Higher June 2018 Paper 2 Q20
20 Martin did this question.
| Rationalise the denominator of \(\dfrac{14}{2 + \sqrt{3}}\) |
Here is how he answered the question.
\(\begin{aligned} \frac{14}{2 + \sqrt{3}} &= \frac{14 \times (2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})} \\[4pt] &= \frac{28 - 14\sqrt{3}}{4 + 2\sqrt{3} - 2\sqrt{3} + 3} \\[4pt] &= \frac{28 - 14\sqrt{3}}{7} \\[4pt] &= 4 - 2\sqrt{3} \end{aligned}\)
Martin’s answer is wrong.
(a) Find Martin’s mistake. (1)
Sian did this question.
| Rationalise the denominator of \(\dfrac{5}{\sqrt{12}}\) |
Here is how she answered the question.
\(\begin{aligned} \frac{5}{\sqrt{12}} &= \frac{5\sqrt{12}}{\sqrt{12} \times \sqrt{12}} \\[4pt] &= \frac{5 \times 3\sqrt{2}}{12} \\[4pt] &= \frac{5\sqrt{2}}{4} \end{aligned}\)
Sian’s answer is wrong.
(b) Find Sian’s mistake. (1)
| Answer | Mark | Mark scheme |
|---|---|---|
| explanation | C1 | for a correct explanation, eg \(\sqrt{3} \times -\sqrt{3} = -3\), not 3 |
| Answer | Mark | Mark scheme |
|---|---|---|
| explanation | C1 | for correct explanation, eg \(\sqrt{12} = 2\sqrt{3}\), not \(3\sqrt{2}\) |