C4 June 2014 (R) Q1

Notes

B1: Writes down \((9 - 10x)^{-\frac{1}{2}}\) or uses power of \(-\dfrac{1}{2}\).
This mark can be implied by a constant term of \((9)^{-\frac{1}{2}}\) or \(\dfrac{1}{3}\).

B1: \(\underline{(9)^{-\frac{1}{2}}}\) or \(\underline{\dfrac{1}{3}}\) outside brackets or \(\dfrac{1}{3}\) 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 an un-simplified,
\(1 + \left(-\tfrac{1}{2}\right)(kx)\) or \(\left(-\tfrac{1}{2}\right)(kx) + \dfrac{(-\frac{1}{2})(-\frac{3}{2})}{2!}(kx)^2\) or \(1 + \ldots\ldots + \dfrac{(-\frac{1}{2})(-\frac{3}{2})}{2!}(kx)^2\), where \(k \neq 1\).

A1: \(\dfrac{1}{3} + \dfrac{5}{27}x\) (simplified fractions)

A1: Accept only \(\dfrac{25}{162}x^2\)

Note: You cannot recover correct work for part (a) in part (b). i.e. if the correct answer to (a) appears as part of their solution in part (b), it cannot be credited in part (a).

SC: If a candidate would otherwise score A0A0 then allow Special Case 1^st A1 for either
SC: \(\dfrac{1}{3}\left[1 + \dfrac{5}{9}x;\ \ldots\right]\) or SC: \(\lambda\left[1 + \dfrac{5}{9}x + \dfrac{25}{54}x^2 + \ldots\right]\) or SC: \(\left[\lambda + \dfrac{5\lambda}{9}x + \dfrac{25\lambda}{54}x^2 + \ldots\right]\)
(where \(\lambda\) can be 1 or omitted), with each term in the \(\left[\ldots\ldots\ldots\right]\) is a simplified fraction

SC: Special case for the M1 mark
Award Special Case M1 for a correct simplified or un-simplified \(1 + n(kx) + \dfrac{n(n - 1)}{2!}(kx)^2\)
expansion with a value of \(n \neq -\dfrac{1}{2}\), \(n \neq\) positive integer and a consistent \((kx)\). Note that \((kx)\) must be consistent (on the RHS, not necessarily the LHS) in a candidate’s expansion.
Note that \(k \neq 1\).

Note: Candidates who write \(\left\{\dfrac{1}{3}\right\}\left[1 + \left(-\dfrac{1}{2}\right)\left(\dfrac{10x}{9}\right) + \dfrac{(-\frac{1}{2})(-\frac{3}{2})}{2!}\left(\dfrac{10x}{9}\right)^2 + \ldots\right]\)
where \(k = \dfrac{10}{9}\) and not \(-\dfrac{10}{9}\) and achieve \(\dfrac{1}{3} - \dfrac{5}{27}x;\ + \dfrac{25}{162}x^2 + \ldots\) will get B1B1M1A0A1.

Alternative Methods for part (a)

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

\(\left\{\dfrac{1}{\sqrt{(9 - 10x)}} =\right\}\ (9 - 10x)^{-\frac{1}{2}} = (9)^{-\frac{1}{2}} + \left(-\dfrac{1}{2}\right)(9)^{-\frac{3}{2}}(-10x) + \dfrac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}(9)^{-\frac{5}{2}}(-10x)^2\)

B1: Writes down \((9 - 10x)^{-\frac{1}{2}}\) or uses power of \(-\dfrac{1}{2}\).

B1: \(9^{-\frac{1}{2}}\) or \(\dfrac{1}{3}\)

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

A1: \(\dfrac{1}{3} + \dfrac{5}{27}x\)

A1: \(\dfrac{25}{162}x^2\)

Note: The terms in C need to be evaluated, so \({}^{-\frac{1}{2}}C_0(9)^{-\frac{1}{2}} + {}^{-\frac{1}{2}}C_1(9)^{-\frac{3}{2}}(-10x) + {}^{-\frac{1}{2}}C_2(9)^{-\frac{5}{2}}(-10x)^2\)
without further working is B1B0M0A0A0.

Alternative Method 2: Maclaurin Expansion

Notes

M1: Writes down \((3 + x)\)(their part (a) answer, at least 2 of the 3 terms.)

Note: \((3 + x)\left(\dfrac{1}{4} + \dfrac{5}{4}x + \ldots\right)\) or \((3 + x)\left(\dfrac{1}{3} + \dfrac{5}{27}x + \dfrac{25}{162}x^2 + \ldots\right)\) are fine for M1.

Note: This mark can also be implied by candidate multiplying out to find two terms (or coefficients) in \(x\).

M1: Multiplies out to give exactly one constant term, exactly 2 terms in \(x\) and exactly 2 terms in \(x^2\).

Note: This M1 mark can be implied. You can also ignore \(x^3\) terms.

A1: \(1 + \dfrac{8}{9}x + \dfrac{35}{54}x^2 + \ldots\)