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If f(x)=[\sqrt{2} \sin x], where [x] denotes the greatest integer function, then

Option: 1

f(x) is continuous at x=0


Option: 2

maximum value of f(x) is 1 in interval [-2 \pi, 2 \pi]


Option: 3

f(x) is discontinuous at x=\frac{n \pi}{2}+\frac{\pi}{4}, n \in I


Option: 4

f(x) is differentiable at x=n \pi, n \in I


Answers (1)

best_answer

 f(x)=[\sqrt{2} \sin x] \\

\therefore \quad f(x)=-2,-\frac{\pi}{2} \leq x<-\frac{\pi}{4}=-1,-\frac{\pi}{4} \leq x<0=0,0 \leq x<\frac{\pi}{4} \\

 =1, \frac{\pi}{4} \leq x \leq \frac{\pi}{2}=1, \frac{\pi}{2} \leq x \leq \frac{3 \pi}{4}=0, \frac{3 \pi}{4}<x \leq \piClearly f(x) is discontinuous at

x=-\frac{\pi}{4}, 0, \frac{\pi}{4}, \frac{3 \pi}{4} etc.

General value corresponding to

x=-\frac{\pi}{4}, \frac{\pi}{4}, \frac{3 \pi}{4} \text { is } \frac{n \pi}{2}+\frac{\pi}{4}

maximum value of f(x) in [-2 \pi, 2 \pi]  is 1 at x=\frac{\pi}{2}.

f(x) is discontinuous and non differentiable at x=0. therefore choice (d) is not correct.

Hence (b) is the correct answer.

Posted by

himanshu.meshram

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