OCR MEI C1 2008 January — Question 6 3 marks

Exam BoardOCR MEI
ModuleC1 (Core Mathematics 1)
Year2008
SessionJanuary
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFactor & Remainder Theorem
TypeSingle remainder condition to find constant
DifficultyModerate -0.8 This is a straightforward application of the Remainder Theorem requiring only substitution and basic algebra. Students substitute x=2, set the result equal to 3, and solve for k in one step. It's simpler than average A-level questions as it involves minimal calculation and no problem-solving insight.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem
6 When \(x ^ { 3 } + k x + 7\) is divided by \(( x - 2 )\), the remainder is 3 . Find the value of \(k\).
AnswerMarks Guidance
Answer/Working: \(f(2)\) used; \(2^3 + 2k + 7 = 3\); \(k = -6\)Marks: M1, M1, A1 Guidance: or division by \(x - 2\) as far as \(x^2 + 2x\) obtained correctly; or remainder \(3 = 2(4 + k) + 7\) o.e. 2nd M1 dep on first 
**Answer/Working:** $f(2)$ used; $2^3 + 2k + 7 = 3$; $k = -6$ | **Marks:** M1, M1, A1 | **Guidance:** or division by $x - 2$ as far as $x^2 + 2x$ obtained correctly; or remainder $3 = 2(4 + k) + 7$ o.e. 2nd M1 dep on first | **Total:** 3
6 When $x ^ { 3 } + k x + 7$ is divided by $( x - 2 )$, the remainder is 3 . Find the value of $k$.

\hfill \mbox{\textit{OCR MEI C1 2008 Q6 [3]}}
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