| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Definite integral with exponentials |
| Difficulty | Standard +0.3 This is a straightforward C3 integration question requiring students to find an intersection point by solving e^(2x) = 8e^(-x) (which simplifies nicely to e^(3x) = 8), then integrate two standard exponential functions and subtract. The algebra is clean with ln 2 appearing naturally, and the integration uses standard results for e^(kx). Slightly easier than average due to the guided structure and standard techniques, though the manipulation of exponentials and exact form answer adds minor challenge. |
| Spec | 1.06g Equations with exponentials: solve a^x = b1.08e Area between curve and x-axis: using definite integrals1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| Either State \(e^{2x} = 8e^{-x}\) and so \(e^{3x} = 8\) | B1 | |
| Obtain \(e^x = 2\) and hence \(x = \ln 2\) | B1 | AG; necessary detail needed |
| Or 1 State \(e^{2x} = 8e^{-x}\) and so \(e^{3x} = 8\) | B1 | |
| State \(3x = \ln 8\), \(x = \ln 8^{\frac{1}{3}}\) and hence \(x = \ln 2\) | B1 | AG; necessary detail needed |
| Or 2 State \(e^{2x} = 8e^{-x}\) and \(2x = \ln 8 - x\) | B1 | |
| State \(3x = \ln 8\), \(x = \ln 8^{\frac{1}{3}}\) and hence \(x = \ln 2\) | B1 | AG; necessary detail needed |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Integrate to obtain \(k_1 e^{-x}\) and \(k_2 e^{2x}\) | M1 | Any non-zero constants \(k_1\) and \(k_2\) |
| Obtain correct \(-8e^{-x} - \frac{1}{2}e^{2x}\) or, if done separately, \(-8e^{-x}\) and \(\frac{1}{2}e^{2x}\) | A1 | |
| Apply limits 0 and \(\ln 2\) correctly to their integral(s) | M1 | Condone one sign slip; earned by sight of \(-8e^{-\ln 2} - \frac{1}{2}e^{2\ln 2} + 8 + \frac{1}{2}\) (or equivs if integrals treated separately) |
| Obtain at least \(-4 - 2 + 8 + \frac{1}{2}\) (or equivs) | *A1 | Final A1 dependent on *A1 |
| Obtain \(\frac{5}{2}\) or equiv | A1 | |
| [5] |
## Part i
Either State $e^{2x} = 8e^{-x}$ and so $e^{3x} = 8$ | B1 |
Obtain $e^x = 2$ and hence $x = \ln 2$ | B1 | AG; necessary detail needed
Or 1 State $e^{2x} = 8e^{-x}$ and so $e^{3x} = 8$ | B1 |
State $3x = \ln 8$, $x = \ln 8^{\frac{1}{3}}$ and hence $x = \ln 2$ | B1 | AG; necessary detail needed | Going immediately from $x = \frac{1}{3}\ln 8$ to $x = \ln 2$ does not earn the second B1
Or 2 State $e^{2x} = 8e^{-x}$ and $2x = \ln 8 - x$ | B1 |
State $3x = \ln 8$, $x = \ln 8^{\frac{1}{3}}$ and hence $x = \ln 2$ | B1 | AG; necessary detail needed | Going immediately from $x = \frac{1}{3}\ln 8$ to $x = \ln 2$ does not earn the second B1
| [2] |
## Part ii
Integrate to obtain $k_1 e^{-x}$ and $k_2 e^{2x}$ | M1 | Any non-zero constants $k_1$ and $k_2$
Obtain correct $-8e^{-x} - \frac{1}{2}e^{2x}$ or, if done separately, $-8e^{-x}$ and $\frac{1}{2}e^{2x}$ | A1 |
Apply limits 0 and $\ln 2$ correctly to their integral(s) | M1 | Condone one sign slip; earned by sight of $-8e^{-\ln 2} - \frac{1}{2}e^{2\ln 2} + 8 + \frac{1}{2}$ (or equivs if integrals treated separately) | M1 also implied by sight only of $-4 - 2 + 8 + \frac{1}{2}$ (or equivs ...)
Obtain at least $-4 - 2 + 8 + \frac{1}{2}$ (or equivs) | *A1 | Final A1 dependent on *A1
Obtain $\frac{5}{2}$ or equiv | A1 |
| [5] |5\\
\includegraphics[max width=\textwidth, alt={}, center]{6d15cb4d-f540-488b-b94e-7a494f192ba5-2_469_721_1932_662}
The diagram shows the curves $y = \mathrm { e } ^ { 2 x }$ and $y = 8 \mathrm { e } ^ { - x }$. The shaded region is bounded by the curves and the $y$-axis. Without using a calculator,\\
(i) solve an appropriate equation to show that the curves intersect at a point for which $x = \ln 2$,\\
(ii) find the area of the shaded region, giving your answer in simplified form.
\hfill \mbox{\textit{OCR C3 2016 Q5 [7]}}