OCR MEI C1 2016 June — Question 8 5 marks

Exam BoardOCR MEI
ModuleC1 (Core Mathematics 1)
Year2016
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFactor & Remainder Theorem
TypeSingle polynomial, two remainder/factor conditions
DifficultyModerate -0.8 This is a straightforward application of the remainder theorem requiring students to set up two simple linear equations from the given conditions and solve simultaneously. The polynomial structure is simple (only two unknowns), the arithmetic is clean, and the method is direct with no conceptual challenges beyond basic remainder theorem recall.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem

8 You are given that \(\mathrm { f } ( x ) = x ^ { 3 } + a x + c\) and that \(\mathrm { f } ( 2 ) = 11\). The remainder when \(\mathrm { f } ( x )\) is divided by \(( x + 1 )\) is 8 . Find the values of \(a\) and \(c\).

Question 8:
AnswerMarks Guidance
\(8+2a+c=11\)B1 accept \(2^3\) instead of 8
\(-1-a+c=8\)B1 or \(c-(a+1)=8\) oe; accept \((-1)^3\) instead of \(-1\)
Correct method for eliminating one variable, condoning one further errorM1 dep on two equations in \(a\) and \(c\) and at least B1 earned
\(a=-2,\ c=7\)A2 (A1 for one correct) [5]
# Question 8:
$8+2a+c=11$ | B1 | accept $2^3$ instead of 8
$-1-a+c=8$ | B1 | or $c-(a+1)=8$ oe; accept $(-1)^3$ instead of $-1$
Correct method for eliminating one variable, condoning one further error | M1 | dep on two equations in $a$ and $c$ and at least B1 earned
$a=-2,\ c=7$ | A2 (**A1** for one correct) | [5]

---
8 You are given that $\mathrm { f } ( x ) = x ^ { 3 } + a x + c$ and that $\mathrm { f } ( 2 ) = 11$. The remainder when $\mathrm { f } ( x )$ is divided by $( x + 1 )$ is 8 . Find the values of $a$ and $c$.

\hfill \mbox{\textit{OCR MEI C1 2016 Q8 [5]}}