| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2006 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Unknown constant, verify then factorise |
| Difficulty | Moderate -0.3 This is a straightforward C2 question testing standard factor theorem application to find k, followed by routine factorisation, integration, and interpretation of signed area. All parts are textbook exercises requiring no novel insight, though the multi-part structure and integration component make it slightly more substantial than the most basic factor theorem questions. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| \(-2 + k + 1 + 6 = 0 \Rightarrow k = -5\) | M1 | For attempting f(−1) |
| A1 | For equating f(-1) to 0 and deducing the correct value of \(k\) AG | |
| OR: Match coefficients and attempt \(k\) | M1 | |
| A1 | Show \(k = -5\) | |
| B2 | Following division, state remainder is 0, hence (x + 1) is a factor | |
| OR: \((x + 1)(2x^2 - 7x + 6)\) | B1 | For correct leading term \(2x^2\) |
| M1 | For attempt at complete division by f(x) by (x + 1) or equiv. | |
| A1 | For completely correct quadratic factor | |
| A1 | For all three factors correct | |
| OR: \(f(2) = 16 - 20 - 2 + 6 = 0\) | M1 | For further relevant use of the factor theorem |
| Hence \((x - 2)\) is a factor | A1 | For correct identification of factor \((x - 2)\) |
| Third factor is \((2x - 3)\) | M1 | For any method for the remaining factor |
| \(f(x) = (x + 1)(x - 2)(2x - 3)\) | A1 | For all three factors correct |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int f(x)\text{d}x = \left[\frac{1}{2}x^4 - \frac{5}{3}x^3 - \frac{1}{3}x^2 + 6x\right]_{-1}^2\) | B1√ | For any two terms integrated correctly |
| B1√ | For all four terms integrated correctly | |
| \(= \left(8 - \frac{40}{3} - 2 + 12\right) - \left(-\frac{1}{2} + \frac{5}{3} - \frac{1}{2} - 6\right) = 9\) | M1 | For evaluation of F(2) − F(−1) |
| A1 | For correct value 9 |
| Answer | Marks | Guidance |
|---|---|---|
| [Sketch showing positive cubic with three distinct, non-zero roots] | B1 | For sketch of positive cubic, with three distinct, non-zero roots |
| B1 | For correct explanation that some of the area is below the axis |
**(i)**
$-2 + k + 1 + 6 = 0 \Rightarrow k = -5$ | M1 | For attempting f(−1)
| A1 | For equating f(-1) to 0 and deducing the correct value of $k$ AG
**OR:** Match coefficients and attempt $k$ | M1 |
| A1 | Show $k = -5$
| B2 | Following division, state remainder is 0, hence (x + 1) is a factor
**OR:** $(x + 1)(2x^2 - 7x + 6)$ | B1 | For correct leading term $2x^2$
| M1 | For attempt at complete division by f(x) by (x + 1) or equiv.
| A1 | For completely correct quadratic factor
| A1 | For all three factors correct
**OR:** $f(2) = 16 - 20 - 2 + 6 = 0$ | M1 | For further relevant use of the factor theorem
Hence $(x - 2)$ is a factor | A1 | For correct identification of factor $(x - 2)$
Third factor is $(2x - 3)$ | M1 | For any method for the remaining factor
$f(x) = (x + 1)(x - 2)(2x - 3)$ | A1 | For all three factors correct
**(ii)**
$\int f(x)\text{d}x = \left[\frac{1}{2}x^4 - \frac{5}{3}x^3 - \frac{1}{3}x^2 + 6x\right]_{-1}^2$ | B1√ | For any two terms integrated correctly
| B1√ | For all four terms integrated correctly
$= \left(8 - \frac{40}{3} - 2 + 12\right) - \left(-\frac{1}{2} + \frac{5}{3} - \frac{1}{2} - 6\right) = 9$ | M1 | For evaluation of F(2) − F(−1)
| A1 | For correct value 9
**(iii)**
[Sketch showing positive cubic with three distinct, non-zero roots] | B1 | For sketch of positive cubic, with three distinct, non-zero roots
| B1 | For correct explanation that some of the area is below the axis
---8 The cubic polynomial $2 x ^ { 3 } + k x ^ { 2 } - x + 6$ is denoted by $\mathrm { f } ( x )$. It is given that $( x + 1 )$ is a factor of $\mathrm { f } ( x )$.\\
(i) Show that $k = - 5$, and factorise $\mathrm { f } ( x )$ completely.\\
(ii) Find $\int _ { - 1 } ^ { 2 } f ( x ) \mathrm { d } x$.\\
(iii) Explain with the aid of a sketch why the answer to part (ii) does not give the area of the region between the curve $y = \mathrm { f } ( x )$ and the $x$-axis for $- 1 \leqslant x \leqslant 2$.
\section*{[Question 9 is printed overleaf.]}
\hfill \mbox{\textit{OCR C2 2006 Q8 [12]}}