| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Determine if inverse exists |
| Difficulty | Standard +0.3 This is a structured multi-part question covering standard C3 topics (even functions, differentiation of ln, inverse functions) with clear scaffolding. Parts (i)-(iii) are routine bookwork, part (iv) requires standard inverse function manipulation, and part (v) applies the chain rule then verifies the inverse function derivative relationship. While comprehensive, each step follows predictable techniques without requiring novel insight, making it slightly easier than average. |
| 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 functions1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(f(-x) = \ln[1+(-x)^2]\) | M1 | If verifies \(f(-x)=f(x)\) using a particular point, allow SCB1; for \(f(-x)=\ln(1+x^2)=f(x)\) allow M1E0 |
| \(= \ln[1+x^2] = f(x)\) | E1 | |
| Symmetrical about \(Oy\) | B1 [3] | or 'reflects in \(Oy\)', etc |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y = \ln(1+x^2)\), let \(u = 1+x^2\) | M1 | Chain rule |
| \(\frac{dy}{du} = \frac{1}{u}\), \(\frac{du}{dx} = 2x\) | B1 | \(\frac{1}{u}\) soi |
| \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{u} \cdot 2x = \frac{2x}{1+x^2}\) | A1 A1cao [4] | |
| When \(x=2\), \(\frac{dy}{dx} = \frac{4}{5}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| The function is not one to one for this domain | B1 [1] | Or many to one |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Graph of \(g(x)\) as reflection of \(f(x)\) in \(y=x\) | M1 | \(g(x)\) is \(f(x)\) reflected in \(y=x\) |
| Reasonable shape and domain | A1 | No \(-ve\) \(x\) values, inflection shown, does not cross \(y=x\) line |
| Domain for \(g(x)\): \(0 \leq x \leq \ln 10\) | B1 | Condone \(y\) instead of \(x\) |
| \(y = \ln(1+x^2)\), swap \(x \leftrightarrow y\) | M1 | Attempt to invert function |
| \(x = \ln(1+y^2) \Rightarrow e^x = 1+y^2\) | M1 | Taking exponentials |
| \(\Rightarrow y^2 = e^x - 1 \Rightarrow y = \sqrt{e^x-1}\) | E1 | \(g(x) = \sqrt{e^x-1}\) www |
| so \(g(x) = \sqrt{e^x - 1}\)* | ||
| Or: \(g\,f(x) = g[\ln(1+x^2)]\) | M1 | forming \(g\,f(x)\) or \(f\,g(x)\) |
| \(= \sqrt{e^{\ln(1+x^2)}-1}\) | M1 | \(e^{\ln(1+x^2)} = 1+x^2\) or \(\ln(1+e^x-1)=x\) |
| \(= \sqrt{(1+x^2)-1} = \sqrt{x^2} = x\) | E1 [6] | www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(g'(x) = \frac{1}{2}(e^x-1)^{-1/2} \cdot e^x\) | B1 B1 | \(\frac{1}{2}u^{-1/2}\) soi; \(\times e^x\) |
| \(\Rightarrow g'(\ln 5) = \frac{1}{2}(e^{\ln 5}-1)^{-1/2} \cdot e^{\ln 5}\) | M1 | substituting \(\ln 5\) into \(g'\) — must be some evidence of substitution |
| \(= \frac{1}{2}(5-1)^{-1/2} \cdot 5 = \frac{5}{4}\) | E1cao | |
| Reciprocal of gradient at \(P\) as tangents are reflections in \(y=x\) | B1 [5] | Must have idea of reciprocal. Not 'inverse'. |
## Question 3:
### Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $f(-x) = \ln[1+(-x)^2]$ | M1 | If verifies $f(-x)=f(x)$ using a particular point, allow SCB1; for $f(-x)=\ln(1+x^2)=f(x)$ allow M1E0 |
| $= \ln[1+x^2] = f(x)$ | E1 | |
| Symmetrical about $Oy$ | B1 [3] | or 'reflects in $Oy$', etc |
### Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $y = \ln(1+x^2)$, let $u = 1+x^2$ | M1 | Chain rule |
| $\frac{dy}{du} = \frac{1}{u}$, $\frac{du}{dx} = 2x$ | B1 | $\frac{1}{u}$ soi |
| $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{u} \cdot 2x = \frac{2x}{1+x^2}$ | A1 A1cao [4] | |
| When $x=2$, $\frac{dy}{dx} = \frac{4}{5}$ | | |
### Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| The function is not one to one for this domain | B1 [1] | Or many to one |
### Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| Graph of $g(x)$ as reflection of $f(x)$ in $y=x$ | M1 | $g(x)$ is $f(x)$ reflected in $y=x$ |
| Reasonable shape and domain | A1 | No $-ve$ $x$ values, inflection shown, does not cross $y=x$ line |
| Domain for $g(x)$: $0 \leq x \leq \ln 10$ | B1 | Condone $y$ instead of $x$ |
| $y = \ln(1+x^2)$, swap $x \leftrightarrow y$ | M1 | Attempt to invert function |
| $x = \ln(1+y^2) \Rightarrow e^x = 1+y^2$ | M1 | Taking exponentials |
| $\Rightarrow y^2 = e^x - 1 \Rightarrow y = \sqrt{e^x-1}$ | E1 | $g(x) = \sqrt{e^x-1}$ www |
| so $g(x) = \sqrt{e^x - 1}$* | | |
| Or: $g\,f(x) = g[\ln(1+x^2)]$ | M1 | forming $g\,f(x)$ or $f\,g(x)$ |
| $= \sqrt{e^{\ln(1+x^2)}-1}$ | M1 | $e^{\ln(1+x^2)} = 1+x^2$ or $\ln(1+e^x-1)=x$ |
| $= \sqrt{(1+x^2)-1} = \sqrt{x^2} = x$ | E1 [6] | www |
### Part (v)
| Answer | Mark | Guidance |
|--------|------|----------|
| $g'(x) = \frac{1}{2}(e^x-1)^{-1/2} \cdot e^x$ | B1 B1 | $\frac{1}{2}u^{-1/2}$ soi; $\times e^x$ |
| $\Rightarrow g'(\ln 5) = \frac{1}{2}(e^{\ln 5}-1)^{-1/2} \cdot e^{\ln 5}$ | M1 | substituting $\ln 5$ into $g'$ — must be some evidence of substitution |
| $= \frac{1}{2}(5-1)^{-1/2} \cdot 5 = \frac{5}{4}$ | E1cao | |
| Reciprocal of gradient at $P$ as tangents are reflections in $y=x$ | B1 [5] | Must have idea of reciprocal. Not 'inverse'. |3 The function $f ( x ) = \ln \left( 1 + x ^ { 2 } \right)$ has domain $- 3 \leqslant x \leqslant 3$.\\
Fig. 9 shows the graph of $y = f ( x )$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6555136d-0444-41f6-9063-21960352089d-3_495_867_519_607}
\captionsetup{labelformat=empty}
\caption{Fig. 9}
\end{center}
\end{figure}
(1) 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 $f ( x )$ is now restricted to $0 \leqslant x \leqslant 3$. The inverse of $f ( x )$ is the function $g ( x )$,\\
(iv) Sketch the curves $y = \mathrm { f } ( x )$ and $y = \mathrm { g } ( x )$ on the same axes.
State the domain of the function $g ( x )$.\\
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 commection between this result and your answer to part (ii).
\hfill \mbox{\textit{OCR MEI C3 Q3 [19]}}