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MathHardAdv 2020 • Paper 1

Difference of Interval Endpoints for Nonnegative Trigonometric Function

Consider the function f:[0,2]→Rf: [0, 2] \rightarrow \mathbb{R} defined by f(x)=(3−sin⁡(2πx))sin⁡(πx−π4)−sin⁡(3πx+π4).f(x) = (3 - \sin(2\pi x)) \sin\left(\pi x - \frac{\pi}{4}\right) - \sin\left(3\pi x + \frac{\pi}{4}\right). If α,β∈[0,2]\alpha, \beta \in [0, 2] are such that {x∈[0,2]:f(x)≥0}=[α,β]\{x \in [0, 2] : f(x) \ge 0\} = [\alpha, \beta], then the value of β−α\beta - \alpha is _____

Official Numerical Answer1

Topics & Concepts

Trigonometric FunctionsTrigonometric Ratios and IdentitiesTrigonometric Equations

Step-by-Step Solution

To find the set of all x∈[0,2]x \in [0, 2] for which f(x)≥0f(x) \ge 0, we first simplify the expression for f(x)f(x).

Let t=πx−π4t = \pi x - \frac{\pi}{4}. Then we have: x=tπ+14x = \frac{t}{\pi} + \frac{1}{4}

Since x∈[0,2]x \in [0, 2], the corresponding range for tt is: t∈[−π4,7π4]t \in \left[-\frac{\pi}{4}, \frac{7\pi}{4}\right]

Now, express the arguments of the trigonometric terms in f(x)f(x) in terms of tt:

  1. 2πx=2t+π22\pi x = 2t + \frac{\pi}{2}, so: sin⁡(2πx)=sin⁡(2t+π2)=cos⁡(2t)\sin(2\pi x) = \sin\left(2t + \frac{\pi}{2}\right) = \cos(2t)

  2. 3πx+π4=3(t+π4)+π4=3t+π3\pi x + \frac{\pi}{4} = 3\left(t + \frac{\pi}{4}\right) + \frac{\pi}{4} = 3t + \pi, so: sin⁡(3πx+π4)=sin⁡(3t+π)=−sin⁡(3t)\sin\left(3\pi x + \frac{\pi}{4}\right) = \sin(3t + \pi) = -\sin(3t)

Substituting these identities back into the definition of f(x)f(x): f(x)=(3−cos⁡(2t))sin⁡(t)−(−sin⁡(3t))=(3−cos⁡(2t))sin⁡(t)+sin⁡(3t)f(x) = (3 - \cos(2t)) \sin(t) - (-\sin(3t)) = (3 - \cos(2t))\sin(t) + \sin(3t)

Using the standard trigonometric identities cos⁡(2t)=1−2sin⁡2(t)\cos(2t) = 1 - 2\sin^2(t) and sin⁡(3t)=3sin⁡(t)−4sin⁡3(t)\sin(3t) = 3\sin(t) - 4\sin^3(t): f(x)=(3−(1−2sin⁡2(t)))sin⁡(t)+(3sin⁡(t)−4sin⁡3(t))f(x) = \left(3 - (1 - 2\sin^2(t))\right)\sin(t) + \left(3\sin(t) - 4\sin^3(t)\right) f(x)=(2+2sin⁡2(t))sin⁡(t)+3sin⁡(t)−4sin⁡3(t)f(x) = (2 + 2\sin^2(t))\sin(t) + 3\sin(t) - 4\sin^3(t) f(x)=2sin⁡(t)+2sin⁡3(t)+3sin⁡(t)−4sin⁡3(t)f(x) = 2\sin(t) + 2\sin^3(t) + 3\sin(t) - 4\sin^3(t) f(x)=5sin⁡(t)−2sin⁡3(t)f(x) = 5\sin(t) - 2\sin^3(t) f(x)=sin⁡(t)(5−2sin⁡2(t))f(x) = \sin(t)\left(5 - 2\sin^2(t)\right)

Since −1≤sin⁡(t)≤1-1 \le \sin(t) \le 1, we have 0≤sin⁡2(t)≤10 \le \sin^2(t) \le 1, which implies: 3≤5−2sin⁡2(t)≤53 \le 5 - 2\sin^2(t) \le 5

Since 5−2sin⁡2(t)>05 - 2\sin^2(t) > 0 for all t∈Rt \in \mathbb{R}, the sign of f(x)f(x) depends entirely on the sign of sin⁡(t)\sin(t): f(x)≥0  ⟺  sin⁡(t)≥0f(x) \ge 0 \iff \sin(t) \ge 0

For t∈[−π4,7π4]t \in \left[-\frac{\pi}{4}, \frac{7\pi}{4}\right], sin⁡(t)≥0\sin(t) \ge 0 if and only if: 0≤t≤π0 \le t \le \pi

Substituting t=πx−π4t = \pi x - \frac{\pi}{4} back into the inequality: 0≤πx−π4≤π0 \le \pi x - \frac{\pi}{4} \le \pi π4≤πx≤5π4\frac{\pi}{4} \le \pi x \le \frac{5\pi}{4} 14≤x≤54\frac{1}{4} \le x \le \frac{5}{4}

Thus, the set {x∈[0,2]:f(x)≥0}\{x \in [0, 2] : f(x) \ge 0\} is the closed interval [14,54]\left[\frac{1}{4}, \frac{5}{4}\right].

Comparing this with [α,β][\alpha, \beta], we get: α=14andβ=54\alpha = \frac{1}{4} \quad \text{and} \quad \beta = \frac{5}{4}

Therefore, the value of β−α\beta - \alpha is: β−α=54−14=1\beta - \alpha = \frac{5}{4} - \frac{1}{4} = 1

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