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If f:R\rightarrow R is a function defined by f(x)=\left [ x-1 \right ] \cos \left ( \frac{2x-1}{2} \right )\pi, where \left [ . \right ] denotes the greatese integer function, then f is :
Option: 1 discontinuous at all integral values of x except a x=1
Option: 2 Continuous only at x=1
Option: 3 disontinuous only at x=1
Option: 4 Continuous for every real x

Answers (1)

best_answer

Doubtful points are x = n, n \in I

\text { L.H.L }=\lim _{x \rightarrow n^{-}}[x-1] \cos \left(\frac{2 x-1}{2}\right) \pi=(n-2) \cos \left(\frac{2 n-1}{2}\right) \pi=0

\text { R.H.L. }=\lim _{x \rightarrow n^{-}}[x-1] \cos \left(\frac{2 x-1}{2}\right) \pi=(n-1) \cos \left(\frac{2 n-1}{2}\right) \pi=0

\mathrm{f}(\mathrm{n})=0

Hence f(x) is continous for all values of x

Posted by

Suraj Bhandari

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