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(B) Given $|f(x)| \leq x^2$ for all $x \in R$.At $x = 0$,$|f(0)| \leq 0^2 = 0$,which implies $f(0) = 0$.To check differentiability at $x = 0$,we evalu

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continuous as well as differentiable

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(B) Given $|f(x)| \leq x^2$ for all $x \in R$.At $x = 0$,$|f(0)| \leq 0^2 = 0$,which implies $f(0) = 0$.To check differentiability at $x = 0$,we evaluate the limit of the derivative:$f'(0) = \lim_{h 	o 0} \frac{f(h) - f(0)}{h - 0} = \lim_{h 	o 0} \frac{f(h)}{h}$.From the given inequality,$|f(h)| \leq h^2$,so $|\frac{f(h)}{h}| \leq |h|$.By the Sandwich Theorem,as $h 	o 0$,$|h| 	o 0$,therefore $\lim_{h 	o 0} \frac{f(h)}{h} = 0$.Thus,$f'(0) = 0$,which means $f$ is differentiable at $x = 0$.Since differentiability implies continuity,$f$ is also continuous at $x = 0$.Therefore,$f$ is continuous as well as differentiable at $x = 0$.

Let $f : R \to R$ be a function such that $|f(x)| \leq x^2$,for all $x \in R$. Then,at $x = 0$,$f$ is

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