A2 June 2022 Paper 2 Q5

EdexcelCurrent spec6 marksDifferentiation & Maclaurin

(a) Given that\[y = \arcsin x \qquad\qquad {-1} \leqslant x \leqslant 1\]show that\[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{\sqrt{1 - x^2}}\] (3)

(b) \[\mathrm{f}(x) = \arcsin\left(\mathrm{e}^x\right) \qquad\qquad x \leqslant 0\]Prove that \(\mathrm{f}(x)\) has no stationary points. (3)

Mark scheme (a)

Scheme	Marks	AO
\(\sin y = x \Rightarrow \cos y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1\) or \(\sin y = x \Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}y} = \cos y\)	M1	1.1b
Uses \(\sin^2 y + \cos^2 y = 1 \Rightarrow \cos y = \sqrt{1 - \sin^2 y} \Rightarrow \sqrt{1 - x^2}\)	M1	2.1
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{\sqrt{1 - x^2}}\) * cso	A1*	1.1b
	(3)

Notes

M1: Finds \(x\) in terms of \(y\) and differentiates

M1: Uses the trig identity \(\sin^2 y + \cos^2 y = 1\) to express \(\cos y\) in terms of \(x\). This may be seen in their derivative or stated on the side

A1*: Correctly achieves the printed answer \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{\sqrt{1 - x^2}}\). cso

Mark scheme (b)

Scheme	Marks	AO
Using the answer to (a) \(\mathrm{f}^\prime(x) = \dfrac{1}{\sqrt{1 - \mathrm{e}^{2x}}} \times \ldots\) or Restart \(\sin y = \mathrm{e}^x \Rightarrow \cos y\dfrac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^x\)	M1	3.1a
\(\mathrm{f}^\prime(x) = \dfrac{1}{\sqrt{1 - \mathrm{e}^{2x}}} \times \mathrm{e}^x\) or \(\mathrm{f}^\prime(x) = \dfrac{\mathrm{e}^x}{\cos y}\)	A1	1.1b
\(\mathrm{e}^x \neq 0\) (or \(\mathrm{e}^x \gt 0\)) therefore, there are no stationary points Alternatively, \(\mathrm{e}^x = 0\) leading to \(x = \ln 0\) which is impossible/undefined therefore there are no stationary points.	A1	2.4
	(3)
	(6 marks)

Notes

M1: Differentiates using the chain rule to achieve the correct form, condone \(\mathrm{f}^\prime(x) = \dfrac{1}{\sqrt{1 - \mathrm{e}^{2x}}}\)
Note \(\mathrm{f}^\prime(x) = \dfrac{1}{\sqrt{1 - \mathrm{e}^x}}\) is B0 for incorrect form
Alternatively restart, finds \(x\) in terms of \(y\) and differentiates

A1: Correct differentiation

A1: Follows correct differentiation. States that as \(\mathrm{e}^x \neq 0\) (or \(\mathrm{e}^x \gt 0\)) or no solutions to \(\mathrm{e}^x = 0\) therefore there are no stationary points.
Alternatively, \(\mathrm{e}^x = 0\) leading to \(x = \ln 0\) which is impossible/undefined/error therefore there are no stationary points. Ignore any reference to the denominator \(= 0\)