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  • For a real number y, let  denotes the greatest integer less than or equal to y. Then

For a real number y, let \mathrm{[y]} denotes the greatest integer less than or equal to y. Then the function \mathrm{f(x)=\frac{\tan \pi[(x-\pi)]}{1+[x]^2}}

Option: 1

discontinuous at some \mathrm{x}


Option: 2

continuous at all x , but the derivative \mathrm{f^{\prime}(x)} does not exist for some x


Option: 3

\mathrm{f^{\prime}(x)} exist for all x


Option: 4

\mathrm{f^{\prime}(x)} exists for all x, but the derivative \mathrm{f^{\prime}(x)} does not exist for some x


Answers (1)

best_answer

Given, f(x)=\frac{\tan \pi[(x-\pi)]}{1+[x]^2}

We know that, {\pi[(x-\pi)]=n \pi \text { and } \tan n \pi=0}

{\begin{aligned} \because & & 1+\left[x^2\right] & \neq 0 \\ & \therefore & f(x) & =0, \vee x \end{aligned}}

So f(x) is a continuous function.

Thus, {f^{\prime \prime}(x), f^{\prime \prime}(x)}....all exist for every x, their value being 0.

Hence, f'(x) exists for all x.

Posted by

Ajit Kumar Dubey

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