| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Second derivative test justification |
| Difficulty | Moderate -0.3 This is a standard C2 stationary points question with routine differentiation of x^2 + 16/x, solving dy/dx = 0, and applying the second derivative test. Part (iv) adds trapezium rule which is also standard bookwork. All parts follow predictable procedures with no problem-solving insight required, making it slightly easier than average but not trivial due to the multi-part nature and algebraic manipulation needed. |
| Spec | 1.07d Second derivatives: d^2y/dx^2 notation1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{dy}{dx} = 2x - \frac{16}{x^2}\) | M1 A1 | 2 marks |
| (ii) \(\frac{dy}{dx} = 0\) when \(2x - \frac{16}{x^2} = 0 \Rightarrow x^3 = 8\) \(\Rightarrow x = 2, y = 12\) | M1 A1 | 2 marks |
| (iii) \(\frac{d^2y}{dx^2} = 2 + \frac{32}{x^3}\) When \(x = 2\), \(\frac{d^2y}{dx^2} > 0\) i.e. minimum. | M1 A1 E1 | 3 marks |
| (iv) Values of \(y\) are: 12, 12.65, 14.33, 16.82, 20 \(A = \frac{1}{2} \times 0.5 \times (12 + 2(12.65 + 14.33 + 16.82) + 20) = 29.9\) | B1 M1 A1 | 3 marks |
| (v) Over-estimates because the curve is concave. | B1 B1 | 2 marks |
**(i)** $\frac{dy}{dx} = 2x - \frac{16}{x^2}$ | M1 A1 | 2 marks
**(ii)** $\frac{dy}{dx} = 0$ when $2x - \frac{16}{x^2} = 0 \Rightarrow x^3 = 8$ $\Rightarrow x = 2, y = 12$ | M1 A1 | 2 marks
**(iii)** $\frac{d^2y}{dx^2} = 2 + \frac{32}{x^3}$ When $x = 2$, $\frac{d^2y}{dx^2} > 0$ i.e. minimum. | M1 A1 E1 | 3 marks
**(iv)** Values of $y$ are: 12, 12.65, 14.33, 16.82, 20 $A = \frac{1}{2} \times 0.5 \times (12 + 2(12.65 + 14.33 + 16.82) + 20) = 29.9$ | B1 M1 A1 | 3 marks
**(v)** Over-estimates because the curve is concave. | B1 B1 | 2 marks10 Fig. 10 shows the curve with equation $y = x ^ { 2 } + \frac { 16 } { x }$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c55a5f04-3573-4f36-a12c-3755bdd4a45b-5_522_1019_403_394}
\captionsetup{labelformat=empty}
\caption{Fig. 10}
\end{center}
\end{figure}
(i) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.\\
(ii) Hence calculate the coordinates of the stationary point on the curve.\\
(iii) Find $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ and explain why this confirms that he stationary point is a minimum.\\
(iv) Using the trapezium rule with 4 intervals, estimate the area between the curve and the $x$ axis between $x = 2$ and $x = 4$.\\
(v) State, giving a reason, whether this estimate of the area under-estimates or over-estimates the true area beneath the curve.
\hfill \mbox{\textit{OCR MEI C2 Q10 [12]}}