If −$$1$$≤x≤$$0$$, then $$cos$$−$$1$$⁡$$2x2$$−$$1$$ equals

Correct option is 32π−2cos−1⁡x

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If $- 1 \leq x \leq 0,$ then $\cos^{- 1} (2 x^{2} - 1)$ equals

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2cos−1⁡x

π−2cos−1⁡x

2π−2cos−1⁡x

−2cos−1⁡x

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detailed solution

Correct option is C

Let cos−1⁡x=θ. Then, x=cos⁡θAlso, −1≤x≤0⇒−1≤cos⁡θ≤0⇒π2≤θ≤πNow, cos−1⁡2x2−1=cos−1⁡(cos⁡2θ)=cos−1⁡(2π−2θ)=2π−2θ ∵π2≤θ≤π⇒π≤2θ≤2π⇒2π−2cos−1⁡x

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If −1≤x≤0, then cos−1⁡2x2−1 equals

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If $- 1 \leq x \leq 0,$ then $\cos^{- 1} (2 x^{2} - 1)$ equals