A downward absolute-value graph with parameters · source reference

Complete question

(i) For a<0 and b<−1 solve $b|x-a|=x-ab$. [2] (ii) Given a<b<0, (a) state the condition on b for a negative real root [1]; (b) under that condition sketch y=x−ab and y=b|x−a| together [2], solve $b|x-a|\ge x-ab$ [1], and deduce the inputs satisfying $-b|e^x+a|\le e^x+ab$. [2]

[2020/ACJC/Prelim/I/4]

Printed working

Part (i)

$$b(x-a)=x-ab\quad\text{or}\quad-b(x-a)=x-ab.$$

Therefore $x(b-1)=0$ or $x+bx=2ab$, giving $x=0$ or $x=2ab/(b+1)$.

Part (ii)(a)

For a negative root, $x=2ab/(b+1)<0$. Since $ab>0$, this requires $b+1<0\Rightarrow b<-1$.

Part (ii)(b)

The source graph labels the downward V $y=b|x-a|$, its vertex at $x=a$, the line $y=x-ab$, common intercept $-ab$, origin O, and second intersection at $x=2ab/(b+1)$. The V is on or above the line between the intersections, giving

$$\frac{2ab}{b+1}\le x\le0.$$

For $-b|e^x+a|\le e^x+ab$:

$$\begin{aligned} b|e^x+a|&\ge-e^x-ab\\ b|-(e^x+a)|&\ge-e^x-ab\\ b|-e^x-a|&\ge-e^x-ab. \end{aligned}$$

Replace $x$ in the preceding result by $-e^x$:

$$\frac{2ab}{b+1}\le-e^x<0\Rightarrow0<e^x\le-\frac{2ab}{b+1} \Rightarrow x\le\ln\left(-\frac{2ab}{b+1}\right).$$

Teacher margin notes

Note: $|f(x)|=p\Rightarrow f(x)=-p$ or $f(x)=p$.

$e^x>0$ for all $x\in\mathbb R$.

Source diagrams

Provenance

Question page references: 3. Solution page references: 14, 15. 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.