| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2008 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polynomial Division & Manipulation |
| Type | Sketching Polynomial Curves |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question testing standard C1 techniques: verifying a root by substitution, polynomial division to factorise, sketching a cubic with known roots, and function transformation. All parts are routine procedures with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02n Sketch curves: simple equations including polynomials1.02w Graph transformations: simple transformations of f(x) |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(-4)\) used | M1 | |
| \(-128 + 112 + 28 - 12 [= 0]\) | A1 | or B2 for \((x+4)(2x^2-x-3)\); or correct division with no remainder |
| Answer | Marks | Guidance |
|---|---|---|
| division of \(f(x)\) by \((x+4)\) | M1 | as far as \(2x^3 + 8x^2\) in working, or two terms of \(2x^2 - x - 3\) obtained by inspection |
| \(2x^2 - x - 3\) | A1 | \(2x^2 - x - 3\) seen implies M1A1 |
| \((x+1)(2x-3)\) | A1 | |
| \([f(x) =] (x+4)(x+1)(2x-3)\) | A1 | or B4; allow final A1 ft their factors if M1A1A0 earned |
| Answer | Marks | Guidance |
|---|---|---|
| sketch of cubic correct way up | G1 | ignore any graph of \(y = f(x-4)\) |
| through \(-12\) shown on \(y\) axis | G1 | or coords stated near graph |
| roots \(-4\), \(-1\), \(1.5\) or ft shown on \(x\) axis | G1 | or coords stated near graph; if no curve drawn, but intercepts marked on axes, can earn max of G0G1G1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(x(x-3)(2[x]-3)\) o.e. or \(x(x-3)(x-5.5)\) or ft their factors | M1 | or \(2(x-4)^3 + 7(x-4)^2 - 7(x-4) - 12\); or stating roots are 0, 3 and 5.5 or ft; condone one error |
| correct expansion of one pair of brackets ft from their factors | M1 | or for correct expn of \((x-4)^3\); or for showing \(g(0) = g(3) = g(5.5) = 0\) in given ans \(g(x)\) |
| correct completion to given answer | M1 | allow M2 for working backwards from given answer to \(x(x-3)(2x-11)\) and M1 for full completion with factors or roots |
# Question 11:
## Part (i)
| $f(-4)$ used | M1 | |
| $-128 + 112 + 28 - 12 [= 0]$ | A1 | or B2 for $(x+4)(2x^2-x-3)$; or correct division with no remainder |
## Part (ii)
| division of $f(x)$ by $(x+4)$ | M1 | as far as $2x^3 + 8x^2$ in working, or two terms of $2x^2 - x - 3$ obtained by inspection |
| $2x^2 - x - 3$ | A1 | $2x^2 - x - 3$ seen implies M1A1 |
| $(x+1)(2x-3)$ | A1 | |
| $[f(x) =] (x+4)(x+1)(2x-3)$ | A1 | or B4; allow final A1 ft their factors if M1A1A0 earned |
## Part (iii)
| sketch of cubic correct way up | G1 | ignore any graph of $y = f(x-4)$ |
| through $-12$ shown on $y$ axis | G1 | or coords stated near graph |
| roots $-4$, $-1$, $1.5$ or ft shown on $x$ axis | G1 | or coords stated near graph; if no curve drawn, but intercepts marked on axes, can earn max of G0G1G1 |
## Part (iv)
| $x(x-3)(2[x]-3)$ o.e. or $x(x-3)(x-5.5)$ or ft their factors | M1 | or $2(x-4)^3 + 7(x-4)^2 - 7(x-4) - 12$; or stating roots are 0, 3 and 5.5 or ft; condone one error |
| correct expansion of one pair of brackets ft from their factors | M1 | or for correct expn of $(x-4)^3$; or for showing $g(0) = g(3) = g(5.5) = 0$ in given ans $g(x)$ |
| correct completion to given answer | M1 | allow M2 for working backwards from given answer to $x(x-3)(2x-11)$ and M1 for full completion with factors or roots |
---11 You are given that $\mathrm { f } ( x ) = 2 x ^ { 3 } + 7 x ^ { 2 } - 7 x - 12$.\\
(i) Verify that $x = - 4$ is a root of $\mathrm { f } ( x ) = 0$.\\
(ii) Hence express $\mathrm { f } ( x )$ in fully factorised form.\\
(iii) Sketch the graph of $y = \mathrm { f } ( x )$.\\
(iv) Show that $\mathrm { f } ( x - 4 ) = 2 x ^ { 3 } - 17 x ^ { 2 } + 33 x$.
\hfill \mbox{\textit{OCR MEI C1 2008 Q11 [12]}}