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Worked solution

Transfer a rational inequality to logarithmic inputs

(i) Without using a calculator, solve the inequality 5x−1x2+x−6≤1\dfrac{5x-1}{x^2+x-6}\leq1. marks

(ii) Hence solve the inequality 1+1−5ln⁡x(ln⁡x)2+ln⁡x−6≥01+\dfrac{1-5\ln x}{(\ln x)^2+\ln x-6}\geq0. marks

[2013/YJC/I/4]

CJC · Review Problems · Question 3

Answer

(i) (−∞,−3)∪[−1,2)∪[5,∞)(-\infty,-3)\cup[-1,2)\cup[5,\infty); (ii) (0,e−3)∪[e−1,e2)∪[e5,∞)(0,e^{-3})\cup[e^{-1},e^2)\cup[e^5,\infty).

(i)

Write 1−5x−1x2+x−6=x2−4x−5x2+x−6=(x+1)(x−5)(x+3)(x−2)≥01-\dfrac{5x-1}{x^2+x-6}=\dfrac{x^2-4x-5}{x^2+x-6}=\dfrac{(x+1)(x-5)}{(x+3)(x-2)}\geq0. The signs across −3,−1,2,5-3,-1,2,5 are +,−,+,−,++,-,+,-,+. Thus x<−3x<-3 or −1≤x<2-1\leq x<2 or x≥5x\geq5.

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(ii)

The displayed expression is 1−5ln⁡x−1(ln⁡x)2+ln⁡x−61-\dfrac{5\ln x-1}{(\ln x)^2+\ln x-6}. Set the input in (i) to ln⁡x\ln x, where x>0x>0. Then ln⁡x<−3\ln x<-3, or −1≤ln⁡x<2-1\leq\ln x<2, or ln⁡x≥5\ln x\geq5. Exponentiation preserves order, giving 0<x<e−30<x<e^{-3} or e−1≤x<e2e^{-1}\leq x<e^2 or x≥e5x\geq e^5.

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The nonnegative rational expression has successive signs positive, negative, positive, negative, positive across −3, −1, 2 and 5. Include −1 and 5; exclude the poles −3 and 2.
The nonnegative rational expression has successive signs positive, negative, positive, negative, positive across −3, −1, 2 and 5. Include −1 and 5; exclude the poles −3 and 2.
The nonnegative rational expression has successive signs positive, negative, positive, negative, positive across −3, −1, 2 and 5. Include −1 and 5; exclude the poles −3 and 2.

Common mistakes

The logarithm requires positive x; the original poles become x=e^(−3) and x=e².

Solution notes

The sign chart is independently drawn from the printed solution. Spacing is illustrative; labelled critical values and open/closed endpoints determine the set.

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