| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polynomial Division & Manipulation |
| Type | Sketching Polynomial Curves |
| Difficulty | Moderate -0.3 This is a structured multi-part question with clear signposting through each step. Part (i) is routine factor theorem application, (ii) guides students to complete factorization with a given form, (iii) is standard curve sketching from factored form, and (iv) is straightforward definite integration. While it covers multiple techniques, each part is scaffolded and represents standard C2 material with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02n Sketch curves: simple equations including polynomials1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks |
|---|---|
| \(f(1) = 1 - 9 + 24 - 16 = 0\) | B1 |
| \(\therefore (x-1)\) is a factor of \(f(x)\) | B1 |
| Answer | Marks |
|---|---|
| \(x - 1 \overline{)\, x^3 - 9x^2 + 24x - 16}\) giving \(x^2 - 8x + 16\) | M1 A1 |
| \(f(x) = (x-1)(x^2 - 8x + 16)\) | |
| \(f(x) = (x-1)(x-4)^2 \quad [p = -1,\ q = -4]\) | M1 A1 |
| Answer | Marks |
|---|---|
| Graph showing correct shape with roots at \(x=1\) (crossing) and \(x=4\) (touching), crossing \(y\)-axis below origin | B2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \int_1^4 (x^3 - 9x^2 + 24x - 16)\, dx\) | ||
| \(= \left[\frac{1}{4}x^4 - 3x^3 + 12x^2 - 16x\right]_1^4\) | M1 A2 | |
| \(= [(64 - 192 + 192 - 64) - (\frac{1}{4} - 3 + 12 - 16)]\) | M1 | |
| \(= 6\frac{3}{4}\) | A1 | (13) |
# Question 9:
## Part (i):
| $f(1) = 1 - 9 + 24 - 16 = 0$ | B1 | |
| $\therefore (x-1)$ is a factor of $f(x)$ | B1 | |
## Part (ii):
| $x - 1 \overline{)\, x^3 - 9x^2 + 24x - 16}$ giving $x^2 - 8x + 16$ | M1 A1 | |
| $f(x) = (x-1)(x^2 - 8x + 16)$ | | |
| $f(x) = (x-1)(x-4)^2 \quad [p = -1,\ q = -4]$ | M1 A1 | |
## Part (iii):
| Graph showing correct shape with roots at $x=1$ (crossing) and $x=4$ (touching), crossing $y$-axis below origin | B2 | |
## Part (iv):
| $= \int_1^4 (x^3 - 9x^2 + 24x - 16)\, dx$ | | |
| $= \left[\frac{1}{4}x^4 - 3x^3 + 12x^2 - 16x\right]_1^4$ | M1 A2 | |
| $= [(64 - 192 + 192 - 64) - (\frac{1}{4} - 3 + 12 - 16)]$ | M1 | |
| $= 6\frac{3}{4}$ | A1 | **(13)** |
**Total: (72)**9.
$$f ( x ) = x ^ { 3 } - 9 x ^ { 2 } + 24 x - 16$$
(i) Evaluate $\mathrm { f } ( 1 )$ and hence state a linear factor of $\mathrm { f } ( x )$.\\
(ii) Show that $\mathrm { f } ( x )$ can be expressed in the form
$$\mathrm { f } ( x ) = ( x + p ) ( x + q ) ^ { 2 }$$
where $p$ and $q$ are integers to be found.\\
(iii) Sketch the curve $y = \mathrm { f } ( x )$.\\
(iv) Using integration, find the area of the region enclosed by the curve $y = \mathrm { f } ( x )$ and the $x$-axis.
\hfill \mbox{\textit{OCR C2 Q9 [13]}}