Higher November 2021 Paper 2 Q21
21 The curve C has equation \(y = \mathrm{f}(x)\) where \(\mathrm{f}(x) = 9 - 3(x + 2)^2\)
The point \(A\) is the maximum point on C.
The curve C is transformed to the curve S by a translation of \(\begin{pmatrix} 4 \\ 0 \end{pmatrix}\)
The curve C is transformed to the curve T.
The curve T has equation \(y = 3(x + 2)^2 - 9\)
The graph of \(y = a\cos(x - b)° + c\) for \(-180 \leqslant x \leqslant 360\) is drawn on the grid below.

| Scheme | Marks |
|---|---|
| (−2, 9) | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \((y =)\ 9 - 3(x - 4 + 2)^2\) | B1 |
| (1) |
Notes
B1: oe eg \((y =) -3x^2 + 12x - 3\)
accept \(\mathrm{f}(x - 4)\)
| Scheme | Marks |
|---|---|
| Reflection in the line \(y = 0\) or \(x\)-axis | B1 |
| (1) |
Notes
B1: with no mention of another transformation
| Scheme | Marks |
|---|---|
| (3, −90, 2) (−3, 90, 2) (3, 270, 2) (−3, 450, 2) etc Answer: eg \(a = 3\) \(b = -90\) \(c = 2\) | B3 |
| (3) | |
| (6 marks) |
Notes
B3: for all 3 correct values
eg 3, −90, 2 or −3, 90, 2
(If not B3 then B2 for any 2 correct values
NB.
2 values from 3, −90, 2 or
2 values from −3, 90, 2
NB: accept a value of (90 + 360n) in place of 90 or (−90 + 360n) in place of −90 where n is an integer (could be negative)
If not B2 then
B1 for any 1 correct value or the graph of \(y = \cos x°\) for \(0 \leqslant x \leqslant 360\))