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]