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Range of the function $f (x) = \cos (K \sin x)$ is [-1,1], then the least positive integral value of K will be

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Correct option is D

y=cosKsinx cos-1yK=sinx ⇒ -1≤cos-1yK≤1 ⇒ -K≤cos-1y≤K Now cos-1y∈0,π ⇒K=4

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Range of the function fx=cosKsinx is [-1,1], then the least positive integral value of K will be

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Range of the function $f (x) = \cos (K \sin x)$ is [-1,1], then the least positive integral value of K will be