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Exercise 3.12 · Q15

Q.If f(θ)=∣sin⁡θ∣+∣cos⁡θ∣, θ∈Rf(\theta)=|\sin\theta|+|\cos\theta|,\ \theta\in\mathbb R, then f(θ)f(\theta) is in the interval

(1) [0,2][0,2]
(2) [1,2]\left[1,\sqrt2\right]
(3) [1,2][1,2]
(4) [0,1][0,1]
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Square f(θ)f(\theta) to turn it into 1+∣sin⁡2θ∣1+|\sin2\theta|, whose range is immediate.

Step 1. f(θ)2=(∣sin⁡θ∣+∣cos⁡θ∣)2=sin⁡2θ+cos⁡2θ+2∣sin⁡θ∣∣cos⁡θ∣=1+2∣sin⁡θcos⁡θ∣=1+∣sin⁡2θ∣f(\theta)^2=(|\sin\theta|+|\cos\theta|)^2=\sin^2\theta+\cos^2\theta+2|\sin\theta||\cos\theta|=1+2|\sin\theta\cos\theta|=1+|\sin2\theta|.

Step 2. Since ∣sin⁡2θ∣∈[0,1]|\sin2\theta|\in[0,1] for all real θ\theta, f(θ)2∈[1,2]f(\theta)^2\in[1,2]. …

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