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If \( \pi \leq x \leq 2 \pi \), then the value of \( \cos ^{-3}(\cos x) \) is (A) \( x \) (B) \( -x \) (C) \( 2 \pi+x \) D \( 2 \pi-x \) |
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Answer» Correct option is (D) 2π - x cos-1(cos x) = 2x But π < x 2π and principal range of cos-1x is (0, π) Now, cos-1(cos x) = cos-1(-(-cos x)) = π - cos-1(-cos x) = π - cos-1(cos(π - x)) (cos(π - θ) = - cos θ) But π - (π - x) = x \(\notin\) [0, π] = π - cos-1(cos(-(x - π))) = π - cos-1(cos(x - π)) ( ∵ cos-1(cos θ) = θ) = π - x + π = 2π - x and 2π - 2π \(\leq\) 2π - x \(\leq\) 2π - π ⇒ 0 \(\leq\) 2π - x \(\leq\)π |
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