| Answer | Marks | Guidance |
|---|---|---|
| \(p(-1) = -1 + 7 - 6 = 0\) therefore \(x + 1\) is a factor | M1, A1 | Finding \(p(-1)\); Shown to \(= 0\) plus statement |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(p(x) = (x+1)(x^2 - x - 6)\) \(p(x) = (x+1)(x+2)(x-3)\) | M1, A1, A1 | Long division/inspection (2 terms correct); Quadratic factor correct; May earn M1,A1 for correct second factor then A1 for \((x+1)(x+2)(x-3)\) |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(A(-2,0)\) | B1 | Condone \(x = -2\) |
| 1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{x^4}{4} - \frac{7x^2}{2} - 6x\) (\(+c\)) (may have \(+c\) or not) | M1, A1, A1 | One term correct; Another term correct; All correct unsimplified |
| 3 | ||
| \(\left[\frac{81}{4} - \frac{63}{2} - 18\right] - \left[-\frac{1}{4} - \frac{7}{2} + 6\right] = -32\) | m1, A1 | \(F(3) - F(-1)\) attempted in correct order; CSO; OE |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Area of shaded region \(= 32\) | B1\(\uparrow\) | FT their (b)(ii) but positive value needed |
| 1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = 3x^2 - 7\) When \(x = -1\), gradient \(= -4\) | M1, A1, A1 | One term correct; All correct (no \(+ c\) etc); CSO |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Gradient of normal \(= \frac{1}{4}\) \(y = \) "their gradient"\((x \pm 1)\) \(y = \frac{1}{4}(x+1)\) | B1\(\uparrow\), M1, A1 | Must be finding normal, not tangent; CSO; any correct form eg \(4y - x = 1\) |
| 3 | ||
| 18 |
## Part (a)(i)
$p(-1) = -1 + 7 - 6 = 0$ therefore $x + 1$ is a factor | M1, A1 | Finding $p(-1)$; Shown to $= 0$ **plus statement**
| | 2 |
## Part (a)(ii)
$p(x) = (x+1)(x^2 - x - 6)$ $p(x) = (x+1)(x+2)(x-3)$ | M1, A1, A1 | Long division/inspection (2 terms correct); Quadratic factor correct; May earn M1,A1 for correct second factor then A1 for $(x+1)(x+2)(x-3)$
| | 3 |
## Part (b)(i)
$A(-2,0)$ | B1 | Condone $x = -2$
| | 1 |
## Part (b)(ii)
$\frac{x^4}{4} - \frac{7x^2}{2} - 6x$ ($+c$) (may have $+c$ or not) | M1, A1, A1 | One term correct; Another term correct; All correct unsimplified
| | 3 |
$\left[\frac{81}{4} - \frac{63}{2} - 18\right] - \left[-\frac{1}{4} - \frac{7}{2} + 6\right] = -32$ | m1, A1 | $F(3) - F(-1)$ attempted in correct order; CSO; OE
| | 5 |
## Part (b)(iii)
Area of shaded region $= 32$ | B1$\uparrow$ | FT their (b)(ii) but positive value needed
| | 1 |
## Part (b)(iv)
$\frac{dy}{dx} = 3x^2 - 7$ When $x = -1$, gradient $= -4$ | M1, A1, A1 | One term correct; All correct (no $+ c$ etc); CSO
| | 3 |
## Part (b)(v)
Gradient of normal $= \frac{1}{4}$ $y = $ "their gradient"$(x \pm 1)$ $y = \frac{1}{4}(x+1)$ | B1$\uparrow$, M1, A1 | Must be finding **normal**, not tangent; CSO; any correct form eg $4y - x = 1$
| | 3 |
| | 18 |
---6
\begin{enumerate}[label=(\alph*)]
\item The polynomial $\mathrm { p } ( x )$ is given by $\mathrm { p } ( x ) = x ^ { 3 } - 7 x - 6$.
\begin{enumerate}[label=(\roman*)]
\item Use the Factor Theorem to show that $x + 1$ is a factor of $\mathrm { p } ( x )$.
\item Express $\mathrm { p } ( x ) = x ^ { 3 } - 7 x - 6$ as the product of three linear factors.
\end{enumerate}\item The curve with equation $y = x ^ { 3 } - 7 x - 6$ is sketched below.\\
\includegraphics[max width=\textwidth, alt={}, center]{de4f827d-f237-488a-9177-3d85d0cb1771-4_403_762_651_641}
The curve cuts the $x$-axis at the point $A$ and the points $B ( - 1,0 )$ and $C ( 3,0 )$.
\begin{enumerate}[label=(\roman*)]
\item State the coordinates of the point $A$.
\item Find $\int _ { - 1 } ^ { 3 } \left( x ^ { 3 } - 7 x - 6 \right) \mathrm { d } x$.
\item Hence find the area of the shaded region bounded by the curve $y = x ^ { 3 } - 7 x - 6$ and the $x$-axis between $B$ and $C$.
\item Find the gradient of the curve $y = x ^ { 3 } - 7 x - 6$ at the point $B$.
\item Hence find an equation of the normal to the curve at the point $B$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2008 Q6 [18]}}