A) continuous but not differentiable.
B) continuous as well as differentiable.
C) neither continuous nor differentiable.
D) differentiable but not continuous.
Correct Answer: B
Solution :
Let \[|f(x)|\le {{x}^{2}},\forall x\in R\] Now, at \[x=0,|f(0)|\le 0\]\[\Rightarrow \]\[f(0)=0\] \[\therefore \]\[f'(0)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(h)-f(0)}{h-0}=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(h)}{h}\]?.(1) Now, \[\left| \frac{f(h)}{h} \right|\le |h|\] \[(\because |f(x)|\le {{x}^{2}})\] \[\Rightarrow \]\[-|h|\le \frac{f(h)}{h}\le |h|\]\[\Rightarrow \]\[\underset{h\to 0}{\mathop{\lim }}\,\frac{f(h)}{h}\to 0\]?(2) (using sandwich Theorem) \[\therefore \]from (1) and (2), we get f? (0) = 0, i.e. - f (x) is differentiable, at x = 0 Since, differentiability \[\Rightarrow \] Continutity \[\therefore \]\[|f(x)|\le {{x}^{2}},\] for all \[x\in R\] is continuous as well as differentiable at x = 0.
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