| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2009 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Prove root count with given polynomial |
| Difficulty | Moderate -0.3 This is a structured multi-part question covering standard C1 techniques: Factor Theorem verification (routine substitution), polynomial division, differentiation, tangent equations, and definite integration. While it requires multiple steps across different topics, each individual part follows textbook procedures with no novel problem-solving required. The integration to find area is slightly more sophisticated than basic exercises, bringing it just below average difficulty. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.07m 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 |
|---|---|---|
| 6(a)(i) | \(p(2) = 8 + 2 - 10 \Rightarrow p(2) = 0 \Rightarrow (x-2)\) is factor | M1 A1 |
| 6(a)(ii) | Attempt at long division (generous) | M1 |
| \(p(x) = (x-2)(x^2 + 2x + 5)\) | A1 | \(a = 2, b = 5\) by inspection B1, B1 |
| 6(b)(i) | \(\frac{dy}{dx} = 3x^2 + 1\) | M1 A1 |
| When \(x = 2\): \(\frac{dy}{dx} = 3 \times 4 + 1\) | m1 | Sub \(x = 2\) into their \(\frac{dy}{dx}\) |
| Therefore gradient at \(Q\) is 13 | A1 | CSO |
| 6(b)(ii) | \(y = 13(x-2)\) | M1 |
| A1 | CSO; correct in any form | |
| 6(b)(iii) | \(\int ... dx = \frac{x^4}{4} + \frac{x^2}{2} - 10x (+c)\) | M1 A1 A1 |
| 6(b)(iv) | \([4 + 2 - 20] - [0] = -14\) | M1 |
| Area of shaded region \(= 14\) | A1 | CSO; separate statement following correct evaluation of limits |
**6(a)(i)** | $p(2) = 8 + 2 - 10 \Rightarrow p(2) = 0 \Rightarrow (x-2)$ is factor | M1 A1 | M1: Must find p(2) NOT long division. A1: Shown $= 0$ plus a statement
**6(a)(ii)** | Attempt at long division (generous) | M1 | Obtaining a quotient $x^2 + cx + d$ or equating coefficients (full method)
| $p(x) = (x-2)(x^2 + 2x + 5)$ | A1 | $a = 2, b = 5$ by inspection B1, B1
**6(b)(i)** | $\frac{dy}{dx} = 3x^2 + 1$ | M1 A1 | M1: One term correct. A1: All correct – no +c etc
| When $x = 2$: $\frac{dy}{dx} = 3 \times 4 + 1$ | m1 | Sub $x = 2$ into their $\frac{dy}{dx}$
| Therefore gradient at $Q$ is 13 | A1 | CSO
**6(b)(ii)** | $y = 13(x-2)$ | M1 | Tangent (NOT normal) attempted. ft their gradient answer from (b)(i)
| | A1 | CSO; correct in any form
**6(b)(iii)** | $\int ... dx = \frac{x^4}{4} + \frac{x^2}{2} - 10x (+c)$ | M1 A1 A1 | M1: one term correct. A1: second term correct. A1: all correct (condone no +c)
**6(b)(iv)** | $[4 + 2 - 20] - [0] = -14$ | M1 | F(2) attempted and possibly F(0). Must have earned M1 in (b)(iii)
| Area of shaded region $= 14$ | A1 | CSO; separate statement following correct evaluation of limits
**Total: 15 marks**
---6
\begin{enumerate}[label=(\alph*)]
\item The polynomial $\mathrm { p } ( x )$ is given by $\mathrm { p } ( x ) = x ^ { 3 } + x - 10$.
\begin{enumerate}[label=(\roman*)]
\item Use the Factor Theorem to show that $x - 2$ is a factor of $\mathrm { p } ( x )$.
\item Express $\mathrm { p } ( x )$ in the form $( x - 2 ) \left( x ^ { 2 } + a x + b \right)$, where $a$ and $b$ are constants.
\end{enumerate}\item The curve $C$ with equation $y = x ^ { 3 } + x - 10$, sketched below, crosses the $x$-axis at the point $Q ( 2,0 )$.\\
\includegraphics[max width=\textwidth, alt={}, center]{22c93dd5-d96a-4e31-8507-9c802e386231-3_444_547_1781_756}
\begin{enumerate}[label=(\roman*)]
\item Find the gradient of the curve $C$ at the point $Q$.
\item Hence find an equation of the tangent to the curve $C$ at the point $Q$.
\item Find $\int \left( x ^ { 3 } + x - 10 \right) \mathrm { d } x$.
\item Hence find the area of the shaded region bounded by the curve $C$ and the coordinate axes.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2009 Q6 [15]}}