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If the period of the function $f (x) = \sin (\sqrt{[n]} x),$ where [n] denotes the greatest integer less than or equal to n, is 2 $π$ , then

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1 ≤ n < 2

1 < n < 2

1 ≤ n ≤ 2

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Correct option is A

Sin x is a periodic function with period 2π, therefore sin⁡([n]x) is a periodic function with period 2π[n].But the period of f (x) is 2π (given).∴2π[n]=2π⇒[n]=1⇒[n]=1⇒1≤n<2

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If the period of the function f(x)=sin⁡([n]x), where [n] denotes the greatest integer less than or equal to n, is 2π, then

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Ready to Test Your Skills?

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If the period of the function $f (x) = \sin (\sqrt{[n]} x),$ where [n] denotes the greatest integer less than or equal to n, is 2 $π$ , then