| Exam Board | AQA |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Normal meets curve/axis — further geometry |
| Difficulty | Standard +0.3 Part (a) requires setting up two simultaneous equations using the point and gradient conditions, then solving—a standard technique. Part (b) involves finding the normal line equation (negative reciprocal gradient) and calculating a triangle area, which is routine coordinate geometry. This is slightly easier than average as it follows well-practiced procedures without requiring novel insight. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(3 = 2^3 + 2^2p + 2q - 45\); \(40 = 4p + 2q\); substitutes coordinates of \(R\) into \(y = x^3 + px^2 + qx - 45\) to form correct equation in \(p\) and \(q\) | B1 (AO 1.1b) | ACF |
| \(\frac{dy}{dx} = 3x^2 + 2px + q\); differentiates with at least two terms correct | M1 (AO 1.1a) | |
| Obtains fully correct derivative | A1 (AO 1.1b) | |
| \(8 = 3\times 2^2 + 4p + q\); \(-4 = 4p+q\); substitutes \(x=2\) and \(\frac{dy}{dx}=8\) into differential equation to give correct equation | A1 (AO 1.1b) | ACF |
| \(p = -12\), \(q = 44\) | A1 (AO 1.1b) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Gradient of normal is \(-\frac{1}{8}\) | B1 (AO 1.2) | PI |
| \((y-3) = -\frac{1}{8}(x-2)\); \(y = -\frac{1}{8}x + \frac{13}{4}\); writes equation of line through \((2,3)\) with 'their' gradient of normal | M1 (AO 1.1a) | ACF |
| Meets \(x\)-axis at \((26, 0)\); meets \(y\)-axis at \(\left(0, 3\frac{1}{4}\right)\); substitutes \(x=0\) or \(y=0\) into 'their' straight line equation to find at least one intercept | M1 (AO 1.1a) | M1M1 PI by \(x=26\) or \(y=3\frac{1}{4}\) |
| Area \(= \frac{1}{2} \times 26 \times 3\frac{1}{4} = \frac{169}{4}\); calculates area of triangle using both 'their' intercepts | M1 (AO 1.1a) | Or calculates area by integration of 'their' line between \(x=0\) and 'their' \(x\) intercept |
| \(\frac{169}{4}\) or \(42\frac{1}{4}\) or \(42.25\) CAO | A1 (AO 1.1b) |
## Question 8(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $3 = 2^3 + 2^2p + 2q - 45$; $40 = 4p + 2q$; substitutes coordinates of $R$ into $y = x^3 + px^2 + qx - 45$ to form correct equation in $p$ and $q$ | B1 (AO 1.1b) | ACF |
| $\frac{dy}{dx} = 3x^2 + 2px + q$; differentiates with at least two terms correct | M1 (AO 1.1a) | |
| Obtains fully correct derivative | A1 (AO 1.1b) | |
| $8 = 3\times 2^2 + 4p + q$; $-4 = 4p+q$; substitutes $x=2$ and $\frac{dy}{dx}=8$ into differential equation to give correct equation | A1 (AO 1.1b) | ACF |
| $p = -12$, $q = 44$ | A1 (AO 1.1b) | |
## Question 8(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Gradient of normal is $-\frac{1}{8}$ | B1 (AO 1.2) | PI |
| $(y-3) = -\frac{1}{8}(x-2)$; $y = -\frac{1}{8}x + \frac{13}{4}$; writes equation of line through $(2,3)$ with 'their' gradient of normal | M1 (AO 1.1a) | ACF |
| Meets $x$-axis at $(26, 0)$; meets $y$-axis at $\left(0, 3\frac{1}{4}\right)$; substitutes $x=0$ or $y=0$ into 'their' straight line equation to find at least one intercept | M1 (AO 1.1a) | M1M1 PI by $x=26$ or $y=3\frac{1}{4}$ |
| Area $= \frac{1}{2} \times 26 \times 3\frac{1}{4} = \frac{169}{4}$; calculates area of triangle using both 'their' intercepts | M1 (AO 1.1a) | Or calculates area by integration of 'their' line between $x=0$ and 'their' $x$ intercept |
| $\frac{169}{4}$ or $42\frac{1}{4}$ or $42.25$ CAO | A1 (AO 1.1b) | |8 A curve has equation
$$y = x ^ { 3 } + p x ^ { 2 } + q x - 45$$
The curve passes through point $R ( 2,3 )$\\
The gradient of the curve at $R$ is 8\\
8
\begin{enumerate}[label=(\alph*)]
\item Find the value of $p$ and the value of $q$.\\
8
\item Calculate the area enclosed between the normal to the curve at $R$ and the coordinate
8 (b) axes.\\
$9 \quad$ A curve $C$ has equation $y = \mathrm { f } ( x )$ where
$$f ( x ) = ( x - 2 ) ( x - 3 ) ^ { 2 }$$
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 2 2019 Q8 [10]}}