| Exam Board | OCR MEI |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | First-order integration |
| Difficulty | Moderate -0.8 This is a straightforward two-part question requiring basic proportionality setup and integration of a power function. Part (a) involves finding a constant of proportionality using given values, and part (b) requires integrating x^(1/2) and applying an initial condition—both routine A-level techniques with no problem-solving insight needed. |
| Spec | 1.02r Proportional relationships: and their graphs1.07b Gradient as rate of change: dy/dx notation1.08a Fundamental theorem of calculus: integration as reverse of differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = k\sqrt{x}\) | B1 (2.1) | may be implied by final answer |
| \(3 = k \times \sqrt{4}\) | M1 (1.1) | |
| \(k = \frac{3}{2}\) or \(\frac{dy}{dx} = \frac{3}{2}\sqrt{x}\) isw | A1 (2.2a) | if B0M0 allow SC1 for \(\frac{dy}{dx} = \frac{k}{\sqrt{x}}\) and \(k=6\) as final answer or \(\frac{dy}{dx} = \frac{6}{\sqrt{x}}\) as final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| their \(k \times \frac{x^{\frac{3}{2}}}{\frac{3}{2}}\) oe | B1 | FT their \(k\) |
| \(10 = (\sqrt{4})^3 + c\) | M1 | FT their integration, one term in \(x\) with index 1.5 |
| \(c = 2\) or \(y = x^{\frac{3}{2}} + 2\) or \(y = x\sqrt{x} + 2\) isw | A1 | must see '\(y=\)' at some point; if B0M0 allow SC2 for \(y = 12x^{\frac{1}{2}} - 14\) or \(y = 12\sqrt{x} - 14\) or \(y = 12x^{\frac{1}{2}} + c\) and \(c = -14\) |
## Question 11(a):
$\frac{dy}{dx} = k\sqrt{x}$ | B1 (2.1) | may be implied by final answer
$3 = k \times \sqrt{4}$ | M1 (1.1) |
$k = \frac{3}{2}$ or $\frac{dy}{dx} = \frac{3}{2}\sqrt{x}$ isw | A1 (2.2a) | if B0M0 allow SC1 for $\frac{dy}{dx} = \frac{k}{\sqrt{x}}$ and $k=6$ as final answer or $\frac{dy}{dx} = \frac{6}{\sqrt{x}}$ as final answer
## Question 11(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| their $k \times \frac{x^{\frac{3}{2}}}{\frac{3}{2}}$ oe | B1 | FT their $k$ |
| $10 = (\sqrt{4})^3 + c$ | M1 | FT their integration, one term in $x$ with index 1.5 |
| $c = 2$ or $y = x^{\frac{3}{2}} + 2$ or $y = x\sqrt{x} + 2$ isw | A1 | must see '$y=$' at some point; if **B0M0** allow **SC2** for $y = 12x^{\frac{1}{2}} - 14$ or $y = 12\sqrt{x} - 14$ or $y = 12x^{\frac{1}{2}} + c$ and $c = -14$ |
---11 In this question you must show detailed reasoning.\\
The variables $x$ and $y$ are such that $\frac { \mathrm { dy } } { \mathrm { dx } }$ is directly proportional to the square root of $x$.\\
When $x = 4 , \frac { d y } { d x } = 3$.
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { dy } } { \mathrm { dx } }$ in terms of $x$.
When $\mathrm { x } = 4 , \mathrm { y } = 10$.
\item Find $y$ in terms of $x$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 2 2023 Q11 [6]}}