Foundation June 2019 Paper 3 Q22
22 Here is a formula.
\[T = n^2 - \frac{12}{n}\](a) Work out \(T\) when \(\quad n = 5\) [1 mark]
(b) Why is \(T\) always positive when \(n\) is negative? [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| 22.6 or \(\dfrac{113}{5}\) or \(22\dfrac{3}{5}\) | B1 |
Additional guidance
| Condone \(22\dfrac{6}{10}\) | B1 |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(n^2\) will be positive and \(\dfrac{12}{n}\) will be negative and positive – negative = positive | B2 | oe B1 \(n^2\) will be positive or \(\dfrac{12}{n}\) will be negative |
| Alternative method 2 | ||
| \(n^2\) will be positive and \(-\dfrac{12}{n}\) will be positive and positive + positive = positive | B2 | oe B1 \(n^2\) will be positive or \(-\dfrac{12}{n}\) will be positive |
Additional guidance
For ‘\(n^2\) will be positive’ accept the square of a negative number is a positive
For ‘\(n^2\) will be positive’ condone square or squared numbers are positive
For ‘positive – negative = positive’ condone +(ve) \(-\) \(-\)(ve) = +(ve)