\cos(-180) = -1\) | B1 |
| She has done \(-y\) instead of \(-x\) | B1 |
| It can’t start as a negative | B1 |
| It should go down not up | B0 |
| She shouldn’t have flipped it | B0 |
| Ignore non-contradictory statements alongside a correct statement | B1 |