OCR C4 2013 January — Question 2 5 marks

Exam BoardOCR
ModuleC4 (Core Mathematics 4)
Year2013
SessionJanuary
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeneralised Binomial Theorem
TypeFactoring out constants first
DifficultyModerate -0.3 This is a straightforward application of the binomial expansion requiring students to factor out 9^(3/2) = 27, then expand (1 - 16x/9)^(3/2) using the standard formula. While it involves fractional powers and finding the validity condition |16x/9| < 1, it's a routine textbook exercise with clear steps and no novel problem-solving required, making it slightly easier than average.
Spec1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions
2 Find the first three terms in the expansion of \(( 9 - 16 x ) ^ { \frac { 3 } { 2 } }\) in ascending powers of \(x\), and state the set of values for which this expansion is valid.
Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
The first 3 marks refer to the expansion of \(\left(1 - \frac{16x}{9}\right)^{1/2}\) only...
First 2 terms \(= 1 - \frac{8}{3}x\)B1 Allow any equiv fraction for \(-\frac{8}{3}\); \(\frac{3}{2} \cdot -\frac{16}{9}\) is not an equiv fraction
\(3^{\text{rd}}\) term \(= \frac{\frac{3}{2} \cdot \frac{1}{2}}{1.2}\left(-\frac{16x}{9}\right)^2\)M1 Allow clear evidence of intention, e.g. \(\frac{\frac{3}{2}\cdot\frac{1}{2}}{1.2} \cdot \frac{-16x^2}{9}\)
\(= \frac{32}{27}x^2\)A1 Allow any equiv fraction for \(\frac{32}{27}\)
Complete expansion \(\approx 27 - 72x + 32x^2\)A1 cao; no equivalents; ignore further terms; if \((a+b)^n\) used, award B1,B1,B1 for \(27, -72x, 32x^2\)
Valid for \(\frac{-9}{16} < x < \frac{9}{16}\) or \(x < \frac{9}{16}\)
[5] 
# Question 2:

| Answer | Marks | Guidance |
|--------|-------|----------|
| The first 3 marks refer to the expansion of $\left(1 - \frac{16x}{9}\right)^{1/2}$ only | ... | |
| First 2 terms $= 1 - \frac{8}{3}x$ | B1 | Allow any equiv fraction for $-\frac{8}{3}$; $\frac{3}{2} \cdot -\frac{16}{9}$ is not an equiv fraction |
| $3^{\text{rd}}$ term $= \frac{\frac{3}{2} \cdot \frac{1}{2}}{1.2}\left(-\frac{16x}{9}\right)^2$ | M1 | Allow clear evidence of intention, e.g. $\frac{\frac{3}{2}\cdot\frac{1}{2}}{1.2} \cdot \frac{-16x^2}{9}$ |
| $= \frac{32}{27}x^2$ | A1 | Allow any equiv fraction for $\frac{32}{27}$ |
| Complete expansion $\approx 27 - 72x + 32x^2$ | A1 | cao; no equivalents; ignore further terms; if $(a+b)^n$ used, award B1,B1,B1 for $27, -72x, 32x^2$ |
| Valid for $\frac{-9}{16} < x < \frac{9}{16}$ or $|x| < \frac{9}{16}$ | B1 | oe; beware e.g. $x < \left|\frac{9}{16}\right|$; condone $\leq$ instead of $<$ |
| **[5]** | | |

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2 Find the first three terms in the expansion of $( 9 - 16 x ) ^ { \frac { 3 } { 2 } }$ in ascending powers of $x$, and state the set of values for which this expansion is valid.

\hfill \mbox{\textit{OCR C4 2013 Q2 [5]}}
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