If m≤3sin−1x+2cos−1x≤n, then the value of m and n are, respectively,
A
π2,3π2
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B
−π2,3π2
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C
π2,−3π2
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D
−π,π
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Solution
The correct option is Aπ2,3π2 m≤3sin−1x+2cos−1x≤n, We know sin−1x+cos−1x=π2 ⇒2sin−1x+2cos−1x=π⋯(1) Also, −π2≤sin1x≤π2⋯(2) Adding(1)&(2), we get π2≤3sin−1x+2cos−1x≤3π2