| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| 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 is a multi-part question requiring standard differentiation (quotient rule and exponential), sign-change verification for a stationary point, showing a tangent passes through the origin (routine substitution), and applying an iterative formula. All techniques are standard C3 material with no novel insight required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.09b Sign change methods: understand failure cases1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks |
|---|---|
| (a) \(\frac{dy}{dx} = -e^x x^2 + e^x\) | M1 A1 |
| Answer | Marks |
|---|---|
| Sign change, \(f(x)\) continuous \(\therefore\) root | M1 M1 A1 |
| (c) \(x = 2, y = \frac{3}{4}e^2\), grad \(= \frac{3}{4}e^2\) | M1 |
| Answer | Marks |
|---|---|
| \(y = \frac{3}{4}e^2 x\) | M1 A1 |
| \(\therefore x = 0 \Rightarrow y = 0\) so passes through origin | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\therefore\) x-coordinate of \(B = -1.1258\) (5sf) | M1 A2 A1 | (13) |
**(a)** $\frac{dy}{dx} = -e^x x^2 + e^x$ | M1 A1
**(b)** SP: $-e^x x^2 + e^x = 0$
Let $f(x) = -e^x x^2 + e^x$
$f(1.3) = -0.70$, $f(1.4) = 0.29$
Sign change, $f(x)$ continuous $\therefore$ root | M1 M1 A1
**(c)** $x = 2, y = \frac{3}{4}e^2$, grad $= \frac{3}{4}e^2$ | M1
$\therefore y - \frac{3}{4}e^2 = \frac{3}{4}e^2(x-2)$
$y = \frac{3}{4}e^2 x$ | M1 A1
$\therefore x = 0 \Rightarrow y = 0$ so passes through origin | A1
**(d)** $x_1 = -1.125589, x_2 = -1.125803, x_3 = -1.125804$ (7sf)
$\therefore$ x-coordinate of $B = -1.1258$ (5sf) | M1 A2 A1 | **(13)**
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**Total (75)**8. A curve has the equation $y = \frac { \mathrm { e } ^ { 2 } } { x } + \mathrm { e } ^ { x } , \quad x \neq 0$.
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.\\[0pt]
\item Show that the curve has a stationary point in the interval [1.3,1.4].
The point $A$ on the curve has $x$-coordinate 2 .
\item Show that the tangent to the curve at $A$ passes through the origin.
The tangent to the curve at $A$ intersects the curve again at the point $B$.\\
The $x$-coordinate of $B$ is to be estimated using the iterative formula
$$x _ { n + 1 } = - \frac { 2 } { 3 } \sqrt { 3 + 3 x _ { n } \mathrm { e } ^ { x _ { n } - 2 } }$$
with $x _ { 0 } = - 1$.
\item Find $x _ { 1 } , x _ { 2 }$ and $x _ { 3 }$ to 7 significant figures and hence state the $x$-coordinate of $B$ to 5 significant figures.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 Q8 [13]}}