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Let \mathrm{f(x)=\frac{\tan (\pi[x-\pi])}{1+[x]^2}} where [ ] denotes G.I.F., then find correct statements.

Option: 1

Discontinuous at same x 


Option: 2

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


Option: 3

 \mathrm{ f^{\prime}(x) } exists for all \mathrm{ x }, but the derivative \mathrm{ f^{\prime}\left(x_0\right) } does not exist second for same \mathrm{ x }.


Option: 4

\mathrm{ f^{\prime}(x) \text { exists for all } x \text {. }}


Answers (1)

best_answer

\mathrm{ \text { } f(x)=\frac{\tan (\pi[x-\pi])}{1+[x]^2}}
\mathrm{ By\, \, definition, [x-\pi] \, \, is\, \, an \, \, integer \, \, whatever\, \, be\, \, the\, \, value\, \, of \, \, x \, \, and \, \, \, so}\\\mathrm{ \pi[x-\pi] \, \, is \, \, an \, \, integral \, \, multiple \, \, of \, \, \pi.}
\mathrm{ Consequently, \tan (\pi[x-\pi])=0 \forall x and since, 1+[x]^2 \neq 0 for any x, we conclude that f(x)=0}\\\mathrm{ Consequently, \tan (\pi[x-\pi])=0 \forall x \, \, \, and \, \, since, 1+[x]^2 \neq 0 for \, \, any x, we\, \, conclude \, \, that \, \, f(x)=0}\mathrm{ Thus, f(x) is\, a\, constant\, function \, and\, so\, it \, \, is \, \, continuous \, \, and \, \, \, differentiable\, \, any \, \, number\, \, of\, \, times\, \, , that \, \, is,\, \, f^{\prime}(x), f^{\prime \prime}(x), f^{\prime \prime \prime}(x), \ldots }
\mathrm{ \text { all exist for every } x \text {, their value being }} 
\mathrm{ 0 \text { (zero) at every point } x \text {. }}

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manish painkra

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