| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2010 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find normal line equation at given point |
| Difficulty | Moderate -0.8 This is a straightforward C1 question requiring routine differentiation to find a gradient, calculating the perpendicular gradient for the normal, and using point-slope form. The integration part uses standard power rule. All techniques are basic with no problem-solving insight needed, making it easier than average but not trivial due to multiple parts. |
| Spec | 1.07b Gradient as rate of change: dy/dx notation1.07m Tangents and normals: gradient and equations1.08b Integrate x^n: where n != -1 and sums1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| [Sketch of \(y = \ln(2x+3)\)] | B2,1,0 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Range of f: \(f(x) \geq \ln 3\) | M1, A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = f^{-1}(x) \Rightarrow f(y) = x\), \(\Rightarrow \ln(2y+3) = x\), \(\Rightarrow 2y + 3 = e^x\) | M1, m1 | \(x \Leftrightarrow y\) at any stage; Use of \(\ln m = N \Rightarrow m = e^N\) |
| \(f^{-1}(x) = \frac{e^x - 3}{2}\) | A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Domain of \(f^{-1}\) is: \(x \geq \ln 3\) | B1F | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d}{dx}[\ln(2x+3)] = \frac{1}{(2x+3)} \times 2\) | M1, A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P\), the point of intersection of \(y = f(x)\) and \(y = f^{-1}(x)\), must lie on the line \(y = x\); so \(P\) has coordinates \((\alpha, \alpha)\) | M1 | |
| \(f(\alpha) = \alpha\) | M1 | OE eg \(f^{-1}(\alpha) = \alpha\) |
| \(\ln(2\alpha + 3) = \alpha \Rightarrow 2\alpha + 3 = e^\alpha\) | A1 | 3 |
# Question 6(a)(i) [XMCA2]:
[Sketch of $y = \ln(2x+3)$] | B2,1,0 | **2** | B2 correct sketch — no part of curve in 2nd, 3rd or 4th quadrants and 'ln3'; B1 for general shape in 1st quadrant, ignore other quadrants; ln3 not required
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# Question 6(a)(ii) [XMCA2]:
Range of f: $f(x) \geq \ln 3$ | M1, A1 | **2** | $\geq \ln 3$ or $> \ln 3$ or $f \geq \ln 3$; Allow $y$ for $f(x)$
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# Question 6(b)(i) [XMCA2]:
$y = f^{-1}(x) \Rightarrow f(y) = x$, $\Rightarrow \ln(2y+3) = x$, $\Rightarrow 2y + 3 = e^x$ | M1, m1 | $x \Leftrightarrow y$ at any stage; Use of $\ln m = N \Rightarrow m = e^N$
$f^{-1}(x) = \frac{e^x - 3}{2}$ | A1 | **3** | ACF — Accept $y$ in place of $f^{-1}(x)$
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# Question 6(b)(ii) [XMCA2]:
Domain of $f^{-1}$ is: $x \geq \ln 3$ | B1F | **1** | ft on (a)(ii) for RHS
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# Question 6(c) [XMCA2]:
$\frac{d}{dx}[\ln(2x+3)] = \frac{1}{(2x+3)} \times 2$ | M1, A1 | **2** | $1/(2x+3)$
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# Question 6(d)(i) [XMCA2]:
$P$, the point of intersection of $y = f(x)$ and $y = f^{-1}(x)$, must lie on the line $y = x$; so $P$ has coordinates $(\alpha, \alpha)$ | M1 |
$f(\alpha) = \alpha$ | M1 | OE eg $f^{-1}(\alpha) = \alpha$
$\ln(2\alpha + 3) = \alpha \Rightarrow 2\alpha + 3 = e^\alpha$ | A1 | **3** | A.G. CSO
# Mark Scheme Extraction
---6 The curve with equation $y = 12 x ^ { 2 } - 19 x - 2 x ^ { 3 }$ is sketched below.\\
\includegraphics[max width=\textwidth, alt={}, center]{2f7a8e95-4994-4732-a9a4-306c7b6cad92-3_444_819_1434_609}
The curve crosses the $x$-axis at the origin $O$, and the point $A ( 2 , - 6 )$ lies on the curve.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the gradient of the curve with equation $y = 12 x ^ { 2 } - 19 x - 2 x ^ { 3 }$ at the point $A$.
\item Hence find the equation of the normal to the curve at the point $A$, giving your answer in the form $x + p y + q = 0$, where $p$ and $q$ are integers.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Find the value of $\int _ { 0 } ^ { 2 } \left( 12 x ^ { 2 } - 19 x - 2 x ^ { 3 } \right) \mathrm { d } x$.
\item Hence determine the area of the shaded region bounded by the curve and the line $O A$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2010 Q6 [15]}}