OCR C1 2006 June — Question 4 8 marks

Exam BoardOCR
ModuleC1 (Core Mathematics 1)
Year2006
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCurve Sketching
TypeReflection or vertical transformation
DifficultyEasy -1.2 This is a straightforward C1 question requiring basic algebraic expansion, identification of roots from factored form, and sketching cubic curves with known intercepts. All steps are routine with no problem-solving or novel insight required—easier than average A-level questions.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02n Sketch curves: simple equations including polynomials

  1. By expanding the brackets, show that $$(x - 4)(x - 3)(x + 1) = x^3 - 6x^2 + 5x + 12.$$ [3]
  2. Sketch the curve $$y = x^3 - 6x^2 + 5x + 12,$$ giving the coordinates of the points where the curve meets the axes. Label the curve \(C_1\). [3]
  3. On the same diagram as in part (ii), sketch the curve $$y = -x^3 + 6x^2 - 5x - 12.$$ Label this curve \(C_2\). [2]

(i)
AnswerMarks Guidance
\((x-4)(x-3)(x+1) = (x^2 - 7x + 12)(x+1)\)B1
\(= x^3 + x^2 - 7x^2 - 7x + 12x + 12\)M1
\(= x^3 - 6x^2 + 5x + 12\)A1 3 marks; \(x^2 - 7x + 12\) or \(x^2 - 2x - 3\) or \(x^2 - 3x - 4\) seen; Attempt to multiply a quadratic by a linear factor or attempt to list an 8 term expansion of all 3 brackets; \(x^3 - 6x^2 + 5x + 12\) (AG) obtained (no wrong working seen)
(ii)-(iii)
AnswerMarks Guidance
Graph with +ve cubic with 3 roots (not 3 line segments)B1
\((0, 12)\) labelled or indicated on \(y\)-axisB1
\((-1, 0)\), \((3, 0)\), \((4, 0)\) labelled or indicated on \(x\)-axisB1 3 marks total
Reflect their (ii) in either \(x\)- or \(y\)-axisM1 2 marks
Reflect their (ii) in \(x\)-axisA1
## (i)
$(x-4)(x-3)(x+1) = (x^2 - 7x + 12)(x+1)$ | B1 |
$= x^3 + x^2 - 7x^2 - 7x + 12x + 12$ | M1 |
$= x^3 - 6x^2 + 5x + 12$ | A1 | 3 marks; $x^2 - 7x + 12$ or $x^2 - 2x - 3$ or $x^2 - 3x - 4$ seen; Attempt to multiply a quadratic by a linear factor or attempt to list an 8 term expansion of all 3 brackets; $x^3 - 6x^2 + 5x + 12$ (AG) obtained (no wrong working seen)

## (ii)-(iii)
Graph with +ve cubic with 3 roots (not 3 line segments) | B1 |
$(0, 12)$ labelled or indicated on $y$-axis | B1 |
$(-1, 0)$, $(3, 0)$, $(4, 0)$ labelled or indicated on $x$-axis | B1 | 3 marks total

Reflect their (ii) in either $x$- or $y$-axis | M1 | 2 marks

Reflect their (ii) in $x$-axis | A1 |

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\begin{enumerate}[label=(\roman*)]
\item By expanding the brackets, show that
$$(x - 4)(x - 3)(x + 1) = x^3 - 6x^2 + 5x + 12.$$ [3]

\item Sketch the curve
$$y = x^3 - 6x^2 + 5x + 12,$$
giving the coordinates of the points where the curve meets the axes. Label the curve $C_1$. [3]

\item On the same diagram as in part (ii), sketch the curve
$$y = -x^3 + 6x^2 - 5x - 12.$$
Label this curve $C_2$. [2]
\end{enumerate}

\hfill \mbox{\textit{OCR C1 2006 Q4 [8]}}