| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2006 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Prove curve has no turning points |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question requiring standard techniques: solving a quadratic in disguise (x² substitution), differentiating a polynomial, and connecting the derivative to stationary points. The 'hence' structure guides students through the solution. Slightly easier than average due to the scaffolding and routine methods, though the connection between parts (i) and (iii) requires some insight. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Let \(y = x^2\) | *M1 | Use a substitution to obtain a quadratic or \((x^2 - 5)(x^2 - 5) = 0\) |
| \(y^2 - 10y + 25 = 0\) | dep*M1 | |
| \((y-5)^2 = 0\) | A1 | Correct method to solve a quadratic; 5 (not \(x = 5\) with no subsequent working) |
| \(y = 5\) | A1 | |
| \(x^2 = 5\) | A1 | 4 marks |
| \(x = \pm\sqrt{5}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = \frac{2x^3}{5} - \frac{20x^3}{3} + 50x + 3\) | B1 | \(2x^4\) or – 20x³ oe seen |
| \(\frac{dy}{dx} = 2x^4 - 20x^2 + 50\) | B1 | 2 marks; \(2x^4\) – 20x² + 50 (integers required) |
| Answer | Marks | Guidance |
|---|---|---|
| \(2x^4 - 20x^2 + 50 = 0\) | M1 | their \(\frac{dy}{dx} = 0\) seen (or implied by correct answer) |
| \(x^4 - 10x^2 + 25 = 0\) | A1 | 2 marks; which has 2 roots; 2 stationary points www in any part |
| which has 2 roots | A1 |
## (i)
Let $y = x^2$ | *M1 | Use a substitution to obtain a quadratic or $(x^2 - 5)(x^2 - 5) = 0$
$y^2 - 10y + 25 = 0$ | dep*M1 |
$(y-5)^2 = 0$ | A1 | Correct method to solve a quadratic; 5 (not $x = 5$ with no subsequent working)
$y = 5$ | A1 |
$x^2 = 5$ | A1 | 4 marks
$x = \pm\sqrt{5}$ | A1 |
## (ii)
$y = \frac{2x^3}{5} - \frac{20x^3}{3} + 50x + 3$ | B1 | $2x^4$ or – 20x³ oe seen
$\frac{dy}{dx} = 2x^4 - 20x^2 + 50$ | B1 | 2 marks; $2x^4$ – 20x² + 50 (integers required)
## (iii)
$2x^4 - 20x^2 + 50 = 0$ | M1 | their $\frac{dy}{dx} = 0$ seen (or implied by correct answer)
$x^4 - 10x^2 + 25 = 0$ | A1 | 2 marks; which has 2 roots; 2 stationary points www in any part
which has 2 roots | A1 |
---\begin{enumerate}[label=(\roman*)]
\item Solve the equation $x^4 - 10x^2 + 25 = 0$. [4]
\item Given that $y = \frac{2}{5}x^5 - \frac{20}{3}x^3 + 50x + 3$, find $\frac{dy}{dx}$. [2]
\item Hence find the number of stationary points on the curve $y = \frac{2}{5}x^5 - \frac{20}{3}x^3 + 50x + 3$. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR C1 2006 Q6 [8]}}