Higher June 2017 Unit 2 Q19
19.
(a) Give one example to show that the square of an irrational number is not always rational. [1]
Number = ……………………… Square of the number = ………………………
(b) Find two different irrational numbers to make the answer to the calculation below rational.
Complete the calculation by filling in the three boxes. [1]\[\boxed{\rule{0pt}{1.8em}\hspace{3.5em}} \times \boxed{\rule{0pt}{1.8em}\hspace{3.5em}} = \boxed{\rule{0pt}{1.8em}\hspace{3.5em}}\]
Complete the calculation by filling in the three boxes. [1]\[\boxed{\rule{0pt}{1.8em}\hspace{3.5em}} \times \boxed{\rule{0pt}{1.8em}\hspace{3.5em}} = \boxed{\rule{0pt}{1.8em}\hspace{3.5em}}\]
| Scheme | Marks | Comments |
|---|---|---|
| Appropriate example: E.g. π × π = π2, \((1 + \sqrt{3})^2 = 4 + 2\sqrt{3}\) \((\sqrt[3]{2})^2 = \sqrt[3]{4}\) OR \(2^{\frac{2}{3}}\) | B1 | The following can be applied if sight of π in the working lines or answer space: If π or 3.141… (with or without the ‘…’) used AND either π2 or 9.8696… (with or without the ‘…’) seen in the answer space, this will gain the B1. However, watch out for π seen, and e.g. 3.141 and 9.8658 offered in the answer spaces. This gains B0 because 3.1412 has been evaluated (not π2). |
| Scheme | Marks | Comments |
|---|---|---|
| Two different irrational numbers and the correct rational number as the answer. Examples: \(\sqrt{2} \times \sqrt{8} = \sqrt{16}\) (or simplified to 4) \(\sqrt{12} \times \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{12}}{\sqrt{3}}\) (or simplified to 2) \(\pi \times \dfrac{1}{\pi} = 1\) \(2^{\frac{1}{2}} \times 2^{\frac{3}{2}} = 2^2\) (answer can be simplified to 4) | B1 | Answers in the boxes take precedence. |