June 2019 Paper 1 Q4

EdexcelCurrent spec6 marksBinomial Expansion

(a) Find the first three terms, in ascending powers of \(x\), of the binomial expansion of\[\frac{1}{\sqrt{4-x}}\]giving each coefficient in its simplest form. (4)

The expansion can be used to find an approximation to \(\sqrt{2}\)
Possible values of \(x\) that could be substituted into this expansion are:

\(x = -14\) because \(\dfrac{1}{\sqrt{4-x}} = \dfrac{1}{\sqrt{18}} = \dfrac{\sqrt{2}}{6}\)
\(x = 2\) because \(\dfrac{1}{\sqrt{4-x}} = \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}\)
\(x = -\dfrac{1}{2}\) because \(\dfrac{1}{\sqrt{4-x}} = \dfrac{1}{\sqrt{\dfrac{9}{2}}} = \dfrac{\sqrt{2}}{3}\)

(b) Without evaluating your expansion,

(i) state, giving a reason, which of the three values of \(x\) should not be used (1)

(ii) state, giving a reason, which of the three values of \(x\) would lead to the most accurate approximation to \(\sqrt{2}\) (1)

Mark scheme (a)

Scheme	Marks	AO
\(\dfrac{1}{\sqrt{4-x}} = (4-x)^{-\frac{1}{2}} = 4^{-\frac{1}{2}} \times (1 \pm \ldots\)	M1	2.1
Uses a "correct" binomial expansion for their \((1 + ax)^n = 1 + nax + \dfrac{n(n-1)}{2}a^2x^2 +\)	M1	1.1b
\(\left(1 - \dfrac{x}{4}\right)^{-\frac{1}{2}} = 1 + \left(-\dfrac{1}{2}\right)\left(-\dfrac{x}{4}\right) + \dfrac{\left(-\dfrac{1}{2}\right) \times \left(-\dfrac{3}{2}\right)}{2}\left(-\dfrac{x}{4}\right)^2\)	A1	1.1b
\(\dfrac{1}{\sqrt{4-x}} = \dfrac{1}{2} + \dfrac{1}{16}x + \dfrac{3}{256}x^2\)	A1	1.1b
	(4)

Notes

M1: For the strategy of expanding \(\dfrac{1}{\sqrt{4-x}}\) using the binomial expansion.
You must see \(4^{-\frac{1}{2}}\) oe and an expansion which may or may not be combined.

M1: Uses a correct binomial expansion for their \((1 \pm ax)^n = 1 \pm nax \pm \dfrac{n(n-1)}{2}a^2x^2 +\)
Condone sign slips and the "\(a\)" not being squared in term 3. Condone \(a = \pm 1\)
Look for an attempt at the correct binomial coefficient for their \(n\), being combined with the correct power of \(ax\)

A1: \(\left(1 - \dfrac{x}{4}\right)^{-\frac{1}{2}} = 1 + \left(-\dfrac{1}{2}\right)\left(-\dfrac{x}{4}\right) + \dfrac{\left(-\dfrac{1}{2}\right) \times \left(-\dfrac{3}{2}\right)}{2}\left(-\dfrac{x}{4}\right)^2\) unsimplified
FYI the simplified form is \(1 + \dfrac{x}{8} + \dfrac{3x^2}{128}\) Accept the terms with commas between.

A1: \(\dfrac{1}{\sqrt{4-x}} = \dfrac{1}{2} + \dfrac{1}{16}x + \dfrac{3}{256}x^2\) Ignore subsequent terms. Allow with commas between.

Note: Alternatively \((4-x)^{-\frac{1}{2}} = 4^{-\frac{1}{2}} + \left(-\dfrac{1}{2}\right)4^{-\frac{3}{2}}(-x) + \dfrac{\left(-\dfrac{1}{2}\right)\left(-\dfrac{3}{2}\right)}{2}4^{-\frac{5}{2}}(-x)^2 + \ldots\)

M1: For \(4^{-\frac{1}{2}} + \ldots\) M1: As above but allow slips on the sign of \(x\) and the value of \(n\) A1: Correct unsimplified (as above) A1: As main scheme

Mark scheme (b)

Scheme	Marks	AO
(i) States \(x = -14\) and gives a valid reason. Eg explains that the expansion is not valid for \(\|x\| \gt 4\)	B1	2.4
	(1)
(ii) States \(x = -\dfrac{1}{2}\) and gives a valid reason. Eg. explains that it is closest to zero	B1	2.4
	(1)
	(6 marks)

Notes

(b) Any evaluations of the expansions are irrelevant.
Look for a suitable value and a suitable reason for both parts.

(b)(i) B1: Requires \(x = -14\) with a suitable reason.
Eg. \(x = -14\) as the expansion is only valid for \(|x| \lt 4\) or equivalent.
Eg ‘\(x = -14\) as \(\lvert -14 \rvert \gt 4\)’ or ‘I cannot use \(x = -14\) as \(\left|\dfrac{-14}{4}\right| \gt 1\)’
Eg. ‘\(x = -14\) as is outside the range \(|x| \lt 4\)’
Do not allow ‘\(-14\) is too big’ or ‘\(x = -14, |x| \lt 4\)’ either way around without some reference to the validity of the expansion.

(b)(ii) B1: Requires \(x = -\dfrac{1}{2}\) with a suitable reason.
Eg. \(x = -\dfrac{1}{2}\) as it is ‘the smallest/smaller value’ or ‘\(x = -\dfrac{1}{2}\) as the value closest to zero’ (that will give the more accurate approximation). The bracketed statement is not required.