| Exam Board | Edexcel |
|---|---|
| Module | AEA (Advanced Extension Award) |
| Year | 2004 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Implicit differentiation |
| Difficulty | Challenging +1.8 This AEA question requires multiple sophisticated techniques: differentiating inverse hyperbolic functions, parametric differentiation, error analysis in a proof, and proving logarithmic identities. While each individual part is manageable, the combination of proof verification, the non-standard parametric curve, and the symmetry deduction requires mathematical maturity beyond typical A-level. The error-spotting in part (b)(i) demands careful attention to the invalid step of differentiating y with respect to x while treating t as constant. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.03g Parametric equations: of curves and conversion to cartesian1.06d Natural logarithm: ln(x) function and properties1.07l Derivative of ln(x): and related functions1.07s Parametric and implicit differentiation |
\begin{enumerate}[label=(\alph*)]
\item Given that $y = \ln [t + \sqrt{(1 + t^2)}]$, show that $\frac{dy}{dt} = \frac{1}{\sqrt{(1+t^2)}}$. [3]
\end{enumerate}
The curve $C$ has parametric equations
$$x = \frac{1}{\sqrt{(1+t^2)}}, \quad y = \ln [t + \sqrt{(1 + t^2)}], \quad t \in \mathbb{R}.$$
A student was asked to prove that, for $t > 0$, the gradient of the tangent to $C$ is negative.
The attempted proof was as follows:
$$y = \ln \left(t + \frac{1}{x}\right)$$
$$= \ln \left(\frac{tx + 1}{x}\right)$$
$$= \ln (tx + 1) - \ln x$$
$$\therefore \frac{dy}{dx} = \frac{t}{tx + 1} - \frac{1}{x}$$
$$= \frac{\frac{t}{x}}{t + \frac{1}{x}} - \frac{1}{x}$$
$$= \frac{t\sqrt{(1+t^2)}}{t + \sqrt{(1+t^2)}} - \sqrt{(1 + t^2)}$$
$$= -\frac{(1+t^2)}{t + \sqrt{(1+t^2)}}$$
As $(1 + t^2) > 0$, and $t + \sqrt{(1 + t^2)} > 0$ for $t > 0$, $\frac{dy}{dx} < 0$ for $t > 0$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item \begin{enumerate}[label=(\roman*)]
\item Identify the error in this attempt.
\item Give a correct version of the proof. [6]
\end{enumerate}
\item Prove that $\ln [-t + \sqrt{(1 + t^2)}] = -\ln [t + \sqrt{(1 + t^2)}]$. [3]
\item Deduce that $C$ is symmetric about the $x$-axis and sketch the graph of $C$. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel AEA 2004 Q5 [15]}}