JEE MAIN - Mathematics (2012 - No. 3)
If $$f:R \to R$$ is a function defined by
$$f\left( x \right) = \left[ x \right]\cos \left( {{{2x - 1} \over 2}} \right)\pi $$,
where [x] denotes the greatest integer function, then $$f$$ is
$$f\left( x \right) = \left[ x \right]\cos \left( {{{2x - 1} \over 2}} \right)\pi $$,
where [x] denotes the greatest integer function, then $$f$$ is
continuous for every real $$x$$
discontinuous only at $$x=0$$
discontinuous only at non-zero integral values of $$x$$
continuous only at $$x=0$$
Explanation
Let $$f\left( x \right) = \left[ x \right]\cos \left( {{{2x - 1} \over 2}} \right)$$
Doubtful points are $$x = n,n \in I$$
$$L.H.L$$ $$ = \mathop {\lim }\limits_{x \to {n^ - }} \left[ x \right]\cos \left( {{{2x - 1} \over 2}} \right)\pi $$
$$ = \left( {n - 1} \right)\cos \left( {{{2n - 1} \over 2}} \right)\pi = 0$$
( As $$\left[ x \right]$$ is the greatest integer function )
$$R.H.L. = \mathop {\lim }\limits_{x \to {n^ + }} \,\left[ x \right]\cos \left( {{{2x - 1} \over 2}} \right)\pi = n\cos \left( {{{2n - 1} \over 2}} \right)\pi = 0$$
Now, value of the function at $$x = n$$ is $$f(n)=0$$
Since, $$L.H.L.$$ $$ = R.H.L. = f\left( n \right)$$
$$\therefore$$ $$f\left( x \right) = \left[ x \right]\cos \left( {{{2x - 1} \over 2}} \right)$$ is continuous for every real $$x.$$
Doubtful points are $$x = n,n \in I$$
$$L.H.L$$ $$ = \mathop {\lim }\limits_{x \to {n^ - }} \left[ x \right]\cos \left( {{{2x - 1} \over 2}} \right)\pi $$
$$ = \left( {n - 1} \right)\cos \left( {{{2n - 1} \over 2}} \right)\pi = 0$$
( As $$\left[ x \right]$$ is the greatest integer function )
$$R.H.L. = \mathop {\lim }\limits_{x \to {n^ + }} \,\left[ x \right]\cos \left( {{{2x - 1} \over 2}} \right)\pi = n\cos \left( {{{2n - 1} \over 2}} \right)\pi = 0$$
Now, value of the function at $$x = n$$ is $$f(n)=0$$
Since, $$L.H.L.$$ $$ = R.H.L. = f\left( n \right)$$
$$\therefore$$ $$f\left( x \right) = \left[ x \right]\cos \left( {{{2x - 1} \over 2}} \right)$$ is continuous for every real $$x.$$
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