Question
If $$f:R \to R$$ is a function defined by $$f\left( x \right) = \left[ x \right]\cos \left( {\frac{{2x - 1}}{2}} \right)\pi ,$$ where $$\left[ x \right]$$ denotes the greatest integer function, then $$f$$ is,
A.
continuous for every real $$x.$$
B.
discontinuous only at $$x = 0$$
C.
discontinuous only at non-zero integral values of $$x.$$
D.
continuous only at $$x = 0.$$
Answer :
continuous for every real $$x.$$
Solution :
$${\text{Let }}f\left( x \right) = \left[ x \right]\cos \left( {\frac{{2x - 1}}{2}} \right)\pi $$
Doubtful points are $$x = n,\,\,n \in I$$
$$\eqalign{
& {\text{L}}{\text{.H}}{\text{.L}}{\text{.}} = \mathop {\lim }\limits_{x \to {n^ - }} \left[ x \right]\cos \left( {\frac{{2x - 1}}{2}} \right)\pi \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {n - 1} \right)\cos \left( {\frac{{2n - 1}}{2}} \right)\pi = 0 \cr} $$
($$\because \,\left[ x \right]$$ is the greatest integer function)
$$\eqalign{
& {\text{R}}{\text{.H}}{\text{.L}}{\text{.}} = \mathop {\lim }\limits_{x \to {n^ + }} \left[ x \right]\cos \left( {\frac{{2x - 1}}{2}} \right)\pi \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = n\,\cos \left( {\frac{{2n - 1}}{2}} \right)\pi = 0 \cr} $$
Now, value of the function at $$x=n$$ is $$f\left( n \right) = 0$$
Since, L.H.L. $$=$$ R.H.L. $$ = f\left( n \right)$$
$$\therefore f\left( x \right) = \left[ x \right]\cos \left( {\frac{{2x - 1}}{2}} \right)$$ is continuous for every real $$x.$$