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Hint: In this question first we will check the continuity of the given function at a given point, if it is discontinuous at that given point, it will also be non-differentiable at that point.
Complete step-by-step answer:
As we are given with a function,
\[ \Rightarrow f(x) = [x],0 < x < 3\]
\[ \Rightarrow \]As, we know that, if \[{\text{ }}f(x)\] is not continuous at a point,
Then it will not be differentiable at that point too.
\[ \Rightarrow \]So, let us check for continuity of \[f(x){\text{ }} = {\text{ }}\left[ x \right]\] at \[x{\text{ }} = {\text{ }}1\] and \[x{\text{ }} = {\text{ 2}}\].
Checking continuity at \[x{\text{ }} = {\text{ }}1\]
\[ \Rightarrow \mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} [x] = 0\]
\[ \Rightarrow \]And, \[\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} [x] = 1\]
As, we have seen above that \[\mathop {\lim }\limits_{x \to {1^ - }} f(x) \ne \mathop {\lim }\limits_{x \to {1^ + }} f(x)\]
\[ \Rightarrow \]Therefore, \[{\text{ }}f(x)\] is neither continuous nor differentiable at \[x{\text{ }} = {\text{ }}1\].
Now, checking continuity at \[{\text{x}} = 2\].
\[ \Rightarrow \mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{x \to {2^ - }} [x] = 1\]
\[ \Rightarrow \]And, \[\mathop {\lim }\limits_{x \to {2^ + }} f(x) = \mathop {\lim }\limits_{x \to {2^ + }} [x] = 2\]
As, we have seen above that, \[\mathop {\lim }\limits_{x \to {2^ - }} f(x) \ne \mathop {\lim }\limits_{x \to {2^ + }} f(x)\].
\[ \Rightarrow \]Therefore, \[{\text{ }}f(x)\] is neither continuous nor differentiable at \[{\text{x}} = 2\].
\[ \Rightarrow \]Hence, \[{\text{ }}f(x)\] is neither differentiable at \[x{\text{ }} = {\text{ }}1\]
nor differentiable at \[{\text{x}} = 2\].
Note: Whenever we come up with this type of problem we are asked to check whether the function, \[{\text{ }}f(x)\] is differentiable or not. Then first we should check the continuity of the given function, if it is continuous then we have to check whether the function is differentiable or not.
Complete step-by-step answer:
As we are given with a function,
\[ \Rightarrow f(x) = [x],0 < x < 3\]
\[ \Rightarrow \]As, we know that, if \[{\text{ }}f(x)\] is not continuous at a point,
Then it will not be differentiable at that point too.
\[ \Rightarrow \]So, let us check for continuity of \[f(x){\text{ }} = {\text{ }}\left[ x \right]\] at \[x{\text{ }} = {\text{ }}1\] and \[x{\text{ }} = {\text{ 2}}\].
Checking continuity at \[x{\text{ }} = {\text{ }}1\]
\[ \Rightarrow \mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} [x] = 0\]
\[ \Rightarrow \]And, \[\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} [x] = 1\]
As, we have seen above that \[\mathop {\lim }\limits_{x \to {1^ - }} f(x) \ne \mathop {\lim }\limits_{x \to {1^ + }} f(x)\]
\[ \Rightarrow \]Therefore, \[{\text{ }}f(x)\] is neither continuous nor differentiable at \[x{\text{ }} = {\text{ }}1\].
Now, checking continuity at \[{\text{x}} = 2\].
\[ \Rightarrow \mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{x \to {2^ - }} [x] = 1\]
\[ \Rightarrow \]And, \[\mathop {\lim }\limits_{x \to {2^ + }} f(x) = \mathop {\lim }\limits_{x \to {2^ + }} [x] = 2\]
As, we have seen above that, \[\mathop {\lim }\limits_{x \to {2^ - }} f(x) \ne \mathop {\lim }\limits_{x \to {2^ + }} f(x)\].
\[ \Rightarrow \]Therefore, \[{\text{ }}f(x)\] is neither continuous nor differentiable at \[{\text{x}} = 2\].
\[ \Rightarrow \]Hence, \[{\text{ }}f(x)\] is neither differentiable at \[x{\text{ }} = {\text{ }}1\]
nor differentiable at \[{\text{x}} = 2\].
Note: Whenever we come up with this type of problem we are asked to check whether the function, \[{\text{ }}f(x)\] is differentiable or not. Then first we should check the continuity of the given function, if it is continuous then we have to check whether the function is differentiable or not.
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