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The function \mathrm{ f(x)=[x] \cos \left(\frac{2 x-1}{2}\right) \pi} , where [ ] denotes the greatest integer function, is discontinuous at

Option: 1

all x


Option: 2

all integer points
 


Option: 3

no x


Option: 4

x which is not an integer.
 


Answers (1)

best_answer

We have [x]=k for \mathrm{k \leq x<k+1}  where k is an integer. Therefore
\mathrm{f(x)=k \cos \left(\frac{2 x-1}{2}\right) \pi}  for \mathrm{k \leq x<k+1}  where \mathrm{k \in Z} 
So, $\mathrm{\lim _{x \rightarrow k^{-}} f(x)=\lim _{h \rightarrow 0} f(k-h)}

\mathrm{ = \lim _{h \rightarrow 0}(k-1) \cos \left(\frac{2(k-h)-1}{2}\right) \pi \cdots }
\mathrm{ {\left[\because f(x)=(k-1) \cos \left(\frac{2 x-1}{2}\right) \pi \text { for } k-15 x<k\right] } }
\mathrm{ = (k-1) \cos \left(\frac{2 k-1}{2}\right) \pi=0}
and 
\mathrm{ \lim _{x \rightarrow k^{+}} f(x)= \lim _{h \rightarrow 0} f(k+h) }
\mathrm{ = \lim _{h \rightarrow 0} k \cos \left(\frac{2(k+h)-1}{2}\right) \pi \\ }
\mathrm{ {\left[\because f(x)=k \cos \left(\frac{2 x-1}{2}\right) \pi \text { for } k \leq x<k+1\right] } }
\mathrm{ =k \cos \left(\frac{2 k-1}{2}\right) \pi=0 }
Also, \mathrm{ f(k)=k \cos \left(\frac{2 k-1}{2}\right) \pi=0 }

Hence f(x) is continuous at all integer points. Since f(x) is a cosine function between any two integer points. Hence, f(x) is everywhere continuous.

Posted by

manish painkra

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