June 2023 Paper 1 Q5

OCR ACurrent spec8 marksFunctions (including |mod|)

(a) The function \(\mathrm{f}(x)\) is defined for all values of \(x\) as \(\mathrm{f}(x) = |ax - b|\), where \(a\) and \(b\) are positive constants.

(i) The graph of \(y = \mathrm{f}(x) + c\), where \(c\) is a constant, has a vertex at (3, 1) and crosses the \(y\)-axis at (0, 7).
Find the values of \(a\), \(b\) and \(c\). [3]

(ii) Explain why \(\mathrm{f}^{-1}(x)\) does not exist. [1]

(b) The function \(\mathrm{g}(x)\) is defined for \(x \geqslant \dfrac{q}{p}\) as \(\mathrm{g}(x) = |px - q|\), where \(p\) and \(q\) are positive constants.

(i) Find, in terms of \(p\) and \(q\), an expression for \(\mathrm{g}^{-1}(x)\), stating the domain of \(\mathrm{g}^{-1}(x)\). [3]

(ii) State the set of values of \(p\) for which the equation \(\mathrm{g}(x) = \mathrm{g}^{-1}(x)\) has no solutions. [1]

Mark scheme (a)

Scheme	Marks	AO
(i) \(a = 2\)	B1	1.1
\(b = 6\)	B1	1.1
\(c = 1\)	B1	1.1
	[3]
(ii) Because f is a many to one function eg \(\mathrm{f}(0) = \mathrm{f}(6)\)	B1	1.2
	[1]

Notes

(a)(i)

B1: (\(a = 2\)) Either stated or embedded in equation
eg \(|2x - b|\) seen
ignore any other values seen
B0 for \(a = -2\), unless subsequently corrected

B1: (\(b = 6\)) Either stated or embedded in equation
eg \(|ax - 6|\) seen
ignore any other values seen

B1: (\(c = 1\)) Either stated or embedded in equation
eg \(|ax - b| + 1\) seen
ignore any other values seen

(a)(ii)

B1: Any correct reason
Condone no explicit example
Could also say ‘because f is not one to one’
B1 BOD for ‘it is not one to one’
If referring to ‘one to many’ or ‘many to one’ it must be clear whether this is f or \(\mathrm{f}^{-1}\) (just ‘it’ or ‘the function’ is not enough)
Allow implication of function eg ‘as it is a many to one function there is no inverse function’
May also refer to the ‘horizontal line test’, but need to state outcome eg horizontal line would cross graph of \(y = \mathrm{f}(x)\) twice

Mark scheme (b)

Scheme	Marks	AO
(i) \(y = px - q\) \(px = y + q\) \(x = \frac{1}{p}(y + q)\)	M1	3.1a
\(\mathrm{g}^{-1}(x) = \frac{1}{p}x + \frac{q}{p}\)	A1	1.1
\(x \geqslant 0\)	B1	1.2
	[3]
(ii) \(0 \lt p \leqslant 1\)	B1	3.1a
	[1]

Notes

(b)(i)

M1: Complete attempt to find inverse function of \(\mathrm{f}(x) = px - q\)
Correct order of operations, allow sign error only
Could use coordinate geometry and reflection in \(y = x\)
Allow M1 BOD if more than one function is being considered

A1: Obtain correct inverse, in terms of \(x\)
Could be single term ie \(\mathrm{g}^{-1}(x) = \frac{x + q}{p}\)
A1 for just \(\frac{1}{p}x + \frac{q}{p}\), ie \(\mathrm{g}^{-1}(x)\) can be omitted
If LHS seen, it must be \(\mathrm{g}^{-1}(x)\) or \(y\) (allow BOD for \(\mathrm{g}^{-1}\), or using f not g)
BOD if modulus sign included
A0 if additional equations given

B1: Correct domain
B0 for \(x \gt 0\)
Independent of the first two marks
If in words then must be correct, so B1 for ‘any non-negative \(x\)’ but B0 for ‘any positive \(x\)’
\(\mathrm{g}^{-1}(x) \geqslant 0\) is B0
Condone incorrect set notation as long as intention is clear

(b)(ii)

B1: Correct set of values, any notation
No need for \(0 \lt p\) as specified in question, so B1 for \(p \leqslant 1\)
B0 for \(p \lt 1\)
B0 for any additional incorrect values
B0 if just single example and not set of values
Condone incorrect set notation as long as intention is clear