C3 January 2006 Q6
6. \[\mathrm{f}(x) = 12\cos x - 4\sin x.\]
Given that \(\mathrm{f}(x) = R\cos(x + \alpha)\), where \(R \geqslant 0\) and \(0 \leqslant \alpha \leqslant 90^\circ\),
(a) find the value of \(R\) and the value of \(\alpha\). (4)
(b) Hence solve the equation\[12\cos x - 4\sin x = 7\]for \(0 \leqslant x \lt 360^\circ\), giving your answers to one decimal place. (5)
(c)
(i) Write down the minimum value of \(12\cos x - 4\sin x\). (1)
(ii) Find, to 2 decimal places, the smallest positive value of \(x\) for which this minimum value occurs. (2)
| Scheme | Marks |
|---|---|
| \(R\cos\alpha = 12,\quad R\sin\alpha = 4\) | |
| \(R = \sqrt{(12^2 + 4^2)} = \sqrt{160}\) Accept if just written down, awrt 12.6 | M1 A1 |
| \(\tan\alpha = \dfrac{4}{12}, \Rightarrow \alpha \approx 18.43^\circ\) awrt \(18.4^\circ\) | M1, A1 |
| (4) |
| Scheme | Marks |
|---|---|
| \(\cos(x + \text{their } \alpha) = \dfrac{7}{\text{their } R}\) \((\approx 0.5534)\) | M1 |
| \(x + \text{their } \alpha = 56.4^\circ\) awrt \(56^\circ\) | A1 |
| \(= \ldots,\ 303.6^\circ\) \(360^\circ -\) their principal value | M1 |
| \(x = 38.0^\circ,\ 285.2^\circ\) Ignore solutions out of range | A1, A1 |
| (5) |
Notes
If answers given to more than 1 dp, penalise first time then accept awrt above.
| Scheme | Marks |
|---|---|
| (i) minimum value is \(-\sqrt{160}\) ft their \(R\) | B1ft |
| (ii) \(\cos(x + \text{their } \alpha) = -1\) | M1 |
| \(x \approx 161.57^\circ\) cao | A1 |
| (3) | |
| (12 marks) |