OCR MEI C1 2013 June — Question 11 12 marks

Exam BoardOCR MEI
ModuleC1 (Core Mathematics 1)
Year2013
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCurve Sketching
TypeVertical translation of cubic with factorisation
DifficultyModerate -0.8 This is a straightforward C1 curve sketching question requiring routine skills: plotting a cubic from factored form, applying a horizontal translation to find roots, expanding brackets, and factorising using the factor theorem. All parts follow standard procedures with no problem-solving insight needed, making it easier than average but not trivial due to the multi-step algebraic manipulation required.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02n Sketch curves: simple equations including polynomials1.02w Graph transformations: simple transformations of f(x)

You are given that \(\text{f}(x) = (2x - 3)(x + 2)(x + 4)\).
  1. Sketch the graph of \(y = \text{f}(x)\). [3]
  2. State the roots of \(\text{f}(x - 2) = 0\). [2]
  3. You are also given that \(\text{g}(x) = \text{f}(x) + 15\).
    1. Show that \(\text{g}(x) = 2x^3 + 9x^2 - 2x - 9\). [2]
    2. Show that \(\text{g}(1) = 0\) and hence factorise \(\text{g}(x)\) completely. [5]

You are given that $\text{f}(x) = (2x - 3)(x + 2)(x + 4)$.
\begin{enumerate}[label=(\roman*)]
\item Sketch the graph of $y = \text{f}(x)$. [3]
\item State the roots of $\text{f}(x - 2) = 0$. [2]
\item You are also given that $\text{g}(x) = \text{f}(x) + 15$.
\begin{enumerate}[label=(\Alph*)]
\item Show that $\text{g}(x) = 2x^3 + 9x^2 - 2x - 9$. [2]
\item Show that $\text{g}(1) = 0$ and hence factorise $\text{g}(x)$ completely. [5]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{OCR MEI C1 2013 Q11 [12]}}