| Date | May 2016 | Marks available | 5 | Reference code | 16M.2.hl.TZ1.8 |
| Level | HL only | Paper | 2 | Time zone | TZ1 |
| Command term | Find | Question number | 8 | Adapted from | N/A |
Question
When \({x^2} + 4x - b\) is divided by \(x - a\) the remainder is 2.
Given that \(a,{\text{ }}b \in \mathbb{R}\), find the smallest possible value for \(b\).
Markscheme
\({a^2} + 4a - b = 2\) M1A1
EITHER
\({a^2} + 4a - (b + 2) = 0\)
as \(a\) is real \( \Rightarrow 16 + 4(b + 2) \geqslant 0\) M1A1
OR
\(b = {a^2} + 4a - 2\) M1
\( = {(a + 2)^2} - 6\) (A1)
THEN
\(b \geqslant - 6\)
hence smallest possible value for \(b\) is \( - 6\) A1
[5 marks]
Examiners report
For quite a difficult question, there were many good solutions for this, including many different methods. It was disturbing to see how many students did not seem to be aware of the remainder theorem, instead choosing to divide the polynomial.
