| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2005 |
| Session | June |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Sketch function and inverse graphs |
| Difficulty | Standard +0.3 This is a comprehensive but standard inverse function question covering routine techniques: proving a function is even (substitution), finding a gradient (chain rule differentiation), explaining why an inverse doesn't exist (horizontal line test), sketching inverse graphs (reflection in y=x), finding an inverse algebraically (rearranging), and verifying the derivative relationship between inverse functions. While multi-part with several marks, each component uses well-practiced C3 techniques without requiring novel insight or complex problem-solving. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence1.06d Natural logarithm: ln(x) function and properties1.07l Derivative of ln(x): and related functions |
| Answer | Marks | Guidance |
|---|---|---|
| Symmetrical about Oy | M1, E1, B1 [3] | If verifies that \(f(-x) = f(x)\) using a particular point, allow SCB1; For \(f(-x) = \ln(1 + x^2) = f(x)\) allow M1E0; For \(f(-x) = \ln(1 + x^2) = f(x)\) allow M1E0; or 'reflects in Oy', etc |
| Answer | Marks | Guidance |
|---|---|---|
| When \(x = 2, dy/dx = 4/5\) | M1, B1, A1, A1cao [4] | Chain rule; \(1/u\) soi; \(\frac{2x}{1+x^2}\) |
| (iii) The function is not one to one for this domain | B1 [1] | Or many to one |
| Answer | Marks | Guidance |
|---|---|---|
| or \(g(f(x)) = g[\ln(1 + x^2)] = \sqrt{e^{\ln(1+x^2)} - 1} = \sqrt{(1 + x^2) - 1} = x\) | M1, A1, B1, M1, M1, E1 [6] | g(x) is f(x) reflected in \(y = x\); Reasonable shape and domain, i.e. no -ve x values, inflection shown, does not cross \(y = x\) line; Condone y instead of x; Attempt to invert function; Taking exponentials; \(g(x) = \sqrt{(e^x - 1)}*\) www; forming g f(x) or f g(x); \(e^{\ln(1+x^2)} = 1 + x^2\) or \(\ln(1 + e^x - 1) = x\) wwww |
| Answer | Marks | Guidance |
|---|---|---|
| Reciprocal of gradient at P as tangents are reflections in \(y = x\) | B1, B1, M1, E1cao [5] | \(1/2 u^{-1/2}\) soi; \(\times e^x\); substituting ln 5 into g' – must be some evidence of substitution; Must have idea of reciprocal. Not 'inverse'. |
**(i)** $f(-x) = \ln[1 + (-x)^2] = \ln[1 + x^2] = f(x)$
Symmetrical about Oy | M1, E1, B1 [3] | If verifies that $f(-x) = f(x)$ using a particular point, allow SCB1; For $f(-x) = \ln(1 + x^2) = f(x)$ allow M1E0; For $f(-x) = \ln(1 + x^2) = f(x)$ allow M1E0; or 'reflects in Oy', etc
**(ii)** $y = \ln(1 + x^2)$, let $u = 1 + x^2$
$dy/du = 1/u, du/dx = 2x$
$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{u} \cdot 2x = \frac{2x}{1 + x^2}$
When $x = 2, dy/dx = 4/5$ | M1, B1, A1, A1cao [4] | Chain rule; $1/u$ soi; $\frac{2x}{1+x^2}$
**(iii)** The function is not one to one for this domain | B1 [1] | Or many to one
**(iv)** Domain for $g(x) = 0 \leq x \leq \ln 10$
$y = \ln(1 + x^2) \rightarrow x \leftrightarrow y$
$x = \ln(1 + y^2) \Rightarrow e^x = 1 + y^2 \Rightarrow e^x - 1 = y^2 \Rightarrow y = \sqrt{e^x - 1}$
so $g(x) = \sqrt{(e^x - 1)}*$
or $g(f(x)) = g[\ln(1 + x^2)] = \sqrt{e^{\ln(1+x^2)} - 1} = \sqrt{(1 + x^2) - 1} = x$ | M1, A1, B1, M1, M1, E1 [6] | g(x) is f(x) reflected in $y = x$; Reasonable shape and domain, i.e. no -ve x values, inflection shown, does not cross $y = x$ line; Condone y instead of x; Attempt to invert function; Taking exponentials; $g(x) = \sqrt{(e^x - 1)}*$ www; forming g f(x) or f g(x); $e^{\ln(1+x^2)} = 1 + x^2$ or $\ln(1 + e^x - 1) = x$ wwww
**(v)** $g'(x) = 1/2(e^x - 1)^{-1/2} \cdot e^x \Rightarrow g'(\ln 5) = 1/2(e^{\ln 5} - 1)^{-1/2} \cdot e^{\ln 5} = 1/2(5-1)^{-1/2} \cdot 5 = 5/4$
Reciprocal of gradient at P as tangents are reflections in $y = x$ | B1, B1, M1, E1cao [5] | $1/2 u^{-1/2}$ soi; $\times e^x$; substituting ln 5 into g' – must be some evidence of substitution; Must have idea of reciprocal. Not 'inverse'.9 The function $\mathrm { f } ( x ) = \ln \left( 1 + x ^ { 2 } \right)$ has domain $- 3 \leqslant x \leqslant 3$.\\
Fig. 9 shows the graph of $y = \mathrm { f } ( x )$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3efea8db-9fa1-47a8-89b8-e4888f87a313-4_540_943_477_550}
\captionsetup{labelformat=empty}
\caption{Fig. 9}
\end{center}
\end{figure}
(i) Show algebraically that the function is even. State how this property relates to the shape of the curve.\\
(ii) Find the gradient of the curve at the point $\mathrm { P } ( 2 , \ln 5 )$.\\
(iii) Explain why the function does not have an inverse for the domain $- 3 \leqslant x \leqslant 3$.
The domain of $\mathrm { f } ( x )$ is now restricted to $0 \leqslant x \leqslant 3$. The inverse of $\mathrm { f } ( x )$ is the function $\mathrm { g } ( x )$.\\
(iv) Sketch the curves $y = \mathrm { f } ( x )$ and $y = \mathrm { g } ( x )$ on the same axes.
State the domain of the function $\mathrm { g } ( x )$.
Show that $\mathrm { g } ( x ) = \sqrt { \mathrm { e } ^ { x } - 1 }$.\\
(v) Differentiate $\mathrm { g } ( x )$. Hence verify that $\mathrm { g } ^ { \prime } ( \ln 5 ) = 1 \frac { 1 } { 4 }$. Explain the connection between this result and your answer to part (ii).
\hfill \mbox{\textit{OCR MEI C3 2005 Q9 [19]}}