Find the set of values of \(k\) that satisfy the inequality \({k^2} - k - 12 < 0\).

The triangle ABC is shown in the following diagram. Given that \(\cos B < \frac{1}{4}\), find the range of possible values for AB.

\({k^2} - k - 12 < 0\)

\((k - 4)(k + 3) < 0\)     (M1)

\( - 3 < k < 4\)     A1

[2 marks]

\(\cos B = \frac{{{2^2} + {c^2} - {4^2}}}{{4c}}{\text{ }}({\text{or }}16 = {2^2} + {c^2} - 4c\cos B)\)     M1

\( \Rightarrow \frac{{{c^2} - 12}}{{4c}} < \frac{1}{4}\)     A1

\( \Rightarrow {c^2} - c - 12 < 0\)

from result in (a)

\(0 < {\text{AB}} < 4\) or \( - 3 < {\text{AB}} < 4\)     (A1)

but AB must be at least 2

\( \Rightarrow 2 < {\text{AB}} < 4\)     A1

Note:     Allow \( \leqslant {\text{AB}}\) for either of the final two A marks.

[4 marks]