October 2021 Paper 1 Q6
6
(a) The diagram shows part of the graph of \(y = \operatorname{cosec} x\), where \(x\) is in radians.
State the equations of the three vertical asymptotes that can be seen. [1]
State the equations of the three vertical asymptotes that can be seen. [1]
The tangent to the graph at the point P with \(x\)-coordinate \(\dfrac{\pi}{3}\) meets the \(x\)-axis at Q.
(b) Show that the \(x\)-coordinate of Q is \(\dfrac{\pi}{3} + \sqrt{3}\). (You may use without proof the result that the derivative of \(\operatorname{cosec} x\) is \(-\operatorname{cosec} x \cot x\).) [6]
| Scheme | Marks | AO |
|---|---|---|
| Asymptotes \(x = 0\) (or \(y\)-axis) \(x = \pi\) and \(x = 2\pi\) | B1 | 1.2 |
| [1] |
Notes
B1: Must have all three
| Scheme | Marks | AO |
|---|---|---|
| When \(x = \dfrac{\pi}{3},\ y = \dfrac{2\sqrt{3}}{3}\) | B1 | 1.1b |
| When \(x = \dfrac{\pi}{3},\ \dfrac{\mathrm{d}y}{\mathrm{d}x} = -\operatorname{cosec}\dfrac{\pi}{3}\cot\dfrac{\pi}{3} = -\dfrac{2}{3}\) | M1 A1 | 3.1a 1.1b |
| Equation of the tangent is \(y - \dfrac{2\sqrt{3}}{3} = -\dfrac{2}{3}\left(x - \dfrac{\pi}{3}\right)\) | M1 | 2.1 |
| When \(y = 0\), \(-\dfrac{2\sqrt{3}}{3} = -\dfrac{2}{3}\left(x - \dfrac{\pi}{3}\right)\) | M1 | 2.1 |
| giving \(x = \dfrac{\pi}{3} + \sqrt{3}\) (AG) | A1 | 2.1 |
| [6] |
Notes
B1: soi; any exact form eg \(\dfrac{2}{\sqrt{3}}\)
M1: Uses the derivative when \(x = \dfrac{\pi}{3}\)
A1: May be embedded in the tangent equation
M1: Uses both their coordinates and their gradient to find the equation of the tangent
If \(y = -\dfrac{2}{3}x + c\) used, there must be an attempt to find \(c\) using both their coordinates and their gradient
M1: Substituting \(y = 0\) into their tangent
A1: Working must be correct and exact throughout