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(D) Given $f(x^2) = f(x^3)$ for all $x \in R$.For any $x > 0$,let $x = t^6$. Then $f(t^{12}) = f(t^{18})$.More generally,for any $x > 0$,we can write

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Both $II$ and $III$ are true

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(D) Given $f(x^2) = f(x^3)$ for all $x \in R$.For any $x > 0$,let $x = t^6$. Then $f(t^{12}) = f(t^{18})$.More generally,for any $x > 0$,we can write $x = t^{6^n}$.By repeating the substitution,$f(x) = f(x^{2/3}) = f(x^{(2/3)^2}) = \dots = f(x^{(2/3)^n})$.As $n 	o \infty$,$(2/3)^n 	o 0$,so $f(x) = f(x^0) = f(1)$ for all $x > 0$.Since $f$ is continuous at $x = 0$,$f(0) = \lim_{x 	o 0^+} f(x) = f(1)$.For $x  0$. Then $f((-t)^2) = f((-t)^3) \implies f(t^2) = f(-t^3)$.Since $f(t^2) = f(1)$,we have $f(-t^3) = f(1)$.As $t^3$ covers all positive values,$f(x) = f(1)$ for all $x Thus,$f(x) = c$ (a constant) for all $x \in R$.$A$ constant function is an even function because $f(-x) = c = f(x)$.$A$ constant function is also differentiable everywhere with $f'(x) = 0$.Therefore,both $II$ and $III$ are true.

Let $f: R \rightarrow R$ be a continuous function such that $f(x^2) = f(x^3)$ for all $x \in R$. Consider the following statements:
$I.$ $f$ is an odd function.
$II.$ $f$ is an even function.
$III.$ $f$ is differentiable everywhere.
Then,

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Let $f: R \rightarrow R$ be a continuous function such that $f(x^2) = f(x^3)$ for all $x \in R$. Consider the following statements:$I.$ $f$ is an odd function.$II.$ $f$ is an even function.$III.$ $f$ is differentiable everywhere.Then,