| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2012 |
| Session | January |
| Marks | 16 |
| 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 testing routine differentiation (power rule), tangent line equation, second derivative test for minimum, and basic integration. All parts follow standard procedures with no problem-solving insight required—easier than average A-level questions. |
| Spec | 1.07b Gradient as rate of change: dy/dx notation1.07d Second derivatives: d^2y/dx^2 notation1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx1.08b Integrate x^n: where n != -1 and sums1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = 5x^4 - 6x + 1\) | M1, A1, A1 (3 marks) | M1 one term correct; A1 another term correct; A1 all correct (no \(+ c\) etc) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d^2y}{dx^2} = 20x^3 - 6\) | B1\(\checkmark\) (1 mark) | FT 'their' \(\frac{dy}{dx}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = -1 \Rightarrow \frac{dy}{dx} = 5(-1)^4 - 6(-1) + 1 = 12\) | M1 | must sub \(x = -1\) into 'their' \(\frac{dy}{dx}\) |
| \(\Rightarrow y = 12(x+1)\) | A1cso (2 marks) | any correct form with \((x - -1)\) simplified; condone \(y = 12x + c\), \(c = 12\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = 1 \Rightarrow \frac{dy}{dx} = 5 - 6 + 1\) | M1 | sub \(x = 1\) into their \(\frac{dy}{dx}\) |
| \(\frac{dy}{dx} = 0 \Rightarrow\) stationary point | A1cso | shown \(= 0\) plus correct statement; or \(\frac{d^2y}{dx^2} = 20 - 6 > 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow (B\) is a) minimum (point) | E1 (3 marks) | must have correct \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}\) for E1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{x^6}{6} - \frac{3x^3}{3} + \frac{x^2}{2} + 5x\) | M1, A1, A1 | one term correct; another term correct; all correct (may have \(+ c\)) |
| \(\left[\frac{1}{6} - 1 + \frac{1}{2} + 5\right] - \left[\frac{1}{6} + 1 + \frac{1}{2} - 5\right]\) | m1 | 'their' \(F(1) - F(-1)\) with powers of 1 and \(-1\) evaluated correctly |
| \(= 8\) | A1cso (5 marks) |
| Answer | Marks |
|---|---|
| 'their answer to part (i)' \(- 2\) | M1 |
| \(\Rightarrow\) Area \(= 6\) | A1cso (2 marks) |
# Question 4:
## Part 4(a)(i):
$\frac{dy}{dx} = 5x^4 - 6x + 1$ | M1, A1, A1 (3 marks) | M1 one term correct; A1 another term correct; A1 all correct (no $+ c$ etc)
## Part 4(a)(ii):
$\frac{d^2y}{dx^2} = 20x^3 - 6$ | B1$\checkmark$ (1 mark) | FT 'their' $\frac{dy}{dx}$
## Part 4(b):
$x = -1 \Rightarrow \frac{dy}{dx} = 5(-1)^4 - 6(-1) + 1 = 12$ | M1 | must sub $x = -1$ into 'their' $\frac{dy}{dx}$
$\Rightarrow y = 12(x+1)$ | A1cso (2 marks) | any correct form with $(x - -1)$ simplified; condone $y = 12x + c$, $c = 12$
## Part 4(c):
$x = 1 \Rightarrow \frac{dy}{dx} = 5 - 6 + 1$ | M1 | sub $x = 1$ into their $\frac{dy}{dx}$
$\frac{dy}{dx} = 0 \Rightarrow$ stationary point | A1cso | shown $= 0$ plus correct statement; or $\frac{d^2y}{dx^2} = 20 - 6 > 0$
when $x = 1$, $\frac{d^2y}{dx^2} = 14$
$\Rightarrow (B$ is a) minimum (point) | E1 (3 marks) | must have correct $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ for E1
## Part 4(d)(i):
$\frac{x^6}{6} - \frac{3x^3}{3} + \frac{x^2}{2} + 5x$ | M1, A1, A1 | one term correct; another term correct; all correct (may have $+ c$)
$\left[\frac{1}{6} - 1 + \frac{1}{2} + 5\right] - \left[\frac{1}{6} + 1 + \frac{1}{2} - 5\right]$ | m1 | 'their' $F(1) - F(-1)$ with powers of 1 and $-1$ evaluated correctly
$= 8$ | A1cso (5 marks) |
## Part 4(d)(ii):
'their answer to part (i)' $- 2$ | M1 |
$\Rightarrow$ Area $= 6$ | A1cso (2 marks) |
---4 The curve with equation $y = x ^ { 5 } - 3 x ^ { 2 } + x + 5$ is sketched below. The point $O$ is at the origin and the curve passes through the points $A ( - 1,0 )$ and $B ( 1,4 )$.\\
\includegraphics[max width=\textwidth, alt={}, center]{91170a77-e266-4c81-89ee-1fc29a538485-3_447_752_438_653}
\begin{enumerate}[label=(\alph*)]
\item Given that $y = x ^ { 5 } - 3 x ^ { 2 } + x + 5$, find:
\begin{enumerate}[label=(\roman*)]
\item $\frac { \mathrm { d } y } { \mathrm {~d} x }$;
\item $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$.
\end{enumerate}\item Find an equation of the tangent to the curve at the point $A ( - 1,0 )$.
\item Verify that the point $B$, where $x = 1$, is a minimum point of the curve.
\item The curve with equation $y = x ^ { 5 } - 3 x ^ { 2 } + x + 5$ is sketched below. The point $O$ is at the origin and the curve passes through the points $A ( - 1,0 )$ and $B ( 1,4 )$.\\
\includegraphics[max width=\textwidth, alt={}, center]{91170a77-e266-4c81-89ee-1fc29a538485-3_451_757_1736_648}
\begin{enumerate}[label=(\roman*)]
\item Find $\int _ { - 1 } ^ { 1 } \left( x ^ { 5 } - 3 x ^ { 2 } + x + 5 \right) \mathrm { d } x$.
\item Hence find the area of the shaded region bounded by the curve between $A$ and $B$ and the line segments $A O$ and $O B$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2012 Q4 [16]}}