June 2024 Paper 1 Q11
11 A curve has equation \(y = 5\ln(1 - \cos 2x)\), where \(x\) is in radians.
Determine the exact coordinates of \(P\). [4]
| Scheme | Marks | AO |
|---|---|---|
| \(\cos 2x = 1\) | M1 | 3.1a |
| \(x = 0, \pm\pi, \ldots\) \(x = k\pi\) for \(k \in \mathbb{Z}\) | A1 | 2.5 |
| [2] |
Notes
M1: Set \(\cos 2x = 1\) soi
\(1 - \cos 2x = 0\), then \(x = 0\), would imply M1
Allow \(\cos 2x \geqslant 1\), but not \(\cos 2x \leqslant 1\) (unless recovered by final answer)
Allow \(\cos 2x \neq 1\) if considering the values that \(x\) cannot take
A1: Identify all multiples of \(\pi\), including negatives
Allow any clear notation, but must include negative integers as well
eg \(x = 0, \pm\pi, \pm 2\pi \ldots\) (allow ‘etc’ for ‘…’)
Condone working in degrees, as long as final answer is in radians
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{10\sin 2x}{1 - \cos 2x}\) | M1 | 1.1 |
| \(\dfrac{10\sin 2x}{1 - \cos 2x} = 0\) \(\sin 2x = 0\) | M1 | 1.1 |
| \(x = \tfrac{1}{2}\pi\) | A1 | 1.1 |
| \(y = 5\ln 2\) | A1 | 1.1 |
| [4] |
Notes
M1: Attempt to differentiate
Obtain \(\dfrac{k\sin 2x}{1 - \cos 2x}\), or unsimplified equiv
Other derivatives may be seen if trig identities, or log laws, used before differentiation; allow coefficient errors only
Could use implicit differentiation on \(1 - \cos 2x = \mathrm{e}^{\frac{1}{5}y}\) oe to obtain \(a\sin 2x = b\mathrm{e}^{\frac{1}{5}y}\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
M1: Equate their derivative to 0 and attempt to solve for a non-zero \(x\) value
Allow M1 as long as their numerator involves a trig term
Allow M1 if working in degrees
A1: Obtain \(x = \tfrac{1}{2}\pi\) only
Must be exact
A0 if \(x = 0\) also given in final answer
A1: Obtain \(y = 5\ln 2\) (or \(\ln 32\))
Must be exact, simplified, and from \(x = \tfrac{1}{2}\pi\)
Allow A1 if \(5\ln 2\) comes from \(90^\circ\)
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = \dfrac{(20\cos 2x)(1 - \cos 2x) - (10\sin 2x)(2\sin 2x)}{(1 - \cos 2x)^2}\) OR \((20\cos 2x)\mathrm{e}^{-\frac{1}{5}y} + (10\sin 2x)\left(-\dfrac{1}{5}\mathrm{e}^{-\frac{1}{5}y}\right)\dfrac{\mathrm{d}y}{\mathrm{d}x}\) | M1* | 3.1a |
| Any correct derivative, including unsimplified | A1 | 2.1 |
| \(= \dfrac{20\cos 2x - 20\cos^2 2x - 20\sin^2 2x}{(1 - \cos 2x)^2}\) \(= \dfrac{20\cos 2x - 20}{(1 - \cos 2x)^2}\) \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = \dfrac{-20(1 - \cos 2x)}{(1 - \cos 2x)^2} = \dfrac{-20}{1 - \cos 2x}\) | M1d* | 2.1 |
| \(\dfrac{-20}{1 - \cos 2x} = \dfrac{-20}{\mathrm{e}^{\frac{1}{5}y}} = -20\mathrm{e}^{-\frac{1}{5}y}\) OR \(20\mathrm{e}^{-\frac{1}{5}y} = \dfrac{20}{\mathrm{e}^{\frac{1}{5}y}} = \dfrac{20}{1 - \cos 2x}\) | M1 | 2.4 |
| \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = -20\mathrm{e}^{-\frac{1}{5}y}\) \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} + 20\mathrm{e}^{-\frac{1}{5}y} = 0.\quad\) A.G. | A1 | 2.1 |
| [5] | ||
| (ii) \(20\mathrm{e}^{-\frac{1}{5}y} \gt 0\) for all \(y\), so \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} \lt 0\) for all \(x\), hence stationary points are all maxima | B1 | 2.2a |
| [1] |
Notes
(c)(i)
M1*: Attempt differentiation using an appropriate method on their first derivative
Starting with \(\dfrac{k\sin 2x}{1 - \cos 2x}\), or a multiple of any other correct first derivative, including eg \(k\cot x\)
Must be correct structure for the differentiation method being attempted, allowing coefficient errors only
Could use implicit differentiation on \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 10\sin 2x \times \mathrm{e}^{-\frac{1}{5}y}\)
A1: If using implicit differentiation then A1 can be awarded if \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) is still present
M1d*: Attempt to simplify their second derivative using at least one trigonometric identity correctly
Only award M1 for trig identities used after differentiation
M1: Correctly replace \(1 - \cos 2x\) with \(\mathrm{e}^{\frac{1}{5}y}\) or vice versa
Used either in their second derivative or in the given answer
If using implicit differentiation then the M1 will be awarded before the differentiation attempt
A1: Obtain / confirm given answer www
Penalise any clearly incorrect equations, but allow BOD if denominator disappears (eg when using trig identities) but then reappears when relevant
(c)(ii)
B1: Correct conclusion, with justification
Refer to the exponential term being positive, hence second derivative must be negative, hence maxima
Could refer to \(\mathrm{e}^k\) not \(\mathrm{e}^{-\frac{1}{5}y}\)
Could refer to the correct second derivative of \(\dfrac{-20}{1 - \cos 2x}\) and explain why this is always negative, hence maxima (so no need to refer to exponential term with this approach)