OCR C1 — Question 8 12 marks

Exam BoardOCR
ModuleC1 (Core Mathematics 1)
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCompleting the square and sketching
TypeSketch quadratic curve
DifficultyModerate -0.3 This is a standard C1 completing the square question with routine sketching and distance calculation. Part (i) is textbook completing the square, part (ii) is direct application of the result, and part (iii) requires finding roots using the quadratic formula and calculating distance—all standard techniques with no novel insight required. Slightly easier than average due to being highly procedural.
Spec1.02e Complete the square: quadratic polynomials and turning points1.02f Solve quadratic equations: including in a function of unknown1.02n Sketch curves: simple equations including polynomials

  1. Express \(3x^2 - 12x + 11\) in the form \(a(x + b)^2 + c\). [4]
  2. Sketch the curve with equation \(y = 3x^2 - 12x + 11\), showing the coordinates of the minimum point of the curve. [3]
Given that the curve \(y = 3x^2 - 12x + 11\) crosses the \(x\)-axis at the points \(A\) and \(B\), \begin{enumerate}[label=(\roman*)] \setcounter{enumi}{2} \item find the length \(AB\) in the form \(k\sqrt{3}\). [5]
AnswerMarks Guidance
(i) \(= 3[x^2 - 4x] + 11\)M1
\(= 3[(x-2)^2 - 4] + 11\)M1
\(= 3(x-2)^2 - 1\)A2
(ii) [Graph showing parabola with vertex at \((2, -1)\), opening upwards, crossing y-axis at approximately \((0, 11)\)]B3
(iii) \(3(x-2)^2 - 1 = 0\)M1
\((x-2)^2 = \frac{1}{3}\)
\(x = 2 \pm \frac{1}{\sqrt{3}} = 2 \pm \frac{1}{\sqrt{3}}\)M1, A1
\(AB = (2 + \frac{1}{\sqrt{3}}) - (2 - \frac{1}{\sqrt{3}}) = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}\)M1, A1 (12)
**(i)** $= 3[x^2 - 4x] + 11$ | M1 |
$= 3[(x-2)^2 - 4] + 11$ | M1 |
$= 3(x-2)^2 - 1$ | A2 |

**(ii)** [Graph showing parabola with vertex at $(2, -1)$, opening upwards, crossing y-axis at approximately $(0, 11)$] | B3 |

**(iii)** $3(x-2)^2 - 1 = 0$ | M1 |
$(x-2)^2 = \frac{1}{3}$ | |
$x = 2 \pm \frac{1}{\sqrt{3}} = 2 \pm \frac{1}{\sqrt{3}}$ | M1, A1 |
$AB = (2 + \frac{1}{\sqrt{3}}) - (2 - \frac{1}{\sqrt{3}}) = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$ | M1, A1 | **(12)** |

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\begin{enumerate}[label=(\roman*)]
\item Express $3x^2 - 12x + 11$ in the form $a(x + b)^2 + c$. [4]
\item Sketch the curve with equation $y = 3x^2 - 12x + 11$, showing the coordinates of the minimum point of the curve. [3]
\end{enumerate}

Given that the curve $y = 3x^2 - 12x + 11$ crosses the $x$-axis at the points $A$ and $B$,

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item find the length $AB$ in the form $k\sqrt{3}$. [5]
</enumerate}

\hfill \mbox{\textit{OCR C1  Q8 [12]}}