| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Differentiate exponential functions |
| Difficulty | Moderate -0.8 This question tests routine differentiation of exponential functions and standard applications. Part (a) is direct recall of a standard derivative, part (b) requires finding a tangent equation and verifying a point lies on it (straightforward substitution), and part (c) involves solving a simple exponential equation. All parts are textbook exercises requiring no problem-solving insight, making this easier than average. |
| Spec | 1.06g Equations with exponentials: solve a^x = b1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| \(ke^{kx}\) | B1 [1] | Ignore "\(y=\)" if seen |
| Answer | Marks | Guidance |
|---|---|---|
| Gradient of tangent is \(\frac{1}{2}e^{\frac{1}{2}\times2}\), ignore eg "\(y=\)" | M1 | Subst \(k=\frac{1}{2}\), \(x=2\) into their (a); Allow decimals throughout; May be implied by \(1.36\) or \(1.35\) ft (a) |
| \(= \frac{1}{2}e\) | A1f | ft (a); ft their gradient so long as it involves \(e\) (possibly implied by decimal) |
| Equation of tangent is \(y-e=\frac{1}{2}e(x-2)\) or \(y=\frac{1}{2}ex+c\) AND sub \((2,e)\) | M1 | M1 for attempt find equation of tangent by correct method |
| \(y=\frac{1}{2}ex\) (cao), Passes through \((0,0)\) or \(x=0\Rightarrow y=0\) or \(y\)-int is \(0\) oe | A1 [4] | A1 for obtain correct equation and state or show or verify that it passes through \((0,0)\). No ft. |
| Answer | Marks | Guidance |
|---|---|---|
| \(3e^x = 1-2e^{\frac{1}{2}x}\) | M1 | oe |
| \(3(e^{\frac{1}{2}x})^2 + 2e^{\frac{1}{2}x} - 1 = 0\) or eg \(3u^2+2u-1=0\) or \(3y+2\sqrt{y}-1=0\) | M1 | Attempt write quadratic equation in \(e^{\frac{1}{2}x}\) or attempt substitution and form QE; Allow one sign error; Allow substitute \(x\) or \(y=e^{\frac{1}{2}x}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(e^{\frac{1}{2}x}=-1\): no solutions stated | B1 | eg "Cannot be negative" or "Impossible" or just "X" |
| \(e^{\frac{1}{2}x}=\frac{1}{3}\) oe | A1 | or \(\frac{1}{2}x=\ln\frac{1}{3}\) or \(\frac{1}{2}x=-1.10\); not just eg \(u=\frac{1}{3}\) |
| \(x=2\ln\frac{1}{3}\) or \(\ln\frac{1}{9}\) or \(-2\ln3\) or \(-\ln9\) or \(-2.20\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Point of intersection is \(\left(\ln\frac{1}{9}, \frac{1}{3}\right)\) | A1 [6] | or equivalent exact or \((-2.20, 0.333)\) (3 sf); If any extra points of intersection: A0; Correct ans with no working or irrelevant working: SC B2 |
## Question 7:
### Part (a):
$ke^{kx}$ | **B1** [1] | Ignore "$y=$" if seen
### Part (b):
Gradient of tangent is $\frac{1}{2}e^{\frac{1}{2}\times2}$, ignore eg "$y=$" | **M1** | Subst $k=\frac{1}{2}$, $x=2$ into their (a); Allow decimals throughout; May be implied by $1.36$ or $1.35$ ft (a)
$= \frac{1}{2}e$ | **A1f** | ft (a); ft their gradient so long as it involves $e$ (possibly implied by decimal)
Equation of tangent is $y-e=\frac{1}{2}e(x-2)$ or $y=\frac{1}{2}ex+c$ AND sub $(2,e)$ | **M1** | M1 for attempt find equation of tangent by correct method
$y=\frac{1}{2}ex$ (cao), Passes through $(0,0)$ or $x=0\Rightarrow y=0$ or $y$-int is $0$ oe | **A1** [4] | A1 for obtain **correct** equation and state or show or verify that it passes through $(0,0)$. **No ft.**
### Part (c):
$3e^x = 1-2e^{\frac{1}{2}x}$ | **M1** | oe
$3(e^{\frac{1}{2}x})^2 + 2e^{\frac{1}{2}x} - 1 = 0$ or eg $3u^2+2u-1=0$ or $3y+2\sqrt{y}-1=0$ | **M1** | Attempt write quadratic equation in $e^{\frac{1}{2}x}$ or attempt substitution and form QE; Allow one sign error; Allow substitute $x$ or $y=e^{\frac{1}{2}x}$
$((3e^{\frac{1}{2}x}-1)(e^{\frac{1}{2}x}+1)=0)$
$e^{\frac{1}{2}x}=-1$: no solutions stated | **B1** | eg "Cannot be negative" or "Impossible" or just "X"
$e^{\frac{1}{2}x}=\frac{1}{3}$ oe | **A1** | or $\frac{1}{2}x=\ln\frac{1}{3}$ or $\frac{1}{2}x=-1.10$; not just eg $u=\frac{1}{3}$
$x=2\ln\frac{1}{3}$ or $\ln\frac{1}{9}$ or $-2\ln3$ or $-\ln9$ or $-2.20$ | **A1** |
$y=3\exp(\ln\frac{1}{9})=\frac{1}{3}$ or $0.333$
Point of intersection is $\left(\ln\frac{1}{9}, \frac{1}{3}\right)$ | **A1** [6] | or equivalent exact or $(-2.20, 0.333)$ (3 sf); If any extra points of intersection: A0; Correct ans with no working or irrelevant working: SC B2
---7
\begin{enumerate}[label=(\alph*)]
\item Write down an expression for the gradient of the curve $y = \mathrm { e } ^ { k x }$.
\item The line L is a tangent to the curve $y = \mathrm { e } ^ { \frac { 1 } { 2 } x }$ at the point where $x = 2$.
Show that L passes through the point $( 0,0 )$.
\item Find the coordinates of the point of intersection of the curves $y = 3 \mathrm { e } ^ { x }$ and $y = 1 - 2 \mathrm { e } ^ { \frac { 1 } { 2 } x }$.
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q7 [11]}}