Keep every absolute value in a logarithmic inequality · source reference

Printed question

Sketch the graphs of $y=\ln(1-|x|)$ and $y=-|x-1|$ on the same diagram, showing clearly the equations of any asymptotes and any intersections with the axes.

Hence, solve the inequality $\ln(1-|x|)+|x-1|\le0$.

Printed working

$y=\ln(1-|x|)$.

Asymptotes: $1-|x|=0\Rightarrow|x|=1\therefore x=-1,x=1$.

From GC, the curves intersect at $x=-0.841$ and $x=0.433$.

$\ln(1-|x|)+|x-1|\le0$.

$\Rightarrow\ln(1-|x|)\le-|x-1|$.

$\Rightarrow-1<x\le-0.841$ or $0.433\le x<1$.

Printed answer key

Native graph above; −1<x≤−0.841 or 0.433≤x<1.

Source provenance

2025 TJC H2 Math (9758) JC2 MSM: GRAPHS AND CURVE SKETCHING. Question/answer booklet printed pages 1–14; solutions printed pages 1–28. Running metadata and repeated headings are transcribed in the topic metadata article.