| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Find gradient at a point - direct evaluation |
| Difficulty | Moderate -0.3 This is a multi-part question covering standard C3 techniques: proving a function is even (routine substitution), finding a gradient using chain rule (straightforward differentiation), explaining why a function lacks an inverse (symmetry argument), finding an inverse function (algebraic manipulation), and verifying the inverse function derivative relationship. While comprehensive, each part uses well-practiced methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.07l Derivative of ln(x): and related functions1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(-x) = \ln[1+(-x)^2] = \ln[1+x^2] = f(x)\) | M1, E1 | If verifies \(f(-x)=f(x)\) using a particular point, allow SCB1; For \(f(-x)=\ln(1+x^2)=f(x)\) allow M1E0 |
| Symmetrical about \(Oy\) | B1 [3] | or 'reflects in \(Oy\)', etc |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = \ln(1+x^2)\), let \(u = 1+x^2\); \(dy/du = 1/u\), \(du/dx = 2x\) | M1, B1 | Chain rule; \(1/u\) soi |
| \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{u} \cdot 2x = \frac{2x}{1+x^2}\) | A1 | |
| When \(x=2\), \(dy/dx = 4/5\) | A1cao [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| The function is not one to one for this domain | B1 [1] | Or many to one |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | 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, no \(-ve\) \(x\) values, inflection shown, does not cross \(y=x\) line | A1 | |
| Domain for \(g(x)\): \(0 \leq x \leq \ln 10\) | B1 | Condone \(y\) instead of \(x\) |
| \(y = \ln(1+x^2)\), swap: \(x = \ln(1+y^2) \Rightarrow e^x = 1+y^2 \Rightarrow y^2 = e^x - 1\) | M1, M1 | Attempt to invert function; taking exponentials |
| \(\Rightarrow y = \sqrt{e^x-1}\), so \(g(x) = \sqrt{e^x-1}\) | E1 | \(g(x) = \sqrt{(e^x-1)}\)* www |
| Or: \(g[f(x)] = g[\ln(1+x^2)] = \sqrt{e^{\ln(1+x^2)}-1} = \sqrt{(1+x^2)-1} = x\) | M1, M1, E1 [6] | Forming \(g\,f(x)\) or \(f\,g(x)\); \(e^{\ln(1+x^2)}=1+x^2\) or \(\ln(1+e^x-1)=x\); www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(g'(x) = \frac{1}{2}(e^x-1)^{-1/2} \cdot e^x\) | B1, B1 | \(\frac{1}{2}u^{-1/2}\) soi; \(\times e^x\) |
| \(\Rightarrow g'(\ln 5) = \frac{1}{2}(e^{\ln 5}-1)^{-1/2} \cdot e^{\ln 5} = \frac{1}{2}(5-1)^{-1/2} \cdot 5 = \frac{5}{4}\) | M1, E1cao | Substituting \(\ln 5\) into \(g'\); must be some evidence of substitution |
| 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/Working | Marks | Guidance |
|---|---|---|
| $f(-x) = \ln[1+(-x)^2] = \ln[1+x^2] = f(x)$ | M1, E1 | If verifies $f(-x)=f(x)$ using a particular point, allow SCB1; For $f(-x)=\ln(1+x^2)=f(x)$ allow M1E0 |
| Symmetrical about $Oy$ | B1 [3] | or 'reflects in $Oy$', etc |
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = \ln(1+x^2)$, let $u = 1+x^2$; $dy/du = 1/u$, $du/dx = 2x$ | M1, B1 | Chain rule; $1/u$ soi |
| $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{u} \cdot 2x = \frac{2x}{1+x^2}$ | A1 | |
| When $x=2$, $dy/dx = 4/5$ | A1cao [4] | |
## Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| The function is not one to one for this domain | B1 [1] | Or many to one |
## Part (iv):
| Answer/Working | Marks | 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, no $-ve$ $x$ values, inflection shown, does not cross $y=x$ line | A1 | |
| Domain for $g(x)$: $0 \leq x \leq \ln 10$ | B1 | Condone $y$ instead of $x$ |
| $y = \ln(1+x^2)$, swap: $x = \ln(1+y^2) \Rightarrow e^x = 1+y^2 \Rightarrow y^2 = e^x - 1$ | M1, M1 | Attempt to invert function; taking exponentials |
| $\Rightarrow y = \sqrt{e^x-1}$, so $g(x) = \sqrt{e^x-1}$ | E1 | $g(x) = \sqrt{(e^x-1)}$* www |
| Or: $g[f(x)] = g[\ln(1+x^2)] = \sqrt{e^{\ln(1+x^2)}-1} = \sqrt{(1+x^2)-1} = x$ | M1, M1, E1 [6] | Forming $g\,f(x)$ or $f\,g(x)$; $e^{\ln(1+x^2)}=1+x^2$ or $\ln(1+e^x-1)=x$; www |
## Part (v):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $g'(x) = \frac{1}{2}(e^x-1)^{-1/2} \cdot e^x$ | B1, B1 | $\frac{1}{2}u^{-1/2}$ soi; $\times e^x$ |
| $\Rightarrow g'(\ln 5) = \frac{1}{2}(e^{\ln 5}-1)^{-1/2} \cdot e^{\ln 5} = \frac{1}{2}(5-1)^{-1/2} \cdot 5 = \frac{5}{4}$ | M1, E1cao | Substituting $\ln 5$ into $g'$; must be some evidence of substitution |
| Reciprocal of gradient at P as tangents are reflections in $y=x$ | B1 [5] | Must have idea of reciprocal. Not 'inverse'. |3 The function $f ( x ) = \ln \left( t + x ^ { 2 } \right)$ 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]{aee8da6a-7d5c-442f-9729-55d81d9a606f-3_510_895_523_604}
\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 $f ( x )$ is now restricted to $0 \leqslant x \leqslant 3$. The inverse of $f ( x )$ is the function $g ( x )$.\\
(iv) Sketch the curves $y = f ( x )$ and $y = g ( x )$ on the same axes.
State the domain of the function $g ( x )$,\\
Show that $\mathrm { g } ( x ) = \sqrt { \mathrm { e } ^ { x } - 1 }$.\\
(v) Differentiate $\mathrm { g } ( \mathrm { x } )$. 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 Q3 [19]}}