FP1 June 2017 Q1

1. \[\mathrm{f}(x) = \frac{1}{3}x^2 + \frac{4}{x^2} - 2x - 1, \quad x > 0\]

(a) Show that the equation \(\mathrm{f}(x) = 0\) has a root \(\alpha\) in the interval \([6, 7]\) (2)

(b) Taking 6 as a first approximation to \(\alpha\), apply the Newton-Raphson process once to \(\mathrm{f}(x)\) to obtain a second approximation to \(\alpha\). Give your answer to 2 decimal places. (5)

Mark scheme (a)

Scheme	Marks
\(\mathrm{f}(x) = \dfrac{1}{3}x^2 + \dfrac{4}{x^2} - 2x - 1,\ x > 0\)
\(\mathrm{f}(6) = -0.88888888\ldots\) \(\mathrm{f}(7) = 1.414965986\ldots\)	M1
Sign change or \(\mathrm{f}(6) = -\text{ve}\) and \(\mathrm{f}(7) = +\text{ve}\) or \(\mathrm{f}(6)\times\mathrm{f}(7) = -\text{ve}\) o.e. (and \(\mathrm{f}(x)\) is continuous) therefore a root / \(\alpha\) (exists between \(x = 6\) and \(x = 7\)) o.e.	A1
	(2)

Notes

M1: Either any one of \(\mathrm{f}(6) =\) awrt \(-0.9\) or \(\mathrm{f}(7) =\) awrt 1.4

A1: Both \(\mathrm{f}(6) =\) awrt \(-0.9\) and \(\mathrm{f}(7) =\) awrt 1.4, sign change and conclusion. Allow \(\mathrm{f}(6) = -\dfrac{8}{9}\) and \(\mathrm{f}(7) = \dfrac{208}{147}\).

Note: Accept at least ‘sign change therefore root’ o.e. for A1.
Any incorrect statements made in the conclusion award A0.

Mark scheme (b)

Scheme	Marks
\(\mathrm{f}'(x) = \dfrac{2}{3}x - \dfrac{8}{x^3} - 2\)	M1 A1 A1
\(\left\{\mathrm{f}'(6) = 1.962962963\ldots\right\}\)
\(\alpha \simeq 6 - \left(\dfrac{\text{"}-0.88888888\ldots\text{"}}{\text{"}1.962962963\ldots\text{"}}\right)\)	M1
\(= 6.452830189\ldots\)
\(= 6.45\ (2\text{ dp})\)	A1 cso
	(5)
	(7 marks)

Notes

M1: \(\dfrac{1}{3}x^2 \to \pm Ax\) or \(\dfrac{4}{x^2} \to \pm Bx^{-3}\) or \(-2x - 1 \to -2\)

A1: At least two of these terms differentiated correctly.

A1: Correct derivative. \(\mathrm{f}'(6) = \dfrac{53}{27}\)

M1: Correct application of Newton-Raphson using their values.

A1 cso: 6.45. Exact form of \(\alpha\) is \(\dfrac{342}{53}\)

Note: Denominator in NR calculation may contain evidence for first 3 marks.
Correct answer of 6.45 with minimal working will imply earlier marks for elements not explicitly stated. However, incorrect values leading to a correct final answer should be marked accordingly.