1.

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 \( 2 \pi-x \)

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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