17M.2.AHL.TZ2.H_4

pestleMathematicsAAHLPaper 217M· sl-3-2-2d-and-3d-trig-sine-rule-cosine-rule-areasource ↗

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

[2]

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

M17/5/MATHL/HP2/ENG/TZ2/04.b

[4]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

${k^2} - k - 12 < 0$

$(k - 4)(k + 3) < 0$ (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]

Examiners’ report

[N/A]