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Worked solution

A parameter in the denominator

Given that aa is a positive real number, solve the inequality x2−x+2x2+(1−a)x−a≤0\dfrac{x^2-x+2}{x^2+(1-a)x-a}\leq0, leaving your answer in terms of aa.

[2013/HCI/I/1]

CJC · Discussion · Question 7

Answer

−1<x<a-1<x<a.

Complete the square and factor

The numerator is (x−1/2)2+7/4(x-1/2)^2+7/4, strictly positive for every real input. The denominator factors as (x−a)(x+1)(x-a)(x+1). It therefore suffices to solve 1(x−a)(x+1)≤0\dfrac1{(x-a)(x+1)}\leq0.

How did this part go?

Sign chart

Since a>0a>0, the poles are ordered −1<a-1<a. The signs are positive, negative, positive in the three intervals. The numerator never vanishes and both poles are excluded. Hence −1<x<a-1<x<a.

How did this part go?

Compare your sketch

Expand diagram

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Since a>0, −1 precedes a. The reciprocal of (x−a)(x+1) has signs positive, negative, positive across these two poles. Both endpoints are excluded and the solution is −1<x<a.
Since a>0, −1 precedes a. The reciprocal of (x−a)(x+1) has signs positive, negative, positive across these two poles. Both endpoints are excluded and the solution is −1<x<a.
Since a>0, −1 precedes a. The reciprocal of (x−a)(x+1) has signs positive, negative, positive across these two poles. Both endpoints are excluded and the solution is −1<x<a.

Common mistakes

The non-strict inequality still gives open endpoints because neither pole is in the domain.

Solution notes

The solution diagrams are independently drawn from the printed mathematical relationships. Geometry is illustrative; use the labelled critical values and endpoint symbols.

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