OCR MEI C3 2012 January — Question 9 18 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
Year2012
SessionJanuary
Marks18
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDifferentiating Transcendental Functions
TypeInverse function properties
DifficultyChallenging +1.2 This is a structured multi-part question covering logarithms, differentiation, inverse functions, and integration with substitution. While it requires multiple C3 techniques (log laws, chain rule, inverse function properties, integration by substitution), each part is well-scaffolded with hints and follows standard patterns. The most challenging aspect is part (iv) requiring reflection of a region and using the inverse function relationship for areas, but this is a known technique at this level. Overall, slightly above average difficulty due to length and the integration/area component, but not requiring exceptional insight.
Spec1.02v Inverse and composite functions: graphs and conditions for existence1.06d Natural logarithm: ln(x) function and properties1.07l Derivative of ln(x): and related functions1.08h Integration by substitution

Fig. 9 shows the curves \(y = \text{f}(x)\) and \(y = \text{g}(x)\). The function \(y = \text{f}(x)\) is given by $$\text{f}(x) = \ln \left( \frac{2x}{1+x} \right), \quad x > 0.$$ The curve \(y = \text{f}(x)\) crosses the \(x\)-axis at P, and the line \(x = 2\) at Q. \includegraphics{figure_9}
  1. Verify that the \(x\)-coordinate of P is 1. Find the exact \(y\)-coordinate of Q. [2]
  2. Find the gradient of the curve at P. [Hint: use \(\frac{a}{b} = \ln a - \ln b\).] [4]
The function \(\text{g}(x)\) is given by $$\text{g}(x) = \frac{e^x}{2-e^x}, \quad x < \ln 2.$$ The curve \(y = \text{g}(x)\) crosses the \(y\)-axis at the point R.
  1. Show that \(\text{g}(x)\) is the inverse function of \(\text{f}(x)\). Write down the gradient of \(y = \text{g}(x)\) at R. [5]
  2. Show, using the substitution \(u = 2 - e^x\) or otherwise, that \(\int_0^{\ln \frac{4}{3}} \text{g}(x) dx = \ln \frac{3}{2}\). Using this result, show that the exact area of the shaded region shown in Fig. 9 is \(\ln \frac{32}{27}\). [Hint: consider its reflection in \(y = x\).] [7]

Fig. 9 shows the curves $y = \text{f}(x)$ and $y = \text{g}(x)$. The function $y = \text{f}(x)$ is given by
$$\text{f}(x) = \ln \left( \frac{2x}{1+x} \right), \quad x > 0.$$

The curve $y = \text{f}(x)$ crosses the $x$-axis at P, and the line $x = 2$ at Q.

\includegraphics{figure_9}

\begin{enumerate}[label=(\roman*)]
\item Verify that the $x$-coordinate of P is 1.

Find the exact $y$-coordinate of Q. [2]

\item Find the gradient of the curve at P. [Hint: use $\frac{a}{b} = \ln a - \ln b$.] [4]
\end{enumerate}

The function $\text{g}(x)$ is given by
$$\text{g}(x) = \frac{e^x}{2-e^x}, \quad x < \ln 2.$$

The curve $y = \text{g}(x)$ crosses the $y$-axis at the point R.

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Show that $\text{g}(x)$ is the inverse function of $\text{f}(x)$.

Write down the gradient of $y = \text{g}(x)$ at R. [5]

\item Show, using the substitution $u = 2 - e^x$ or otherwise, that $\int_0^{\ln \frac{4}{3}} \text{g}(x) dx = \ln \frac{3}{2}$.

Using this result, show that the exact area of the shaded region shown in Fig. 9 is $\ln \frac{32}{27}$.
[Hint: consider its reflection in $y = x$.] [7]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI C3 2012 Q9 [18]}}