Let f(x)=sec−1[1+cos2x], where [.] denotes the greatest integer function, then the range of f(x) is
A
[1, 2]
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B
[0, 2]
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C
[sec−11,sec−12]
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D
None of these
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Solution
The correct option is C[sec−11,sec−12] f(x)=sec−1{1+cos2x}0≤cos2x≤10+1≤1+cos2x≤1+11≤1+cos2x≤2 [1+cos2x]=1 ([] is the greatest integer function) (for1≤(1+cos2x)<2) [1+cos2x]=2 (for(1+cos2x)=2) ∴ the values of y=sec−1[1+cos2x]aresec−11 and sec−12