OCR MEI C1 2007 June — Question 4 3 marks

Exam BoardOCR MEI
ModuleC1 (Core Mathematics 1)
Year2007
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFactor & Remainder Theorem
TypeTwo factors given
DifficultyModerate -0.8 This is a straightforward application of the factor theorem requiring two simple substitutions: f(0)=6 immediately gives c=6, then f(2)=0 gives a linear equation in k. The algebra is minimal and the method is direct textbook application with no problem-solving insight needed.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem

4 You are given that \(\mathrm { f } ( x ) = x ^ { 3 } + k x + c\). The value of \(\mathrm { f } ( 0 )\) is 6, and \(x - 2\) is a factor of \(\mathrm { f } ( x )\).
Find the values of \(k\) and \(c\).

Question 4:
AnswerMarks Guidance
\(f(0) = c = 6\)B1 \(c = 6\)
\(f(2) = 8 + 2k + 6 = 0\)M1 Substituting \(x = 2\)
\(2k = -14\), \(k = -7\)A1 \(k = -7\)
## Question 4:
$f(0) = c = 6$ | B1 | $c = 6$
$f(2) = 8 + 2k + 6 = 0$ | M1 | Substituting $x = 2$
$2k = -14$, $k = -7$ | A1 | $k = -7$

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4 You are given that $\mathrm { f } ( x ) = x ^ { 3 } + k x + c$. The value of $\mathrm { f } ( 0 )$ is 6, and $x - 2$ is a factor of $\mathrm { f } ( x )$.\\
Find the values of $k$ and $c$.

\hfill \mbox{\textit{OCR MEI C1 2007 Q4 [3]}}