C4 June 2013 (R) Q4
4.
| Scheme | Marks |
|---|---|
| \(\left\{\sqrt[3]{(8 - 9x)}\right\} = (8 - 9x)^{\frac{1}{3}}\) Power of \(\dfrac{1}{3}\) | M1 |
| \(= \underline{(8)^{\frac{1}{3}}}\left(1 - \dfrac{9x}{8}\right)^{\frac{1}{3}} = \underline{2}\left(1 - \dfrac{9x}{8}\right)^{\frac{1}{3}}\) \(\underline{(8)^{\frac{1}{3}}}\) or \(\underline{2}\) | B1 |
| \(= \{2\}\left[1 + \left(\dfrac{1}{3}\right)(kx) + \dfrac{(\frac{1}{3})(-\frac{2}{3})}{2!}(kx)^2 + \dfrac{(\frac{1}{3})(-\frac{2}{3})(-\frac{5}{3})}{3!}(kx)^3 + \ldots\right]\) see notes | M1 A1 |
| \(= \{2\}\left[\underline{1 + \left(\dfrac{1}{3}\right)\left(\dfrac{-9x}{8}\right) + \dfrac{(\frac{1}{3})(-\frac{2}{3})}{2!}\left(\dfrac{-9x}{8}\right)^2 + \dfrac{(\frac{1}{3})(-\frac{2}{3})(-\frac{5}{3})}{3!}\left(\dfrac{-9x}{8}\right)^3 + \ldots}\right]\) | |
| \(= 2\left[1 - \dfrac{3}{8}x;\ - \dfrac{9}{64}x^2 - \dfrac{45}{512}x^3 + \ldots\right]\) See notes below! | |
| \(= 2 - \dfrac{3}{4}x;\ - \dfrac{9}{32}x^2 - \dfrac{45}{256}x^3 + \ldots\) | A1; A1 |
| (6) |
Notes
M1: Writes or uses \(\dfrac{1}{3}\). This mark can be implied by a constant term of \(8^{\frac{1}{3}}\) or 2.
B1: \(\underline{(8)^{\frac{1}{3}}}\) or \(\underline{2}\) outside brackets or \(\underline{2}\) as candidate’s constant term in their binomial expansion.
M1: Expands \((\ldots + kx)^{\frac{1}{3}}\) to give any 2 terms out of 4 terms simplified or un-simplified,
Eg: \(1 + \left(\dfrac{1}{3}\right)(kx)\) or \(\dfrac{(\frac{1}{3})(-\frac{2}{3})}{2!}(kx)^2 + \dfrac{(\frac{1}{3})(-\frac{2}{3})(-\frac{5}{3})}{3!}(kx)^3\) or \(1 + \ldots\ldots + \dfrac{(\frac{1}{3})(-\frac{2}{3})}{2!}(kx)^2\)
or \(\dfrac{(\frac{1}{3})(-\frac{2}{3})}{2!}(kx)^2 + \dfrac{(\frac{1}{3})(-\frac{2}{3})(-\frac{5}{3})}{3!}(kx)^3\) where \(k \neq 1\) are fine for M1.
A1: A correct simplified or un-simplified \(1 + \left(\dfrac{1}{3}\right)(kx) + \dfrac{(\frac{1}{3})(-\frac{2}{3})}{2!}(kx)^2 + \dfrac{(\frac{1}{3})(-\frac{2}{3})(-\frac{5}{3})}{3!}(kx)^3\)
expansion with consistent \((kx)\). Note that \((kx)\) must be consistent (on the RHS, not necessarily the LHS) in a candidate’s expansion. Note that \(k \neq 1\).
You would award B1M1A0 for \(2\left[\underline{1 + \left(\dfrac{1}{3}\right)\left(\dfrac{-9x}{8}\right) + \dfrac{(\frac{1}{3})(-\frac{2}{3})}{2!}(-9x)^2 + \dfrac{(\frac{1}{3})(-\frac{2}{3})(-\frac{5}{3})}{3!}\left(\dfrac{-9x}{8}\right)^3 + \ldots}\right]\)
because \((kx)\) is not consistent.
“Incorrect bracketing” \(= \{2\}\left[\underline{1 + \left(\dfrac{1}{3}\right)\left(\dfrac{-9x}{8}\right) + \dfrac{(\frac{1}{3})(-\frac{2}{3})}{2!}\left(\dfrac{-9x^2}{8}\right) + \dfrac{(\frac{1}{3})(-\frac{2}{3})(-\frac{5}{3})}{3!}\left(\dfrac{-9x^3}{8}\right) + \ldots}\right]\)
is M1A0 unless recovered.
A1: For \(2 - \dfrac{3}{4}x\) (simplified please) or also allow \(2 - 0.75x\).
Allow Special Case A1A0 for either SC: \(= 2\left[1 - \dfrac{3}{8}x;\ \ldots\right]\) or SC: \(K\left[1 - \dfrac{3}{8}x - \dfrac{9}{64}x^2 - \dfrac{45}{512}x^3 + \ldots\right]\)
(where \(K\) can be 1 or omitted), with each term in the \(\left[\ldots\ldots\ldots\right]\) either a simplified fraction or a decimal.
A1: Accept only \(-\dfrac{9}{32}x^2 - \dfrac{45}{256}x^3\) or \(-0.28125x^2 - 0.17578125x^3\)
Candidates who write \(= 2\left[\underline{1 + \left(\dfrac{1}{3}\right)\left(\dfrac{9x}{8}\right) + \dfrac{(\frac{1}{3})(-\frac{2}{3})}{2!}\left(\dfrac{9x}{8}\right)^2 + \dfrac{(\frac{1}{3})(-\frac{2}{3})(-\frac{5}{3})}{3!}\left(\dfrac{9x}{8}\right)^3 + \ldots}\right]\) where \(k = \dfrac{9}{8}\)
and not \(-\dfrac{9}{8}\) and achieve \(2 + \dfrac{3}{4}x;\ - \dfrac{9}{32}x^2 + \dfrac{45}{256}x^3 + \ldots\) will get B1M1A1A0A0.
Note for final two marks:
\(2\left[1 - \dfrac{3}{8}x;\ - \dfrac{9}{64}x^2 - \dfrac{45}{512}x^3 + \ldots\right] = 2 + \dfrac{3}{4}x - \dfrac{9}{32}x^2 - \dfrac{45}{256}x^3 + \ldots\) scores final A0A1.
\(2\left[1 - \dfrac{3}{8}x;\ - \dfrac{9}{64}x^2 - \dfrac{45}{512}x^3 + \ldots\right] = 2 - \dfrac{3}{4} - \dfrac{9}{32}x^2 - \dfrac{45}{256}x^3 + \ldots\) scores final A0A1
Alternative method: Candidates can apply an alternative form of the binomial expansion.
\(\left\{\sqrt[3]{(8 - 9x)}\right\} = (8 - 9x)^{\frac{1}{3}} = (8)^{\frac{1}{3}} + \left(\dfrac{1}{3}\right)(8)^{-\frac{2}{3}}(-9x) + \dfrac{(\frac{1}{3})(-\frac{2}{3})}{2!}(8)^{-\frac{5}{3}}(-9x)^2 + \dfrac{(\frac{1}{3})(-\frac{2}{3})(-\frac{5}{3})}{3!}(8)^{-\frac{8}{3}}(-9x)^3\)
B1: \((8)^{\frac{1}{3}}\) or 2
M1: Any two of four (un-simplified or simplified) terms correct.
A1: All four (un-simplified or simplified) terms correct.
A1: \(2 - \dfrac{3}{4}x\)
A1: \(-\dfrac{9}{32}x^2 - \dfrac{45}{256}x^3\)
Note: The terms in C need to be evaluated,
so \({}^{\frac{1}{3}}C_0(8)^{\frac{1}{3}} + {}^{\frac{1}{3}}C_1(8)^{-\frac{2}{3}}(-9x) + {}^{\frac{1}{3}}C_2(8)^{-\frac{5}{3}}(-9x)^2 + {}^{\frac{1}{3}}C_3(8)^{-\frac{8}{3}}(-9x)^3\) without further working is B0M0A0.
| Scheme | Marks |
|---|---|
| \(\left\{\sqrt[3]{7100} = 10\sqrt[3]{71} = 10\sqrt[3]{(8 - 9x)},\right\}\) so \(x = 0.1\) Writes down or uses \(x = 0.1\) | B1 |
| When \(x = 0.1\), \(\sqrt[3]{(8 - 9x)} \approx 2 - \dfrac{3}{4}(0.1) - \dfrac{9}{32}(0.1)^2 - \dfrac{45}{256}(0.1)^3 + \ldots\) \(= 2 - 0.075 - 0.0028125 - 0.00017578125\) \(= 1.922011719\) | M1 |
| So, \(\sqrt[3]{7100} = 19.22011719... = \underline{19.2201}\) (4 dp) 19.2201 cso | A1 cao |
| (3) | |
| (9 marks) |
Notes
B1: Writes down or uses \(x = 0.1\)
M1: Substitutes their \(x\), where \(|x| < \dfrac{8}{9}\) into at least two terms of their binomial expansion.
A1: 19.2201 cao
Be Careful!: The binomial answer is 19.22011719
and the calculated \(\sqrt[3]{7100}\) is 19.21997343... which is 19.2200 to 4 decimal places.
(corrected from the printed mark scheme: the line \(\sqrt[3]{7100} = 19.22011719...\) is printed as \(19.220117919...\); \(10 \times 1.922011719 = 19.22011719\), as the “Be Careful!” note says.)