| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Find stationary points - polynomial/exponential products |
| Difficulty | Standard +0.3 This is a straightforward C3 question requiring product rule differentiation of exponentials, solving a simple exponential equation for stationary points, and integration of exponentials. All techniques are standard with no novel insight required, making it slightly easier than average. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.06b Gradient of e^(kx): derivative and exponential model1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State first derivative is \(3e^{3x}-12e^{2x}\) | B1 | Or equiv |
| Equate first derivative to zero and attempt solution of equation of form \(k_1e^{3x}-k_2e^{2x}=0\) | M1 | At least as far as \(e^x=c\); M0 for false method such as \(\ln(3e^{3x})-\ln(12e^{2x})=0\) |
| Obtain \(\ln 4\) or exact equiv and no other | A1 | Obtained by legitimate method |
| Substitute \(x=\ln 4\) or \(e^x=4\) to confirm \(y=0\) | A1 | AG; using exact working with all detail present: needs sight of \(4^3-6\times4^2+32\) or similar equiv |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Integrate to obtain \(k_3e^{3x}+k_4e^{2x}+32x\) | M1 | For non-zero constants |
| Obtain \(\frac{1}{3}e^{3x}-3e^{2x}+32x\) or equiv | A1 | |
| Apply limits correctly to expression of form \(k_3e^{3x}+k_4e^{2x}+32x\) | M1 | Using limits 0 and their answer from part (i) |
| Simplify to obtain \(32\ln 4-24\) or \(64\ln 2-24\) | A1 | Or suitably simplified equiv |
| [4] |
# Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State first derivative is $3e^{3x}-12e^{2x}$ | B1 | Or equiv |
| Equate first derivative to zero and attempt solution of equation of form $k_1e^{3x}-k_2e^{2x}=0$ | M1 | At least as far as $e^x=c$; M0 for false method such as $\ln(3e^{3x})-\ln(12e^{2x})=0$ |
| Obtain $\ln 4$ or exact equiv and no other | A1 | Obtained by legitimate method |
| Substitute $x=\ln 4$ or $e^x=4$ to confirm $y=0$ | A1 | AG; using exact working with all detail present: needs sight of $4^3-6\times4^2+32$ or similar equiv |
| **[4]** | | |
---
# Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Integrate to obtain $k_3e^{3x}+k_4e^{2x}+32x$ | M1 | For non-zero constants |
| Obtain $\frac{1}{3}e^{3x}-3e^{2x}+32x$ or equiv | A1 | |
| Apply limits correctly to expression of form $k_3e^{3x}+k_4e^{2x}+32x$ | M1 | Using limits 0 and their answer from part (i) |
| Simplify to obtain $32\ln 4-24$ or $64\ln 2-24$ | A1 | Or suitably simplified equiv |
| **[4]** | | |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{00a4be37-c095-4d9c-a1cd-d03b8ab1d411-2_455_643_1327_694}
The diagram shows the curve $y = \mathrm { e } ^ { 3 x } - 6 \mathrm { e } ^ { 2 x } + 32$.\\
(i) Find the exact $x$-coordinate of the minimum point and verify that the $y$-coordinate of the minimum point is 0 .\\
(ii) Find the exact area of the region (shaded in the diagram) enclosed by the curve and the axes.
\hfill \mbox{\textit{OCR C3 2015 Q5 [8]}}