FP1 June 2010 Q3

EdexcelOld spec10 marksDifferentiation Numerical Methods

3. \[\mathrm{f}(x) = x^3 - \frac{7}{x} + 2, \quad x > 0\]

(a) Show that \(\mathrm{f}(x) = 0\) has a root \(\alpha\) between 1.4 and 1.5 (2)

(b) Starting with the interval \([1.4,\ 1.5]\), use interval bisection twice to find an interval of width 0.025 that contains \(\alpha\). (3)

(c) Taking 1.45 as a first approximation to \(\alpha\), apply the Newton-Raphson procedure once to \(\mathrm{f}(x) = x^3 - \dfrac{7}{x} + 2\) to obtain a second approximation to \(\alpha\), giving your answer to 3 decimal places. (5)

Mark scheme (a)

Scheme	Marks
\(\mathrm{f}(1.4) = \ldots\) and \(\mathrm{f}(1.5) = \ldots\) Evaluate both	M1
\(\mathrm{f}(1.4) = -0.256\) (or \(-\dfrac{32}{125}\)), \(\mathrm{f}(1.5) = 0.708\ldots\) (or \(\dfrac{17}{24}\)) Change of sign, \(\therefore\) root	A1
	(2)

Alternative method

Graphical method could earn M1 if 1.4 and 1.5 are both indicated

A1 then needs correct graph and conclusion, i.e. change of sign \(\therefore\) root

Notes

(a) M1: Some attempt at two evaluations

A1: needs accuracy to 1 figure truncated or rounded and conclusion including sign change indicated (One figure accuracy sufficient)

Mark scheme (b)

Scheme	Marks
\(\mathrm{f}(1.45) = 0.221\ldots\) or 0.2 [\(\therefore\) root is in [1.4, 1.45] ]	M1
\(\mathrm{f}(1.425) = -0.018\ldots\) or −0.019 or −0.02	M1
\(\therefore\) root is in [1.425, 1.45]	A1cso
	(3)

Notes

(b) M1: See f(1.45) attempted and positive

M1: See f(1.425) attempted and negative

A1: is cso – any slips in numerical work are penalised here even if correct region found.

Answer may be written as \(1.425 \leqslant \alpha \leqslant 1.45\) or \(1.425 \lt \alpha \lt 1.45\) or (1.425, 1.45) must be correct way round. Between is sufficient.

There is no credit for linear interpolation. This is M0 M0 A0Answer with no working is also M0M0A0

Mark scheme (c)

Scheme	Marks
\(\mathrm{f}'(x) = 3x^2 + 7x^{-2}\)	M1 A1
\(\mathrm{f}'(1.45) = 9.636\ldots\) (Special case: \(\mathrm{f}'(x) = 3x^2 + 7x^{-2} + 2\) then \(\mathrm{f}'(1.45) = 11.636\ldots\))	A1ft
\(x_1 = 1.45 - \dfrac{\mathrm{f}(1.45)}{\mathrm{f}'(1.45)} = 1.45 - \dfrac{0.221\ldots}{9.636\ldots} = 1.427\)	M1 A1cao
	(5)
	10 marks

Notes

Second A1 may be implied by correct answer (do not need to see it)

ft is limited to special case given.

2^nd M1: for attempt at Newton Raphson with their values for f(1.45) and \(\mathrm{f}'(1.45)\).

A1: is cao and needs to be correct to 3dp

Newton Raphson used more than once – isw.

Special case: \(\mathrm{f}'(x) = 3x^2 + 7x^{-2} + 2\) then \(\mathrm{f}'(1.45) = 11.636\ldots\)) is M1 A0 A1ft M1 A0 This mark can also be given by implication from final answer of 1.43