C4 June 2015 Q1
1.
Give each coefficient in its simplest form. (5)
Give your answer in the form \(k\sqrt{2}\), where \(k\) is a constant to be determined. (1)
Give your answer in the form \(\dfrac{p}{q}\) where \(p\) and \(q\) are integers. (2)
| Scheme | Marks |
|---|---|
| \((4 + 5x)^{\frac{1}{2}} = \underline{(4)^{\frac{1}{2}}}\left(1 + \dfrac{5x}{4}\right)^{\frac{1}{2}} = \underline{2}\left(1 + \dfrac{5x}{4}\right)^{\frac{1}{2}}\) \(\underline{(4)^{\frac{1}{2}}}\) or \(\underline{2}\) | B1 |
| \(= \{2\}\left[1 + \left(\tfrac{1}{2}\right)(kx) + \dfrac{(\frac{1}{2})(-\frac{1}{2})}{2!}(kx)^2 + \ldots\right]\) see notes | M1 A1ft |
| \(= \{2\}\left[1 + \left(\tfrac{1}{2}\right)\left(\dfrac{5x}{4}\right) + \dfrac{(\frac{1}{2})(-\frac{1}{2})}{2!}\left(\dfrac{5x}{4}\right)^2 + \ldots\right]\) | |
| \(= 2\left[1 + \dfrac{5}{8}x - \dfrac{25}{128}x^2 + \ldots\right]\) See notes below! | |
| \(= 2 + \dfrac{5}{4}x;\ - \dfrac{25}{64}x^2 + \ldots\) isw | A1; A1 |
| (5) |
Notes
B1: \((4)^{\frac{1}{2}}\) or \(\underline{2}\) outside brackets or \(\underline{2}\) as candidate’s constant term in their binomial expansion.
M1: Expands \((\ldots + kx)^{\frac{1}{2}}\) to give any 2 terms out of 3 terms simplified or un-simplified,
Eg: \(1 + \left(\tfrac{1}{2}\right)(kx)\) or \(\left(\tfrac{1}{2}\right)(kx) + \dfrac{(\frac{1}{2})(-\frac{1}{2})}{2!}(kx)^2\) or \(1 + \ldots + \dfrac{(\frac{1}{2})(-\frac{1}{2})}{2!}(kx)^2\)
where \(k\) is a numerical value and where \(k \neq 1\).
A1: A correct simplified or un-simplified \(1 + \left(\tfrac{1}{2}\right)(kx) + \dfrac{(\frac{1}{2})(-\frac{1}{2})}{2!}(kx)^2\) expansion with consistent \((kx)\).
Note: \((kx)\), \(k \neq 1\), must be consistent (on the RHS, not necessarily on the LHS) in a candidate’s expansion.
Note: Award B1M1A0 for \(2\left[1 + \left(\tfrac{1}{2}\right)(5x) + \dfrac{(\frac{1}{2})(-\frac{1}{2})}{2!}\left(\dfrac{5x}{4}\right)^2 + \ldots\right]\) because \((kx)\) is not consistent.
Note: Incorrect bracketing: \(2\left[1 + \left(\tfrac{1}{2}\right)\left(\dfrac{5x}{4}\right) + \dfrac{(\frac{1}{2})(-\frac{1}{2})}{2!}\left(\dfrac{5x^2}{4}\right) + \ldots\right]\) is B1M1A0 unless recovered.
A1: \(2 + \dfrac{5}{4}x\) (simplified fractions) or allow \(2 + 1.25x\) or \(2 + 1\tfrac{1}{4}x\)
A1: Accept only \(-\dfrac{25}{64}x^2\) or \(-0.390625x^2\)
SC: If a candidate would otherwise score 2nd A0, 3rd A0 then allow Special Case 2nd A1 for either
SC: \(2\left[1 + \dfrac{5}{8}x;\ \ldots\right]\) or SC: \(2\left[1 + \ldots - \dfrac{25}{128}x^2 + \ldots\right]\) or SC: \(\lambda\left[1 + \dfrac{5}{8}x - \dfrac{25}{128}x^2 + \ldots\right]\)
or SC: \(\left[\lambda + \dfrac{5\lambda}{8}x - \dfrac{25\lambda}{128}x^2 + \ldots\right]\) (where \(\lambda\) can be 1 or omitted), where each term in the \(\left[\ldots\ldots\right]\) is a simplified fraction or a decimal,
OR SC: for \(2 + \dfrac{10}{8}x - \dfrac{50}{128}x^2 + \ldots\) (i.e. for not simplifying their correct coefficients.)
Note: Candidates who write \(2\left[1 + \left(\tfrac{1}{2}\right)\left(-\dfrac{5x}{4}\right) + \dfrac{(\frac{1}{2})(-\frac{1}{2})}{2!}\left(-\dfrac{5x}{4}\right)^2 + \ldots\right]\), where \(k = -\dfrac{5}{4}\) and not \(\dfrac{5}{4}\) and achieve \(2 - \dfrac{5}{4}x - \dfrac{25}{64}x^2 + \ldots\) will get B1M1A1A0A1
Note: Ignore extra terms beyond the term in \(x^2\).
Note: You can ignore subsequent working following a correct answer.
Alternative method 1
Candidates can apply an alternative form of the binomial expansion.
| Scheme | Marks |
|---|---|
| \(\left\{(4 + 5x)^{\frac{1}{2}}\right\} = (4)^{\frac{1}{2}} + \left(\tfrac{1}{2}\right)(4)^{-\frac{1}{2}}(5x) + \dfrac{(\frac{1}{2})(-\frac{1}{2})}{2!}(4)^{-\frac{3}{2}}(5x)^2\) |
B1: \((4)^{\frac{1}{2}}\) or 2
M1: Any two of three (un-simplified) terms correct.
A1: All three (un-simplified) terms correct.
A1: \(2 + \dfrac{5}{4}x\) (simplified fractions) or allow \(2 + 1.25x\) or \(2 + 1\tfrac{1}{4}x\)
A1: Accept only \(-\dfrac{25}{64}x^2\) or \(-0.390625x^2\)
Note: The terms in C need to be evaluated.
So \({}^{\frac{1}{2}}C_0(4)^{\frac{1}{2}} + {}^{\frac{1}{2}}C_1(4)^{-\frac{1}{2}}(5x);\ + {}^{\frac{1}{2}}C_2(4)^{-\frac{3}{2}}(5x)^2\) without further working is B0M0A0.
Alternative Method 2: Maclaurin Expansion \(\mathrm{f}(x) = (4 + 5x)^{\frac{1}{2}}\)
| Scheme | Marks |
|---|---|
| \(\mathrm{f}''(x) = -\dfrac{25}{4}(4 + 5x)^{-\frac{3}{2}}\) Correct \(\mathrm{f}''(x)\) | B1 |
| \(\mathrm{f}'(x) = \dfrac{1}{2}(4 + 5x)^{-\frac{1}{2}}(5)\) \(\pm a(4 + 5x)^{-\frac{1}{2}};\ a \neq \pm 1\) \(\dfrac{1}{2}(4 + 5x)^{-\frac{1}{2}}(5)\) | M1 A1 oe |
| \(\left\{\therefore \mathrm{f}(0) = 2,\ \mathrm{f}'(0) = \dfrac{5}{4} \text{ and } \mathrm{f}''(0) = -\dfrac{25}{32}\right\}\) | |
| So, \(\mathrm{f}(x) = 2 + \dfrac{5}{4}x;\ - \dfrac{25}{64}x^2 + \ldots\) | A1; A1 |
| Scheme | Marks |
|---|---|
| \(\left\{x = \dfrac{1}{10} \Rightarrow (4 + 5(0.1))^{\frac{1}{2}} = \sqrt{4.5} = \sqrt{\dfrac{9}{2}} = \dfrac{3}{\underline{\underline{\sqrt{2}}}} = \dfrac{3}{\sqrt{2}}\dfrac{\sqrt{2}}{\sqrt{2}}\right\}\) | |
| \(= \dfrac{3}{2}\sqrt{2}\) \(\dfrac{3}{2}\sqrt{2}\) or \(k = \dfrac{3}{2}\) or 1.5 o.e. | B1 |
| (1) |
Notes
B1: \(\dfrac{3}{2}\sqrt{2}\) or \(1.5\sqrt{2}\) or \(k = \dfrac{3}{2}\) or 1.5 o.e. (Ignore how \(k = \dfrac{3}{2}\) is found.)
| Scheme | Marks |
|---|---|
| \(\dfrac{3}{2}\sqrt{2}\) or \(1.5\sqrt{2}\) or \(\dfrac{3}{\underline{\underline{\sqrt{2}}}} = 2 + \dfrac{5}{4}\left(\dfrac{1}{10}\right) - \dfrac{25}{64}\left(\dfrac{1}{10}\right)^2 + \ldots\ \{= 2.121\ldots\}\) See notes | M1 |
| So, \(\dfrac{3}{2}\sqrt{2} = \dfrac{543}{256}\) or \(\dfrac{3}{\underline{\underline{\sqrt{2}}}} = \dfrac{543}{256}\) | |
| yields, \(\sqrt{2} = \dfrac{181}{128}\) or \(\underline{\underline{\sqrt{2}}} = \dfrac{256}{181}\) \(\dfrac{181}{128}\) or \(\dfrac{362}{256}\) or \(\dfrac{543}{384}\) or \(\dfrac{256}{181}\) etc. | A1 oe |
| (2) | |
| (8 marks) |
Notes
M1: Substitutes \(x = \dfrac{1}{10}\) or 0.1 into their binomial expansion found in part (a) which must contain both an \(x\) term and an \(x^2\) term (or even an \(x^3\) term) and equates this to either \(\dfrac{3}{\sqrt{2}}\) or their \(k\sqrt{2}\) from (b), where \(k\) is a numerical value.
Note: M1 can be implied by \(\dfrac{3}{2}\sqrt{2}\) or \(1.5\sqrt{2}\) or \(\dfrac{3}{\underline{\underline{\sqrt{2}}}}\) = awrt 2.121
Note: M1 can be implied by \(\dfrac{1}{k}\left(\text{their } \dfrac{543}{256}\right)\), with their \(k\) found in part (b).
Note: M1 cannot be implied by \((k)\left(\text{their } \dfrac{543}{256}\right)\), with their \(k\) found in part (b).
A1: \(\dfrac{181}{128}\) or any equivalent fraction, eg: \(\dfrac{362}{256}\) or \(\dfrac{543}{384}\). Also allow \(\dfrac{256}{181}\) or any equivalent fraction.
Note: Also allow A1 for \(p = 181, q = 128\) or \(p = 181\lambda, q = 128\lambda\) or \(p = 256, q = 181\) or \(p = 256\lambda, q = 181\lambda\), where \(\lambda \in \mathbb{R}^{+}\)
Note: You can recover work for part (c) in part (b). You cannot recover part (b) work in part (c).
Note: Candidates are allowed to restart and gain all 2 marks in part (c) from an incorrect part (b).
Note: Award M1 A1 for the correct answer from no working.