| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2011 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find tangent at given point (polynomial/algebraic) |
| Difficulty | Moderate -0.8 This is a straightforward C1 question requiring routine differentiation to find a gradient, then using point-slope form for the tangent equation. Both steps are standard textbook exercises with no problem-solving required, making it easier than average but not trivial since it requires correct application of basic techniques. |
| 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 |
|---|---|---|
| \(\frac{dy}{dx} = -1 - 4x^3\) | M1, A1 | one of these terms correct; all correct (no + c) |
| (When \(x = 1\), grad =) \(-5\) | A1cso | (Check that \(\frac{dy}{dx}\) is actually correct!) |
| Answer | Marks | Guidance |
|---|---|---|
| \(y - 12 =\) their grad\((x - 1)\) | M1 | any form of equation through (1, 12) and attempt at c if using \(y = mx + c\) |
| \(y = -5x + 17\) (or \(y = 17 - 5x\)) | A1√ | FT their gradient; Condone \(y = -5x + c, c = 17\) etc |
| Answer | Marks | Guidance |
|---|---|---|
| \(14x - \frac{x^2}{2} - \frac{x^5}{5}\) | M1, A1, A1 | one of these terms correct; another term correct; all correct (may have + c) |
| \([]_{-2}^1 =\) | ||
| \(\left(14 - \frac{1}{2} - \frac{1}{5}\right) - \left(-28 - 2 + \frac{32}{5}\right)\) | m1 | F(1) and F(–2) attempted |
| \(= 36.9\) OE | A1 | Condone recovery to this value |
| Answer | Marks | Guidance |
|---|---|---|
| Area \(\Delta = \frac{1}{2} \times 3 \times 12\) | M1 | Correct area of triangle unsimplified |
| \(= 18\) | ||
| \(\Rightarrow\) shaded area \(= 18.9\) | A1cso | 2 |
| Total | 12 |
**4(a)(i)**
$\frac{dy}{dx} = -1 - 4x^3$ | M1, A1 | one of these terms correct; all correct (no + c)
(When $x = 1$, grad =) $-5$ | A1cso | (Check that $\frac{dy}{dx}$ is actually correct!) | 3
**4(a)(ii)**
$y - 12 =$ their grad$(x - 1)$ | M1 | any form of equation through (1, 12) and attempt at c if using $y = mx + c$
$y = -5x + 17$ (or $y = 17 - 5x$) | A1√ | FT their gradient; Condone $y = -5x + c, c = 17$ etc | 2
**4(b)(i)**
$14x - \frac{x^2}{2} - \frac{x^5}{5}$ | M1, A1, A1 | one of these terms correct; another term correct; all correct (may have + c)
$[]_{-2}^1 =$ | |
$\left(14 - \frac{1}{2} - \frac{1}{5}\right) - \left(-28 - 2 + \frac{32}{5}\right)$ | m1 | F(1) and F(–2) attempted
$= 36.9$ OE | A1 | Condone recovery to this value | 5
**4(b)(ii)**
Area $\Delta = \frac{1}{2} \times 3 \times 12$ | M1 | Correct area of triangle unsimplified
$= 18$ | |
$\Rightarrow$ shaded area $= 18.9$ | A1cso | 2
| | Total | 12
---4 The curve sketched below passes through the point $A ( - 2,0 )$.\\
\includegraphics[max width=\textwidth, alt={}, center]{889639d6-0a31-4569-8370-1e72291a0c47-3_538_734_365_662}
The curve has equation $y = 14 - x - x ^ { 4 }$ and the point $P ( 1,12 )$ lies on the curve.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the gradient of the curve at the point $P$.
\item Hence find the equation of the tangent to the curve at the point $P$, giving your answer in the form $y = m x + c$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Find $\int _ { - 2 } ^ { 1 } \left( 14 - x - x ^ { 4 } \right) \mathrm { d } x$.
\item Hence find the area of the shaded region bounded by the curve $y = 14 - x - x ^ { 4 }$ and the line $A P$.\\
(2 marks)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2011 Q4 [12]}}