| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2011 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Show stationary point exists or gradient has specific property |
| Difficulty | Standard +0.3 This question involves routine differentiation of exponential functions and straightforward algebraic manipulation. Part (i)(a) requires basic chain rule application and factoring to show positivity. Part (i)(b) involves computing the second derivative and comparing expressions—mechanical but requiring care. Part (ii) requires finding a minimum using calculus (setting g'(x)=0) and evaluating, which is standard optimization. All techniques are core C3 material with no novel insight required, making it slightly easier than average. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.06a Exponential function: a^x and e^x graphs and properties1.07e Second derivative: as rate of change of gradient1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| (i)(a) Differentiate to obtain \(k_1 e^{2x} + k_2 e^{-2x}\) | M1 | any constants \(k_1\) and \(k_2\), but derivative must be different from \(f(x)\); condone presence of \(+ c\) |
| Obtain \(2e^{2x} + 6e^{-2x}\) | A1 | or unsimplified equiv; no \(+ c\) now |
| Refer to \(e^{2x} > 0\) and \(e^{-2x} > 0\) or to more general comment about exponential functions | A1 | 3 or equiv (which might be sketch of \(y = f(x)\) with comment that gradient is positive or might be sketch of \(y = f'(x)\) with comment that \(y > 0\); AG |
| (i)(b) Differentiate to obtain \(k_3 e^{2x} + k_4 e^{-2x}\) | M1 | any constants \(k_3\) and \(k_4\) but second derivative must be different from their first derivative; condone presence of \(+ c\) |
| Obtain \(4e^{2x} - 12e^{-2x}\) | A1 | or unsimplified equiv; no \(+ c\) now |
| Attempt solution of \(f''(x) > 0\) or of \(f(x) > 0\) or of corresponding eqn | M1 | at least as far as term involving \(e^{4x}\) or \(e^{-4x}\) |
| Obtain \(x > \frac{1}{4}\ln 3\) | A1 | |
| Confirm both give same result | B1 | 5 AG; necessary detail needed; either by solving the other or by observing that same inequality involved (just noting that \(f''(x) = 4f(x)\) is sufficient) |
| (ii) Differentiate to obtain \(2e^{2x} - 2ke^{-2x}\) | B1 | or unsimplified equiv |
| Attempt to find x-coordinate of stationary pt | M1 | equating to 0 and reaching \(e^{4x} = ...\) or equiv |
| Obtain \(e^{4x} = k\) and hence \(\frac{1}{4}\ln k\) or equiv | A1 | or equiv such as \(e^{2x} = \sqrt{k}\) |
| Substitute and attempt simplification | M1 | using valid processes but allow if only limited progress [note that question can be successfully concluded (without actually finding x) by substitution of \(e^{2x} = \sqrt{k}\) and \(e^{-2x} = \frac{1}{\sqrt{k}}\)] |
| Obtain \(g(x) \geq 2\sqrt{k}\) or \(y \geq 2\sqrt{k}\) | A1 | 5 or similarly simplified equiv with \(\geq\) not \(>\) |
**(i)(a)** Differentiate to obtain $k_1 e^{2x} + k_2 e^{-2x}$ | M1 | any constants $k_1$ and $k_2$, but derivative must be different from $f(x)$; condone presence of $+ c$
Obtain $2e^{2x} + 6e^{-2x}$ | A1 | or unsimplified equiv; no $+ c$ now
Refer to $e^{2x} > 0$ and $e^{-2x} > 0$ or to more general comment about exponential functions | A1 | 3 or equiv (which might be sketch of $y = f(x)$ with comment that gradient is positive or might be sketch of $y = f'(x)$ with comment that $y > 0$; AG
**(i)(b)** Differentiate to obtain $k_3 e^{2x} + k_4 e^{-2x}$ | M1 | any constants $k_3$ and $k_4$ but second derivative must be different from their first derivative; condone presence of $+ c$
Obtain $4e^{2x} - 12e^{-2x}$ | A1 | or unsimplified equiv; no $+ c$ now
Attempt solution of $f''(x) > 0$ or of $f(x) > 0$ or of corresponding eqn | M1 | at least as far as term involving $e^{4x}$ or $e^{-4x}$
Obtain $x > \frac{1}{4}\ln 3$ | A1 |
Confirm both give same result | B1 | 5 AG; necessary detail needed; either by solving the other or by observing that same inequality involved (just noting that $f''(x) = 4f(x)$ is sufficient)
**(ii)** Differentiate to obtain $2e^{2x} - 2ke^{-2x}$ | B1 | or unsimplified equiv
Attempt to find x-coordinate of stationary pt | M1 | equating to 0 and reaching $e^{4x} = ...$ or equiv
Obtain $e^{4x} = k$ and hence $\frac{1}{4}\ln k$ or equiv | A1 | or equiv such as $e^{2x} = \sqrt{k}$
Substitute and attempt simplification | M1 | using valid processes but allow if only limited progress [note that question can be successfully concluded (without actually finding x) by substitution of $e^{2x} = \sqrt{k}$ and $e^{-2x} = \frac{1}{\sqrt{k}}$]
Obtain $g(x) \geq 2\sqrt{k}$ or $y \geq 2\sqrt{k}$ | A1 | 5 or similarly simplified equiv with $\geq$ not $>$
---9 (i) The function f is defined for all real values of $x$ by
$$f ( x ) = e ^ { 2 x } - 3 e ^ { - 2 x } .$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { f } ^ { \prime } ( x ) > 0$ for all $x$.
\item Show that the set of values of $x$ for which $\mathrm { f } ^ { \prime \prime } ( x ) > 0$ is the same as the set of values of $x$ for which $\mathrm { f } ( x ) > 0$, and state what this set of values is.\\
(ii)\\
\includegraphics[max width=\textwidth, alt={}, center]{774bb427-5392-45d3-8e4e-47d08fb8a792-04_634_830_641_699}
The function g is defined for all real values of $x$ by
$$\mathrm { g } ( x ) = \mathrm { e } ^ { 2 x } + k \mathrm { e } ^ { - 2 x } ,$$
where $k$ is a constant greater than 1 . The graph of $y = \mathrm { g } ( x )$ is shown above. Find the range of g , giving your answer in simplified form.
\end{enumerate}
\hfill \mbox{\textit{OCR C3 2011 Q9 [13]}}