Functions
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Higher June 2025 Paper 2R Q19
19 The functions f and g are such that
\[\begin{aligned} \mathrm{f}(x) &= 3x - 4 \quad \text{where } x \gt 2 \\ \mathrm{g}(x) &= \dfrac{x}{2x + 1} \end{aligned}\]
(a) State the value of \(x\) that cannot be included in any domain of g (1)
(b) Find \(\mathrm{gf}(x)\)
Give your answer in its simplest form. (2)
Mark scheme (a)| Scheme | Marks |
|---|
| \(-\dfrac{1}{2}\) | B1 |
| (1) |
Notes
B1: oe
Accept \(x = -\dfrac{1}{2}\), accept \(x \ne -\dfrac{1}{2}\)
Do not allow inequalities, eg \(x \gt -\dfrac{1}{2}\) or \(x \leqslant -\dfrac{1}{2}\)
Mark scheme (b)| Scheme | Marks |
|---|
| \(\dfrac{3x - 4}{2(3x - 4) + 1}\) | M1 |
Correct answer only scores full marks (unless from obviously incorrect working) Answer: \(\dfrac{3x - 4}{6x - 7}\) | A1 |
| (2) |
| (3 marks) |
Notes
M1: for a correct unsimplified expression for \(\mathrm{gf}(x)\)
A1: oe eg \(\dfrac{4 - 3x}{7 - 6x}\)
Correct answer seen followed by incorrect subsequent working scores M1A0
Higher June 2025 Paper 1 Q16
16 \(\mathrm{f}(x) = 9 - \sqrt{x}\) where \(x \geqslant 0\)
\(\mathrm{g}(x) = 4x^2\)
(a) Find \(\mathrm{f}(9)\) (1)
(b) Solve \(\mathrm{fg}(x) \lt 0\)
Show clear algebraic working. (3)
Mark scheme (a)Mark scheme (b)| Scheme | Marks |
|---|
| \(9 - \sqrt{4x^2}\;(\lt 0)\) | M1 |
| eg \(9 - 2x\;(\lt 0)\) or \(9 \lt 2x\) or \(81 \lt 4x^2\) or \(9^2 \lt 4x^2\) | M1 |
Working required Answer: \(x \gt \dfrac{9}{2}\) | A1 |
| (3) |
| (4 marks) |
Notes
M1: for substituting \(\mathrm{g}(x)\) in \(\mathrm{f}(x)\), allow incorrect inequality sign or = sign
M1: for removing the square root, allow incorrect inequality sign or = sign
A1: (dep on M1)
oe eg \(x \gt 4.5\), \(\dfrac{9}{2} \lt x\), \(4.5 \lt x\)
Higher November 2024 Paper 1 Q25
25 The function f is such that \(\mathrm{f}(x) = 2x^2 - 24x + 7\) where \(x \geqslant 6\)
Find the inverse function \(\mathrm{f}^{-1}(x)\)
(4)
Mark scheme| Scheme | Marks |
|---|
\((y =)\;2(x^2 - 12x) + 7\) or \((y =)\;2\left(x^2 - 12x + \dfrac{7}{2}\right)\) or \(\dfrac{y - 7}{2} = x^2 - 12x\)
or\((x =)\;2(y^2 - 12y) + 7\) or \((x =)\;2\left(y^2 - 12y + \dfrac{7}{2}\right)\) or \(\dfrac{x - 7}{2} = y^2 - 12y\) | M1 |
eg \((y =)\;2\left((x - 6)^2 - 6^2\right) + 7\) or \((y =)\;2\left((x - 6)^2 - 6^2 + \dfrac{7}{2}\right)\) or \((y =)\;2(x - 6)^2 - 65\) oe or \(\dfrac{y - 7}{2} = (x - 6)^2 - 6^2\) oe
oreg \((x =)\;2\left((y - 6)^2 - 6^2\right) + 7\) or \((x =)\;2\left((y - 6)^2 - 6^2 + \dfrac{7}{2}\right)\) or \((x =)\;2(y - 6)^2 - 65\) oe or \(\dfrac{x - 7}{2} = (y - 6)^2 - 6^2\) oe | M1 |
\((x - 6)^2 = \dfrac{y + 65}{2}\) oe or \((x - 6)^2 = \dfrac{y - 7}{2} + 6^2\) oe
or\((y - 6)^2 = \dfrac{x + 65}{2}\) oe or \((y - 6)^2 = \dfrac{x - 7}{2} + 6^2\) oe | M1 |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(6 + \sqrt{\dfrac{x + 65}{2}}\) | A1 |
| (4) |
| (4 marks) |
Notes
M1: for a correct first step in order to complete the square
For each method mark the function must be correct
M1: dep on M1
A1: oe eg \(6 + \sqrt{\dfrac{x - 7}{2} + 36}\)
Must be in terms of \(x\)
M3A0 for \(6 \pm \sqrt{\dfrac{x + 65}{2}}\) or \(6 \pm \sqrt{\dfrac{y + 65}{2}}\)
or \(6 + \sqrt{\dfrac{y + 65}{2}}\)
Note: Allow candidates to swap \(x\) and \(y\) when finding the inverse
Mark scheme (ALT)| Scheme | Marks |
|---|
| \(2x^2 - 24x + 7 - y\;(= 0)\) | M1 |
\((x =)\;\dfrac{24 \pm \sqrt{576 - 8(7 - y)}}{4}\) or \((x =)\;\dfrac{24 + \sqrt{576 - 8(7 - y)}}{4}\)
or\(2\left((x - 6)^2 - 6^2\right) + 7 - y\;(= 0)\) or \(2\left((x - 6)^2 - 6^2 + \dfrac{7}{2}\right) - y\;(= 0)\) | M1 |
\((x =)\;6 \pm \sqrt{\dfrac{y + 65}{2}}\)
or\((x - 6)^2 = \dfrac{y + 65}{2}\) | M1 |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(6 + \sqrt{\dfrac{x + 65}{2}}\) | A1 |
Notes
M1: for a correct first step
M1: dep on M1
A1: oe eg \(6 + \sqrt{\dfrac{x - 7}{2} + 36}\)
Must be in terms of \(x\)
M3A0 for \(6 \pm \sqrt{\dfrac{x + 65}{2}}\) or \(6 \pm \sqrt{\dfrac{y + 65}{2}}\)
or \(6 + \sqrt{\dfrac{y + 65}{2}}\)
Note: Allow candidates to swap \(x\) and \(y\) when finding the inverse
No questions match these filters.
