FP1 June 2014 Q2

2. \[\mathrm{f}(x) = x^3 - \frac{5}{2x^{\frac{3}{2}}} + 2x - 3, \quad x > 0\]

(a) Show that the equation \(\mathrm{f}(x) = 0\) has a root \(\alpha\) in the interval \([1.1, 1.5]\). (2)

(b) Find \(\mathrm{f}^{\prime}(x)\). (2)

(c) Using \(x_0 = 1.1\) as a first approximation to \(\alpha\), apply the Newton-Raphson procedure once to \(\mathrm{f}(x)\) to find a second approximation to \(\alpha\), giving your answer to 3 decimal places. (3)

Mark scheme (a)

Scheme	Marks
\(\mathrm{f}(x) = x^3 - \dfrac{5}{2x^{\frac{3}{2}}} + 2x - 3\)
\(\mathrm{f}(1.1) = -1.6359604,\) \(\mathrm{f}(1.5) = 2.0141723\) Attempts to evaluate both \(\mathrm{f}(1.1)\) and \(\mathrm{f}(1.5)\) and evaluates at least one of them correctly to awrt (or trunc.) 2 sf.	M1
Sign change (and \(\mathrm{f}(x)\) is continuous) therefore a root / \(\boldsymbol{\alpha}\) is between \(x = 1.1\) and \(x = 1.5\) Both values correct to awrt (or trunc.) 2 sf, sign change (or a statement which implies this e.g. \(-1.63.. < 0 < 2.014..\)) and conclusion.	A1
	(2)

Mark scheme (b)

Scheme	Marks
\(\mathrm{f}(x) = x^3 - \dfrac{5}{2}x^{-\frac{3}{2}} + 2x - 3\) \(\Rightarrow \mathrm{f}^{\prime}(x) = 3x^2 + \dfrac{15}{4}x^{-\frac{5}{2}} + 2\) M1: \(x^n \to x^{n-1}\) for at least one term A1: Correct derivative oe	M1A1
	(2)

Mark scheme (c)

Scheme	Marks
\(\mathrm{f}^{\prime}(1.1) = 3(1.1)^2 + \dfrac{15}{4}(1.1)^{-\frac{5}{2}} + 2\ (= 8.585)\) Attempt to find \(\mathrm{f}^{\prime}(1.1)\). Accept \(\mathrm{f}^{\prime}(1.1)\) seen and their value.	M1
\(\alpha_2 = 1.1 - \left(\dfrac{\text{"}-1.6359604\text{"}}{\text{"}8.585\text{"}}\right)\) Correct application of N-R	M1
\(\alpha_2 = 1.291\) cao	A1
	(3)
	(7 marks)