FP1 June 2012 Q6

6. \[\mathrm{f}(x) = \tan\left(\frac{x}{2}\right) + 3x - 6, \quad -\pi < x < \pi\]

(a) Show that the equation \(\mathrm{f}(x) = 0\) has a root \(\alpha\) in the interval \([1,\ 2]\). (2)

(b) Use linear interpolation once on the interval \([1,\ 2]\) to find an approximation to \(\alpha\). Give your answer to 2 decimal places. (3)

Mark scheme (a)

Scheme	Marks
\(\mathrm{f}(x) = \tan\left(\dfrac{x}{2}\right) + 3x - 6,\quad -\pi < x < \pi\)
\(\mathrm{f}(1) = -2.45369751\ldots\) \(\mathrm{f}(2) = 1.557407725\ldots\) Attempts to evaluate both \(\mathrm{f}(1)\) and \(\mathrm{f}(2)\) and evaluates at least one of them correctly to awrt (or trunc.) 2 sf. Nm	M1
Sign change (and \(\mathrm{f}(x)\) is continuous) therefore a root \(\alpha\) is between \(x = 1\) and \(x = 2\). Both values correct to awrt (or trunc.) 2 sf, sign change (or a statement which implies this e.g. \(-2.453.. < 0 < 1.5574..\)) and conclusion.	A1
	[2]

Mark scheme (b)

Scheme	Marks
\(\dfrac{\alpha - 1}{\text{"}2.45369751\ldots\text{"}} = \dfrac{2 - \alpha}{\text{"}1.557407725\ldots\text{"}}\) or \(\dfrac{\text{"}2.45369751\ldots\text{"} + \text{"}1.557407725\text{"}}{1} = \dfrac{\text{"}2.45369751\ldots\text{"}}{\alpha - 1}\) Correct linear interpolation method. It must be a correct statement using their f(2) and f(1). Can be implied by working below.	M1
\(\alpha = 1 + \left(\dfrac{\text{"}2.45369751\ldots\text{"}}{\text{"}1.557407725\ldots\text{"} + \text{"}2.45369751\ldots\text{"}}\right)1\) \(= \dfrac{6.464802745}{4.011105235}\) Correct follow through expression to find \(\alpha\). Method can be implied here. (Can be implied by awrt 1.61.)	A1ft
\(= 1.611726037\ldots\) awrt 1.61	A1
	[3]
	5 marks

If any “negative lengths” are used, score M0

Scheme	Marks
\(\mathrm{f}(1) = -2.991273132\ldots\) \(\mathrm{f}(2) = 0.017455064\ldots\) Attempts to evaluate both \(\mathrm{f}(1)\) and \(\mathrm{f}(2)\) and evaluates at least one of them correctly to awrt (or trunc.) 2 sf.	M1A0
\(\dfrac{\alpha - 1}{\text{"}2.991273132\ldots\text{"}} = \dfrac{2 - \alpha}{\text{"}0.017455064\ldots\text{"}}\) Correct linear interpolation method. It must be a correct statement using their f(2) and f(1). Can be implied by working below.	M1
If any “negative lengths” are used, score M0
\(\alpha = 1 + \left(\dfrac{\text{"}2.99127123\ldots\text{"}}{\text{"}0.017455064\ldots\text{"} + \text{"}2.99127123\ldots\text{"}}\right)1\) Correct follow through expression to find \(\alpha\). Method can be implied here. (Can be implied by awrt 1.99.)	A1ft
\(= 1.994198523\ldots\)	A0

Scheme	Marks
\(y - f(2) = \dfrac{f(2) - f(1)}{2 - 1}(x - 2)\) or \(y - f(1) = \dfrac{f(2) - f(1)}{2 - 1}(x - 1)\) or \(y = \dfrac{f(2) - f(1)}{2 - 1}x + c\) with an attempt to find \(c\) Correct straight line method. It must be a correct statement using their f(2) and f(1). Can be implied by working below.	M1
NB ‘\(m\)’ \(= 4.011105235\)
\(y = 0 \Rightarrow \alpha = \dfrac{f(2)}{f(1) - f(2)} + 2\) or \(\alpha = \dfrac{f(1)}{f(1) - f(2)} + 1\) Correct follow through expression to find \(\alpha\). Method can be implied here. (Can be implied by awrt 1.61.)	A1ft
\(= 1.611726037\ldots\) awrt 1.61	A1
	[3]