OCR C3 2013 June — Question 9 11 marks

Exam BoardOCR
ModuleC3 (Core Mathematics 3)
Year2013
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAreas Between Curves
TypeArea with Exponential or Logarithmic Curves
DifficultyStandard +0.8 This question requires finding a minimum by differentiation (standard), but then computing the area between a curve and a slant line requires setting up a definite integral with the difference of functions. The exponential integration and exact answer manipulation (involving e^(2ln3) = 9) elevates this above routine area questions, requiring careful algebraic work across multiple steps.
Spec1.06a 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 integrals1.08f Area between two curves: using integration

9 \includegraphics[max width=\textwidth, alt={}, center]{71e01d8f-d0ed-4f17-b7cd-6f5a93bbe329-4_661_915_269_557} The diagram shows the curve $$y = \mathrm { e } ^ { 2 x } - 18 x + 15 .$$ The curve crosses the \(y\)-axis at \(P\) and the minimum point is \(Q\). The shaded region is bounded by the curve and the line \(P Q\).
  1. Show that the \(x\)-coordinate of \(Q\) is \(\ln 3\).
  2. Find the exact area of the shaded region.

Question 9(i):
AnswerMarks Guidance
AnswerMark Guidance
Differentiate to obtain \(2e^{2x} - 18\)B1
Equate first derivative to zero and use legitimate method to reach equation without e involvedM1
Confirm \(x = \ln 3\)A1 AG; necessary detail needed (in particular, for solutions concluding \(x = \frac{1}{2}\ln 9 = \ln 3\) or equiv — award A0)
[3]
Question 9(ii):
AnswerMarks Guidance
AnswerMark Guidance
Attempt integration*M1 confirmed by at least one correct term
Obtain \(\frac{1}{2}e^{2x} - 9x^2 + 15x\)A1 or equiv
Apply limits 0 and \(\ln 3\) to obtain exact unsimplified expressionM1 dep *M
Obtain \(4 - 9(\ln 3)^2 + 15\ln 3\)A1 or exact (maybe unsimplified) equiv perhaps still involving e
Attempt area of trapezium or equiv, retaining exactness throughoutM1 using \(\frac{1}{2}\ln 3 \times (y_1 + y_2)\) where \(y_1\) is 15 or 16 and \(y_2\) is attempt at \(y\)-coordinate of \(Q\); if using alternative approach involving rectangle and triangle, complete attempt needs to be seen for M1; another alternative approach involves equation of \(PQ\) (\(y = \frac{8-18\ln 3}{\ln 3}\,x + 16\)) with integration: M1 for attempting equation and integration, A1 for correct answer
Obtain \(\frac{1}{2}\ln 3 \times (16 + 24 - 18\ln 3)\)A1 or equiv perhaps still including e
Subtract areas the right way round, retaining exactnessM1 dep on award of all three M marks
Obtain \(5\ln 3 - 4\)A1 or similarly simplified exact equiv
[8]
## Question 9(i):

| Answer | Mark | Guidance |
|--------|------|----------|
| Differentiate to obtain $2e^{2x} - 18$ | B1 | |
| Equate first derivative to zero and use legitimate method to reach equation without e involved | M1 | |
| Confirm $x = \ln 3$ | A1 | AG; necessary detail needed (in particular, for solutions concluding $x = \frac{1}{2}\ln 9 = \ln 3$ or equiv — award A0) |
| **[3]** | | |

---

## Question 9(ii):

| Answer | Mark | Guidance |
|--------|------|----------|
| Attempt integration | *M1 | confirmed by at least one correct term |
| Obtain $\frac{1}{2}e^{2x} - 9x^2 + 15x$ | A1 | or equiv |
| Apply limits 0 and $\ln 3$ to obtain exact unsimplified expression | M1 | dep *M |
| Obtain $4 - 9(\ln 3)^2 + 15\ln 3$ | A1 | or exact (maybe unsimplified) equiv perhaps still involving e |
| Attempt area of trapezium or equiv, retaining exactness throughout | M1 | using $\frac{1}{2}\ln 3 \times (y_1 + y_2)$ where $y_1$ is 15 or 16 and $y_2$ is attempt at $y$-coordinate of $Q$; if using alternative approach involving rectangle and triangle, complete attempt needs to be seen for M1; another alternative approach involves equation of $PQ$ ($y = \frac{8-18\ln 3}{\ln 3}\,x + 16$) with integration: M1 for attempting equation and integration, A1 for correct answer |
| Obtain $\frac{1}{2}\ln 3 \times (16 + 24 - 18\ln 3)$ | A1 | or equiv perhaps still including e |
| Subtract areas the right way round, retaining exactness | M1 | dep on award of all three M marks |
| Obtain $5\ln 3 - 4$ | A1 | or similarly simplified exact equiv |
| **[8]** | | |
9\\
\includegraphics[max width=\textwidth, alt={}, center]{71e01d8f-d0ed-4f17-b7cd-6f5a93bbe329-4_661_915_269_557}

The diagram shows the curve

$$y = \mathrm { e } ^ { 2 x } - 18 x + 15 .$$

The curve crosses the $y$-axis at $P$ and the minimum point is $Q$. The shaded region is bounded by the curve and the line $P Q$.\\
(i) Show that the $x$-coordinate of $Q$ is $\ln 3$.\\
(ii) Find the exact area of the shaded region.

\hfill \mbox{\textit{OCR C3 2013 Q9 [11]}}