Solve this following


If $f: \mathrm{R} \rightarrow \mathrm{R}$ is a function defined by

$f(x)=[x-1] \cos \left(\frac{2 x-1}{2}\right) \pi$, where [.] denotes

the greatest integer function, then $f$ is :


  1. discontinuous at all integral values of $x$ except at $x=1$

  2. continuous only at $x=1$

  3. continuous for every real $x$

  4. discontinuous only at $x=1$

Correct Option: , 3


For $x=n, n \in Z$

$\mathrm{LHL}=\lim _{x \rightarrow \mathrm{n}^{-}} f(\mathrm{x})=\lim _{\mathrm{x} \rightarrow \mathrm{n}^{-}}[\mathrm{x}-1] \cos \left(\frac{2 \mathrm{x}-1}{2}\right) \pi$


$\mathrm{RHL}=\lim _{x \rightarrow \mathrm{n}^{+}} f(\mathrm{x})=\lim _{x \rightarrow \mathrm{n}^{+}}[\mathrm{x}-1] \cos \left(\frac{2 \mathrm{x}-1}{2}\right) \pi$



$\Rightarrow \quad \mathrm{LHL}=\mathrm{RHL}=f(\mathrm{n})$

$\Rightarrow f(\mathrm{x})$ is continuous for every real $\mathrm{x}$.


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