Porisma

Worked solution

Use a nonpositive squared input

(a) Using an algebraic method, solve the inequality x+43+2x−x2<1\dfrac{x+4}{3+2x-x^2}<1. marks

(b) Hence solve the inequality x2−4x4+2x2−3<1\dfrac{x^2-4}{x^4+2x^2-3}<1. marks

[2013/CJC/I/4]

CJC · Review Problems · Question 8

Answer

(a) x<−1x<-1 or x>3x>3; (b) x<−1x<-1 or x>1x>1.

(a)

Subtract 1 to obtain x2−x+1−(x+1)(x−3)<0\dfrac{x^2-x+1}{-(x+1)(x-3)}<0. The numerator x2−x+1=(x−12)2+34x^2-x+1=(x-\frac12)^2+\frac34 is always positive. Alternatively, its discriminant is 1−4=−3<01-4=-3<0 and its leading coefficient is positive. Thus 1(x+1)(x−3)>0\dfrac1{(x+1)(x-3)}>0, with signs +,−,++,-,+ across the excluded poles −1,3-1,3. Hence x<−1x<-1 or x>3x>3.

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(b)

Replace the original input by −x2-x^2: −x2+43−2x2−x4=x2−4x4+2x2−3\dfrac{-x^2+4}{3-2x^2-x^4}=\dfrac{x^2-4}{x^4+2x^2-3}. The result in (a) gives −x2<−1-x^2<-1 or −x2>3-x^2>3. The second branch is impossible because −x2≤0-x^2\leq0; the first gives x2>1x^2>1, hence x<−1x<-1 or x>1x>1.

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Compare your sketch

Expand diagram

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The sign chart of 1/((x+1)(x−3)) is positive, negative, positive across the excluded poles −1 and 3.
The sign chart of 1/((x+1)(x−3)) is positive, negative, positive across the excluded poles −1 and 3.
The sign chart of 1/((x+1)(x−3)) is positive, negative, positive across the excluded poles −1 and 3.

Common mistakes

Use −x², rather than x², and simplify both numerator and denominator before transferring the solution.

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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