| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2014 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find tangent equation at point |
| Difficulty | Standard +0.3 Part (a) is a standard proof of exponential differentiation using logarithms (routine A-level technique). Part (b) requires implicit differentiation of a moderately complex equation and substituting a point to find the tangent—this is a standard C3/C4 exercise with multiple terms but no novel insight required. Slightly above average due to the algebraic complexity of the implicit differentiation. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.06b Gradient of e^(kx): derivative and exponential model1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Set \(y = 2^x\), take \(\ln\) of both sides to get \(\ln y = x \ln 2\) | M1 | |
| Differentiate wrt \(x\): \(\frac{1}{y}\frac{dy}{dx} = \ln 2 \Rightarrow \frac{dy}{dx} = \ldots\) | dM1 | |
| \(\frac{dy}{dx} = 2^x \ln 2\) | A1* | cao; given answer — all aspects of proof must be present |
| *Alt 1:* Write \(2^x = e^{x\ln 2}\), differentiate to get \(\frac{d}{dx}(e^{x\ln 2}) = e^{x\ln 2}\ln 2 = 2^x \ln 2\) | M1, dM1 A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Differentiate wrt \(x\): \(2 + 6y\frac{dy}{dx} + 6xy + 3x^2\frac{dy}{dx} = 4 \times 2^x \ln 2\) | M1, B1, A1 | M1 for product rule on \(3x^2y\); B1 for \(2x + 3y^2 \rightarrow 2 + 6y\frac{dy}{dx}\); A1 for completely correct differential |
| Substitute \((2, 0)\) and rearrange: \(2 + 12\frac{dy}{dx} = 16\ln 2 \Rightarrow \frac{dy}{dx} = \frac{16\ln 2 - 2}{12}\) \((= 0.758)\) | M1 | Substitutes \(x=2\), \(y=0\) and rearranges to numerical \(\frac{dy}{dx}\) |
| Use \((2, 0)\) and numerical \(\frac{dy}{dx}\) to find tangent equation | dM1 | Accept \(\frac{y-0}{x-2} = \) numerical \(\frac{dy}{dx}\) |
| \(y = \frac{(16\ln 2 - 2)(x-2)}{12}\) | A1 | oe; accept \(y = 0.76x - 1.52\) (awrt 2dp) |
# Question 5:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Set $y = 2^x$, take $\ln$ of both sides to get $\ln y = x \ln 2$ | M1 | |
| Differentiate wrt $x$: $\frac{1}{y}\frac{dy}{dx} = \ln 2 \Rightarrow \frac{dy}{dx} = \ldots$ | dM1 | |
| $\frac{dy}{dx} = 2^x \ln 2$ | A1* | cao; given answer — all aspects of proof must be present |
*Alt 1:* Write $2^x = e^{x\ln 2}$, differentiate to get $\frac{d}{dx}(e^{x\ln 2}) = e^{x\ln 2}\ln 2 = 2^x \ln 2$ | M1, dM1 A1* |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Differentiate wrt $x$: $2 + 6y\frac{dy}{dx} + 6xy + 3x^2\frac{dy}{dx} = 4 \times 2^x \ln 2$ | M1, B1, A1 | M1 for product rule on $3x^2y$; B1 for $2x + 3y^2 \rightarrow 2 + 6y\frac{dy}{dx}$; A1 for completely correct differential |
| Substitute $(2, 0)$ and rearrange: $2 + 12\frac{dy}{dx} = 16\ln 2 \Rightarrow \frac{dy}{dx} = \frac{16\ln 2 - 2}{12}$ $(= 0.758)$ | M1 | Substitutes $x=2$, $y=0$ and rearranges to numerical $\frac{dy}{dx}$ |
| Use $(2, 0)$ and numerical $\frac{dy}{dx}$ to find tangent equation | dM1 | Accept $\frac{y-0}{x-2} = $ numerical $\frac{dy}{dx}$ |
| $y = \frac{(16\ln 2 - 2)(x-2)}{12}$ | A1 | oe; accept $y = 0.76x - 1.52$ (awrt 2dp) |
---\begin{enumerate}
\item (a) Prove, by using logarithms, that
\end{enumerate}
$$\frac { \mathrm { d } } { \mathrm {~d} x } \left( 2 ^ { x } \right) = 2 ^ { x } \ln 2$$
The curve $C$ has the equation
$$2 x + 3 y ^ { 2 } + 3 x ^ { 2 } y + 12 = 4 \times 2 ^ { x }$$
The point $P$, with coordinates $( 2,0 )$, lies on $C$.\\
(b) Find an equation of the tangent to $C$ at $P$.\\
\hfill \mbox{\textit{Edexcel C34 2014 Q5 [9]}}