Mathematics - Continuity - Differentiability Question with Solution | TestHub

Doubtful points are $x=n, n\in I$

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

$R.H.L =\underset{x\to n^{+}}{\lim }\left[x-1\right]\cos \left(\frac{2x-1}{2}\right)\pi =\left(n-1\right)\cos \left(\frac{2n-1}{2}\right)\pi =0$

$f\left(n\right)=0$

Hence continuous.