Porisma

Original question & working

Reciprocals reverse negative interval order · source reference

[2014/ACJC/Promo/2] · Question pp. 2 · Working pp. 12, 13

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Complete question

(i) Without a calculator solve 4/(x+3)≤x4/(x+3)\le x. [3] (ii) Hence solve 4x/(1+3x)≤1/x4x/(1+3x)\le1/x. [3]

[2014/ACJC/Promo/2]

Printed working

Part (i)

4x+3≤x,x≠−34−x(x+3)x+3≤04−x2−3xx+3≤0x2+3x−4x+3≥0(x+4)(x−1)x+3≥0(x+4)(x+3)(x−1)≥0.\begin{aligned} \frac4{x+3}&\le x,\qquad x\ne-3\\ \frac{4-x(x+3)}{x+3}&\le0\\ \frac{4-x^2-3x}{x+3}&\le0\\ \frac{x^2+3x-4}{x+3}&\ge0\\ \frac{(x+4)(x-1)}{x+3}&\ge0\\ (x+4)(x+3)(x-1)&\ge0. \end{aligned}

Hence −4≤x<−3-4\le x<-3 or x≥1x\ge1; retain x≠−3x\ne-3.

Part (ii)

4x1+3x≤1x⟹41x+3≤1x.\frac{4x}{1+3x}\le\frac1x\quad\Longrightarrow\quad\frac4{\frac1x+3}\le\frac1x.

Replacing xx in (i) by 1/x1/x:

−4≤1x<−3or1x≥1,-4\le\frac1x<-3\quad\text{or}\quad\frac1x\ge1,

so −13<x≤−14-\tfrac13<x\le-\tfrac14 or 0<x≤10<x\le1.

Teacher margin notes

Can sketch the graph of y=1/xy=1/x.

Source diagrams

Printed cubic sign graph with roots−4,−3,1; closed endpoints−4 and1 and excluded open input−3. Printed reciprocal graph: levels−4,−3,1 correspond to−1/4,−1/3,1, with correct open/closed boundaries and the resulting number-line intervals.

Provenance

Question page references: 2. Solution page references: 12, 13. The original compilation header and printed page numbers are preserved in the edition notes. Original PDFs and page images remain only in the private QA archive.