OCR MEI C1 2013 June — Question 6 4 marks

Exam BoardOCR MEI
ModuleC1 (Core Mathematics 1)
Year2013
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicBinomial Theorem (positive integer n)
TypeStandard binomial expansion coefficient
DifficultyModerate -0.8 This is a straightforward application of the binomial theorem requiring identification of the correct term and calculation using the binomial coefficient formula. It's slightly easier than average because it's a single-step problem with clear methodology, though students must handle the negative coefficient and power of 2 carefully in the arithmetic.
Spec1.04a Binomial expansion: (a+b)^n for positive integer n

Find the coefficient of \(x^3\) in the binomial expansion of \((2 - 4x)^5\). [4]

AnswerMarks Guidance
\(-2560\) www4 B3 for 2560 from correct term (NB coefficient of \(x^3\) is 2560); or B3 for neg answer following \(10 \times 4 \times -64\) and then an error in multiplication; or M2 for \(10 \times 2^2 \times (-4)^3\) oe; must have multn signs or be followed by a clear attempt at multn; or M1 for \(2^2 \times (-4)^3\) oe (condone missing brackets) or for 10 used or for 1 5 10 10 5 1 seen; for those who find the coefft of \(x^2\) instead: allow M1 for 10 used or for 1 5 10 10 5 1 seen; and a further SC1 if they get 1280, similarly for finding coefficient of \(x^4\) as 2560
[4]
Question 7 (i)
AnswerMarks Guidance
\(5^{\frac{1}{3}}\) oe or \(k = 7/2\) oe2 M1 for \(125 = 5^3\) or \(\sqrt{5} = 5^{\frac{1}{2}}\) soi
[2]
Question 7 (ii)
AnswerMarks Guidance
attempting to multiply numerator and denominator of fraction by \(1 + 2\sqrt{5}\)M1 some cands are incorporating the \(10 + 7\sqrt{5}\) into the fraction. The M1s are available even if this is done wrongly or if \(10 + 7\sqrt{5}\) is also multiplied by \(1 + 2\sqrt{5}\)
denominator \(= -19\) soiM1 must be obtained correctly, but independent of first M1; eg M1 for denominator of 19 with a minus sign in front of whole expression or with attempt to change signs in numerator
\(8 + 3\sqrt{5}\)A1 [3]
$-2560$ www | 4 | B3 for 2560 from correct term (NB coefficient of $x^3$ is 2560); or B3 for neg answer following $10 \times 4 \times -64$ and then an error in multiplication; or M2 for $10 \times 2^2 \times (-4)^3$ oe; must have multn signs or be followed by a clear attempt at multn; or M1 for $2^2 \times (-4)^3$ oe (condone missing brackets) or for 10 used or for 1 5 10 10 5 1 seen; for those who find the coefft of $x^2$ instead: allow M1 for 10 used or for 1 5 10 10 5 1 seen; and a further SC1 if they get 1280, similarly for finding coefficient of $x^4$ as 2560 | ignore terms for other powers; condone $x^3$ included; but eg $10 \times 4 \times -64 = 40 - 64 = -24$ gets M2 only; condone missing brackets eg allow M2 for $10 \times 2^2 \times -4x^3$ or $C_3$ or factorial notation is not sufficient but accept $\frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1 \times 3 \times 2 \times 1}$ oe; 10 may be unsimplified, as above; M1 only for eg 10, $2^2$ and $-4x^3$ seen in table with no multn signs or evidence of attempt at multn [lack of neg sign in the $x^2$ or $x^4$ terms means that these are easier and so not eligible for just a 1 mark MR penalty]

[4]

## Question 7 (i)

$5^{\frac{1}{3}}$ oe or $k = 7/2$ oe | 2 | M1 for $125 = 5^3$ or $\sqrt{5} = 5^{\frac{1}{2}}$ soi | M0 for just answer of 5' with no reference to 125

[2]

## Question 7 (ii)

attempting to multiply numerator and denominator of fraction by $1 + 2\sqrt{5}$ | M1 | some cands are incorporating the $10 + 7\sqrt{5}$ into the fraction. The M1s are available even if this is done wrongly or if $10 + 7\sqrt{5}$ is also multiplied by $1 + 2\sqrt{5}$

denominator $= -19$ soi | M1 | must be obtained correctly, but independent of first M1; eg M1 for denominator of 19 with a minus sign in front of whole expression or with attempt to change signs in numerator

$8 + 3\sqrt{5}$ | A1 [3]
Find the coefficient of $x^3$ in the binomial expansion of $(2 - 4x)^5$. [4]

\hfill \mbox{\textit{OCR MEI C1 2013 Q6 [4]}}