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Question 59 of 71

Q.If 3cos⁡−1x=cos⁡−1(4x3−3x)3\cos^{-1}x=\cos^{-1}(4x^3-3x),

(a) x∈(12,1)x\in\left(\dfrac12, 1\right)
(b) x∈[12,1]x\in\left[\dfrac12, 1\right]
(c) x∈(−∞,1]x\in(-\infty, 1]
(d) x∈[12,∞)x\in\left[\dfrac12, \infty\right)
Puducherry TnboardTamil Nadu HSC (DGE) Board 2023MCQ· 1mImportance★★★★★
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The triple-angle identity cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos3\theta=4\cos^3\theta-3\cos\theta only matches cos⁡−1\cos^{-1}'s principal branch when 3θ∈[0,π]3\theta\in[0,\pi].

  1. Let x=cos⁡θx=\cos\theta with θ=cos⁡−1x∈[0,π]\theta=\cos^{-1}x\in[0,\pi] (the principal branch of cos⁡−1\cos^{-1}).
  2. The identity cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos3\theta=4\cos^3\theta-3\cos\theta gives 4x3−3x=cos⁡3θ4x^3-3x=\cos3\theta.
  3. For the given equation 3cos⁡−1x=cos⁡−1(4x3−3x)3\cos^{-1}x=\cos^{-1}(4x^3-3x) to hold, we need cos⁡−1(cos⁡3θ)=3θ\cos^{-1}(\cos3\theta)=3\theta, which is true only when 3θ3\theta itself lies in [0,π][0,\pi] (the range of cos⁡−1\cos^{-1}). …

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