FP1 June 2013 Q2
2. \[\mathrm{f}(x) = \cos(x^2) - x + 3, \qquad 0 < x < \pi\]
(a) Show that the equation \(\mathrm{f}(x) = 0\) has a root \(\alpha\) in the interval \([2.5,\ 3]\). (2)
(b) Use linear interpolation once on the interval \([2.5,\ 3]\) to find an approximation for \(\alpha\), giving your answer to 2 decimal places. (3)
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(x) = \cos\left(x^2\right) - x + 3\) | |
| \(\mathrm{f}(2.5) = 1.499\ldots..\) \(\mathrm{f}(3) = -0.9111\ldots..\) Either any one of \(\mathrm{f}(2.5) = \) awrt 1.5 or \(\mathrm{f}(3) = \) awrt \(-0.91\) | M1 |
| Sign change (positive, negative) (and \(\mathrm{f}(x)\) is continuous) therefore root or equivalent. Both \(\mathrm{f}(2.5) = \) awrt 1.5 and \(\mathrm{f}(3) = \) awrt \(-0.91\), sign change and conclusion. | A1 |
| (2) |
Notes
Use of degrees gives \(\mathrm{f}(2.5) = 1.494\) and \(\mathrm{f}(3) = 0.988\) which is awarded M1A0
| Scheme | Marks |
|---|---|
| \(\dfrac{3 - \alpha}{\text{"}0.91113026188\text{"}} = \dfrac{\alpha - 2.5}{\text{"}1.4994494182\text{"}}\) Correct linear interpolation method – accept equivalent equation - ensure signs are correct. | M1 A1ft |
| \(\alpha = \dfrac{3 \times 1.499\ldots + 2.5 \times 0.9111\ldots.}{1.499\ldots + 0.9111\ldots.}\) | |
| \(\alpha = 2.81\ (2\text{d.p.})\) cao | A1 |
| (3) | |
| Total 5 |
Notes
Alternative (b)
Gradient of line is \(-\dfrac{\text{'}1.499\ldots\text{'} + \text{'}0.9111\ldots\text{'}}{0.5}\ (= -4.82)\) (3sf). Attempt to find equation of straight line and equate y to 0 award M1 and A1ft for their gradient awrt 3sf.