C3 June 2010 Q3

EdexcelOld spec9 marksNumerical Methods Trigonometry

\[\mathrm{f}(x) = 4\operatorname{cosec} x - 4x + 1, \quad \text{where } x \text{ is in radians.}\]

(a) Show that there is a root \(\alpha\) of \(\mathrm{f}(x) = 0\) in the interval \([1.2, 1.3]\). (2)

(b) Show that the equation \(\mathrm{f}(x) = 0\) can be written in the form\[x = \frac{1}{\sin x} + \frac{1}{4}\] (2)

(c) Use the iterative formula\[x_{n+1} = \frac{1}{\sin x_n} + \frac{1}{4}, \quad x_0 = 1.25,\]to calculate the values of \(x_1\), \(x_2\) and \(x_3\), giving your answers to 4 decimal places. (3)

(d) By considering the change of sign of \(\mathrm{f}(x)\) in a suitable interval, verify that \(\alpha = 1.291\) correct to 3 decimal places. (2)

Mark scheme (a)

Scheme	Marks
\(\mathrm{f}(1.2) = 0.49166551\ldots,\quad \mathrm{f}(1.3) = -0.048719817\ldots\) Sign change (and as \(\mathrm{f}(x)\) is continuous) therefore a root \(\alpha\) is such that \(\alpha \in [1.2, 1.3]\)	M1A1
	(2)

Notes

(a) M1: Attempts to evaluate both \(\mathrm{f}(1.2)\) and \(\mathrm{f}(1.3)\) and evaluates at least one of them correctly to awrt (or truncated) 1 sf.

A1: both values correct to awrt (or truncated) 1 sf, sign change and conclusion.

Mark scheme (b)

Scheme	Marks
\(4\operatorname{cosec} x - 4x + 1 = 0 \Rightarrow 4x = 4\operatorname{cosec} x + 1\)	M1
\(\Rightarrow x = \operatorname{cosec} x + \dfrac{1}{4} \Rightarrow \underline{x = \dfrac{1}{\sin x} + \dfrac{1}{4}}\)	A1 *
	(2)

Notes

(b) M1: Attempt to make \(4x\) or \(x\) the subject of the equation.

A1: Candidate must then rearrange the equation to give the required result. It must be clear that candidate has made their initial \(\mathrm{f}(x) = 0\).

Mark scheme (c)

Scheme	Marks
\(x_1 = \dfrac{1}{\sin(1.25)} + \dfrac{1}{4}\)	M1
\(x_1 = 1.303757858\ldots,\quad x_2 = 1.286745793\ldots\)	A1
\(x_3 = 1.291744613\ldots\)	A1
	(3)

Notes

(c) M1: An attempt to substitute \(x_0 = 1.25\) into the iterative formula Eg \(= \dfrac{1}{\sin(1.25)} + \dfrac{1}{4}\). Can be implied by \(x_1 = \text{awrt } 1.3\) or \(x_1 = \text{awrt } 46^\circ\).

A1: Both \(x_1 = \text{awrt } 1.3038\) and \(x_2 = \text{awrt } 1.2867\)

A1: \(x_3 = \text{awrt } 1.2917\)

Mark scheme (d)

Scheme	Marks
\(\mathrm{f}(1.2905) = 0.00044566695\ldots,\quad \mathrm{f}(1.2915) = -0.00475017278\ldots\)	M1
Sign change (and as \(\mathrm{f}(x)\) is continuous) therefore a root \(\alpha\) is such that \(\alpha \in (1.2905, 1.2915) \Rightarrow \alpha = 1.291\ (3\text{ dp})\)	A1
	(2)
	(9 marks)

Notes

(d) M1: Choose suitable interval for \(x\), e.g. [1.2905, 1.2915] or tighter and at least one attempt to evaluate \(\mathrm{f}(x)\).

A1: both values correct to awrt (or truncated) 1 sf, sign change and conclusion.