June 2023 Paper 2 Q7

AQACurrent spec5 marksFunctions (including |mod|)

7 The functions \(\mathrm{f}\) and \(\mathrm{g}\) are defined by

\[\mathrm{f}(x) = \sqrt{10 - 2x} \quad \text{for} \quad x \leqslant 5\]\[\mathrm{g}(x) = \frac{1}{x} \quad \text{for} \quad x \neq 0\]

The function \(\mathrm{h}\) has maximum possible domain and is defined by

\[\mathrm{h}(x) = \mathrm{gf}(x)\]

(a) Find an expression for \(\mathrm{h}(x)\) [1 mark]

(b) Find the domain of \(\mathrm{h}\) [1 mark]

Scheme	Marks	AO
Obtains \(\dfrac{1}{\sqrt{10 - 2x}}\) ACF	B1	1.1b
	(1)

Scheme	Marks	AO
Deduces \(x \lt 5\) ACF Condone incorrect set notation	B1	2.2a
	(1)

Mark scheme (c)

Scheme	Marks	AO
Forms the equation \(y\) = their \(\mathrm{h}(x)\) and squares both sides of the equation to remove the square root correctly. or Forms the equation \(y\) = their \(\mathrm{h}(x)\) and rearranges to obtain an expression for \(\sqrt{10 - 2x}\) \(x\) and \(y\) can be switched at any point.	M1	3.1a
Obtains \(10 - 2x = \dfrac{1}{y^2}\) or \(5 - x = \dfrac{1}{2y^2}\) \(x\) and \(y\) can be switched at any point.	A1	1.1b
Completes reasoned argument with no incorrect steps to show the given result. Must use correct notation \(\mathrm{h}^{-1}(x)\) and be consistent with use of variables. AG	R1	2.1
	(3)
	(5 marks)

Typical solution

\[y = \frac{1}{\sqrt{10 - 2x}}\]\[\sqrt{10 - 2x} = \frac{1}{y}\]\[10 - 2x = \frac{1}{y^2}\]\[2x = 10 - \frac{1}{y^2}\]\[x = 5 - \frac{1}{2y^2}\]\[\mathrm{h}^{-1}(x) = 5 - \frac{1}{2x^2}\]