| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Topic | Differential equations |
| Type | Separable variables - standard (requires integration by parts) |
| Difficulty | Standard +0.3 This is a straightforward separable variables question requiring standard integration techniques (polynomial × ln x, and exponential). Part (a) involves routine separation and integration with a boundary condition. Part (b) is algebraic manipulation to express the answer in a given form. While it requires careful execution, it follows standard A-level procedures without requiring problem-solving insight or novel approaches. |
| Spec | 1.07l Derivative of ln(x): and related functions1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\int e^{3y}\,dy = \int 3x^2\ln x\,dx\) | M1 (3.1a) | Separate variables and attempt integration of at least one side; allow \(ke^{3y}\) with \(k\neq 1\) as 'attempt' at LHS |
| \(\int e^{3y}\,dy = \frac{1}{3}e^{3y}\) | B1 (1.1) | Correct LHS; B0 if still part of expression involving \(x\) |
| \(\int 3x^2\ln x\,dx = x^3\ln x - \int x^2\,dx\) | M1 (3.1a) | Attempt integration by parts on RHS; must have correct parts |
| \(= x^3\ln x - \frac{1}{3}x^3 + c\) | A1 (1.1) | Correct RHS (condone no \(+c\)); condone no modulus sign on \(\ln x\) |
| \(\frac{1}{3}e^3 = e^3\ln e - \frac{1}{3}e^3 + c\) so \(c = -\frac{1}{3}e^3\) | M1 (1.1a) | Attempt use of \((e, 1)\) to find \(c\); M1 implied by sight of \(-\frac{1}{3}e^3\) or \(-6.695...\) |
| \(\frac{1}{3}e^{3y} = x^3\ln x - \frac{1}{3}x^3 - \frac{1}{3}e^3\) | A1 (1.1) | Obtain correct equation in required form; must be \(e^{3y} = ...\); A0 if decimal approximation for \(e^3\) |
| \(e^{3y} = 3x^3\ln x - x^3 - e^3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(e^{3y} = 3e^6\ln e^2 - e^6 - e^3 = 6e^6 - e^6 - e^3 = 5e^6 - e^3\) | M1* | 2.1 — Substitute \(x = e^2\) into their integral involving \(\ln x\), and attempt to simplify. \(\ln x\) may be \(\ln x^p\) if any coefficient has been taken into the \(\ln\) term. Must be working exactly, so M0 if decimals seen before \(\ln\) dealt with. |
| \(3y = \ln(e^3(5e^3 - 1)) = 3 + \ln(5e^3 - 1)\) | M1 dep* | 2.1 — Introduce logs correctly, and attempt to rearrange to given form. Their equation must have two terms, possibly more, with terms having a common factor of \(e^k\). Attempt must go as far as splitting into sum of two terms, with \(\ln e^k\) simplified to \(k\). |
| \(y = 1 + \frac{1}{3}\ln(5e^3 - 1)\) | A1 | 2.1 — Obtain \(y = 1 + \frac{1}{3}\ln(5e^3 - 1)\). No need to state \(a\), \(b\) and \(c\) explicitly. |
# Question 11(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\int e^{3y}\,dy = \int 3x^2\ln x\,dx$ | M1 (3.1a) | Separate variables and attempt integration of at least one side; allow $ke^{3y}$ with $k\neq 1$ as 'attempt' at LHS |
| $\int e^{3y}\,dy = \frac{1}{3}e^{3y}$ | B1 (1.1) | Correct LHS; B0 if still part of expression involving $x$ |
| $\int 3x^2\ln x\,dx = x^3\ln x - \int x^2\,dx$ | M1 (3.1a) | Attempt integration by parts on RHS; must have correct parts |
| $= x^3\ln x - \frac{1}{3}x^3 + c$ | A1 (1.1) | Correct RHS (condone no $+c$); condone no modulus sign on $\ln x$ |
| $\frac{1}{3}e^3 = e^3\ln e - \frac{1}{3}e^3 + c$ so $c = -\frac{1}{3}e^3$ | M1 (1.1a) | Attempt use of $(e, 1)$ to find $c$; M1 implied by sight of $-\frac{1}{3}e^3$ or $-6.695...$ |
| $\frac{1}{3}e^{3y} = x^3\ln x - \frac{1}{3}x^3 - \frac{1}{3}e^3$ | A1 (1.1) | Obtain correct equation in required form; must be $e^{3y} = ...$; A0 if decimal approximation for $e^3$ |
| $e^{3y} = 3x^3\ln x - x^3 - e^3$ | | |
## Question 11(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $e^{3y} = 3e^6\ln e^2 - e^6 - e^3 = 6e^6 - e^6 - e^3 = 5e^6 - e^3$ | M1* | 2.1 — Substitute $x = e^2$ into their integral involving $\ln x$, and attempt to simplify. $\ln x$ may be $\ln x^p$ if any coefficient has been taken into the $\ln$ term. Must be working exactly, so M0 if decimals seen before $\ln$ dealt with. |
| $3y = \ln(e^3(5e^3 - 1)) = 3 + \ln(5e^3 - 1)$ | M1 dep* | 2.1 — Introduce logs correctly, and attempt to rearrange to given form. Their equation must have two terms, possibly more, with terms having a common factor of $e^k$. Attempt must go as far as splitting into sum of two terms, with $\ln e^k$ simplified to $k$. |
| $y = 1 + \frac{1}{3}\ln(5e^3 - 1)$ | A1 | 2.1 — Obtain $y = 1 + \frac{1}{3}\ln(5e^3 - 1)$. No need to state $a$, $b$ and $c$ explicitly. |
---11 The gradient function of a curve is given by $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 3 x ^ { 2 } \ln x } { \mathrm { e } ^ { 3 y } }$.\\
The curve passes through the point (e, 1).
\begin{enumerate}[label=(\alph*)]
\item Find the equation of this curve, giving your answer in the form $\mathrm { e } ^ { 3 y } = \mathrm { f } ( x )$.
\item Show that, when $x = \mathrm { e } ^ { 2 }$, the $y$-coordinate of this curve can be written as $y = a + \frac { 1 } { 3 } \ln \left( b \mathrm { e } ^ { 3 } + c \right)$, where $a , b$ and $c$ are constants to be determined.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/01 2022 Q11 [9]}}