| Exam Board | WJEC |
|---|---|
| Module | Unit 3 (Unit 3) |
| Session | Specimen |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Find stationary point then sketch curve |
| Difficulty | Standard +0.8 This question requires understanding of inflection points (second derivative = 0), solving simultaneous equations from multiple conditions, finding and classifying stationary points, and producing an accurate sketch. The multi-step nature, need to work with both first and second derivatives systematically, and the requirement to verify a second inflection point make this more demanding than standard calculus exercises, though the techniques themselves are A-level standard. |
| Spec | 1.07f Convexity/concavity: points of inflection1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d^2y}{dx^2} = 12ax^2 + 6bx + 36\) | M1 | (attempt to find \(\frac{d^2y}{dx^2}\), 2 correct terms) |
| For point of inflection at \((1,11)\): \(12a + 6b + 36 = 0\) So that \(2a + b + 6 = 0\) (1) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Also \(a + b + 18 = 11\) (2) | B1, M1 | (Attempt to solve for \(a, b\)) |
| From (1), (2): \(a = 1, b = -8\) | A1 | |
| \(\therefore \frac{d^2y}{dx^2} = 12x^2 - 48x + 36 = 12(x^2 - 4x + 3) = 12(x-1)(x-3) = 0\) | M1 | |
| \(\therefore \frac{d^2y}{dx^2} = 0\) when \(x = 3\) | A1 | |
| and \(\frac{d^2y}{dx^2}\) changes sign as \(x\) passes through 3 | m1 | |
| \(\therefore\) There is a point of inflection at \(x = 3\), \(y = 3^3 - 8 \times 3^3 + 18 \times 3^2 = 27\) , i.e at \((3,27)\) | A1, A1 | (Only if m1 is awarded) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = 4x^3 - 24x^2 + 36x = 0\) | M2 | (M1 for correct differentiation but not equal to 0) |
| \(\therefore 4x(x^2 - 6x + 9) = 0\) | A1 | |
| giving \(x = 0, x = 3\) | ||
| Then at \(x = 0\), \(y = 0\) and \(\frac{d^2y}{dx^2} = 36\) | A1 | |
| There is a minimum at \(x = 0, y = 0\) | A1 | |
| [Sketch showing general shape with points of inflection at \((1,11)\) and \((3,27)\), and minimum at origin] | G1, G1 | (general shape) (min two points of inflection) |
### Part (a):
$\frac{d^2y}{dx^2} = 12ax^2 + 6bx + 36$ | M1 | (attempt to find $\frac{d^2y}{dx^2}$, 2 correct terms)
For point of inflection at $(1,11)$: $12a + 6b + 36 = 0$ So that $2a + b + 6 = 0$ (1) | A1 |
### Part (b):
Also $a + b + 18 = 11$ (2) | B1, M1 | (Attempt to solve for $a, b$)
From (1), (2): $a = 1, b = -8$ | A1 |
$\therefore \frac{d^2y}{dx^2} = 12x^2 - 48x + 36 = 12(x^2 - 4x + 3) = 12(x-1)(x-3) = 0$ | M1 |
$\therefore \frac{d^2y}{dx^2} = 0$ when $x = 3$ | A1 |
and $\frac{d^2y}{dx^2}$ changes sign as $x$ passes through 3 | m1 |
$\therefore$ There is a point of inflection at $x = 3$, $y = 3^3 - 8 \times 3^3 + 18 \times 3^2 = 27$ , i.e at $(3,27)$ | A1, A1 | (Only if m1 is awarded)
### Part (c):
$\frac{dy}{dx} = 4x^3 - 24x^2 + 36x = 0$ | M2 | (M1 for correct differentiation but not equal to 0)
$\therefore 4x(x^2 - 6x + 9) = 0$ | A1 |
giving $x = 0, x = 3$ | |
Then at $x = 0$, $y = 0$ and $\frac{d^2y}{dx^2} = 36$ | A1 |
There is a minimum at $x = 0, y = 0$ | A1 |
[Sketch showing general shape with points of inflection at $(1,11)$ and $(3,27)$, and minimum at origin] | G1, G1 | (general shape) (min two points of inflection)
**Total: [16]**The curve $y = ax^4 + bx^3 + 18x^2$ has a point of inflection at $(1, 11)$.
\begin{enumerate}[label=(\alph*)]
\item Show that $2a + b + 6 = 0$. [2]
\item Find the values of the constants $a$ and $b$ and show that the curve has another point of inflection at $(3, 27)$. [8]
\item Sketch the curve, identifying all the stationary points including their nature. [6]
\end{enumerate}
\hfill \mbox{\textit{WJEC Unit 3 Q7 [16]}}