4. (a) Find the binomial expansion of
$$\sqrt [ 3 ] { ( 8 - 9 x ) , \quad } \quad | x | < \frac { 8 } { 9 }$$
in ascending powers of \(x\), up to and including the term in \(x ^ { 3 }\). Give each coefficient as a simplified fraction.
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Question 4:
Part (a): Binomial expansion of \(\sqrt[3]{(8-9x)}\)
Answer Marks
Guidance
Working/Answer Mark
Guidance
\(\{\sqrt[3]{(8-9x)}\} = (8-9x)^{\frac{1}{3}}\) M1
Power of \(\frac{1}{3}\)
\(= (8)^{\frac{1}{3}}\left(1-\frac{9x}{8}\right)^{\frac{1}{3}} = 2\left(1-\frac{9x}{8}\right)^{\frac{1}{3}}\) B1
\((8)^{\frac{1}{3}}\) or \(2\) outside brackets
Expands \((\ldots + kx)^{\frac{1}{3}}\) to give any 2 terms out of 4 M1 A1
\(kx\) must be consistent
\(= 2\left[1 - \frac{3}{8}x - \frac{9}{64}x^2 - \frac{45}{512}x^3 + \ldots\right]\) \(= 2 - \frac{3}{4}x - \frac{9}{32}x^2 - \frac{45}{256}x^3 + \ldots\) A1; A1
Two separate A1 marks for final simplified terms
Total: [6] Part (b): Approximation of \(\sqrt[3]{7100}\)
Answer Marks
Guidance
Working/Answer Mark
Guidance
\(\sqrt[3]{7100} = 10\sqrt[3]{71} = 10\sqrt[3]{(8-9x)}\), so \(x = 0.1\) B1
Writes down or uses \(x=0.1\)
\(\sqrt[3]{(8-9x)} \approx 2 - \frac{3}{4}(0.1) - \frac{9}{32}(0.1)^2 - \frac{45}{256}(0.1)^3 + \ldots = 1.922011719\ldots\) M1
\(\sqrt[3]{7100} = 19.2201\) (4 dp) A1 cao
Total: [3] Question 4:
Part (a):
Answer Marks
Guidance
Answer/Working Mark
Guidance
\(2 - \frac{3}{4}x\) A1
Simplified form required; also allow \(2 - 0.75x\)
\(-\frac{9}{32}x^2 - \frac{45}{256}x^3\) A1
Accept only these or decimal equivalents \(-0.28125x^2 - 0.17578125x^3\)
\(2\left[1 - \frac{3}{8}x; -\frac{9}{64}x^2 - \frac{45}{512}x^3 + ...\right] = 2 + \frac{3}{4}x - \frac{9}{32}x^2 - \frac{45}{256}x^3 + ...\) A0A1
Final two marks
\(2\left[1 - \frac{3}{8}x; -\frac{9}{64}x^2 - \frac{45}{512}x^3 + ...\right] = 2 - \frac{3}{4} - \frac{9}{32}x^2 - \frac{45}{256}x^3 + ...\) A0A1
Final two marks
Alternative method:
Answer Marks
Guidance
Answer/Working Mark
Guidance
\((8)^{\frac{1}{3}}\) or \(2\) B1
Any two of four terms correct (un-simplified or simplified) M1
All four terms correct A1
\(2 - \frac{3}{4}x\) A1
\(-\frac{9}{32}x^2 - \frac{45}{256}x^3\) A1
Part (b):
Answer Marks
Guidance
Answer/Working Mark
Guidance
Writes down or uses \(x = 0.1\) B1
Substitutes their \(x\), where \(\ x\
< \frac{8}{9}\), into at least two terms of binomial expansion
\(19.2201\) A1 cao
Binomial answer is \(19.22011719\); actual \(\sqrt[3]{7100} = 19.21997343...\) which is \(19.2200\) to 4 d.p.
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# Question 4:
## Part (a): Binomial expansion of $\sqrt[3]{(8-9x)}$
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\{\sqrt[3]{(8-9x)}\} = (8-9x)^{\frac{1}{3}}$ | M1 | Power of $\frac{1}{3}$ |
| $= (8)^{\frac{1}{3}}\left(1-\frac{9x}{8}\right)^{\frac{1}{3}} = 2\left(1-\frac{9x}{8}\right)^{\frac{1}{3}}$ | B1 | $(8)^{\frac{1}{3}}$ or $2$ outside brackets |
| Expands $(\ldots + kx)^{\frac{1}{3}}$ to give any 2 terms out of 4 | M1 A1 | $kx$ must be consistent |
| $= 2\left[1 - \frac{3}{8}x - \frac{9}{64}x^2 - \frac{45}{512}x^3 + \ldots\right]$ | | |
| $= 2 - \frac{3}{4}x - \frac{9}{32}x^2 - \frac{45}{256}x^3 + \ldots$ | A1; A1 | Two separate A1 marks for final simplified terms |
**Total: [6]**
## Part (b): Approximation of $\sqrt[3]{7100}$
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\sqrt[3]{7100} = 10\sqrt[3]{71} = 10\sqrt[3]{(8-9x)}$, so $x = 0.1$ | B1 | Writes down or uses $x=0.1$ |
| $\sqrt[3]{(8-9x)} \approx 2 - \frac{3}{4}(0.1) - \frac{9}{32}(0.1)^2 - \frac{45}{256}(0.1)^3 + \ldots = 1.922011719\ldots$ | M1 | |
| $\sqrt[3]{7100} = 19.2201$ (4 dp) | A1 cao | |
**Total: [3]**
# Question 4:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2 - \frac{3}{4}x$ | A1 | Simplified form required; also allow $2 - 0.75x$ |
| $-\frac{9}{32}x^2 - \frac{45}{256}x^3$ | A1 | Accept only these or decimal equivalents $-0.28125x^2 - 0.17578125x^3$ |
| $2\left[1 - \frac{3}{8}x; -\frac{9}{64}x^2 - \frac{45}{512}x^3 + ...\right] = 2 + \frac{3}{4}x - \frac{9}{32}x^2 - \frac{45}{256}x^3 + ...$ | A0A1 | Final two marks |
| $2\left[1 - \frac{3}{8}x; -\frac{9}{64}x^2 - \frac{45}{512}x^3 + ...\right] = 2 - \frac{3}{4} - \frac{9}{32}x^2 - \frac{45}{256}x^3 + ...$ | A0A1 | Final two marks |
**Alternative method:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(8)^{\frac{1}{3}}$ or $2$ | B1 | |
| Any two of four terms correct (un-simplified or simplified) | M1 | |
| All four terms correct | A1 | |
| $2 - \frac{3}{4}x$ | A1 | |
| $-\frac{9}{32}x^2 - \frac{45}{256}x^3$ | A1 | |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Writes down or uses $x = 0.1$ | B1 | |
| Substitutes their $x$, where $\|x\| < \frac{8}{9}$, into at least two terms of binomial expansion | M1 | |
| $19.2201$ | A1 cao | Binomial answer is $19.22011719$; actual $\sqrt[3]{7100} = 19.21997343...$ which is $19.2200$ to 4 d.p. |
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4. (a) Find the binomial expansion of
$$\sqrt [ 3 ] { ( 8 - 9 x ) , \quad } \quad | x | < \frac { 8 } { 9 }$$
in ascending powers of $x$, up to and including the term in $x ^ { 3 }$. Give each coefficient as a simplified fraction.\\
(b) Use your expansion to estimate an approximate value for $\sqrt [ 3 ] { 7100 }$, giving your answer to 4 decimal places. State the value of $x$, which you use in your expansion, and show all your working.\\
\hfill \mbox{\textit{Edexcel C4 2013 Q4 [9]}}