June 2023 Paper 3 Q6
6
(a) Sketch the curve with equation\[y = x^2(2x + a)\]
where \(a \gt 0\) [3 marks]
(b) The polynomial \(\mathrm{p}(x)\) is given by\[\mathrm{p}(x) = x^2(2x + a) + 36\]
(i) It is given that \(x + 3\) is a factor of \(\mathrm{p}(x)\)
Use the factor theorem to show \(a = 2\) [2 marks]
(ii) State the transformation which maps the curve with equation\[y = x^2(2x + 2)\]
onto the curve with equation
\[y = x^2(2x + 2) + 36\] [2 marks](iii) The polynomial \(x^2(2x + 2) + 36\) can be written as \((x + 3)(2x^2 + bx + c)\)
Without finding the values of \(b\) and \(c\), use your answers to parts (a) and (b)(ii) to explain why
\[b^2 \lt 8c\] [2 marks]| Scheme | Marks | AO |
|---|---|---|
| Draws cubic curve in the correct orientation | M1 | 1.1a |
| Deduces minimum or maximum at (0,0) on their curve | M1 | 2.2a |
| Draws a fully correct cubic curve with \(x\)-intercept at \(-\dfrac{a}{2}\) shown on the curve | A1 | 2.2a |
| (3) |
Typical solution
| Scheme | Marks | AO |
|---|---|---|
| (i) Substitutes \(x = -3\) into \(\mathrm{p}(x)\) Condone missing bracket for \((-3)^2\) Must see an expression in terms of \(a\) | M1 | 1.1a |
| Completes reasoned argument with at least one correct intermediate step and no error seen to show \(a = 2\) AG Must set an expression for \(\mathrm{p}(-3) = 0\) Condone recovery of missing bracket for \((-3)^2\) to get 9 Do not condone any other missing bracket | R1 | 2.1 |
| (2) | ||
| (ii) States ‘translation’ or ‘translate’ or ‘translated’ Must not have other transformation other than translation | B1 | 1.1b |
| States the vector \(\begin{pmatrix} 0 \\ 36 \end{pmatrix}\) or \(36\mathbf{j}\) | B1 | 1.1b |
| (2) | ||
| (iii) Explains that the translated graph only has one real solution or only has a root at \(-3\) Condone missing ‘real’ | E1 | 2.4 |
| Deduces that the discriminant of \(2x^2 + bx + c\) must be negative and shows the required result Do not allow the use of \(a\) = 2 with reference to part (b)(i) Allow \(b^2 - 8c \lt 0\) following from \(b^2 - 4ac\) seen | E1 | 2.2a |
| (2) | ||
| (9 marks) |
Typical solution
(i)
\[(-3)^2\left(2 \times -3 + a\right) + 36 = 0\]\[-54 + 9a + 36 = 0\]\[9a - 18 = 0\]\[a = 2\](ii)
\[\text{Translation } \begin{pmatrix} 0 \\ 36 \end{pmatrix}\](iii)
The translated graph will only have one real solution.
\[b^2 - 4ac \lt 0\]\[\text{Hence } b^2 - 4 \times 2 \times c \lt 0\]\[b^2 \lt 8c\]