A downward absolute-value graph with parameters

(i) For a<0 and b<−1 solve b∣x−a∣=x−abb|x-a|=x-ab. marks

(ii) Given a<b<0, (a) state the condition on b for a negative real root mark; (b) under that condition sketch y=x−ab and y=b|x−a| together marks, solve b∣x−a∣≥x−abb|x-a|\ge x-ab mark, and deduce the inputs satisfying −b∣ex+a∣≤ex+ab-b|e^x+a|\le e^x+ab. marks

[2020/ACJC/Prelim/I/4]

Total marks: 8

Solution

(i)

Absolute-value equations

For a<0a<0 and b<−1b<-1, solve b∣x−a∣=x−abb|x-a|=x-ab.

Hint

  1. Split into the two cases of the absolute value: b(x−a)=x−abb(x-a)=x-ab and −b(x−a)=x−ab-b(x-a)=x-ab.

  2. Solve each linear equation for xx.

Answer · (i)

x=0x=0 or x=2ab/(b+1)x=2ab/(b+1).

(ii)a

Conditions for a root

Given a<b<0a<b<0, state the condition on bb for the equation in (i) to have a negative real root.

Hint

  1. x=0x=0 is not negative, so the root must be 2abb+1\frac{2ab}{b+1}.

  2. Find the sign of this fraction.

Answer · (ii)a

b<−1b<-1.

(ii)b

Curve sketching

Under this condition, sketch y=x−aby=x-ab and y=b∣x−a∣y=b|x-a| on the same diagram.

Explore this idea

Hint

  1. y=b∣x−a∣y=b|x-a|: a V with vertex (a,0)(a,0), opening downwards since b<0b<0.

  2. y=x−aby=x-ab: a line of gradient 1 with yy-intercept −ab<0-ab<0.

  3. Mark the intersections at x=0x=0 and x=2abb+1x=\frac{2ab}{b+1}.

b|x−a|≥x−ab: representative a<b<−1
b|x−a|≥x−ab: representative a<b<−1Scroll across to see the full diagram.

Answer · (ii)b

Downward V vertex (a,0)(a,0); line y=x−aby=x-ab; intersections at x=0,2ab/(b+1)x=0,2ab/(b+1).

a<b<−1a<b<-1.

Explore the idea

An upside-down V and a line

Drawn with a = −4 and b = −2. The line y = x − 8 meets the V at x = −16 and x = 0.

Switch on a feature to add it to the sketch and see how it is found. A full-mark sketch shows all of them.

  • V y = b|x − a|
  • Y = x − ab
  • Intersections
Ideas to try
Between the two intersections, which graph is higher?
(ii)c

Inequalities from graphs

Hence solve b∣x−a∣≥x−abb|x-a|\ge x-ab.

Hint

  1. The V must be on or above the line.

  2. Read the interval between the two intersections.

Answer · (ii)c

2ab/(b+1)≤x≤02ab/(b+1)\le x\le0.

(ii)d

Exponential substitution

Deduce the values of xx satisfying −b∣ex+a∣≤ex+ab-b|e^x+a|\le e^x+ab.

Explore this idea

Hint

  1. Multiply by −1-1 to get b∣ex+a∣≥−ex−abb|e^x+a|\ge-e^x-ab.

  2. Match with the previous inequality with xx replaced by −ex-e^x: ∣−ex−a∣=∣ex+a∣|{-e^x}-a|=|e^x+a|.

  3. Keep only −ex<0-e^x<0.

Answer · (ii)d

x≤ln⁡(−2ab/(b+1))x\le\ln(-2ab/(b+1)).

Explore the idea

e^x below 16

Here −2ab/(b+1) = 16. The solution is every x where eˣ is at most 16.

Switch on a feature to add it to the sketch and see how it is found. A full-mark sketch shows all of them.

  • Y = eˣ
  • Bound 16
  • Boundary x = ln 16
Ideas to try
What happens to the bound if bb is closer to −1-1 (say b=−1.1b=-1.1)?