Porisma

Worked solution

Rational inequality followed by a sine substitution

Without using a calculator, solve 3x−11−2x2≥1\dfrac{3x-1}{1-2x^2}\geq1.

Hence find the solution of 3sin⁡θ−1cos⁡2θ≥1\dfrac{3\sin\theta-1}{\cos2\theta}\geq1, where −π2≤θ≤π2-\dfrac\pi2\leq\theta\leq\dfrac\pi2.

[2016/AJC/Promo/5]

CJC · Discussion · Question 8

Answer

[−2,−1/2)∪[1/2,1/2)[-2,-1/\sqrt2)\cup[1/2,1/\sqrt2). Hence [−π/2,−π/4)∪[π/6,π/4)[-\pi/2,-\pi/4)\cup[\pi/6,\pi/4).

Rational inequality

Subtract 1: 2x2+3x−21−2x2≥0\dfrac{2x^2+3x-2}{1-2x^2}\geq0, equivalently (2x−1)(x+2)(2x+1)(2x−1)≤0\dfrac{(2x-1)(x+2)}{(\sqrt2x+1)(\sqrt2x-1)}\leq0. The critical inputs are −2,−1/2,1/2,1/2-2,-1/\sqrt2,1/2,1/\sqrt2. Successive signs are positive, negative, positive, negative, positive. Include numerator zeros and exclude denominator zeros: −2≤x<−1/2-2\leq x<-1/\sqrt2 or 1/2≤x<1/21/2\leq x<1/\sqrt2.

How did this part go?

Hence

Use cos⁡2θ=1−2sin⁡2θ\cos2\theta=1-2\sin^2\theta and replace x with sin⁡θ\sin\theta. This gives −2≤sin⁡θ<−1/2-2\leq\sin\theta<-1/\sqrt2 or 1/2≤sin⁡θ<1/21/2\leq\sin\theta<1/\sqrt2. On [−π/2,π/2][-\pi/2,\pi/2], sine increases from −1 to 1. The printed sine graph shows the levels −2, −1/2-1/\sqrt2, 1/21/2 and 1/21/\sqrt2; the −2 lower bound is automatic. Therefore −π/2≤θ<−π/4-\pi/2\leq\theta<-\pi/4 or π/6≤θ<π/4\pi/6\leq\theta<\pi/4.

How did this part go?

Compare your sketch

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Rational signs are positive, negative, positive, negative, positive across −2, −1/√2, 1/2 and 1/√2. Include −2 and 1/2; exclude both poles. The increasing sine graph on [−π/2,π/2] is compared with levels −2, −1/√2, 1/2 and 1/√2. Since sine is at least −1, the lower bound −2 imposes no restriction. Solution: [−π/2,−π/4) union [π/6,π/4).
Rational signs are positive, negative, positive, negative, positive across −2, −1/√2, 1/2 and 1/√2. Include −2 and 1/2; exclude both poles. The increasing sine graph on [−π/2,π/2] is compared with levels −2, −1/√2, 1/2 and 1/√2. Since sine is at least −1, the lower bound −2 imposes no restriction. Solution: [−π/2,−π/4) union [π/6,π/4).
Rational signs are positive, negative, positive, negative, positive across −2, −1/√2, 1/2 and 1/√2. Include −2 and 1/2; exclude both poles. The increasing sine graph on [−π/2,π/2] is compared with levels −2, −1/√2, 1/2 and 1/√2. Since sine is at least −1, the lower bound −2 imposes no restriction. Solution: [−π/2,−π/4) union [π/6,π/4).

Common mistakes

Intersect the algebraic solution with the range of sine; exclude cos 2θ=0.

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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