C4 June 2018 Q1

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,
E.g. \(1 + \left(\dfrac{1}{2}\right)(kx)\) or \(\left(\dfrac{1}{2}\right)(kx) + \dfrac{(\frac{1}{2})(-\frac{1}{2})}{2!}(kx)^2\) or \(1 + \ldots + \dfrac{(\frac{1}{2})(-\frac{1}{2})}{2!}(kx)^2\)
where \(k\) is a numerical value and where \(k \neq 1\)

A1ft: A correct simplified or un-simplified \(1 + \left(\dfrac{1}{2}\right)(kx) + \dfrac{(\frac{1}{2})(-\frac{1}{2})}{2!}(kx)^2\) expansion with consistent \((kx)\)

Note: \((kx)\), \(k \neq 1\) must be consistent (on the RHS, not necessarily on the LHS) in their expansion

Note: Award B1M1A0 for \(2\left[1 + \left(\dfrac{1}{2}\right)(-9x) + \dfrac{(\frac{1}{2})(-\frac{1}{2})}{2!}\left(-\dfrac{9x}{4}\right)^2 + \ldots\right]\) because \((kx)\) is not consistent

Note: Incorrect bracketing: \(2\left[1 + \left(\dfrac{1}{2}\right)\left(-\dfrac{9x}{4}\right) + \dfrac{(\frac{1}{2})(-\frac{1}{2})}{2!}\left(-\dfrac{9x^2}{4}\right) + \ldots\right]\) is B1M1A0 unless recovered

A1: \(2 - \dfrac{9}{4}x\) (simplified fractions) or allow \(2 - 2.25x\) or \(2 - 2\tfrac{1}{4}x\)

A1: Accept only \(-\dfrac{81}{64}x^2\) or \(-1\tfrac{17}{64}x^2\) or \(-1.265625x^2\)

SC: If a candidate would otherwise score 2^nd A0, 3^rd A0 (i.e. scores A0A0 in the final two marks to (a)) then allow Special Case 2^nd A1 for either
SC: \(2\left[1 - \dfrac{9}{8}x;\ \ldots\right]\) or SC: \(2\left[1 + \ldots - \dfrac{81}{128}x^2 + \ldots\right]\) or SC: \(\lambda\left[1 - \dfrac{9}{8}x - \dfrac{81}{128}x^2 + \ldots\right]\)
or SC: \(\left[\lambda - \dfrac{9\lambda}{8}x - \dfrac{81\lambda}{128}x^2 + \ldots\right]\) (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 \(2 - \dfrac{18}{8}x - \dfrac{162}{128}x^2 + \ldots\) (i.e. for not simplifying their correct coefficients)

Note: Candidates who write \(2\left[1 + \left(\dfrac{1}{2}\right)\left(\dfrac{9x}{4}\right) + \dfrac{(\frac{1}{2})(-\frac{1}{2})}{2!}\left(\dfrac{9x}{4}\right)^2 + \ldots\right]\), where \(k = \dfrac{9}{4}\) and not \(-\dfrac{9}{4}\) and achieve \(2 + \dfrac{9}{4}x;\ - \dfrac{81}{64}x^2 + \ldots\) will get B1M1A1A0A1

Note: Ignore extra terms beyond the term in \(x^2\)

Note: You can ignore subsequent working following a correct answer

Note: Allow B1M1A1 for \(2\left[1 + \left(\dfrac{1}{2}\right)\left(-\dfrac{9x}{4}\right) + \dfrac{(\frac{1}{2})(-\frac{1}{2})}{2!}\left(\dfrac{9x}{4}\right)^2 + \ldots\right]\)

Note: Allow B1M1A1A1A1 for \(2\left[1 + \left(\dfrac{1}{2}\right)\left(-\dfrac{9x}{4}\right) + \dfrac{(\frac{1}{2})(-\frac{1}{2})}{2!}\left(\dfrac{9x}{4}\right)^2 + \ldots\right] = 2 - \dfrac{9}{4}x - \dfrac{81}{64}x^2 + \ldots\)

Alternative method 1: Candidates can apply an alternative form of the binomial expansion

B1: \((4)^{\frac{1}{2}}\) or 2

M1: Any two of three (un-simplified) terms correct

A1: All three (un-simplified) terms correct

A1: \(2 - \dfrac{9}{4}x\) (simplified fractions) or allow \(2 - 2.25x\) or \(2 - 2\tfrac{1}{4}x\)

A1: Accept only \(-\dfrac{81}{64}x^2\) or \(-1\tfrac{17}{64}x^2\) or \(-1.265625x^2\)

Note: The terms in C need to be evaluated.
So \({}^{\frac{1}{2}}C_0(4)^{\frac{1}{2}} + {}^{\frac{1}{2}}C_1(4)^{-\frac{1}{2}}(-9x);\ + {}^{\frac{1}{2}}C_2(4)^{-\frac{3}{2}}(-9x)^2\) without further working is B0M0A0

Alternative Method 2: Maclaurin Expansion \(\mathrm{f}(x) = (4 - 9x)^{\frac{1}{2}}\)

Notes

Note: Give B1 M1 for \(\sqrt{310} \approx 10\left(2 - \dfrac{9}{4}(0.1) - \dfrac{81}{64}(0.1)^2\right)\)

Note: Other alternative suitable values for \(x\) for \(\sqrt{310} \approx \beta\sqrt{4 - 9(\text{their } x)}\)

Note: Apply the scheme in the same way for their \(\beta\) and their \(x\)
E.g. Give B1 M1 A1 for \(\sqrt{310} \approx 12\left(2 - \dfrac{9}{4}\left(\dfrac{133}{648}\right) - \dfrac{81}{64}\left(\dfrac{133}{648}\right)^2\right) = 17.819\) (3 dp)

Note: Allow B1 M1 A1 for \(\sqrt{310} \approx 100\left(2 - \dfrac{9}{4}(0.441) - \dfrac{81}{64}(0.441)^2\right) = 76.161\) (3 dp)

Note: Give B1 M1 A0 for \(\sqrt{310} \approx 10\left(2 - \dfrac{9}{4}(0.1) - \dfrac{81}{64}(0.1)^2 - \dfrac{729}{512}(0.1)^3\right) = 17.609\) (3 dp)

Note: Send to review using \(\beta = \sqrt{155}\) and \(x = \dfrac{2}{9}\) (which gives 17.897 (3 dp))

Note: Send to review using \(\beta = \sqrt{1000}\) and \(x = 0.41\) (which gives 27.346 (3 dp))