June 2017 Paper 1 Q7
7 The function f is given by \(\quad \mathrm{f}(x) = \sqrt{2x - 5}\)
(a) Which of these inequalities is a possible domain for \(\mathrm{f}(x)\)?
Circle the inequality. [1 mark]
- \(x \geqslant 0\)
- \(x \geqslant \dfrac{2}{5}\)
- \(x \geqslant 2\)
- \(x \geqslant \dfrac{5}{2}\)
(b) Work out \(x\) when \(\mathrm{f}(x) = 1.2\) [2 marks]
(c) Work out the value of \(\quad \mathrm{f}\!\left(2\dfrac{5}{8}\right)\)
Give your answer as a fraction in its simplest form. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(x \geqslant \dfrac{5}{2}\) | B1 |
| Answer | Mark | Comments |
|---|---|---|
| \(1.2^2 = 2x - 5\) or \(1.44 = 2x - 5\) | M1 | oe |
| (\(x\) =) 3.22 | A1 | oe eg \(\dfrac{161}{50}\) |
| Answer | Mark | Comments |
|---|---|---|
| \(\sqrt{5\frac{1}{4} - 5}\) or \(\sqrt{\dfrac{21}{4} - 5}\) | M1 | oe \(\sqrt{\dfrac{2(21)}{8} - 5}\quad \sqrt{\dfrac{42}{8} - 5}\quad \sqrt{2\left(2\dfrac{5}{8}\right) - 5}\quad \sqrt{5.25 - 5}\quad \sqrt{2(2.625) - 5}\) |
| \(\sqrt{\dfrac{1}{4}}\) or \(\sqrt{(0.25)}\) | A1 | oe |
| \(\dfrac{1}{2}\) or 0.5 | A1 | Condone \(\pm\dfrac{1}{2}\) but not \(-\dfrac{1}{2}\) on its own |
Additional guidance
Condone decimals throughout
An answer of \(\dfrac{\sqrt{1}}{2}\) is M1 M1 A0