| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2013 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Given factor, find all roots |
| Difficulty | Moderate -0.3 This is a standard C1 factor theorem question with clear guidance at each step. Part (i) involves routine verification and polynomial division, part (ii) requires testing simple integer values, and part (iii) involves recognizing an irreducible quadratic. While it requires multiple steps and some algebraic manipulation, the question structure is highly scaffolded with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02f Solve quadratic equations: including in a function of unknown1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(f(1) = 1 - 1 + 1 + 9 - 10 [= 0]\) | B1 | Allow for correct division of \(f(x)\) by \((x-1)\) showing there is no remainder, or for \((x-1)(x^3 + x + 10)\) found, showing it 'works' by multiplying out. Condone \(1^4 - 1^3 + 1^2 + 9 - 10\) |
| Attempt at division by \((x-1)\) as far as \(x^4 - x^3\) in working | M1 | Allow equiv for \((x+2)\) as far as \(x^4 + 2x^3\) in working, or for inspection with at least two terms of cubic factor correct. eg for inspection, M1 for two terms right and two wrong |
| Correctly obtaining \(x^3 + x + 10\) | A1 | Or \(x^3 - 3x^2 + 7x - 5\). If M0 and this division/factorising is done in part (ii) or (iii), allow SC1 if correct cubic obtained there; attach the relevant part to (i) with a formal chain link if not already seen in the image zone for (i) |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([g(-2) =] -8 - 2 + 10\) or \(f(-2) = 16 + 8 + 4 - 18 - 10\) | M1 | [In this scheme \(g(x) = x^3 + x + 10\)] Allow M1 for correct trials with at least two values of \(x\) (other than 1) using \(g(x)\) or \(f(x)\) or \(x^3 - 3x^2 + 7x - 5\). eg \(f(2) = 16-8+4+18-10\) or 20; \(f(3)=81-27+9+27-10\) or 80; \(f(0)=-10\); \(f(-1)=1+1+1-9-10\) or \(-16\). No ft from wrong cubic 'factors' from (i) |
| \(x = -2\) isw | A1 | Allow these marks if already earned in (i). NB factorising of \(x^3 + x + 10\) or \(x^3 - 3x^2 + 7x - 5\) in (ii) earns credit for (iii) |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Attempted division of \(x^3 + x + 10\) by \((x+2)\) as far as \(x^3 + 2x^2\) in working | M1 | Or \(x^3 - 3x^2 + 7x - 5\) by \((x-1)\) as far as \(x^3 - x^2\) in working, or inspection with at least two terms of quadratic factor correct. Alt method: allow M1 for attempted division of quartic by \(x^2 + x - 2\) as far as \(x^4 + x^3 - 2x^2\) in working, or inspection etc |
| Correctly obtaining \(x^2 - 2x + 5\) | A1 | Allow these first 2 marks if this has been done in (ii), even if not used here |
| Use of \(b^2 - 4ac\) with \(x^2 - 2x + 5\) | M1 | May be in attempt at formula (ignore rest of formula), or completing square form attempted, or attempt at calculus or symmetry to find min pt. NB M0 for use of \(b^2 - 4ac\) with cubic factor etc |
| \(b^2 - 4ac = 4 - 20 [= -16]\) | A1 | May be in formula; or \((x-1)^2 + 4\); or \(\min = (1, 4)\) |
| So only two real roots [of \(f(x)\)] [and hence no more linear factors] | A1 | Or no real roots of \(x^2 - 2x + 5 = 0\); allow this last mark if clear use of \(x^2 - 2x + 5 = 0\), even if error in \(b^2 - 4ac\), provided result negative, but no ft from wrong factor. Or \((x-1)^2 + 4\) is always positive so no real roots [of \((x-1)^2 + 4 = 0\)] and hence no linear factors], or similar conclusion from min pt. If last M1 not earned, allow SC1 for stating that the only factors of 5 are 1 and 5 and reasoning eg that \((x-1)(x-5)\) and \((x+1)(x+5)\) do not give \(x^2 - 2x + 5\) [hence \(x^2 - 2x + 5\) does not factorise] |
| [5] |
## Question 12 (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $f(1) = 1 - 1 + 1 + 9 - 10 [= 0]$ | B1 | Allow for correct division of $f(x)$ by $(x-1)$ showing there is no remainder, or for $(x-1)(x^3 + x + 10)$ found, showing it 'works' by multiplying out. Condone $1^4 - 1^3 + 1^2 + 9 - 10$ |
| Attempt at division by $(x-1)$ as far as $x^4 - x^3$ in working | M1 | Allow equiv for $(x+2)$ as far as $x^4 + 2x^3$ in working, or for inspection with at least two terms of cubic factor correct. eg for inspection, M1 for two terms right and two wrong |
| Correctly obtaining $x^3 + x + 10$ | A1 | Or $x^3 - 3x^2 + 7x - 5$. If M0 and this division/factorising is done in part (ii) or (iii), allow SC1 if correct cubic obtained there; attach the relevant part to (i) with a formal chain link if not already seen in the image zone for (i) |
| | **[3]** | |
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## Question 12 (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[g(-2) =] -8 - 2 + 10$ or $f(-2) = 16 + 8 + 4 - 18 - 10$ | M1 | [In this scheme $g(x) = x^3 + x + 10$] Allow M1 for correct trials with at least two values of $x$ (other than 1) using $g(x)$ or $f(x)$ or $x^3 - 3x^2 + 7x - 5$. eg $f(2) = 16-8+4+18-10$ or 20; $f(3)=81-27+9+27-10$ or 80; $f(0)=-10$; $f(-1)=1+1+1-9-10$ or $-16$. No ft from wrong cubic 'factors' from (i) |
| $x = -2$ isw | A1 | Allow these marks if already earned in (i). NB factorising of $x^3 + x + 10$ or $x^3 - 3x^2 + 7x - 5$ in (ii) earns credit for (iii) |
| | **[2]** | |
---
## Question 12 (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempted division of $x^3 + x + 10$ by $(x+2)$ as far as $x^3 + 2x^2$ in working | M1 | Or $x^3 - 3x^2 + 7x - 5$ by $(x-1)$ as far as $x^3 - x^2$ in working, or inspection with at least two terms of quadratic factor correct. Alt method: allow M1 for attempted division of quartic by $x^2 + x - 2$ as far as $x^4 + x^3 - 2x^2$ in working, or inspection etc |
| Correctly obtaining $x^2 - 2x + 5$ | A1 | Allow these first 2 marks if this has been done in (ii), even if not used here |
| Use of $b^2 - 4ac$ with $x^2 - 2x + 5$ | M1 | May be in attempt at formula (ignore rest of formula), or completing square form attempted, or attempt at calculus or symmetry to find min pt. NB M0 for use of $b^2 - 4ac$ with cubic factor etc |
| $b^2 - 4ac = 4 - 20 [= -16]$ | A1 | May be in formula; or $(x-1)^2 + 4$; or $\min = (1, 4)$ |
| So only two real roots [of $f(x)$] [and hence no more linear factors] | A1 | Or no real roots of $x^2 - 2x + 5 = 0$; allow this last mark if clear use of $x^2 - 2x + 5 = 0$, even if error in $b^2 - 4ac$, provided result negative, but no ft from wrong factor. Or $(x-1)^2 + 4$ is always positive so no real roots [of $(x-1)^2 + 4 = 0$] and hence no linear factors], or similar conclusion from min pt. If last M1 not earned, allow SC1 for stating that the only factors of 5 are 1 and 5 and reasoning eg that $(x-1)(x-5)$ and $(x+1)(x+5)$ do not give $x^2 - 2x + 5$ [hence $x^2 - 2x + 5$ does not factorise] |
| | **[5]** | |12 You are given that $\mathrm { f } ( x ) = x ^ { 4 } - x ^ { 3 } + x ^ { 2 } + 9 x - 10$.\\
(i) Show that $x = 1$ is a root of $\mathrm { f } ( x ) = 0$ and hence express $\mathrm { f } ( x )$ as a product of a linear factor and a cubic factor.\\
(ii) Hence or otherwise find another root of $\mathrm { f } ( x ) = 0$.\\
(iii) Factorise $\mathrm { f } ( x )$, showing that it has only two linear factors. Show also that $\mathrm { f } ( x ) = 0$ has only two real roots.
\section*{THERE ARE NO QUESTIONS WRITTEN ON THIS PAGE.}
\hfill \mbox{\textit{OCR MEI C1 2013 Q12 [10]}}