FP1 June 2015 Q2
2. In the interval \(13 < x < 14\), the equation \[3 + x\sin\left(\frac{x}{4}\right) = 0, \text{ where } x \text{ is measured in radians,}\] has exactly one root, \(\alpha\).
Give your answer to 3 decimal places. (4)
| Scheme | Marks |
|---|---|
| Let \(\mathrm{f}(x) = 3 + x\sin\left(\tfrac{x}{4}\right)\) then \(\mathrm{f}(13) = 1.593\) [and \(\mathrm{f}(14) = -1.911\) need not be seen in (a)] | |
| \(\mathrm{f}(13.5) = -0.122\), so root in \([13, 13.5]\) | M1 A1 |
| \(\mathrm{f}(13.25) = 0.746\) so root in \([13.25, 13.5]\) | A1 |
| (3) |
Notes
M1: Evaluate f(13) and f(13.5) giving at least positive, negative OR evaluate f(13.5) and f(13.25) to give at least negative, positive. Do not award if using degrees.
A1: \(\mathrm{f}(13.5) =\) awrt \(-0.1\), \(\mathrm{f}(13.25) =\) awrt 0.7(5).
A1: Correct interval \([13.25, 13.5]\) or equivalent form with or without boundaries.
| Scheme | Marks |
|---|---|
| \(\dfrac{\alpha - 13}{14 - \alpha} = \dfrac{1.593}{1.911}\) or \(\dfrac{\alpha - 13}{1} = \dfrac{1.593}{1.593 + 1.911}\) | M1 A1 |
| So \(\alpha(1.911 + 1.593) = 1.593 \times 14 + 13 \times 1.911\) and \(\alpha = \dfrac{47.145}{3.504} = 13.455\) | dM1 A1 |
| (4) | |
| (7 marks) |
Notes
M1: Attempt at linear interpolation on either side of equation with correct signs.
A1: Correct equivalent statement
dM1: Makes alpha subject of formula
A1: cao. Award A0 for 13.456 and 13.454
ALT (b)
Using equation of line
M1: Attempt to find gradient \(\dfrac{y_1 - y_0}{x_1 - x_0} = \dfrac{-1.911 - 1.593}{14 - 13}\ (= -3.504\ldots)\), attempt to use \(y - y_0 = m(x - x_0)\) with either 13 or 14 (gives \(y = -3.504x + 47.145\)) and substitute \(y = 0\)
A1: Correct statement after substituting \(y = 0\) in their equation i.e. \(0 = -3.504x + 47.145\)
dM1: Makes \(x\) the subject of the formula
A1: cao. Award A0 for 13.456 and 13.454