| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Solve equation involving derivatives |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on exponential differentiation and integration. Part (a) requires simple algebraic manipulation (substituting and multiplying through), part (b) involves solving a quadratic in e^x (standard technique), and part (c) requires basic integration with boundary conditions. All steps are routine A-level techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.06b Gradient of e^(kx): derivative and exponential model1.07j Differentiate exponentials: e^(kx) and a^(kx)1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = e^x - 4e^{-x} = 3\) | M1 | Equate to 3 |
| \((e^x)^2 - 3e^x - 4 = 0\) | E1 | AG Rearrange to quadratic in \(e^x\). Also allow \(e^{2x} - 3e^x - 4 = 0\). Expression \(= 0\) must be seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Solve to give \(e^x = [-1], 4\) | M1 | May be BC giving at least one root of the quadratic equation (a) |
| When \(e^x = 4\), \(x = \ln 4\) | A1 | \(x = \ln 4\) must be seen explicitly |
| When \(e^x = -1\) there are no real values of \(x\), so no other points on the curve | E1 | Must explain why they reject the value \(-1\) for \(e^x\), or state \(e^x + 1\) is never zero |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Equation \(y = \int(e^x - 4e^{-x})\, dx\); \([y =] e^x + 4e^{-x} [+c]\) | B1 | Condone missing \(+ c\) in their integral |
| When \(x = 0\), \(y = 0 = 1 + 4 + c\) | M1 | Attempt to evaluate \(c\) |
| So \(c = -5\); \([y = e^x + 4e^{-x} - 5]\) | A1 | |
| When \(x = 1\), \(y = e^1 + 4e^{-1} - 5\) | M1 | Substituting \(x = 1\) into their expression |
| \(y = -0.810 < 0\) so below the \(x\)-axis | E1 | AG must argue below the axis from correct \(y\) value. Must be clear that \(-0.81\) is a \(y\)-coordinate |
## Question 9(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = e^x - 4e^{-x} = 3$ | M1 | Equate to 3 |
| $(e^x)^2 - 3e^x - 4 = 0$ | E1 | AG Rearrange to quadratic in $e^x$. Also allow $e^{2x} - 3e^x - 4 = 0$. Expression $= 0$ must be seen |
---
## Question 9(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Solve to give $e^x = [-1], 4$ | M1 | May be BC giving at least one root of the quadratic equation (a) |
| When $e^x = 4$, $x = \ln 4$ | A1 | $x = \ln 4$ must be seen explicitly |
| When $e^x = -1$ there are no real values of $x$, so no other points on the curve | E1 | Must explain why they reject the value $-1$ for $e^x$, or state $e^x + 1$ is never zero |
---
## Question 9(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Equation $y = \int(e^x - 4e^{-x})\, dx$; $[y =] e^x + 4e^{-x} [+c]$ | B1 | Condone missing $+ c$ in their integral |
| When $x = 0$, $y = 0 = 1 + 4 + c$ | M1 | Attempt to evaluate $c$ |
| So $c = -5$; $[y = e^x + 4e^{-x} - 5]$ | A1 | |
| When $x = 1$, $y = e^1 + 4e^{-1} - 5$ | M1 | Substituting $x = 1$ into their expression |
| $y = -0.810 < 0$ so below the $x$-axis | E1 | AG must argue below the axis from correct $y$ value. Must be clear that $-0.81$ is a $y$-coordinate |
---9 The gradient of a curve is given by $\frac { d y } { d x } = e ^ { x } - 4 e ^ { - x }$.
\begin{enumerate}[label=(\alph*)]
\item Show that the $x$-coordinate of any point on the curve at which the gradient is 3 satisfies the equation $\left( e ^ { x } \right) ^ { 2 } - 3 e ^ { x } - 4 = 0$.
\item Hence show that there is only one point on the curve at which the gradient is 3 , stating the exact value of its $x$-coordinate.
\item The curve passes through the point $( 0,0 )$.
Show that when $x = 1$ the curve is below the $x$-axis.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2023 Q9 [10]}}