A2 June 2025 Paper 2 Q9
9. Given that
\[y = \arcsin 3x \qquad -\frac{1}{3} \leqslant x \leqslant \frac{1}{3}\]The curve \(C\) has equation
\[y = \cos(\arcsin 3x) \qquad -\frac{1}{3} \leqslant x \leqslant \frac{1}{3}\]| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\lambda}{\sqrt{1 - 9x^2}}\) | M1 | 1.1b |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3}{\sqrt{1 - 9x^2}}\) | A1 | 1.1b |
| (2) |
Notes
M1: Uses the chain rule to differentiate to the correct form. (\(\lambda\) can be 1). Condone a missing bracket on \(3x^2\) for this mark.
A1: Correct answer. Accept \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3}{\sqrt{1 - (3x)^2}}\) oe
Alt
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{1}{3}\sin y = x \Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}y} = \lambda\cos y = \lambda\sqrt{1 - 9x^2}\) | M1 | 1.1b |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3}{\sqrt{1 - 9x^2}}\) | A1 | 1.1b |
| (2) |
Alternatively
M1: Uses implicit differentiation and trig identities to achieve the correct form for \(\dfrac{\mathrm{d}x}{\mathrm{d}y}\) in terms of \(x\)
A1: Correct answer. Accept \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3}{\sqrt{1 - (3x)^2}}\) oe Accept \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3}{\cos(\arcsin 3x)}\)
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots\sin(\arcsin 3x) \times \text{their}\left(\dfrac{3}{\sqrt{1 - 9x^2}}\right)\) | M1 | 1.1b |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{9x}{\sqrt{1 - 9x^2}}\) | A1 | 2.1 |
| (2) |
Notes
M1: Uses the chain rule and their answer to part (a) to differentiate to the correct form. May use \(\arccos y = \arcsin 3x\) – see scheme for form but must get to \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots\)
A1: Correct simplified answer. If using the Alt they must replace the \(y\) and simplify to achieve the same answer. Must be their answer to (b). Do not accept forms in trig functions such as \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -3\tan(\arcsin 3x)\) as these are not simplest form.
Do not allow recovery if they simplify the sin(arcsin..) as part of their working in part (c).
Alt
| Scheme | Marks | AO |
|---|---|---|
| \(\arccos y = \arcsin 3x\left[\Rightarrow -\dfrac{1}{\sqrt{1 - y^2}}\dfrac{\mathrm{d}y}{\mathrm{d}x} = \text{their}\left(\dfrac{3}{\sqrt{1 - 9x^2}}\right)\right]\) \(\Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots\sqrt{1 - y^2} \times \text{their}\left(\dfrac{3}{\sqrt{1 - 9x^2}}\right)\) | M1 | 1.1b |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{9x}{\sqrt{1 - 9x^2}}\) | A1 | 2.1 |
| (2) |
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{9x}{\sqrt{1 - 9x^2}} = 4 \Rightarrow -9x = 4\sqrt{1 - 9x^2} \Rightarrow 81x^2 = 16\left(1 - 9x^2\right)\) Leading to a value for \(x\) or \(x^2\) | M1 | 3.1a |
| \(x^2 = \dfrac{16}{225}\) | A1 | 1.1b |
| \(x = -\dfrac{4}{15}\) | A1 | 2.3 |
| (3) | ||
| (7 marks) |
Notes
M1: A complete method to find a value for at least \(x^2\). Sets their answer to part (b) equal to 4, rearranges, squares both sides to reach a quadratic in \(x\) before achieving a value for \(x\) or \(x^2\). Note that if they never simplify the \(\sin(\arcsin 3x)\) in the derivative they will not be able to solve for \(x^2\) and so will not gain this mark.
A1: Correct value for \(x^2\) from a correct derivative in (b) (may be implied).
A1: Selects the correct value of \(x\). Must have come from a correct derivative in (b).
Caution: Watch out for those who get \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{9x}{\sqrt{1 - 9x^2}}\) in (b) which will give the correct values for part (c) but should score (b) M1A0 (c) M1A0A0.
Alt
M1: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -3\tan(\arcsin 3x) = 4 \Rightarrow \tan(\arcsin 3x) = -\dfrac{4}{3} \Rightarrow \sin(\arcsin 3x) = -\dfrac{4}{5} \Rightarrow 3x = \ldots\)
Uses the alt form in trig functions, set equal to 4 and proceeds to use right angle triangle or trig identities to proceed to a value for \(3x\)
A1: Correct value for \(3x\)
A1: Correct value for \(x\)