| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2005 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Given factor, find all roots |
| Difficulty | Moderate -0.3 This is a comprehensive but routine C1 question covering multiple standard techniques (factor theorem, factorization, differentiation, integration). Part (a) is straightforward substitution and factorization; parts (b-d) involve standard calculus procedures. While multi-part with several marks, each component is textbook-standard with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.07d Second derivatives: d^2y/dx^2 notation1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.08b Integrate x^n: where n != -1 and sums1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(p(3) = 27 - 45 + 21 - 3\) | M1 | Finding \(p(3)\) |
| \(p(3) = 0 \Rightarrow x-3\) is a factor | A1 | Shown \(= 0\) plus a statement; or \((x-3)(x^2 - 2x + 1)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(B\) is point \((3, 0)\) | B1 | Must have coordinates |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\frac{dy}{dx} = 3x^2 - 10x + 7\) | M1 | One term correct |
| A1 | All correct with NO \(+c\) etc |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(3x^2 - 10x + 7 = 0\) | M1 | Putting their \(\frac{dy}{dx} = 0\) |
| \(\Rightarrow (x-1)(3x-7) = 0\) | m1 | Attempt to use quad formula or factorise |
| \(\Rightarrow\) at \(M\), \(x = \frac{7}{3}\) | A1 | CSO factors correct etc |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\frac{d^2y}{dx^2} = 6x - 10\) | B1\(\checkmark\) | ft their \(\frac{dy}{dx}\) |
| sub \(x = 1\), \(\Rightarrow \frac{d^2y}{dx^2} = -4\) | B1\(\checkmark\) | ft their \(\frac{d^2y}{dx^2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\frac{x^4}{4} - \frac{5x^3}{3} + \frac{7x^2}{2} - 3x \quad (+c)\) | M1 | Increase one power by 1 |
| A1 | One term correct | |
| A1 | Two other terms correct | |
| A1 | All correct (condone missing \(+c\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Realisation that limits are 0 and 1 | B1 | Condone wrong way round |
| \(\left[\frac{1}{4} - \frac{5}{3} + \frac{7}{2} - 3\right] - 0\) | M1 | Attempt to sub their limits into their (d)(i) |
| \(= -\frac{11}{12}\) | A1 | CSO Must use \(F(1) - F(0)\) correctly |
| Area \(= \frac{11}{12}\) | E1 | CSO Convincing argument |
## Question 4:
**Part (a)(i)**
| Working | Marks | Guidance |
|---------|-------|----------|
| $p(3) = 27 - 45 + 21 - 3$ | M1 | Finding $p(3)$ |
| $p(3) = 0 \Rightarrow x-3$ is a factor | A1 | Shown $= 0$ plus a statement; or $(x-3)(x^2 - 2x + 1)$ |
**Part (a)(ii)**
| Working | Marks | Guidance |
|---------|-------|----------|
| $B$ is point $(3, 0)$ | B1 | Must have coordinates |
**Part (b)(i)**
| Working | Marks | Guidance |
|---------|-------|----------|
| $\frac{dy}{dx} = 3x^2 - 10x + 7$ | M1 | One term correct |
| | A1 | All correct with NO $+c$ etc |
**Part (b)(ii)**
| Working | Marks | Guidance |
|---------|-------|----------|
| $3x^2 - 10x + 7 = 0$ | M1 | Putting their $\frac{dy}{dx} = 0$ |
| $\Rightarrow (x-1)(3x-7) = 0$ | m1 | Attempt to use quad formula or factorise |
| $\Rightarrow$ at $M$, $x = \frac{7}{3}$ | A1 | **CSO** factors correct etc |
**Part (c)**
| Working | Marks | Guidance |
|---------|-------|----------|
| $\frac{d^2y}{dx^2} = 6x - 10$ | B1$\checkmark$ | ft their $\frac{dy}{dx}$ |
| sub $x = 1$, $\Rightarrow \frac{d^2y}{dx^2} = -4$ | B1$\checkmark$ | ft their $\frac{d^2y}{dx^2}$ |
**Part (d)(i)**
| Working | Marks | Guidance |
|---------|-------|----------|
| $\frac{x^4}{4} - \frac{5x^3}{3} + \frac{7x^2}{2} - 3x \quad (+c)$ | M1 | Increase one power by 1 |
| | A1 | One term correct |
| | A1 | Two other terms correct |
| | A1 | All correct (condone missing $+c$) |
**Part (d)(ii)**
| Working | Marks | Guidance |
|---------|-------|----------|
| Realisation that limits are 0 and 1 | B1 | Condone wrong way round |
| $\left[\frac{1}{4} - \frac{5}{3} + \frac{7}{2} - 3\right] - 0$ | M1 | Attempt to sub their limits into their (d)(i) |
| $= -\frac{11}{12}$ | A1 | **CSO** Must use $F(1) - F(0)$ correctly |
| Area $= \frac{11}{12}$ | E1 | **CSO** Convincing argument |
**Total: 18 marks**
---4 The curve with equation $y = x ^ { 3 } - 5 x ^ { 2 } + 7 x - 3$ is sketched below.\\
\includegraphics[max width=\textwidth, alt={}, center]{3729de55-7139-4f41-8584-640f173c0e09-3_444_588_411_717}
The curve touches the $x$-axis at the point $A ( 1,0 )$ and cuts the $x$-axis at the point $B$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Use the factor theorem to show that $x - 3$ is a factor of
$$\mathrm { p } ( x ) = x ^ { 3 } - 5 x ^ { 2 } + 7 x - 3$$
\item Hence find the coordinates of $B$.
\end{enumerate}\item The point $M$, shown on the diagram, is a minimum point of the curve with equation $y = x ^ { 3 } - 5 x ^ { 2 } + 7 x - 3$.
\begin{enumerate}[label=(\roman*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item Hence determine the $x$-coordinate of $M$.
\end{enumerate}\item Find the value of $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ when $x = 1$.
\item \begin{enumerate}[label=(\roman*)]
\item Find $\int \left( x ^ { 3 } - 5 x ^ { 2 } + 7 x - 3 \right) \mathrm { d } x$.
\item Hence determine the area of the shaded region bounded by the curve and the coordinate axes.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2005 Q4 [15]}}