| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2014 |
| Session | June |
| Marks | 12 |
| 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 standard C3/C4 question on differentiating exponential functions and finding stationary points. Part (a) requires product rule and solving dy/dx=0, part (b) is straightforward equation solving with exponentials, and part (c) is a routine modulus graph sketch. All techniques are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.06a Exponential function: a^x and e^x graphs and properties1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = -3e^{a-3x} + 3e^{-x}\) | M1 A1 | M1: at least one term differentiated correctly; A1: correct differentiation of both terms |
| Sets \(\frac{dy}{dx} = 0\): \(-3e^{a-3x} + 3e^{-x} = 0 \Rightarrow e^{-x} = e^{a-3x} \Rightarrow -x = a-3x\) | M1 | Sets \(\frac{dy}{dx}=0\) and applies correct method to eliminate exponentials to reach \(x=\) |
| \(x = \frac{1}{2}a\) | A1 | After correct work |
| \(y_P = e^{a-3(\frac{a}{2})} - 3e^{-(\frac{a}{2})} = -2e^{-\frac{a}{2}}\) | ddM1; A1 | ddM1: needs both previous M marks; substitutes \(x\)-coordinate into \(y\) (not \(\frac{dy}{dx}\)); A1: \(y_P = -2e^{-\frac{a}{2}}\) as one term |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Method 1: \(0 = e^{a-3x} - 3e^{-x} \Rightarrow e^{a-"2"x} = 3 \Rightarrow a - "2"x = \ln 3\) | M1 | Put \(y=0\) and attempt to obtain \(e^{f(x)} = k\), e.g. \(e^{a \pm \lambda x} = 3\); all \(x\) terms on one side |
| Method 2: \(0 = e^{a-3x} - 3e^{-x} \Rightarrow e^{"2"x} = \frac{e^a}{3}\); \("2"x = a - \ln 3\) | dM1 | Depends on previous M; take logs correctly |
| Method 3: \(0 = e^{a-3x} - 3e^{-x} \Rightarrow 3e^{-2"x"} = e^a\); \(\ln 3 + "2"x = a\) | A1 | \(x_Q = \frac{a - \ln 3}{2}\) (must be exact) |
| \(\Rightarrow x = \frac{a - \ln 3}{2}\) or equivalent e.g. \(\frac{1}{2}\ln\!\left(\frac{e^a}{3}\right)\) or \(-\ln\!\sqrt{\left(\frac{3}{e^a}\right)}\) | ||
| Method 4: \(e^{a-3x} = 3e^{-x}\) so \(a - 3x = \ln 3 - x\); \("2"x = a - \ln 3\); \(x = \frac{a-\ln 3}{2}\) | M1, dM1, A1 | M1: puts \(e^{a-3x} = 3e^{-x}\) then takes lns; dM1: collects \(x\) terms; A1: \(x_Q = \frac{a-\ln 3}{2}\) cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct overall shape: \(y \geq 0\) for all \(x\), curve crossing positive \(y\)-axis, small portion to left of \(y\)-axis, meets \(x\)-axis once, one maximum turning point | B1 | |
| Cusp at \(x = x_Q\) (not zero gradient) and no appearance of curve clearly increasing as \(x\) becomes large | B1 | |
| \((0, e^a - 3)\) labelled | B1 | Either writes full coordinates \((0, e^a-3)\) in text or on graph, or \(e^a - 3\) marked on \(y\)-axis; allow \( |
# Question 11:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = -3e^{a-3x} + 3e^{-x}$ | M1 A1 | M1: at least one term differentiated correctly; A1: correct differentiation of both terms |
| Sets $\frac{dy}{dx} = 0$: $-3e^{a-3x} + 3e^{-x} = 0 \Rightarrow e^{-x} = e^{a-3x} \Rightarrow -x = a-3x$ | M1 | Sets $\frac{dy}{dx}=0$ and applies correct method to eliminate exponentials to reach $x=$ |
| $x = \frac{1}{2}a$ | A1 | After correct work |
| $y_P = e^{a-3(\frac{a}{2})} - 3e^{-(\frac{a}{2})} = -2e^{-\frac{a}{2}}$ | ddM1; A1 | ddM1: needs both previous M marks; substitutes $x$-coordinate into $y$ (not $\frac{dy}{dx}$); A1: $y_P = -2e^{-\frac{a}{2}}$ as one term |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| **Method 1:** $0 = e^{a-3x} - 3e^{-x} \Rightarrow e^{a-"2"x} = 3 \Rightarrow a - "2"x = \ln 3$ | M1 | Put $y=0$ and attempt to obtain $e^{f(x)} = k$, e.g. $e^{a \pm \lambda x} = 3$; all $x$ terms on one side |
| **Method 2:** $0 = e^{a-3x} - 3e^{-x} \Rightarrow e^{"2"x} = \frac{e^a}{3}$; $"2"x = a - \ln 3$ | dM1 | Depends on previous M; take logs correctly |
| **Method 3:** $0 = e^{a-3x} - 3e^{-x} \Rightarrow 3e^{-2"x"} = e^a$; $\ln 3 + "2"x = a$ | A1 | $x_Q = \frac{a - \ln 3}{2}$ (must be exact) |
| $\Rightarrow x = \frac{a - \ln 3}{2}$ or equivalent e.g. $\frac{1}{2}\ln\!\left(\frac{e^a}{3}\right)$ or $-\ln\!\sqrt{\left(\frac{3}{e^a}\right)}$ | | |
| **Method 4:** $e^{a-3x} = 3e^{-x}$ so $a - 3x = \ln 3 - x$; $"2"x = a - \ln 3$; $x = \frac{a-\ln 3}{2}$ | M1, dM1, A1 | M1: puts $e^{a-3x} = 3e^{-x}$ then takes lns; dM1: collects $x$ terms; A1: $x_Q = \frac{a-\ln 3}{2}$ cao |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct overall shape: $y \geq 0$ for all $x$, curve crossing positive $y$-axis, small portion to left of $y$-axis, meets $x$-axis once, one maximum turning point | B1 | |
| Cusp at $x = x_Q$ (not zero gradient) and no appearance of curve clearly increasing as $x$ becomes large | B1 | |
| $(0, e^a - 3)$ labelled | B1 | Either writes full coordinates $(0, e^a-3)$ in text or on graph, or $e^a - 3$ marked on $y$-axis; allow $|e^a - 3|$; can be earned without the graph |
---11.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{423eb549-0873-4185-8faf-12dedafcd108-17_600_1024_221_470}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of part of the curve $C$ with equation
$$y = \mathrm { e } ^ { a - 3 x } - 3 \mathrm { e } ^ { - x } , \quad x \in \mathbb { R }$$
where $a$ is a constant and $a > \ln 4$
The curve $C$ has a turning point $P$ and crosses the $x$-axis at the point $Q$ as shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $a$, the coordinates of the point $P$.
\item Find, in terms of $a$, the $x$ coordinate of the point $Q$.
\item Sketch the curve with equation
$$y = \left| \mathrm { e } ^ { a - 3 x } - 3 \mathrm { e } ^ { - x } \right| , \quad x \in \mathbb { R } , \quad a > \ln 4$$
Show on your sketch the exact coordinates, in terms of $a$, of the points at which the curve meets or cuts the coordinate axes.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2014 Q11 [12]}}