| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 12 |
| 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 involving standard differentiation of exponential functions (quotient/sum rules), locating stationary points by sign change, showing a tangent passes through a point, and applying a given iterative formula. All techniques are routine C3 material with no novel problem-solving required; the iterative formula is provided rather than derived. Slightly easier than average due to straightforward application of standard methods. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07m Tangents and normals: gradient and equations1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = -e^2 x^{-2} + e^x\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| SP: \(-e^2 x^{-2} + e^x = 0\); let \(f(x) = -e^2 x^{-2} + e^x\) | M1 | |
| \(f(1.3) = -0.70\), \(f(1.4) = 0.29\) | M1 | |
| sign change, \(f(x)\) continuous \(\therefore\) root | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = 2\), \(y = \frac{3}{2}e^2\), grad \(= \frac{3}{4}e^2\) | M1 | |
| \(\therefore y - \frac{3}{2}e^2 = \frac{3}{4}e^2(x-2)\) | M1 A1 | |
| \(y = \frac{3}{4}e^2 x\) | ||
| \(\therefore x = 0 \Rightarrow y = 0\) so passes through origin | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x_1 = -1.125589\), \(x_2 = -1.125803\), \(x_3 = -1.125804\) (7sf) | M1 A1 | |
| \(\therefore\) \(x\)-coordinate of \(B = -1.1258\) (5sf) | A1 | (12) |
# Question 8:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = -e^2 x^{-2} + e^x$ | M1 A1 | |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| SP: $-e^2 x^{-2} + e^x = 0$; let $f(x) = -e^2 x^{-2} + e^x$ | M1 | |
| $f(1.3) = -0.70$, $f(1.4) = 0.29$ | M1 | |
| sign change, $f(x)$ continuous $\therefore$ root | A1 | |
## Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 2$, $y = \frac{3}{2}e^2$, grad $= \frac{3}{4}e^2$ | M1 | |
| $\therefore y - \frac{3}{2}e^2 = \frac{3}{4}e^2(x-2)$ | M1 A1 | |
| $y = \frac{3}{4}e^2 x$ | | |
| $\therefore x = 0 \Rightarrow y = 0$ so passes through origin | A1 | |
## Part (iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x_1 = -1.125589$, $x_2 = -1.125803$, $x_3 = -1.125804$ (7sf) | M1 A1 | |
| $\therefore$ $x$-coordinate of $B = -1.1258$ (5sf) | A1 | **(12)** |\begin{enumerate}
\item A curve has the equation $y = \frac { \mathrm { e } ^ { 2 } } { x } + \mathrm { e } ^ { x } , x \neq 0$.\\
(i) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.\\[0pt]
(ii) Show that the curve has a stationary point in the interval [1.3,1.4].
\end{enumerate}
The point $A$ on the curve has $x$-coordinate 2 .\\
(iii) 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 $x _ { 0 } = - 1$.\\
(iv) Find $x _ { 1 } , x _ { 2 }$ and $x _ { 3 }$ to 7 significant figures and hence state the $x$-coordinate of $B$ to 5 significant figures.
\hfill \mbox{\textit{OCR C3 Q8 [12]}}