OCR MEI C1 — Question 2 12 marks

Exam BoardOCR MEI
ModuleC1 (Core Mathematics 1)
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPolynomial Division & Manipulation
TypeSketching Polynomial Curves
DifficultyModerate -0.3 This is a structured multi-part question covering standard C1 topics: forming a cubic from roots, expanding brackets, graphical intersection, and solving a quadratic. Each part guides students through routine techniques with no novel problem-solving required. While it has multiple steps (4 parts), each individual step is straightforward—slightly easier than a typical average A-level question due to the scaffolding provided.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02n Sketch curves: simple equations including polynomials1.02q Use intersection points: of graphs to solve equations

2 \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{a94593ca-d84d-4747-ac19-b05389e85b3c-1_781_1462_1118_342} \captionsetup{labelformat=empty} \caption{Fig. 12}
\end{figure} Fig. 12 shows the graph of a cubic curve. It intersects the axes at \(( - 5,0 ) , ( - 2,0 ) , ( 1.5,0 )\) and \(( 0 , - 30 )\).
  1. Use the intersections with both axes to express the equation of the curve in a factorised form.
  2. Hence show that the equation of the curve may be written as \(y = 2 x ^ { 3 } + 11 x ^ { 2 } - x - 30\).
  3. Draw the line \(y = 5 x + 10\) accurately on the graph. The curve and this line intersect at ( \(- 2,0\) ); find graphically the \(x\)-coordinates of the other points of intersection.
  4. Show algebraically that the \(x\)-coordinates of the other points of intersection satisfy the equation $$2 x ^ { 2 } + 7 x - 20 = 0$$ Hence find the exact values of the \(x\)-coordinates of the other points of intersection.

Question 2:
Part (i)
AnswerMarks Guidance
AnswerMark Guidance
\(y = (x+5)(x+2)(2x-3)\) or \(y = 2(x+5)(x+2)(x-\frac{3}{2})\)2 M1 for \(y=(x+5)(x+2)(x-\frac{3}{2})\) or \((x+5)(x+2)(2x-3)\) with no equation or \((x+5)(x+2)(2x-3)=0\); but M0 for \(y=(x+5)(x+2)(2x-3)-30\) etc; allow "\(f(x)=\)" instead of "\(y=\)"; ignore further work towards (ii)
[2 marks]
Part (ii)
AnswerMarks Guidance
AnswerMark Guidance
Correct expansion of a pair of linear two-term factors ft iswM1 ft their factors from (i); need not be simplified; may be seen in a grid; allow only first M1 for expansion if their (i) has an extra \(-30\) etc
Correct expansion of correct linear and quadratic factors, completion to \(y = 2x^3 + 11x^2 - x - 30\)M1 Must be working for this step before given answer; or for direct expansion of all three factors, allow M2 for \(2x^3 + 10x^2 + 4x^2 - 3x^2 + 20x - 15x - 6x - 30\) oe (M1 if one error); condone omission of "\(y=\)" or "\(=0\)"
[2 marks]
Part (iii)
AnswerMarks Guidance
AnswerMark Guidance
Ruled line drawn through \((-2, 0)\) and \((0, 10)\), long enough to intersect curve at least twiceB1 Tolerance half a small square at \((-2,0)\) and \((0,10)\); accept in coordinate form ignoring any \(y\) coordinates
\(-5.3\) to \(-5.4\) and \(1.8\) to \(1.9\)B2 B1 for one correct; ignore solution \(-2\); allow B1 for both values correct but one extra or wrong "coordinate" form such as \((1.8, -5.3)\)
[3 marks]
Part (iv)
AnswerMarks Guidance
AnswerMark Guidance
\(2x^3 + 11x^2 - x - 30 = 5x + 10\)M1 For equating curve and line; correct equations only
\(2x^3 + 11x^2 - 6x - 40\ [=0]\)M1 For rearrangement to zero, condoning one error
Division by \((x+2)\) and correctly obtaining \(2x^2 + x - 20\)M1 Or showing that \((x+2)(2x^2+7x-20) = 2x^3 + 11x^2 - 6x - 40\) with supporting working
Substitution into quadratic formula or completing the square as far as \(x + \frac{7}{4}^2 = \frac{209}{16}\) oeM1 Condone one error; using given \(2x^2 + 7x - 20 = 0\)
\([x=]\dfrac{-7 \pm \sqrt{209}}{4}\) oe iswA1 Dependent only on 4th M1
[5 marks]
# Question 2:

## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $y = (x+5)(x+2)(2x-3)$ or $y = 2(x+5)(x+2)(x-\frac{3}{2})$ | 2 | M1 for $y=(x+5)(x+2)(x-\frac{3}{2})$ or $(x+5)(x+2)(2x-3)$ with no equation or $(x+5)(x+2)(2x-3)=0$; but M0 for $y=(x+5)(x+2)(2x-3)-30$ etc; allow "$f(x)=$" instead of "$y=$"; ignore further work towards (ii) |

**[2 marks]**

## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Correct expansion of a pair of linear two-term factors ft isw | M1 | ft their factors from (i); need not be simplified; may be seen in a grid; allow only first M1 for expansion if their (i) has an extra $-30$ etc |
| Correct expansion of correct linear and quadratic factors, completion to $y = 2x^3 + 11x^2 - x - 30$ | M1 | Must be working for this step before given answer; or for direct expansion of all three factors, allow M2 for $2x^3 + 10x^2 + 4x^2 - 3x^2 + 20x - 15x - 6x - 30$ oe (M1 if one error); condone omission of "$y=$" or "$=0$" |

**[2 marks]**

## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Ruled line drawn through $(-2, 0)$ and $(0, 10)$, long enough to intersect curve at least twice | B1 | Tolerance half a small square at $(-2,0)$ and $(0,10)$; accept in coordinate form ignoring any $y$ coordinates |
| $-5.3$ to $-5.4$ and $1.8$ to $1.9$ | B2 | B1 for one correct; ignore solution $-2$; allow B1 for both values correct but one extra or wrong "coordinate" form such as $(1.8, -5.3)$ |

**[3 marks]**

## Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| $2x^3 + 11x^2 - x - 30 = 5x + 10$ | M1 | For equating curve and line; correct equations only |
| $2x^3 + 11x^2 - 6x - 40\ [=0]$ | M1 | For rearrangement to zero, condoning one error |
| Division by $(x+2)$ and correctly obtaining $2x^2 + x - 20$ | M1 | Or showing that $(x+2)(2x^2+7x-20) = 2x^3 + 11x^2 - 6x - 40$ with supporting working |
| Substitution into quadratic formula or completing the square as far as $x + \frac{7}{4}^2 = \frac{209}{16}$ oe | M1 | Condone one error; using given $2x^2 + 7x - 20 = 0$ |
| $[x=]\dfrac{-7 \pm \sqrt{209}}{4}$ oe isw | A1 | Dependent only on 4th M1 |

**[5 marks]**

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2

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{a94593ca-d84d-4747-ac19-b05389e85b3c-1_781_1462_1118_342}
\captionsetup{labelformat=empty}
\caption{Fig. 12}
\end{center}
\end{figure}

Fig. 12 shows the graph of a cubic curve. It intersects the axes at $( - 5,0 ) , ( - 2,0 ) , ( 1.5,0 )$ and $( 0 , - 30 )$.\\
(i) Use the intersections with both axes to express the equation of the curve in a factorised form.\\
(ii) Hence show that the equation of the curve may be written as $y = 2 x ^ { 3 } + 11 x ^ { 2 } - x - 30$.\\
(iii) Draw the line $y = 5 x + 10$ accurately on the graph. The curve and this line intersect at ( $- 2,0$ ); find graphically the $x$-coordinates of the other points of intersection.\\
(iv) Show algebraically that the $x$-coordinates of the other points of intersection satisfy the equation

$$2 x ^ { 2 } + 7 x - 20 = 0$$

Hence find the exact values of the $x$-coordinates of the other points of intersection.

\hfill \mbox{\textit{OCR MEI C1  Q2 [12]}}