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Let \mathrm{f}: \mathrm{R} \rightarrow \mathrm{R} satisfying |\mathrm{f}(\mathrm{x})| \leq \mathrm{x}^{2} \forall \mathrm{x} \in \mathrm{R}, then

Option: 1

f is continuous but non-differentiable at x=0


Option: 2

f is discontinuous at x=0


Option: 3

f is differentiable at x=0


Option: 4

none of these.


Answers (1)

best_answer

Putting x=0, we get, |f(x)| \leq 0

\Rightarrow \mathrm{f}(0)=0 \\

\mathrm{f}^{\prime}(0)=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(\mathrm{h})-\mathrm{f}(0)}{\mathrm{h}}=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(\mathrm{h})}{\mathrm{h}}

Now \left|\frac{f(h)}{h}\right| \leq|h|

\Rightarrow-|h| \leq \frac{f(h)}{h} \leq|h| \Rightarrow \lim _{h \rightarrow 0} \frac{f(h)}{h}=0

\Rightarrow {f}(\mathrm{x}) is differentiable at x=0.

Posted by

Nehul

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