FP1 June 2011 Q1

1. \[\mathrm{f}(x) = 3^x + 3x - 7\]

(a) Show that the equation \(\mathrm{f}(x) = 0\) has a root \(\alpha\) between \(x = 1\) and \(x = 2\). (2)

(b) Starting with the interval \([1,\ 2]\), use interval bisection twice to find an interval of width 0.25 which contains \(\alpha\). (3)

Mark scheme (a)

Scheme	Marks
\(\mathrm{f}(x) = 3^x + 3x - 7\)
\(\mathrm{f}(1) = -1\) \(\mathrm{f}(2) = 8\) Either any one of \(\mathrm{f}(1) = -1\) or \(\mathrm{f}(2) = 8\).	M1
Sign change (positive, negative) (and \(\mathrm{f}(x)\) is continuous) therefore (a root) \(\alpha\) is between \(x = 1\) and \(x = 2\). Both values correct, sign change and conclusion	A1
	(2)

Mark scheme (b)

Scheme	Marks
\(\mathrm{f}(1.5) = 2.696152423\ldots\quad \{\Rightarrow 1 \leqslant \alpha \leqslant 1.5\}\) \(\mathrm{f}(1.5) = \) awrt 2.7 (or truncated to 2.6)	B1
\(\mathrm{f}(1.25) = 0.698222038\ldots\) Attempt to find \(\mathrm{f}(1.25)\).	M1
\(\Rightarrow 1 \leqslant \alpha \leqslant 1.25\) \(\mathrm{f}(1.25) = \) awrt 0.7 with \(1 \leqslant \alpha \leqslant 1.25\) or \(1 < \alpha < 1.25\) or \([1,\ 1.25]\) or \((1,\ 1.25)\), or equivalent in words.	A1
	(3)
	(5 marks)

In (b) there is no credit for linear interpolation and a correct answer with no working scores no marks.