OCR MEI C1 2010 June — Question 6 5 marks

Exam BoardOCR MEI
ModuleC1 (Core Mathematics 1)
Year2010
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFactor & Remainder Theorem
TypeFind constants from coefficient conditions
DifficultyModerate -0.3 This is a straightforward two-equation system requiring basic polynomial expansion and the remainder theorem. Students multiply out to find one equation (coefficient of x³), apply remainder theorem for the second equation, then solve simultaneously. While it involves multiple steps, each technique is standard C1 material with no conceptual challenges or novel insights required.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem

You are given that • the coefficient of \(x^3\) in the expansion of \((5 + 2x^2)(x^3 + kx + m)\) is 29, • when \(x^3 + kx + m\) is divided by \((x - 3)\), the remainder is 59. Find the values of \(k\) and \(m\). [5]

AnswerMarks Guidance
\(5 + 2k\) soiM1 allow M1 for expansion with \(5x^3 + 2kx^3\) and no other \(x^3\) terms or M1 for \((29 - 5)/2\) soi
\(k = 12\)A1
attempt at \(f(3)\)M1 must substitute 3 for \(x\) in cubic not product or long division as far as obtaining \(x^2 + 3x\) in quotient
\(27 + 36 + m = 59\) o.e.A1 or from division \(m - (-63) = 59\) o.e. or for \(27 + 3k + m = 59\) or ft their \(k\)
\(m = -4\) caoA1
$5 + 2k$ soi | M1 | allow M1 for expansion with $5x^3 + 2kx^3$ and no other $x^3$ terms or M1 for $(29 - 5)/2$ soi

$k = 12$ | A1 |

attempt at $f(3)$ | M1 | must substitute 3 for $x$ in cubic not product or long division as far as obtaining $x^2 + 3x$ in quotient

$27 + 36 + m = 59$ o.e. | A1 | or from division $m - (-63) = 59$ o.e. or for $27 + 3k + m = 59$ or ft their $k$

$m = -4$ cao | A1 |
You are given that
• the coefficient of $x^3$ in the expansion of $(5 + 2x^2)(x^3 + kx + m)$ is 29,
• when $x^3 + kx + m$ is divided by $(x - 3)$, the remainder is 59.

Find the values of $k$ and $m$. [5]

\hfill \mbox{\textit{OCR MEI C1 2010 Q6 [5]}}