FP1 June 2014 (R) Q2

2. \[\mathrm{f}(x) = 3\cos 2x + x - 2, \qquad -\pi \leqslant x < \pi\]

(a) Show that the equation \(\mathrm{f}(x) = 0\) has a root \(\alpha\) in the interval \([2, 3]\). (2)

(b) Use linear interpolation once on the interval \([2, 3]\) to find an approximation to \(\alpha\).
Give your answer to 3 decimal places. (3)

(c) The equation \(\mathrm{f}(x) = 0\) has another root \(\beta\) in the interval \([-1, 0]\). Starting with this interval, use interval bisection to find an interval of width 0.25 which contains \(\beta\). (4)

Mark scheme (a)

Scheme	Marks
\(\mathrm{f}(x) = 3\cos 2x + x - 2\)
\(\mathrm{f}(2) = -1.9609\ldots\ldots\) \(\mathrm{f}(3) = 3.8805\ldots\ldots\) Attempts to evaluate both \(\mathrm{f}(2)\) and \(\mathrm{f}(3)\) 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 \(\alpha\) is between \(x = 2\) and \(x = 3\) Both values correct to awrt (or trunc.) 2 sf, sign change (or a statement which implies this e.g. \(-1.96.. < 0 < 3.88..\)) and conclusion.	A1
	(2)

Mark scheme (b)

Scheme	Marks
\(\dfrac{\alpha - 2}{\text{"}1.9609\ldots\text{"}} = \dfrac{3 - \alpha}{\text{"}3.8805\ldots\text{"}}\) Correct linear interpolation method. It must be a correct statement using their f(2) and f(3). Can be implied by working below.	M1
If any “negative lengths” are used, score M0
\((3.88\ldots + 1.96\ldots)\alpha = 3 \times 1.96 + 2 \times 3.88\)
\(\alpha_2 = \dfrac{3 \times 1.96.. + 2 \times 3.88..}{1.96\ldots + 3.88\ldots}\) Follow through their values if seen explicitly.	A1ft
\(\alpha_2 = 2.336\) cao	A1
	(3)

Mark scheme (c)

Scheme	Marks
\(\mathrm{f}(0) = +(1)\) or \(\mathrm{f}(-1) = -(4.248)\) Award for correct sign, can be in a table.	B1
\(\mathrm{f}(-0.5)\ (= -0.879\ldots\ldots)\) Attempt \(\mathrm{f}(-0.5)\)	M1
\(\mathrm{f}(-0.25)\ (= 0.382\ldots\ldots)\) Attempt \(\mathrm{f}(-0.25)\)	M1
\(\therefore -0.5 < \beta < -0.25\) oe with no numerical errors seen	A1
	(4)
	(9 marks)