18M.1.AHL.TZ1.H_5

pestleMathematicsAAHLPaper 118M· sl-1-5-intro-to-logssource ↗

Solve ${\left( {{\text{ln}}\,x} \right)^2} - \left( {{\text{ln}}\,2} \right)\left( {{\text{ln}}\,x} \right) < 2{\left( {{\text{ln}}\,2} \right)^2}$ .

Markscheme / solution

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

${\left( {{\text{ln}}\,x} \right)^2} - \left( {{\text{ln}}\,2} \right)\left( {{\text{ln}}\,x} \right) - 2{\left( {{\text{ln}}\,2} \right)^2}\left( { = 0} \right)$

EITHER

${\text{ln}}\,x = \frac{{{\text{ln}}\,2 \pm \sqrt {{{\left( {{\text{ln}}\,2} \right)}^2} + 8{{\left( {{\text{ln}}\,2} \right)}^2}} }}{2}$ M1

$= \frac{{{\text{ln}}\,2 \pm 3\,{\text{ln}}\,2}}{2}$ A1

$\left( {{\text{ln}}\,x - 2\,{\text{ln}}\,2} \right)\left( {{\text{ln}}\,x + 2\,{\text{ln}}\,2} \right)\left( { = 0} \right)$ M1A1

THEN

${\text{ln}}\,x = 2\,{\text{ln}}\,2$ or $- {\text{ln}}\,2$ A1

$\Rightarrow x = 4$ or $x = \frac{1}{2}$ (M1)A1

Note: (M1) is for an appropriate use of a log law in either case, dependent on the previous M1 being awarded, A1 for both correct answers.

solution is $\frac{1}{2} < x < 4$ A1

[6 marks]

Examiners’ report

[N/A]