JEE MAIN - Mathematics (2021 - 24th February Morning Shift - No. 2)
If f : R $$ \to $$ R is a function defined by f(x)= [x - 1] $$\cos \left( {{{2x - 1} \over 2}} \right)\pi $$, where [.] denotes the greatest
integer function, then f is :
continuous for every real x
discontinuous at all integral values of x except at x = 1
discontinuous only at x = 1
continuous only at x = 1
Explanation
Given, $$f(x) = [x - 1]\cos \left( {{{2x - 1} \over 2}} \right)\pi $$ where [ . ] is greatest integer function and f : R $$\to$$ R
$$\because$$ It is a greatest integer function then we need to check its continuity at x $$\in$$ I except these it is continuous.
Let, x = n where n $$\in$$ I
Then
LHL = $$\mathop {\lim }\limits_{x \to {n^ - }} [x - 1]\cos \left( {{{2x - 1} \over 2}} \right)\pi $$
$$ = (n - 2)\cos \left( {{{2x - 1} \over 2}} \right)\pi = 0$$
RHL = $$\mathop {\lim }\limits_{x \to {n^ + }} [x - 1]\cos \left( {{{2x - 1} \over 2}} \right)\pi $$
$$ = (n - 1)\cos \left( {{{2x - 1} \over 2}} \right)\pi = 0$$
and f(n) = 0
Here, $$\mathop {\lim }\limits_{x \to {n^ - }} f(x) = \mathop {\lim }\limits_{x \to {n^ + }} f(x) = f(n)$$
$$\therefore$$ It is continuous at every integers.
Therefore, the given function is continuous for all real x.
$$\because$$ It is a greatest integer function then we need to check its continuity at x $$\in$$ I except these it is continuous.
Let, x = n where n $$\in$$ I
Then
LHL = $$\mathop {\lim }\limits_{x \to {n^ - }} [x - 1]\cos \left( {{{2x - 1} \over 2}} \right)\pi $$
$$ = (n - 2)\cos \left( {{{2x - 1} \over 2}} \right)\pi = 0$$
RHL = $$\mathop {\lim }\limits_{x \to {n^ + }} [x - 1]\cos \left( {{{2x - 1} \over 2}} \right)\pi $$
$$ = (n - 1)\cos \left( {{{2x - 1} \over 2}} \right)\pi = 0$$
and f(n) = 0
Here, $$\mathop {\lim }\limits_{x \to {n^ - }} f(x) = \mathop {\lim }\limits_{x \to {n^ + }} f(x) = f(n)$$
$$\therefore$$ It is continuous at every integers.
Therefore, the given function is continuous for all real x.
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