4MA0 Higher January 2017 Paper 3R Q18
18 Given that \(p\) is a prime number, rationalise the denominator of \(\;\dfrac{7\sqrt{p} - p^2}{\sqrt{p^3}}\)
Simplify your answer.
(3)
| Scheme | Marks |
|---|---|
| \(\dfrac{7\sqrt{p} - p^2}{p\sqrt{p}}\) or \(\dfrac{7\sqrt{p} - p^2}{\sqrt{p^3}} \times \dfrac{\sqrt{p^3}}{\sqrt{p^3}}\) oe | M1 |
\(\dfrac{7\sqrt{p} - p^2}{p\sqrt{p}} \times \dfrac{\sqrt{p}}{\sqrt{p}}\) or \(\dfrac{7\sqrt{p} - p\sqrt{p}\sqrt{p}}{p\sqrt{p}}\) \(\dfrac{7\sqrt{p}\sqrt{p^3} - p^2\sqrt{p^3}}{p^3}\) oe | M1 |
| \(\dfrac{7 - p\sqrt{p}}{p}\) | A1 |
| (3) | |
| (3 marks) |
Notes
M1: e.g. \(\dfrac{7p^{\frac{1}{2}} - p^2}{p^{\frac{3}{2}}}\)
M1: e.g. \(\dfrac{7p^2 - p^{\frac{7}{2}}}{p^3}\) oe
A1: for \(\dfrac{7 - p\sqrt{p}}{p}\) or \(\dfrac{7}{p} - \sqrt{p}\) oe or \(\dfrac{7 - p^{\frac{3}{2}}}{p}\) oe