C3 June 2016 Q5
5.
Write your answer in the form\[\frac{\mathrm{d}y}{\mathrm{d}x} = p\,\mathrm{cosec}(qy), \qquad 0 < y < \frac{\pi}{4}\]where \(p\) and \(q\) are constants to be determined.
(5)| Scheme | Marks |
|---|---|
| \(y = \mathrm{e}^{3x}\cos 4x \Rightarrow \left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right) = \cos 4x \times 3\mathrm{e}^{3x} + \mathrm{e}^{3x} \times -4\sin 4x\) | M1A1 |
| Sets \(\cos 4x \times 3\mathrm{e}^{3x} + \mathrm{e}^{3x} \times -4\sin 4x = 0 \Rightarrow 3\cos 4x - 4\sin 4x = 0\) | M1 |
| \(\Rightarrow x = \dfrac{1}{4}\arctan\dfrac{3}{4}\) | M1 |
| \(\Rightarrow x = \text{awrt } 0.9463\) 4dp | A1 |
| (5) |
Notes
M1: Uses the product rule \(uv' + vu'\) to achieve \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right) = A\mathrm{e}^{3x}\cos 4x \pm B\mathrm{e}^{3x}\sin 4x \quad A, B \neq 0\)
The product rule if stated must be correct
A1: Correct (unsimplified) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \cos 4x \times 3\mathrm{e}^{3x} + \mathrm{e}^{3x} \times -4\sin 4x\)
M1: Sets/implies their \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) factorises/cancels)by \(\mathrm{e}^{3x}\) to form a trig equation in just \(\sin 4x\) and \(\cos 4x\)
M1: Uses the identity \(\dfrac{\sin 4x}{\cos 4x} \equiv \tan 4x\), moves from \(\tan 4x = C, C \neq 0\) using correct order of operations to \(x = \ldots\) Accept \(x = \text{awrt } 0.16\) (radians) \(x = \text{awrt } 9.22\) (degrees) for this mark.
If a candidate elects to pursue a more difficult method using \(R\cos(\theta + \alpha)\), for example, the minimum expectation will be that they get (1) the identity correct, and (2) the values of \(R\) and \(\alpha\) correct to 2dp. So for the correct equation you would only accept \(5\cos(4x + \text{awrt } 0.93)\) or \(5\sin(4x - \text{awrt } 0.64)\) before using the correct order of operations to \(x = \ldots\)
Similarly candidates who square \(3\cos 4x - 4\sin 4x = 0\) then use a Pythagorean identity should proceed from either \(\sin 4x = \dfrac{3}{5}\) or \(\cos 4x = \dfrac{4}{5}\) before using the correct order of operations …
A1: \(\Rightarrow x = \text{awrt } 0.9463\).
Ignore any answers outside the domain. Withhold mark for additional answers inside the domain
| Scheme | Marks |
|---|---|
| \(x = \sin^2 2y \Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}y} = 2\sin 2y \times 2\cos 2y\) | M1A1 |
| Uses \(\sin 4y = 2\sin 2y\cos 2y\) in their expression | M1 |
| \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = 2\sin 4y \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2\sin 4y} = \dfrac{1}{2}\mathrm{cosec}\,4y\) | M1A1 |
| (5) | |
| (10 marks) |
(ii) Alt I
| Scheme | Marks |
|---|---|
| \(x = \sin^2 2y \Rightarrow x = \dfrac{1}{2} - \dfrac{1}{2}\cos 4y\) | 2nd M1 |
| \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = 2\sin 4y\) | 1st M1 A1 |
| \(\Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2\sin 4y} = \dfrac{1}{2}\mathrm{cosec}\,4y\) | M1A1 |
| (5) |
(ii) Alt II
| Scheme | Marks |
|---|---|
| \(x^{\frac{1}{2}} = \sin 2y \Rightarrow \dfrac{1}{2}x^{-\frac{1}{2}} = 2\cos 2y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) | M1A1 |
| Uses \(x^{\frac{1}{2}} = \sin 2y\) AND \(\sin 4y = 2\sin 2y\cos 2y\) in their expression | M1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2\sin 4y} = \dfrac{1}{2}\mathrm{cosec}\,4y\) | M1A1 |
| (5) |
(ii) Alt III
| Scheme | Marks |
|---|---|
| \(x^{\frac{1}{2}} = \sin 2y \Rightarrow 2y = \mathrm{invsin}\,x^{\frac{1}{2}} \Rightarrow 2\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{\sqrt{1 - x}} \times \dfrac{1}{2}x^{-\frac{1}{2}}\) | M1A1 |
| Uses \(x^{\frac{1}{2}} = \sin 2y\), \(\sqrt{1 - x} = \cos 2y\) and \(\sin 4y = 2\sin 2y\cos 2y\) in their expression | M1 |
| \(\Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2\sin 4y} = \dfrac{1}{2}\mathrm{cosec}\,4y\) | M1A1 |
| (5) |
Notes
M1: Uses chain rule (or product rule) to achieve \(\pm P\sin 2y\cos 2y\) as a derivative.
There is no need for lhs to be seen/ correct
If the product rule is used look for \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = \pm A\sin 2y\cos 2y \pm B\sin 2y\cos 2y\),
A1: Both lhs and rhs correct (unsimplified) . \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = 2\sin 2y \times 2\cos 2y = (4\sin 2y\cos 2y)\) or \(1 = 2\sin 2y \times 2\cos 2y\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
M1: Uses \(\sin 4y = 2\sin 2y\cos 2y\) in their expression.
You may just see a statement such as \(4\sin 2y\cos 2y = 2\sin 4y\) which is fine.
Candidates who write \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = A\sin 2x\cos 2x\) can score this for \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = \dfrac{A}{2}\sin 4x\)
M1: Uses \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1\Big/\dfrac{\mathrm{d}x}{\mathrm{d}y}\) for their expression in \(y\). Concentrate on the trig identity rather than the coefficient in awarding this. Eg \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = 2\sin 4y \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = 2\mathrm{cosec}\,4y\) is condoned for the M1
If \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = a + b\) do not allow \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{a} + \dfrac{1}{b}\)
A1: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2}\mathrm{cosec}\,4y\) If a candidate then proceeds to write down incorrect values of \(p\) and \(q\) then do not withhold the mark.
NB: See the three alternatives which may be less common but mark in exactly the same way. If you are uncertain as how to mark these please consult your team leader.
In Alt I the second M is for writing \(x = \sin^2 2y \Rightarrow x = \pm\dfrac{1}{2} \pm \dfrac{1}{2}\cos 4y\) from \(\cos 4y = \pm 1 \pm 2\sin^2 2y\)
In Alt II the first M is for writing \(x^{\frac{1}{2}} = \sin 2y\) and differentiating both sides to \(Px^{-\frac{1}{2}} = Q\cos 2y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) oe
In Alt III the first M is for writing \(2y = \mathrm{invsin}\left(x^{0.5}\right)\) oe and differentiating to \(M\dfrac{\mathrm{d}y}{\mathrm{d}x} = N\dfrac{1}{\sqrt{1 - \left(x^{0.5}\right)^2}} \times x^{-0.5}\)