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Let $f: R \rightarrow R$ be a continuous function such that $f\left(x^2\right)=f\left(x^3\right)$ 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,
$I$ is true and $III$ is false
$II$ is true and $III$ is false
Both $I$ and $III$ are true
Both $II$ and $III$ are true
Solution
(d)
Given function $f: R \longrightarrow R$ be a continuous function such that $f\left(x^2\right)=f\left(x^3\right) \forall x \in R$
then $f(x)=f\left(x^{23}\right) \quad$ [on replacing $x$ by $x^{1 / 3}$ ]
Similarly,
$f(x)=f\left(x^{23}\right)=f\left(x^{4 / 9}\right)=f\left(x^{2 / 27}\right)=$
$\quad \ldots=f\left(x^{(23)^n}\right)$
$=f\left(x^0\right) \text { [as } x \text { tends to infinity] }=f(1)$
$\therefore f(x)=f(1)=\text { constant }$
The function $f(x)=$ constant is even and differentiable everywhere.